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Triangle T(n,k) with the coefficient [x^k] (n+1)!* C(n,x), in row n, column k, where C(.,.) are the Bernoulli twin number polynomials of A129378.
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%I #15 Feb 12 2014 18:09:11

%S 1,-1,2,-2,0,6,-4,-12,12,24,-4,-60,-60,120,120,24,-120,-720,-240,1080,

%T 720,120,840,-2520,-8400,0,10080,5040,-960,6720,20160,-47040,-100800,

%U 20160,100800,40320,-12096,-60480,241920,423360

%N Triangle T(n,k) with the coefficient [x^k] (n+1)!* C(n,x), in row n, column k, where C(.,.) are the Bernoulli twin number polynomials of A129378.

%C The terms at x=0 define the Bernoulli twin numbers, C(n,0)=C(n) = A129826(n)/(n+1)! .

%C Because the C(n,x) are derived from the Bernoulli polynomials B(n,x) via a binomial transformation and because the odd-indexed Bernoulli numbers are (essentially) zero, the following sum rules for the C(n) emerge (partially in Umbral notation):

%C For odd C(n): C(2n)=(C-1)^(2n-1), n > 1, C(2n) disappears; example: C(4)=C(4)-3C(3)+3C(2)-C(1).

%C 0r for C(2n+1): (C-1)^2n=0, n >0; example: C(1)-4C(2)+6C(3)-4C(4)+C(5)=0.

%C With positive coefficients, table

%C 1, 2;

%C 2, 2, 3;

%C 3, 2, 3, 6;

%C 4, 2, 3, 6, 30;

%C 5, 2, 3, 6, 30, -30;

%C 6, 2, 3, 6, 30, -30, -42;

%C gives C(n). Example: 3C(0)+2C(1)+3C(2)+6C(3)=0. See -A051717(n+1), Bernoulli twin numbers denominators, with from 30 opposite twin.

%e 1; C(0,x) = 1

%e -1, 2; C(1,x) = -1/2+x

%e -2, 0, 6; C(2,x) = -1/3+x^2

%e -4, -12, 12, 24; C(3,x) = -1/6 -x/2 +x^2/2 +x^3

%e -4, -60, -60, 120, 120;

%p C := proc(n,x) if n =0 then 1; else add( binomial(n-1,j-1)*bernoulli(j,x),j=1..n) ; expand(%) ; end if; end proc:

%p A140333 := proc(n,k) (n+1)!*C(n,x) ; coeftayl(%,x=0,k) ; end proc: # _R. J. Mathar_, Jun 27 2011

%t c[0, _] = 1; c[n_, x_] := Sum[Binomial[n-1, j-1]*BernoulliB[j, x], {j, 1, n}]; t[n_, k_] := (n+1)!*Coefficient[c[n, x], x, k]; Table[t[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Sep 16 2013 *)

%Y Cf. A129378.

%K sign,less,uned,tabl

%O 0,3

%A _Paul Curtz_, May 28 2008