Surprises in
Experimental Mathematics
Michael I. Shamos
School of Computer Science
Carnegie Mellon University
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Mathematical Discovery
• Where do theorems come from?
• Theorem easy to conjecture, proof is hard
– Fermat’s last theorem, four-color theorem
• Theorem surprising, but results from a logical
investigation
– Fundamental theorem of algebra
• Theorem difficult to invent, straightforward to prove
– Hadamard three-circles theorem
– Where do these come from?
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Hadamard Three-Circles Theorem
• If f(z) is holomorphic (complex differentiable) on the annulus
a  z  b centered at the origin and M (r )  sup f ( z )
z r
• then
b
b


log
log
M
(
r
)

log



 log M (a)
a
r


r

  log  log M (b)
a

b
r
a
•
•
M (b)
MCGILL UNIVERSITY
M (a)
•
How was this
theorem ever
conjectured?
M (r )
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Outline
• The problem

1
– Closed-form expression for  (3)   3  1.2020569031...
k 1 k
• The approach
– Build a catalog of real-valued expressions indexed by first 20
digits
– Equivalent expressions will “collide”
– Look up 1.20205690315959428539
• The discoveries
– The Partial Sum Theorem
– Overcounting functions
– How many ways can n be expressed as an integer power k j ?

1
– Expression for  2 k
k 1 2
– ...
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
The Problem
• Closed form expressions for values of the zeta

1
function
 (s) 
k
k 1
s
• Euler found an expression for all even values of s:
 (1) s 1 22 s 1 B2 s  2 s
 
 (2s)  
(2s)!


1 n 1  n  1

 Bk ;
Bn  

n  1 k 0  k 
  (2s )
B1  
1
2
2s
is a rational multiple of 
• No expression is known for even a single odd value,
e.g.  (3)
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
The Catalog
• Some values of  (n):
2
4
176411 20
 (2) 
;  (4) 
;  (20) 
6
90
1531329465 290625
176411 20
1.0000009539620338727941   (20) 
1531329465 290625
1.0823232337111381915160   (4) 
4
90
1
1.2020224917616113171527 
2Catalan  1
1.2020569031595942853997   (3)  ?
1.6449340668482264364724   (2) 
2
6
(1) k
0.91596559417721901505  Catalan  
2
k 0 (2k  1)

MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Other Catalogs
• Sloane’s Encyclopedia of Integer Sequences
– Terrific, but for integer sequences, not reals
• Plouffe’s Inverter
– Huge (215 million entries), but not “natural”
expressions from actual mathematical work
• Simon Fraser Inverse Symbolic Calculator
– 50 million constants
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Discovery A

 (k )
k 1
2k

 2

1

p
2
p prime
where
 (k ) is the number of primes k
• Is this a coincidence?
• Why the factor of 2?
• Is there a general principle at work?
– In fact,

 (k )
k 1
ak

MCGILL UNIVERSITY

a
1


a  1 p prime a p
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Observation
 (k ) is a partial sum function, i.e.,
 (k ) 
1
p prime
pk
More generally,  (k ) is the partial sum function of the
indicator function of the property “primeness”:
 (k ) 
1, j prime
where I prime ( j )  
0, otherwise
k
 I prime( j ),
j 1
So

 (k )
k 1
2k


 2
1


p
p prime 2
  (k ) f (k )
k 1


1
can be rewritten as:
p 1
p prime 2
 g (k )
p prime


I
k 1
prime
(k ) g (k )
where f and g are “related”
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
The Partial Sum Theorem
(New)
• Given a sequence S of complex numbers s(k), let
t ( n) 
n
 s(k )
be the sequence of partial sums of S.
k 1
• Given a function f, if certain convergence criteria are satisfied,
then
PARTIAL
TAILS OF f ( k )

 t (k ) f (k )

k 1

 s( j ) g ( j )
j 1
PARTIAL
SUMS OF s( j )
where
g ( j) 

 f (k )
k j
(the partial tails of f) is a transform of f independent of s & t
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Partial Sum Functions
• Many sequences are partial sum functions:
H ( n) 
H
(s)
( n) 
n
1

k 1 k
n
1
k
k 1
the harmonic function
n

1

n
k 1 2
n 2  4n  6
6

2n
k2

k
k 1 2
1 2
n
log (n) 
generalized harmonic function
s
n
n
 log k
k 1
• Actually, every sequence is the partial sum function of some other
sequence
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Some Partial Sum Transforms

f (k )  g ( j )   f (k )

f (k )
g ( j)
k j

( j ,1) 

e 1 

(
j
)


