History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Benford’s Law: Tables of Logarithms, TaxCheats, and The Leading Digit
Phenomenon
Michelle Manes ([email protected])
9 September, 2008
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
History
(1881) Simon Newcomb publishes “Note on thefrequency of use of the different digits in naturalnumbers.” The world ignores it.
(1938) Frank Benford (unaware of Newcomb’s work,presumably) publishes “The law of anomalousnumbers.”
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
History
(1881) Simon Newcomb publishes “Note on thefrequency of use of the different digits in naturalnumbers.” The world ignores it.
(1938) Frank Benford (unaware of Newcomb’s work,presumably) publishes “The law of anomalousnumbers.”
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Statement of Benford’s Law
Newcomb noticed that the early pages of the book oftables of logarithms were much dirtier than the laterpages, so were presumably referenced more often.
He stated the rule this way:
Prob(first significant digit = d) = log10
(1 +
1d
).
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Statement of Benford’s Law
Newcomb noticed that the early pages of the book oftables of logarithms were much dirtier than the laterpages, so were presumably referenced more often.
He stated the rule this way:
Prob(first significant digit = d) = log10
(1 +
1d
).
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Benford Distribution
DefinitionA sequence of positive numbers {xn} is Benford if
Prob(first significant digit = d) = log10
(1 +
1d
).
A sequence of positive numbers {xn} is Benford base b if
Prob(first significant digit = d) = logb
(1 +
1d
).
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Benford Distribution
DefinitionA sequence of positive numbers {xn} is Benford if
Prob(first significant digit = d) = log10
(1 +
1d
).
A sequence of positive numbers {xn} is Benford base b if
Prob(first significant digit = d) = logb
(1 +
1d
).
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Benford’s Law
Base 10 Predictionsdigit probability it occurs as a leading digit
1 30.1%2 17.6%3 12.5%4 9.7%5 7.9%6 6.7%7 5.8%8 5.1%9 4.6%
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Benford’s Data
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
More Data
Benford’s Law compared with: numbers from the frontpages of newspapers, U.S. county populations, and theDow Jones Industrial Average.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Dow Illustrates Benford’s Law
Suppose the Dow Jones average is about $1,000. If theaverage goes up at a rate of about 20% a year, it wouldtake four years to get from 1 to 2 as a first digit.
If we start with a first digit 5, it only requires a 20%increase to get from $5,000 to $6,000, and that isachieved in one year.
When the Dow reaches $9,000, it takes only an 11%increase and just seven months to reach the $10,000mark. This again has first digit 1, so it will take anotherdoubling (and four more years) to get back to first digit 2.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Dow Illustrates Benford’s Law
Suppose the Dow Jones average is about $1,000. If theaverage goes up at a rate of about 20% a year, it wouldtake four years to get from 1 to 2 as a first digit.
If we start with a first digit 5, it only requires a 20%increase to get from $5,000 to $6,000, and that isachieved in one year.
When the Dow reaches $9,000, it takes only an 11%increase and just seven months to reach the $10,000mark. This again has first digit 1, so it will take anotherdoubling (and four more years) to get back to first digit 2.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Dow Illustrates Benford’s Law
Suppose the Dow Jones average is about $1,000. If theaverage goes up at a rate of about 20% a year, it wouldtake four years to get from 1 to 2 as a first digit.
If we start with a first digit 5, it only requires a 20%increase to get from $5,000 to $6,000, and that isachieved in one year.
When the Dow reaches $9,000, it takes only an 11%increase and just seven months to reach the $10,000mark. This again has first digit 1, so it will take anotherdoubling (and four more years) to get back to first digit 2.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Dow Illustrates Benford’s Law
Suppose the Dow Jones average is about $1,000. If theaverage goes up at a rate of about 20% a year, it wouldtake four years to get from 1 to 2 as a first digit.
If we start with a first digit 5, it only requires a 20%increase to get from $5,000 to $6,000, and that isachieved in one year.
