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%I #26 Jul 02 2015 10:17:23
%S 1,0,-5,3,-61,5,-125,7,-121,9,-325,11,-17071,13,-35,15,-7697,17,36685,
%T 19,-177911,21,852995,23,-236396851,25,8553025,27,-23749473209,29,
%U 8615841061175,31,-7709321049377,33,2577687858265,35,-26315271553088022793,37
%N Numerators of the inverse binomial transform of the Bernoulli numbers with B(1)=1.
%C Difference table of 1, 1, 1/6, 0, -1/30, ... :
%C 1, 1, 1/6, 0, -1/30, 0, 1/42, 0, ...
%C 0, -5/6, -1/6, -1/30, 1/30, 1/42, -1/42, ...
%C -5/6, 2/3, 2/15, 1/15, -1/105, -1/21, ...
%C 3/2, -8/15, -1/15, -8/105, -4/105, ...
%C -61/30, 7/15, -1/105, 4/105, ...
%C 5/2, -10/21, 1/21, ...
%C -125/42, 11/21, ...
%C 7/2, ...
%C etc.
%C The inverse binomial transform is the first column. a(n) is the n-th term of the numerators. See A027641(n+1).
%C Denominators: A176591.
%C Is a(4n+2) a multiple of 5? This is true, at least up to 4n+2 = 998. - _Jean-François Alcover_, Jul 02 2015
%H Colin Barker, <a href="/A257935/b257935.txt">Table of n, a(n) for n = 0..629</a>
%F a(n) = numerators of A027641(n)/A027642(n) - (-1)^n*n/2.
%F a(n) = (A176328(n) - (-1)^n*n)*A176591(n).
%F a(n) = 2*A027641(n)*A176591(n)/A027642(n) - A176328(n).
%e By the first formula: numerators of 1-0=1, -1/2+1/2=0, 1/6-1=-5/6, 0+3/2=3/2,....
%t max = 40; B[1] = 1; B[n_] := BernoulliB[n]; BB = Array[B, max, 0]; a[n_] := Differences[BB, n] // First // Numerator; Table[a[n], {n, 0, max-1}] (* _Jean-François Alcover_, May 20 2015 *)
%o (PARI)
%o firstdiff(s) = my(t=vector(#s-1)); for(i=2, #s, t[i-1]=s[i]-s[i-1]); t
%o a257935(k) = {
%o my(s=[], b = concat([1,1], vector(k, n, n++; bernfrac(n))));
%o until(#b<2,
%o s = concat(s, numerator(b[1]));
%o b = firstdiff(b)
%o );
%o s
%o }
%o a257935(50) \\ _Colin Barker_, May 13 2015
%Y Cf. A257106, A027641/A027642, A164555/A027642, A176327/A027642, A176328/A176591, A026741.
%K sign
%O 0,3
%A _Paul Curtz_, May 13 2015
%E More terms from _Colin Barker_, May 13 2015
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