EDITED A097345

[Max Alekseyev]

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

On Jan 25, 2008 12:13 PM, Maximilian Hasler <maximilian.hasler at gmail.com> wrote:
> quite interesting : in addition to the link to Mathar's PDF, the
> formula and most terms of this sequence also were wrong,
> but by correcting it, it seems that again A097345(n)=A097344(n) except
> for very few cases
> (n=59, no other until n=1519,... I suspected a bug in my beta version of PARI,
> but maple confirmed the first value, then crashed  before reaching the
> second...)
>
> Maximilian
>
> %I A097345
> %S A097345 1, 5, 29, 103, 887, 1517, 18239, 63253, 332839, 118127,
> 2331085, 4222975, 100309579, 184649263,
> %T A097345 1710440723, 6372905521, 202804884977, 381240382217,
> 13667257415003, 25872280345103,
> %U A097345 49119954154463, 93501887462903, 4103348710010689,
> 7846225754967739, 75162749477272151
> %N A097345 Numerators of the partial sums of the binomial transform of 1/(n+1).
> %C A097345 Is this identical to A097344? - Aaron Gulliver, Jul 19 2007
> %C A097345 From n=9 on, the formula a(n)=A003418(n+1)*sum{k=0..n,
> (2^(k+1)-1)/(k+1)} is no more correct.
> The least n for which a(n) is different from A097344(n) is n=59, then
> they agree again until n=1519.
> - M. Hasler, Jan 25 2008
> %H A097345 R. J. Mathar, <a
> href="http://www.research.att.com/~njas/sequences/a097345.pdf">
>                Notes on an attempt to prove that A097344 and A097345
> are identical</a>
> %o A097345 (PARI) A097345(n) = numerator(sum(k=0,n,(2^(k+1)-1)/(k+1)))
> %Y A097345 Adjacent sequences: A097342 A097343 A097344 this_sequence
> A097346 A097347 A097348
> %Y A097345 Sequence in context: A050409 A111937 A097344 this_sequence
> A034700 A057721 A085151
> %K A097345 easy,nonn,frac
> %O A097345 0,2
> %A A097345 Paul Barry (pbarry(AT)wit.ie), Aug 06 2004
> %E A097345 Edited and corrected by Maximilian F. Hasler
> (Maximilian.Hasler(AT)gmail.com), Jan 25 2008
>




Dear Seqfans,   I just came across a draft of a list
of open problems from the Western Number Theory Meeting, Asilomar, 1988.

This suggested two possibly new sequences - maybe

1.
Let T(n) := (p, p+2) denote the n-th pair of twin primes.
Let S(n) = 2p+2.

Then a(n) = number of ways of writing S(n) as S(i) + S(j) with i <= j < m.
Sequence begins 0,0,1,1,...

a(4) = 1 because s(4) = 17+19 = (5+7) + (11+13) = S(2)+S(3).


2. 

Consider A001043:
%I A001043 M3780 N0968
%S A001043 5,8,12,18,24,30,36,42,52,60,68,78,84,90,100,112,120,128,138,144,152,
%T A001043 162,172,186,198,204,210,216,222,240,258,268,276,288,300,308,320,330,
%U A001043 340,352,360,372,384,390,396,410,434,450,456,462,472,480,492,508,520
%N A001043 Numbers that are the sum of 2 successive primes.

The new sequence would be:
Let m = A001043(n). Then a(n) = number of ways of writing
m = A001043(i) + A001043(j) with i <= j < n.

This is not always possible, and we have:

%I A134650
%S A134650 5,8,12,18,52,100
%N A134650 Numbers n such that n is the sum of two consecutive primes but is not the sum of tw
o sums of consecutive primes.
%C A134650 Numbers in A001043 but not in A134651.
%C A134650 Conjectured to be finite, may be complete.
%O A134650 1,1
%D A134650 R. K. Guy, ed., Unsolved Problems, Western Number Theory Meeting, Las Vegas, 1988.
%K A134650 nonn,fini,more
%A A134650 njas, Jan 25 2008

Could someone work out these sequences?

Neil