When the Dow reaches $9,000, it takes only an 11%increase and just seven months to reach the $10,000mark. This again has first digit 1, so it will take anotherdoubling (and four more years) to get back to first digit 2.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Benford’s Law and Tax Fraud (Nigrini, 1992)
Most people can’t fake data convincingly.
Many states (including California) and the IRS now usefraud-detection software based on Benford’s Law.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Benford’s Law and Tax Fraud (Nigrini, 1992)
Most people can’t fake data convincingly.
Many states (including California) and the IRS now usefraud-detection software based on Benford’s Law.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Benford’s Law and Tax Fraud (Nigrini, 1992)
Most people can’t fake data convincingly.
Many states (including California) and the IRS now usefraud-detection software based on Benford’s Law.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
True Life Tale
Manager from Arizona State Treasurer wasembezzling funds.
Most amounts were below $100,000 (criticalthreshold for checks that would require morescrutiny).Over 90% of the checks had a first digit 7, 8, or 9.(Trying to get close to the threshold without goingover — artificially changes the data and so breaks fitwith Benford’s law.)
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
True Life Tale
Manager from Arizona State Treasurer wasembezzling funds.Most amounts were below $100,000 (criticalthreshold for checks that would require morescrutiny).
Over 90% of the checks had a first digit 7, 8, or 9.(Trying to get close to the threshold without goingover — artificially changes the data and so breaks fitwith Benford’s law.)
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
True Life Tale
Manager from Arizona State Treasurer wasembezzling funds.Most amounts were below $100,000 (criticalthreshold for checks that would require morescrutiny).Over 90% of the checks had a first digit 7, 8, or 9.(Trying to get close to the threshold without goingover — artificially changes the data and so breaks fitwith Benford’s law.)
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
True Life Tale
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
What types of sequences are Benford?
Real-world data can be a good fit or not, depending onthe type of data. Data that is a good fit is “suitablyrandom” — comes in many different scales, and is a largeand randomly distributed data set, with no artificial orexternal limitations on the range of the numbers.
Some numerical sequences are clearly not Benford:1,2,3,4,5,6,7, . . . (uniform distribution)
1,10,100,1000, . . . (first digit is always 1)
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
What types of sequences are Benford?
Real-world data can be a good fit or not, depending onthe type of data. Data that is a good fit is “suitablyrandom” — comes in many different scales, and is a largeand randomly distributed data set, with no artificial orexternal limitations on the range of the numbers.
Some numerical sequences are clearly not Benford:
1,2,3,4,5,6,7, . . . (uniform distribution)
1,10,100,1000, . . . (first digit is always 1)
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
What types of sequences are Benford?
Real-world data can be a good fit or not, depending onthe type of data. Data that is a good fit is “suitablyrandom” — comes in many different scales, and is a largeand randomly distributed data set, with no artificial orexternal limitations on the range of the numbers.
Some numerical sequences are clearly not Benford:1,2,3,4,5,6,7, . . . (uniform distribution)
1,10,100,1000, . . . (first digit is always 1)
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
What types of sequences are Benford?
Real-world data can be a good fit or not, depending onthe type of data. Data that is a good fit is “suitablyrandom” — comes in many different scales, and is a largeand randomly distributed data set, with no artificial orexternal limitations on the range of the numbers.
Some numerical sequences are clearly not Benford:1,2,3,4,5,6,7, . . . (uniform distribution)
1,10,100,1000, . . . (first digit is always 1)
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Powers of Two
1,2,4,8,16,32,64, . . .
tn = 2n
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Powers of Two
1,2,4,8,16,32,64, . . .
tn = 2n
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Powers of Two
1,2,4,8,16,32,64, . . .
tn = 2n
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Powers of Two
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Fibonacci Numbers
1,1,2,3,5,8,13,21, . . .
Fn+1 = Fn + Fn−1
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Fibonacci Numbers
1,1,2,3,5,8,13,21, . . .