1
ak
1
(a  1)a j 1
1
k2  k
1
j
k 2  k 1
(k  2)!
1
k3  k
1
2 j ( j  1)
sin k
ak
( 1) k
k 2 1
(1) j
2 j ( j  1)
1
k ak
n
1

i 0 (k  i )
1
k!
1  n

  (1)i S1 (n, i) j i 1 
n  1  i 1

MCGILL UNIVERSITY
1
SEPTEMBER 21, 2007
1
 2k 
 
k 
j
( j  1)!
sin( 1  j )  a sin j
a j 1 1  a 2  2 a cos 1


1
a


  , j, 0 
 ( j )
4 j ( j  ½ )
F 1, j  1, j  ½, ¼ 
2 1
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Some Partial Sum Transforms


k 1
j 1
1
t
(
k
)

ak
k 1
k 2  k 1
t (k )

(k  2)!
k 1

(1) k
t (k ) 2

k 1
k 2
(1) j
s( j )

2 j ( j  1)
j 2
MCGILL UNIVERSITY
j 1

 s( j )
j 1

1
t
(
k
)

 2k 
k 1
 
k
SEPTEMBER 21, 2007
sin( 1  j )  a sin j
a j 1 1  a 2  2 a cos1


1

s
(
j
)

 , j, 0 

a

j 1
1
t (k ) k

ka
k 1
1
j
( j  1)!


s( j )  n

  (1)i S1 (n, i) j i 1 

j 1 n  1  i 1


 s( j )
sin k
t (k ) k

a
k 1





1
s
(
j
)

2 j ( j  1)
j 2

1
t
(
k
)

k!
k 1
1
s( j )

j
j 1
1
t (k ) 3

k k
k 2

( j,1) 


s
(
j
)
e
1




(
j
)
j 1





t (k )

k 1 i  0 ( k  i )
j 1
1
s( j )

(a  1)a j 1
j 1
1
t (k ) 2

k k
k 1
n
k 1




 t (k ) f (k )   s( j ) g ( j )
 t (k ) f (k )   s( j ) g ( j )


s ( j )  ( j )

2 F1 1, j  1, j  ½, ¼ 
j
4

(
j

½
)
j 1

COPYRIGHT © 2007 MICHAEL I. SHAMOS
Partial Sum Theorem (Proof)
• Consider the upper triangular matrix mi , j  s (i ) f ( j ),
 s (1) f (1) s (1) f (2) s (1) f (3) s (1) f (4)

s (2) f (2) s (2) f (3) s (2) f (4)
 0
 0
0
s (3) f (3) s (3) f (4)

0
0
s (4) f (4)
 0
 0
0
0
0

 0
0 HEADS 0OF s
0

= t(i)0
0
0
0








Column sums are
MCGILL UNIVERSITY
 t (k ) f (k )
SEPTEMBER 21, 2007
s (1) f (5)
s (2) f (5)
s (3) f (5)
s (4) f (5)
s (5) f (5)
0
0

ij:
... 

TAILS OF f
...  = g(j)
... 
 Row
...  sums
...  are
s( j ) g ( j )

... 

... 



