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%I #16 Jul 12 2015 19:42:45
%S 1,5,29,103,887,1517,18239,63253,332839,118127,2331085,4222975,
%T 100309579,184649263,1710440723,6372905521,202804884977,381240382217,
%U 13667257415003,25872280345103,49119954154463,93501887462903
%N Numerators of the partial sums of the binomial transform of 1/(n+1).
%C Numerators in the expansion of log((1-x)/(1-2x)) / (1-x) are 0,1,5,29,.. - _Paul Barry_, Feb 09 2005
%C Is this identical to A097344? - Aaron Gulliver, Jul 19 2007. The answer turns out to be No - see A134652.
%C From n=9 on, the putative formula a(n)=A003418(n+1)*sum{k=0..n, (2^(k+1)-1)/(k+1)} is false. The least n for which a(n) is different from A097344(n) is n=59, then they agree again until n=1519. - _M. F. Hasler_, Jan 25 2008
%H R. J. Mathar, <a href="/A097345/a097345.pdf">Notes on an attempt to prove that A097344 and A097345 are identical</a>
%t Table[ Sum[(2^(k+1)-1)/(k+1), {k, 0, n}] // Numerator, {n, 0, 21}] (* _Jean-François Alcover_, Oct 14 2013, after Pari *)
%o (PARI) A097345(n) = numerator(sum(k=0,n,(2^(k+1)-1)/(k+1)))
%Y Cf. A097344, A134652.
%K easy,nonn,frac
%O 0,2
%A _Paul Barry_, Aug 06 2004
%E Edited and corrected by Daniel Glasscock (glasscock(AT)rice.edu), Jan 04 2008 and _M. F. Hasler_, Jan 25 2008
%E Moved comment concerning numerators of the logarithm from A097344 to here where it is correct - _R. J. Mathar_, Mar 04 2010
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