Fn+1 = Fn + Fn−1
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Fibonacci Numbers
1,1,2,3,5,8,13,21, . . .
Fn+1 = Fn + Fn−1
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Fibonacci Numbers
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Mantissa Function
DefinitionDefine the mantissa function
M : R+ → [1,10)
x 7→ r
where r is the unique number in [1,10) such thatx = r × 10n for some n ∈ Z.
In other words, write the number in scientific notation.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Mantissa Function
DefinitionDefine the mantissa function
M : R+ → [1,10)
x 7→ r
where r is the unique number in [1,10) such thatx = r × 10n for some n ∈ Z.
In other words, write the number in scientific notation.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Mantissa Function
DefinitionDefine the mantissa function
M : R+ → [1,10)
x 7→ r
where r is the unique number in [1,10) such thatx = r × 10n for some n ∈ Z.
ExamplesM(9,000,001) = 9.000001.M(0.01247) = 1.247.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford if {yi} is equidistributed mod 1,where yi = log10 xi .
Proof:x = M(x) · 10k for some k ∈ Z.First digit of x is d iff d ≤ M(x) < d + 1.log10 d ≤ y < log10(d + 1), wherey = log10(M(x)) = log10 x mod 1.If the distribution is uniform (mod 1), then theprobability y is in this range is
log10(d+1)−log10(d) = log10
(d + 1
d
)= log10
(1 +
1d
).
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford if {yi} is equidistributed mod 1,where yi = log10 xi .
Proof:x = M(x) · 10k for some k ∈ Z.
First digit of x is d iff d ≤ M(x) < d + 1.log10 d ≤ y < log10(d + 1), wherey = log10(M(x)) = log10 x mod 1.If the distribution is uniform (mod 1), then theprobability y is in this range is
log10(d+1)−log10(d) = log10
(d + 1
d
)= log10
(1 +
1d
).
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford if {yi} is equidistributed mod 1,where yi = log10 xi .
Proof:x = M(x) · 10k for some k ∈ Z.First digit of x is d iff d ≤ M(x) < d + 1.
log10 d ≤ y < log10(d + 1), wherey = log10(M(x)) = log10 x mod 1.If the distribution is uniform (mod 1), then theprobability y is in this range is
log10(d+1)−log10(d) = log10
(d + 1
d
)= log10
(1 +
1d
).
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford if {yi} is equidistributed mod 1,where yi = log10 xi .
Proof:x = M(x) · 10k for some k ∈ Z.First digit of x is d iff d ≤ M(x) < d + 1.log10 d ≤ y < log10(d + 1), wherey = log10(M(x)) = log10 x mod 1.
If the distribution is uniform (mod 1), then theprobability y is in this range is
log10(d+1)−log10(d) = log10
(d + 1
d
)= log10
(1 +
1d
).
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford if {yi} is equidistributed mod 1,where yi = log10 xi .
Proof:x = M(x) · 10k for some k ∈ Z.First digit of x is d iff d ≤ M(x) < d + 1.log10 d ≤ y < log10(d + 1), wherey = log10(M(x)) = log10 x mod 1.If the distribution is uniform (mod 1), then theprobability y is in this range is
log10(d+1)−log10(d) = log10
(d + 1
d
)= log10
(1 +
1d
).
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford if {yi} is equidistributed mod 1,where yi = log10 xi .
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford if {yi} is equidistributed mod 1,where yi = log10 xi .
0 1
1 102
log 2 ! log 10
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford if {yi} is equidistributed mod 1,where yi = log10 xi .
Kronecker-Weyl TheoremIf β 6∈ Q then nβ mod 1 is equidistributed.(Thus if log10 α 6∈ Q, then αn is Benford.)
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Powers of 2
TheoremThe sequence {2n} for n ≥ 0 is Benford.