COPYRIGHT © 2007 MICHAEL I. SHAMOS
The Convergence Criteria

 t (k ) f (k )

k 1

 s (k ) g (k )
iff
k 1
1. All sums g(k) converge

2.
 s(k ) g (k ) converges; and
k 1
3. lim t (n) g (n)  0
n 
Proof: By Markoff’s theorem on convergence of double series
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Further Applications
• The number of perfect n th powers  k is n k 
• The number of positive integers powers of a  k is log a k 
• Therefore, by inspection,
log a k  
 2

k 2
k k
1


n
n a positive

1
1


j
a
a 1
j 1
*
power of a
k

k k

k 1
m
2

2 (k )


3
k 2 k  k


n a perfect
m th power

p prime
(2k  1) (k )

2
2
k  2 k k  1

MCGILL UNIVERSITY
1

n


j 1
1
p ( p  1)

*

p prime
1
  ( m)
jm
*
k (k )


(
k

1
)!
k 2

*
1

2
p (Old)
SEPTEMBER 21, 2007

 (k )
k 1
k


p prime
1
p!
*
log  (2k ) 
COPYRIGHT © 2007 MICHAEL I. SHAMOS
The Inverse Transform

 t (k ) f (k )

k 1

 s( j ) g ( j )
j 1
• Given g(j), how can we find f(k)?
• Since g is a sum of f’s, f is the sequence of finite differences of g :

 f (k )
 g ( j)
k j

 f (k )
 g ( j  1)
k  j 1
• Subtracting,
f ( j )  g ( j )  g ( j  1)
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Some Inverse Transforms
g ( j) 
g ( j )  f (k )  g (k )  g (k  1)
f (k )
1
ja
(k  1) a  k a
k a (k  1) a
1
aj
a 1
a k 1
1
j  j!
1
j 1
1
(k  1)( k  2)
1
j! j!
k 2  2k
(k  1)!(k  1)!
(1) j
j ( j  1)
2(1) k
k (k  2)
1
jaj
a  k  ak
k (k  1)a k 1
1
j!
1
Fj
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
k
(k  1)!
k 2  k 1
k (k  1)(k  1)!
Fk 1
Fk Fk 1
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Some Inverse Transforms

 s( j ) g ( j )

j 1

1
s
(
j
)

ja
j 1

1
s( j ) j

a
j 1


 t (k ) f (k )
k 1
 s( j ) g ( j )
(k  1) a  k a
t (k ) a

k (k  1) a
k 1

s( j )

j 1 j  j!


2(1) k t (k )

k 1 k (k  2)
s( j )

j 1 j! j!


s( j )

j
j 1 j a

s( j )

j 1 F j
MCGILL UNIVERSITY
k 1

s( j )

j!
j 1
a 1
t (k ) k 1

a
k 1
(1) j
s( j )

j ( j  1)
j 1
 t (k ) f (k )



t (k )

k 1 ( k  1)( k  2)

j 1

1
s( j )

j 1
j 1


SEPTEMBER 21, 2007
 t (k )
k 1
k
(k  1)!
k 2  k 1
t (k )

k (k  1)( k  1)!
k 1

k 2  2k
t (k )

(k  1)!(k  1)!
k 1

a  k  ak

 t (k ) k (k  1)a
k 1

 t (k )
k 1
k 1
Fk 1
Fk Fk 1
COPYRIGHT © 2007 MICHAEL I. SHAMOS
The Partial Integral Theorem *
• Given a function s(x), let t(y) be the “left integral” of s :
y
t ( y) 
 s( x) dx
0
• Given a function f(y), if certain convergence criteria are satisfied,
RIGHT INTEGRAL
OF f ( y )
then

 t ( y) f ( y) dy


0
 s( x) g ( x) dx
0
LEFT INTEGRAL
OF s(x)
where

g ( x) 
 f ( y) dx
x
(the right integral of f) is a transform of f independent of s & t
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Example
1
1
;
f
(
x
)

x2 1
( x  1) 2
s ( x) 

y
t ( y)   s( x) dx  arctan y ;
g ( x) 
0
x

Therefore,

f ( y) dy 
arctan x
0 ( x  1)2 dx 
1
x 1

dx
0 ( x  1)( x 2  1)
Note: if s(x) is a probability density
then t(y) is its cumulative distribution function
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Example
Consider the special case in which
f ( x)  e
This implies
x
f ( x)  g ( x)

