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 A227500 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). 1
 0, 0, 1, 1, 5, 13, 19, 179, 1028, 1103, 893, 2889, 15445, 249787, 24988, 8494711, 6888613, 7423979, 101535859, 329279361, 1187585188, 128951009, 2513033741, 25007430139, 599126628077, 591141383117, -3361274604, 1470023540617, 22712552603063, 322385807064733, -26340115994784101 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS 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,... . 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, .... 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. This conjecture is confirmed up to 100 terms. [Jean-François Alcover, Jul 19 2013] LINKS EXAMPLE a(2)=1 because r(2)=r(1)+r(0)+A027641(0)/A027642(0)=0+0+1=1; a(3)=1 because r(3)=r(2)+r(1)+A027641(1)/A027642(1)=1+0-1/2=1/2; a(4)=5 because r(4)=r(3)+r(2)+A027641(2)/A027642(2)=1+1/2+1/6=5/3. MATHEMATICA 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 *) CROSSREFS Sequence in context: A154634 A232655 A175866 * A265814 A087714 A055045 Adjacent sequences:  A227497 A227498 A227499 * A227501 A227502 A227503 KEYWORD sign,frac AUTHOR Paul Curtz, Jul 13 2013 EXTENSIONS More terms from Jean-François Alcover, Jul 19 2013 STATUS approved

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Last modified May 7 19:23 EDT 2021. Contains 343652 sequences. (Running on oeis4.)