Proof:Consider the sequence of logarithms {n(log10 2)}.By the Kronecker-Weyl Theorem, this is uniform(mod 1) because log10 2 6∈ Q.Since the sequence of logarithms is uniformlydistributed (mod 1), the original sequence isBenford.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Powers of 2
TheoremThe sequence {2n} for n ≥ 0 is Benford.
Proof:Consider the sequence of logarithms {n(log10 2)}.By the Kronecker-Weyl Theorem, this is uniform(mod 1) because log10 2 6∈ Q.Since the sequence of logarithms is uniformlydistributed (mod 1), the original sequence isBenford.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Fibonacci Numbers
TheoremThe sequence {Fn} of Fibonacci numbers is Benford.
Heuristic Argument:Closed form for Fibonacci numbers:
Fn =1√5
[(1 +√
52
)n
−
(1−√
52
)n].
∣∣∣(1−√
52
)∣∣∣ < 1, so the leading digits are completely
determined by 1√5
(1+√
52
)n.
This sequence is Benford because log10
(1+√
52
)6∈ Q.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Fibonacci Numbers
TheoremThe sequence {Fn} of Fibonacci numbers is Benford.
Heuristic Argument:Closed form for Fibonacci numbers:
Fn =1√5
[(1 +√
52
)n
−
(1−√
52
)n].
∣∣∣(1−√
52
)∣∣∣ < 1, so the leading digits are completely
determined by 1√5
(1+√
52
)n.
This sequence is Benford because log10
(1+√
52
)6∈ Q.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Fibonacci Numbers
TheoremThe sequence {Fn} of Fibonacci numbers is Benford.
Heuristic Argument:Closed form for Fibonacci numbers:
Fn =1√5
[(1 +√
52
)n
−
(1−√
52
)n].
∣∣∣(1−√
52
)∣∣∣ < 1, so the leading digits are completely
determined by 1√5
(1+√
52
)n.
This sequence is Benford because log10
(1+√
52
)6∈ Q.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Fibonacci Numbers
TheoremThe sequence {Fn} of Fibonacci numbers is Benford.
Heuristic Argument:Closed form for Fibonacci numbers:
Fn =1√5
[(1 +√
52
)n
−
(1−√
52
)n].
∣∣∣(1−√
52
)∣∣∣ < 1, so the leading digits are completely
determined by 1√5
(1+√
52
)n.
This sequence is Benford because log10
(1+√
52
)6∈ Q.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Linear Recurrence Sequences
Consider the sequence {an} given by some initialconditions a0,a1, . . . ,ak−1 and then a recurrence relation
an+k = c1an+k−1 + c2an+k−2 + · · ·+ ckan,
with c1, c2, . . . , ck fixed real numbers.
Find the eigenvalues of the recurrence relation and orderthem so that |λ1| ≥ |λ2| ≥ · · · ≥ |λk |.
There exist numbers u1,u2, . . . ,uk (which depend on theinitial conditions) so that an = u1λ
n1 + u2λ
n2 + · · ·+ ukλ
nk .
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Linear Recurrence Sequences
Consider the sequence {an} given by some initialconditions a0,a1, . . . ,ak−1 and then a recurrence relation
an+k = c1an+k−1 + c2an+k−2 + · · ·+ ckan,
with c1, c2, . . . , ck fixed real numbers.
Find the eigenvalues of the recurrence relation and orderthem so that |λ1| ≥ |λ2| ≥ · · · ≥ |λk |.
There exist numbers u1,u2, . . . ,uk (which depend on theinitial conditions) so that an = u1λ
n1 + u2λ
n2 + · · ·+ ukλ
nk .
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Linear Recurrence Sequences
Consider the sequence {an} given by some initialconditions a0,a1, . . . ,ak−1 and then a recurrence relation
an+k = c1an+k−1 + c2an+k−2 + · · ·+ ckan,
with c1, c2, . . . , ck fixed real numbers.
Find the eigenvalues of the recurrence relation and orderthem so that |λ1| ≥ |λ2| ≥ · · · ≥ |λk |.
