EDITED A097345

[Maximilian Hasler]

There's another simple formula : g(n) = sum( k=1,n+1, binomial(n+1,k)/k )

Also, it is easy to prove the formula I gave earlier,
f(n,m) = binomial transform of k -> 1/(k+m)
= m-th differences of f(n,0) = sum( k=1,n, binomial(n,k)/k^2 ),
f(n,1) = f(n) in Richard's notes.

Maximilian

On Jan 25, 2008 7:15 PM, Max Alekseyev <maxale at gmail.com> wrote:
> This is an alternative proof of the formula mentioned in Richard
> Mathar's note that
> g_n = (n+1) f_n
> where
> f_n = sum(k=0,n,binomial(n,k)/(k+1)^2)
> and
> g_n = sum(k=0,n,(2^(k+1)-1)/(k+1)).
>
> First off, it is easy to see that binomial(n,k)/(k+1) =
> binomial(n+1,k+1)/(n+1) and thus
> (n+1) f_n = sum(k=0,n,binomial(n+1,k+1)/(k+1)).
>
> Now all we need is to notice that
> sum(k=0,n,binomial(n+1,k+1)/(k+1)) is simply an integral of
> sum(k=0,n,binomial(n+1,k+1)*x^k) = ((x+1)^(n+1)-1)/x taken over x from 0 to 1;
> while
> sum(k=0,n,(2^(k+1)-1)/(k+1)) is an integral of
> sum(k=0,n,y^k) = (y^(n+1)-1)/(y-1) taken over y from 1 to 2.
>
> These integrals are equal as there is a simple substitution x=y-1 or
> y=x+1 transforming one into the other.
>
> Regards,
> Max




njas> ...
njas> 2.
njas>
njas> Consider A001043:
njas> %I A001043 M3780 N0968
njas> %S A001043
njas> 5,8,12,18,24,30,36,42,52,60,68,78,84,90,100,112,120,128,138,144,152,
njas> %T A001043
njas> 162,172,186,198,204,210,216,222,240,258,268,276,288,300,308,320,330,
njas> %U A001043
njas> 340,352,360,372,384,390,396,410,434,450,456,462,472,480,492,508,520
njas> %N A001043 Numbers that are the sum of 2 successive primes.
njas>
njas> The new sequence would be:
njas> Let m = A001043(n). Then a(n) = number of ways of writing
njas> m = A001043(i) + A001043(j) with i <= j < n.
njas>
njas> This is not always possible, and we have:
njas>
njas> %I A134650
njas> %S A134650 5,8,12,18,52,100
njas> %N A134650 Numbers n such that n is the sum of two consecutive primes but
njas> is not the sum of tw
njas> o sums of consecutive primes.
njas> %C A134650 Numbers in A001043 but not in A134651.
njas> %C A134650 Conjectured to be finite, may be complete.
njas> %O A134650 1,1
njas> %D A134650 R. K. Guy, ed., Unsolved Problems, Western Number Theory
njas> Meeting, Las Vegas, 1988.
njas> %K A134650 nonn,fini,more
njas> %A A134650 njas, Jan 25 2008
njas>...

js> From seqfan-owner at ext.jussieu.fr  Sat Jan 26 02:37:36 2008
js> Date: Fri, 25 Jan 2008 20:36:10 -0500 (EST)
js> Subject: Re: 2 or 3 possibly new sequences from 1998
js> From: sellersj at math.psu.edu
js> To: "N. J. A. Sloane" <njas at research.att.com>
js> Cc: rkg at cpsc.ucalgary.ca, seqfan at ext.jussieu.fr, sellersj at math.psu.edu
js> ...
js> Just calculated the second sequence below for the first 5000 primes and
js> obtained the following:
js> 
js> 5,8,12,18,52,100,946
js> 
js> So I only obtained one more term.
js> ...

Next term in A134650, after 946, is larger than 20100000 if existent.
Straight forward Maple implementation follows, just for reference.

Richard

isA001043 := proc(n)
end:

A001043 := proc(n) option remember ;
end:

isA134651 := proc(n)
end:

isA134650 := proc(n)
end:

for n from 1 do
od:




njas> From seqfan-owner at ext.jussieu.fr  Sat Jan 26 00:25:50 2008
njas> Date: Fri, 25 Jan 2008 18:25:17 -0500
njas> From: "N. J. A. Sloane" <njas at research.att.com>
njas> To: rkg at cpsc.ucalgary.ca, seqfan at ext.jussieu.fr
njas> Subject: 2 or 3 possibly new sequences from 1998
njas> Cc: njas at research.att.com
njas> 
njas> Dear Seqfans,   I just came across a draft of a list
njas> of open problems from the Western Number Theory Meeting, Asilomar, 1988.
njas> 
njas> This suggested two possibly new sequences - maybe
njas> someone would like to look into them?
njas> 
njas> 1.
njas> Let T(n) := (p, p+2) denote the n-th pair of twin primes.
njas> Let S(n) = 2p+2.
njas> 
njas> Then a(n) = number of ways of writing S(n) as S(i) + S(j) with i <= j < m.
njas> Sequence begins 0,0,1,1,...
njas> 
njas> a(4) = 1 because s(4) = 17+19 = (5+7) + (11+13) = S(2)+S(3).
njas> 
njas> ...

S(n) appears as A054735, p as A001359 in the OEIS. The a(n) starts

0,0,1,1,1,1,2,2,2,1,1,2,3,2,3,1,4,3,3,3,2,6,3,5,3,3,3,3,3,8,4,2,3,3,6,4,4,6,7,8,
3,6,3,9,8,7,7,5,8,4,1,6,6,3,7,1,6,6,4,8,1,5,5,8,9,11,10,6,8,16,13,9,12,6,7,8,4,
16,9,6,13,10,9,5,6,6,8,11,16,11,13,6,6,6,17,9,6,6,4,14,12,6,10,12,13,10,9,8,12,
12,13,12,11,16,9,17,13,4,8,6,

according to the highly-non-optimized Maple program which is attached.
Richard

A001359 := proc(n) option remember ;
end:

A054735 := proc(n) option remember ;
end:

A000001 := proc(n)
end:

for n from 1 to 120 do
od: