0,2

Numerators in the expansion of ln((1-x)/(1-2x)) / (1-x) are 0,1,5,29,.. - Paul Barry (pbarry(AT)wit.ie), Feb 09 2005

Is this identical to A097345? - Aaron Gulliver, Jul 19 2007. The answer turns out to be No - see A134652.

If the putative formula a(n)=A081528(n) sum{k=0..n, binomial(n, k)/(k+1)^2} were true, then this sequence coincides with A097345 according to Mathar's notes. However, the term n=9 in the binomial transform of 1/(n+1)^2 has the denominator 5040=A081528(9)/4=A081528(10)/5. So the formula cannot be true. - M. F. Hasler, Jan 25 2008

a(n) is also the numerator of u(n+1), with u(n)=(1/n)*sum((2^k-1)/k,k=1..n)and we have the formula : polylog(2,x/(1-x))=sum(u(n)*x^n, n=1..infinity) on the interval [ -1/2,1/2]. [From Groux Roland (roland.groux(AT)orange.fr), Feb 01 2009]

R. J. Mathar, Notes on an attempt to prove that A097344 and A097345 are identical

The first values of the binomial transform of 1/(n+1)^2 are 1, 5/4, 29/18, 103/48, 887/300, 1517/360, 18239/2940, 63253/6720, 332839/22680, 118127/5040, 2331085/60984, ...

f:=n->add( binomial(n, k)/(k+1)^2, k=0..n);

(PARI) A097344(n)=numerator(sum(k=0, n, binomial(n, k)/(k+1)^2)) \\ - M. F. Hasler, Jan 25 2008

Cf. A097345, A134652.

Sequence in context: A111937 A139856 A097345 this_sequence A153076 A034700 A057721

Adjacent sequences: A097341 A097342 A097343 this_sequence A097345 A097346 A097347

easy,nonn,frac

Paul Barry (pbarry(AT)wit.ie), Aug 06 2004

Edited and corrected by Daniel Glasscock (glasscock(AT)rice.edu), Jan 04 2008 and M. F. Hasler (Maximilian.Hasler(AT)gmail.com), Jan 25 2008