There exist numbers u1,u2, . . . ,uk (which depend on theinitial conditions) so that an = u1λ
n1 + u2λ
n2 + · · ·+ ukλ
nk .
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Linear Recurrence Sequences
TheoremWith a linear recurrence sequence as described, iflog10 |λ1| 6∈ Qand the initial conditions are such thatu1 6= 0, then the sequence {an} is Benford.
Sketch of Proof:
Rewrite the closed form as an = u1λn1
(1 +O
(kuλn
2λn
1
))where u = maxi |ui |+ 1.Some clever algebra using our assumptions to findthat yn = log10(an) = n log10 λ1 + log10 u1 +O(βn) forsome appropriate β.Show in the limit the error term affects a vanishinglysmall portion of the distribution.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Linear Recurrence Sequences
TheoremWith a linear recurrence sequence as described, iflog10 |λ1| 6∈ Qand the initial conditions are such thatu1 6= 0, then the sequence {an} is Benford.
Sketch of Proof:
Rewrite the closed form as an = u1λn1
(1 +O
(kuλn
2λn
1
))where u = maxi |ui |+ 1.
Some clever algebra using our assumptions to findthat yn = log10(an) = n log10 λ1 + log10 u1 +O(βn) forsome appropriate β.Show in the limit the error term affects a vanishinglysmall portion of the distribution.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Linear Recurrence Sequences
TheoremWith a linear recurrence sequence as described, iflog10 |λ1| 6∈ Qand the initial conditions are such thatu1 6= 0, then the sequence {an} is Benford.
Sketch of Proof:
Rewrite the closed form as an = u1λn1
(1 +O
(kuλn
2λn
1
))where u = maxi |ui |+ 1.Some clever algebra using our assumptions to findthat yn = log10(an) = n log10 λ1 + log10 u1 +O(βn) forsome appropriate β.
Show in the limit the error term affects a vanishinglysmall portion of the distribution.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Linear Recurrence Sequences
TheoremWith a linear recurrence sequence as described, iflog10 |λ1| 6∈ Qand the initial conditions are such thatu1 6= 0, then the sequence {an} is Benford.
Sketch of Proof:
Rewrite the closed form as an = u1λn1
(1 +O
(kuλn
2λn
1
))where u = maxi |ui |+ 1.Some clever algebra using our assumptions to findthat yn = log10(an) = n log10 λ1 + log10 u1 +O(βn) forsome appropriate β.Show in the limit the error term affects a vanishinglysmall portion of the distribution.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Elliptic Divisibility Sequences
DefinitionAn integral divisibility sequence is a sequence of integers{un} satisfying
un | um whenever n | m.
An elliptic divisibility sequence is an integral divisibilitysequence which satisfies the following recurrence relationfor all m ≥ n ≥ 1:
um+num−nu21
= um+1um−1u2n − un+1un−1u2
m.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Boring Elliptic Divisibility Sequences
The sequences of integers, where un = n.
The sequence 0,1,−1,0,1,−1, . . ..
The sequence1,3,8,21,55,144,377,987,2584,6765, . . . (this isevery-other Fibonacci number).
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Boring Elliptic Divisibility Sequences
The sequences of integers, where un = n.
The sequence 0,1,−1,0,1,−1, . . ..
The sequence1,3,8,21,55,144,377,987,2584,6765, . . . (this isevery-other Fibonacci number).
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Boring Elliptic Divisibility Sequences
The sequences of integers, where un = n.
The sequence 0,1,−1,0,1,−1, . . ..
The sequence1,3,8,21,55,144,377,987,2584,6765, . . . (this isevery-other Fibonacci number).
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Not-So-Boring Elliptic Divisibility Sequences
The sequences which begins0,1,1,−1,1,2,−1,−3,−5,7,−4,−28,29,59,129,−314,−65,1529,−3689,−8209,−16264,833313,113689,−620297,2382785,7869898,7001471,−126742987,−398035821,168705471, . . .(This is sequence A006769 in the On-LineEncyclopedia of Integer Sequences.)