.
So
 t ( x) e dx 
x
0


0
e x
dx 
2
x 1


e  x arctan x dx 
x
s
(
x
)
e
dx

0
 cos1
2
0

 ci(1) sin 1  si (1) cos1
1
0.8
e x
x2 1
0.6
e  x arctan x
0.4
0.2
1
MCGILL UNIVERSITY
2
3
4
SEPTEMBER 21, 2007
5
6
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Example
1
s ( x) 
x2 1
; t ( x)  arcsinh x ;
sinh x 
1 x x

e e 
2

x
e
 arcsinh x dx  ???
Mathematica gives up.
0


0
e x
x 1
2
dx 

2
H 0 (1)  Y0 (1)
 0.7546100258
Mathematica has no problem
Struve function
Bessel function
What about Risch’s theorem? Risch, R. “The Solution of the Problem of
Integration in Finite Terms.” Bull. Amer. Math. Soc., 1-76, 605-608, 1970.
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Discovery B


1
  (2)   (3) 

3
j 1 k 1 (k  j )

•

1

3 must exceed
j 1 k 1 ( k  j )

 1 1
 2  3

k 
k 1  k

1
  (3)  1 ,

3
k 2 k
but by how much?
• Is there a general principle at work?
– In fact,


1
  (s  1)   (s)

s
j 1 k 1 (k  j )
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Overcounting Functions
Let S 


and
f
(
k
)
S


k 1


 f ( g (k , j))
,
j 1 k 1
where g (k , j ) ranges over the natural numbers
1. Every term of S+ occurs at least once in S.
2. In general, S+ overcounts S, since some terms of S occur many
times in S+
3. If Kg(k) is the number of times f(k) is included in S+, then
S




 f ( g (k , j))
j 1 k 1


K
k 1
g
(k ) f (k )
where Kg(k) depends only on g and not on f .
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Examples
• Let g(k, j) = k + j . How many ordered pairs (k, j) of natural
numbers give k + j = n?
Answer: Kk+j (n) = n - 1
• Therefore, by inspection,




1


s
j 1 k 1 (k  j )

j 1 k 1

 (k  j )
2k  j


1
 1
 s 1  s    ( s  1)   ( s)

n 
n 1  n
n 1
2
  n  ( n) 
8
n 1 2
1


j 1 k 1 (k  j )!
MCGILL UNIVERSITY
n 1


s
n 1 n


n 1
 e  e  1  1

n 1 n !

SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Examples
• Let g(k, j) = k • j . How many ordered pairs (k, j) give k • j = n?
Answer: Kk•j (n) = d(n), the number of divisors of n .
• Therefore, by inspection






1


s
j 1 k 1 (k  j )
1


k j
j 1 k 1 a
d (n)
2


( s)

s
n 1 n


1
d ( n)



j
n
a

1
a
j 1
n 1
1


j 1 k 1 (k  j )!
MCGILL UNIVERSITY


d (n)
 2.4810610197 907626979 ...

n 1 n !
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Enumerating Non-Trivial Powers
• Let g(k, j) = k j. How many ordered pairs (k, j) give k j = n?
Or, how many ways K(n) can n be expressed as a positive
integral power of a positive integer?
• Let n  p1 1 p2 2 ... be the prime factorization of n
e
e
• n can be a non-trivial power of an integer > 1 iff
G(n) = gcd(e1, e2, . . . ) exceeds 1; otherwise K(n) = 1.


• Suppose b > 1 divides G(n). Then n  p1 1 p2 2 ... b ,
where each of the ei /b is a natural number, so n is the b th power
of a natural number
e /b
e /b
• Suppose c > 1 does not divide G(n). Then at least one of the
exponents ei /c is not a natural number and n is not the c th power
of a natural number. Therefore,
K k j (n)  d (gcd( exponents of the prime factorizat ion of n )) *
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
A Remarkable Series
• Let S 


1
  ( s)  1 .

s
k 2 k

1
S   j s 
j 1 k  2 (k )

S  S 
 K

n 1
k
j
( n )


n
n2
1 

k
 js
k  2 j 1
(n)  1k  s 
1
1
1


k s 1 k 2s  k s k s


Then
1
  s

k 2 k  1

k

j
  (n s) 1
n 1
( n) k  s
S S 
yields

the “overcounting” function

n 1

 ( n ) n
n2
d (G (n))  1


n
n2


(Old)
s


1

2s
 ns
n2 n

1
 1 *

2
n2 n  n
1
1 1
2
1
1
1
1
1
3
  






 ...
4
8 9 16 25 27 32 36 49 64
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Goldbach’s Theorem
• In 1729, Christian Goldbach proved that

