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%I #20 Jul 21 2013 18:17:00
%S 0,0,1,1,5,13,19,179,1028,1103,893,2889,15445,249787,24988,8494711,
%T 6888613,7423979,101535859,329279361,1187585188,128951009,2513033741,
%U 25007430139,599126628077,591141383117,-3361274604,1470023540617,22712552603063,322385807064733,-26340115994784101
%N a(0)=a(1)=0; for n>1, a(n) = numerator( r(n) ), where r(n) = r(n-1)+r(n-2)+A027641(n-2)/A027642(n-2) and r(0)=r(1)=a(0).
%C Reduced a(n)/c(n) = 0, 0, 1, 1/2, 5/3, 13/6, 19/5, 179/30, 1028/105, 1103/70, 893/35, 2889/70, 15445/231, 249787/2310,... .
%C After the first Bernoulli numbers we consider the same transform applied to the second Bernoulli numbers A164555(n)/A027642(n). Hence reduced b(n)/c(n) = 0, 0, 1, 3/2, 8/3, 25/6, 34/5, 329/30, 1868/105, 2013/70, 1628/35, 5269/70, 28150/231, 455377/2310, ....
%C Conjecture: (b(n)-a(n))/c(n) = 0, 0, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ..., that is two 0 followed by A000045.
%C This conjecture is confirmed up to 100 terms. [_Jean-François Alcover_, Jul 19 2013]
%e a(2)=1 because r(2)=r(1)+r(0)+A027641(0)/A027642(0)=0+0+1=1;
%e a(3)=1 because r(3)=r(2)+r(1)+A027641(1)/A027642(1)=1+0-1/2=1/2;
%e a(4)=5 because r(4)=r(3)+r(2)+A027641(2)/A027642(2)=1+1/2+1/6=5/3.
%t b1[0] = b1[1] = 0; b1[n_] := b1[n] = b1[n - 1] + b1[n - 2] + BernoulliB[n - 2]; a[n_] := Numerator[b1[n]]; Table[a[n], {n, 0, 30}] (* _Jean-François Alcover_, Jul 19 2013 *)
%K sign,frac
%O 0,5
%A _Paul Curtz_, Jul 13 2013
%E More terms from _Jean-François Alcover_, Jul 19 2013
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