The sequence which begins1,1,−3,11,38,249,−2357,8767,496036,−3769372,−299154043,−12064147359, . . ..
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Not-So-Boring Elliptic Divisibility Sequences
The sequences which begins0,1,1,−1,1,2,−1,−3,−5,7,−4,−28,29,59,129,−314,−65,1529,−3689,−8209,−16264,833313,113689,−620297,2382785,7869898,7001471,−126742987,−398035821,168705471, . . .(This is sequence A006769 in the On-LineEncyclopedia of Integer Sequences.)
The sequence which begins1,1,−3,11,38,249,−2357,8767,496036,−3769372,−299154043,−12064147359, . . ..
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Why We Like Elliptic Divisibility Sequences
Special case of Somos sequences, which areinteresting and an active area of research.
Connection to elliptic curves, also an active area ofresearch.
Elliptic curve
E over Qwith rational point
P on E
↔ EDS: denominators
of the sequenceof points {P,2P,3P, . . .}
Applications to elliptic curve cryptography.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Why We Like Elliptic Divisibility Sequences
Special case of Somos sequences, which areinteresting and an active area of research.
Connection to elliptic curves, also an active area ofresearch.
Elliptic curve
E over Qwith rational point
P on E
↔ EDS: denominators
of the sequenceof points {P,2P,3P, . . .}
Applications to elliptic curve cryptography.
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History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Why We Like Elliptic Divisibility Sequences
Special case of Somos sequences, which areinteresting and an active area of research.
Connection to elliptic curves, also an active area ofresearch.
Elliptic curve
E over Qwith rational point
P on E
↔ EDS: denominators
of the sequenceof points {P,2P,3P, . . .}
Applications to elliptic curve cryptography.66
History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Elliptic Divisibility Sequences are Benford?
67
History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Elliptic Divisibility Sequences are Benford?
68
History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Elliptic Divisibility Sequences are Benford?
69
History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Heuristic Argument
It’s well-known that elliptic divisibility sequencessatisfy a growth condition like un ≈ cn2 where theconstant c depends on the arithmetic height of thepoint P and on the curve E .
Weyl’s theorem tells us that {nkα} is uniformdistributed (mod 1) iff α 6∈ Q.
So we should at least be able to conclude that a givenEDS is Benford base b for almost every b.
But: The argument with the big-O error terms isdelicate, and not enough is known in the case of EDS.
70
History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Heuristic Argument
It’s well-known that elliptic divisibility sequencessatisfy a growth condition like un ≈ cn2 where theconstant c depends on the arithmetic height of thepoint P and on the curve E .
Weyl’s theorem tells us that {nkα} is uniformdistributed (mod 1) iff α 6∈ Q.
So we should at least be able to conclude that a givenEDS is Benford base b for almost every b.
But: The argument with the big-O error terms isdelicate, and not enough is known in the case of EDS.
71
History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Heuristic Argument
It’s well-known that elliptic divisibility sequencessatisfy a growth condition like un ≈ cn2 where theconstant c depends on the arithmetic height of thepoint P and on the curve E .
Weyl’s theorem tells us that {nkα} is uniformdistributed (mod 1) iff α 6∈ Q.
So we should at least be able to conclude that a givenEDS is Benford base b for almost every b.
But: The argument with the big-O error terms isdelicate, and not enough is known in the case of EDS.
72
History Applications Benford and Integer Sequences Formalism Benford and Recurrence Relations
Heuristic Argument
It’s well-known that elliptic divisibility sequencessatisfy a growth condition like un ≈ cn2 where theconstant c depends on the arithmetic height of thepoint P and on the curve E .
Weyl’s theorem tells us that {nkα} is uniformdistributed (mod 1) iff α 6∈ Q.
So we should at least be able to conclude that a givenEDS is Benford base b for almost every b.
But: The argument with the big-O error terms isdelicate, and not enough is known in the case of EDS.
73