1
1 1 1 1
1
1
1    


 ...
q 1
3 7 8 15 24 26

1

q 1
q a non trivial
integerpower
q a non trivial
integerpower
1 

1



k
q a non trivial k 1 q
(k )
1

k
k 1

integerpower
1
1 1
2
1
1
1
1
1
3
  






 ...
4
8 9 16 25 27 32 36 49 64
42
43
82
162
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
3
256
44 . . .
83
163
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Discovery C

q a non trivial
integerpower
1
1 
q 1
What is

q a non trivial
integerpower

q a non trivial
integerpower

q a non trivial
integerpower

  (k )  1
(Old)
k 1
1
???
q

1
    (k )  (k )  1  .87446436840494486669 ... *
q
k 2
k 1
1,

 (k )  (1) n , k a product of n distinct primes
0,
k has a repeated prime factor

1
 1
q(q  1)
MCGILL UNIVERSITY

  (k )  (k )  1
k 2
SEPTEMBER 21, 2007
*
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Discovery D


k 1
 (k ) *

2k
1
,

p
2

1
p prime
where  (k ) is the number of distinct prime factors of k
1
a k (a  1)
• Since the partial sum function of k
is
a 1
a 2 k 1  a k 1  a k  1


k 1
 (k )
a
k
1
  p

p prime a  1
MCGILL UNIVERSITY
 (k ) a k

a
k 1
2 k 1
SEPTEMBER 21, 2007
a
k 1
 a 1
k
,
a 1
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Discovery E
For c > 1 real and p prime,


1
c
pj
j 1
 
k 1
 (k p)
ck p 1
In particular,

2
j 1


j 1
2j

1
2

1
2j
 
k 1

 1 
k 1
MCGILL UNIVERSITY
 
k 1
 ( 2k )
2k  1
 ( 2k )
2 1
k
 ( 2k )
4k  1



k 1
 (2k  1)
22 k 1  1

 1 
k 1
log k
   k2  

k 1
2

1
2
k 0
 ( 2k )
4k  1
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
2k
Results
The counting function Kmax(n) of max(k, j) is 2n-1. So


 2
k 1 j 1




1
max( k , j )

2k  1
 3

k
2
k 1

1


max(
k
,
j
)!
k 1 j 1
1


3
max(
k
,
j
)
k 1 j 1
2k  1
 e 1

k
!
k 1

2k  1
2

  (3)

3
k
3
k 1

The counting function Klcm(n) of lcm(k, j) is d(n 2). So


1


s
lcm
(
k
,
j
)
k 1 j 1
MCGILL UNIVERSITY
d (k 2 )
 3 ( s)


a
k
 ( 2a )
k 1

SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
The First-Digit Phenomenon
• Given a random integer, what is the probability that its
leading digit is a 1?
• Answer: depends on the distribution from which k is
chosen
• If k is chosen uniformly in [1, n], then let p(d, n) be the
probability that the leading digit of k is d
• For n = 19+, 5/9 < p(1,n) < .579; 1/19 p(9,n) < 1/18
• For n = 9+, p(1,n) = 1/9; p(9,n) = 1/9
• The “average” is log10(1+1/d)
• {.301, .176, .124, .097, .079, .066, .058, .051, .046}
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Relative Digit Frequency (Benford’s Law)
0.35
0.3
log10(1+1/d)
0.25
0.2
Benford
0.15
0.1
0.05
0
1
2
MCGILL UNIVERSITY
3
4
5
6
SEPTEMBER 21, 2007
7
8
9
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Relative Digit Frequency
0.35
0.3
0.25
0.2
Benford
Catalog Frac
0.15
0.1
0.05
0
1, 0
2, 1
MCGILL UNIVERSITY
3, 2
4, 3
5, 4
6, 5
7, 6
SEPTEMBER 21, 2007
8, 7
9, 8
9
COPYRIGHT © 2007 MICHAEL I. SHAMOS
First-Digit Phenomenon
0.35
0.3
0.25
0.2
Benford
Log[11,1+1/d]
Catalog Frac
0.15
0.1
0.05
0
0, 1
1, 2
MCGILL UNIVERSITY
2, 3
3, 4
4, 5
5, 6
6, 7
SEPTEMBER 21, 2007
7, 8
8, 9
9
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Major Ideas
• For mathematicians:
– How to populate the catalog
– How to generalize from discoveries
• For computer scientists:
– Use in symbolic manipulation systems
• For data miners:
– How to mine the catalog, i.e. how to find new relations
• For statisticians:


