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A binomial sum of powers related to the Bernoulli numbers, triangular array, read by rows.
3

%I #25 Jan 28 2021 14:01:31

%S -1,-2,2,-4,10,-6,-8,38,-54,24,-16,130,-330,336,-120,-32,422,-1710,

%T 3000,-2400,720,-64,1330,-8106,21840,-29400,19440,-5040,-128,4118,

%U -36414,141624,-285600,312480,-176400,40320

%N A binomial sum of powers related to the Bernoulli numbers, triangular array, read by rows.

%C T(n,k) = sum_{v=0..k} (-1)^v*v*binomial(k,v)*(v+1)^(n-1)

%C for n >= 1, k >= 1; by convention T(0,0) = 1.

%C Gives a representation of the Bernoulli numbers B_{n} = B_{n}(1) (with B_1 = 1/2)

%C B_{n} = sum_{j=0..n} sum_{k=0..j} T(j,k)/(k+1)

%C T(n,1) = -2^(n-1) (n>=1)

%C T(n,n) = (-1)^n*n! (n>=1)

%C sum_{k=0..n} T(n,k) = -A000007(n-1) = -1,0,0,0,0,... (n>=1)

%C sum_{k=0..n} abs(T(n,k)) = A162509(n) = A073146(n,n-1) (n>=1)

%C sum_{k=0..n} T(n,k)/(k+1) = Bernoulli(n,1)-Bernoulli(n-1,1) (n>=1)

%C numer(sum(T(n,k)/(k+1),k=0..n)) = A051716(n) (n>=0)

%C denom(sum(T(n,k)/(k+1),k=0..n)) = A051717(n) (n>=0)

%C Contribution from _Peter Luschny_, Jul 08 2009: (Start)

%C More generally, define the polynomials (assume p[0,0](x)=1 and 0^0=1)

%C p[n,k](x) = sum_{v=0..k} (-1)^v*v*binomial(k,v)*(v+1+x)^(n-1)

%C [1], [0, -1], [0, -2-x, 2], [0, -4-4x-x^2, 10+4x, -6], ...

%C then T(n,k)=p[n,k](0) and (-1)^k*k!*Stirling2(n,k)=p[n,k](-1) (cf. A019538).

%C Assume now k >= 1 and read by rows. