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%I #29 Nov 06 2022 09:10:40
%S 0,1,1,3,2,17,4,23,25,61,18,107,40,421,1363,1103,210,5777,492,7563,
%T 24475,19801,2786,103681,33552,135721,146401,355323,39650,1860497,
%U 97108,2435423,2627065,6376021,20633238,11128427,1459960,43701901
%N a(n) = numerator of B(0,n) where B(n,n) = 0, B(n-1,n) = 1/n, and B(m,n) = B(m-1,n+1) - B(m-1,n).
%C Square array B(m,n) begins:
%C 0, 1/1, 1/1, 3/2, 2/1, 17/6, ...
%C 1/1, 0, 1/2, 1/2, 5/6, 7/6, ...
%C -1/1, 1/2, 0, 1/3, 1/3, 7/12, ...
%C 3/2, -1/2, 1/3, 0, 1/4, 1/4, ...
%C -2/1, 5/6, -1/3, 1/4, 0, 1/5, ...
%C 17/6, -7/6, 7/12, -1/4, 1/5, 0, ...
%C The inverse binomial transform of B(0,n) gives B(n,0) and thus it is an eigensequence in the sense that it remains the same (up to a sign) under inverse binomial transform.
%C The bisection of B(0,n) (odd part) gives A175385/A175386, and thus a(2*n+1) = A175385(n+1).
%H Alois P. Heinz, <a href="/A189731/b189731.txt">Table of n, a(n) for n = 0..1000</a>
%F Numerator of (A000204(n) - 1)/n. - _Artur Jasinski_, Oct 21 2022
%p B:= proc(m, n) option remember;
%p if m=n then 0
%p elif n=m+1 then 1/n
%p elif n>m then B(m, n-1) +B(m+1, n-1)
%p else B(m-1, n+1) -B(m-1, n)
%p fi
%p end:
%p a:= n-> numer(B(0, n)):
%p seq(a(n), n=0..50); # _Alois P. Heinz_, Apr 29 2011
%t Rest[Numerator[Abs[CoefficientList[Normal[Series[Log[1 - x^2/(1 + x)], {x, 0, 40}]], x]]]] (* _Vaclav Kotesovec_, Jul 07 2020 *)
%t Table[Numerator[(LucasL[n]-1)/n],{n,1,38}] (* _Artur Jasinski_, Oct 21 2022 *)
%Y Cf. A000204, A242926 (denominators).
%Y Cf. A174341, A177690, A181722.
%K nonn,easy
%O 0,4
%A _Paul Curtz_, Apr 26 2011
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