0
0
– How to use the fact that  P( y) f ( y) dy 
 p( x) g ( x) dx
where P is the cumulative distribution of density p
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
A Parting Philosophy
“The object of mathematical rigor is to
sanction and legitimize the conquests of
intuition, and there was never any other
object for it.”
Jacques Hadamard
(as quoted by Borel in 1928)
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Q&A
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Results
 (k )

k
k 2

k
k 2
2
(k  1)
 (k )
2
(k  1)

  (k )



1

  (k  1) 
k 2 



k   (k )


(

1
)

1

  (k  1)
k 2



1


k
k  2 2  (k )
 (k )


k  2 k ( k )

 

2


1   (k )
 

2 k 1 k (2k  1)
k 1 2k  1

 (k )


1
 (2) (3)


1


  p( p  1) 
 (6)
p prime 

(k )  1 
k 2

d (k )

k  2 k ( k  1)
n! is the nearest integer to (1)
MCGILL UNIVERSITY

n 1

(n)
log n k
(2)   2
k
k 2
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Hypergometric Functions

(a) k (b) k z k
2 F1 a, b; c; z   
(c ) k
k!
k 0
where (a) k  a(a  1) . . . (a  k  1)  (a  k ) / (a)
2
•
F1 a, b; c; z 
 (c )
b 1
c b
a

t
(
1

t
)
(
1

tz
)
dt

(b)(c  b) 0
1
The 2 F1 a, b; c; z  are solutions of the hypergeometric differential equation:
z (1  z ) y  [c  (a  b  1) z ] y  a b y  0
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Partial Sum Theorem (Proof)
• Consider the upper triangular matrix mi , j  s (i ) f ( j ),
ij
• The sum of row i is

m
i, j
j 1

 s(i) f ( j )  s(i) g (i)
j i
• The sum of column j is

m
i 1
i, j
j
 f ( j ) s(i)  t ( j ) f ( j )
i 1
• The sum of the row sums equals the sum of the columns sums
precisely when the conditions of Markoff’s theorem are satisfied.
QED
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Correspondence with Plouffe
0.7
0.6
0.5
0.4
Catalog Int
Benford
Log[11,1+1/d]
Catalog Frac
0.3
0.2
0.1
0
0, 1
1, 2
MCGILL UNIVERSITY
2, 3
3, 4
4, 5
5, 6
6, 7
SEPTEMBER 21, 2007
7, 8
8, 9
9
COPYRIGHT © 2007 MICHAEL I. SHAMOS
First-Digit Phenomenon
SOURCE: SIMON PLOUFFE
MCGILL UNIVERSITY
SEPTEMBER 21, 2007
COPYRIGHT © 2007 MICHAEL I. SHAMOS
First Digit Frequency
0.7
0.6
0.5
0.4
Catalog Int
Benford
Catalog Frac
0.3
0.2
0.1
0
0, 1
1, 2
MCGILL UNIVERSITY
2, 3
3, 4
4, 5
5, 6
6, 7
SEPTEMBER 21, 2007
7, 8
8, 9
9
COPYRIGHT © 2007 MICHAEL I. SHAMOS
Correspondence with Plouffe
0.7
0.6
0.5
0.4
Catalog Int
0.3
0.2
0.1
0
0
1
MCGILL UNIVERSITY
2
3
4
5
SEPTEMBER 21, 2007
6
7
8
9
COPYRIGHT © 2007 MICHAEL I. SHAMOS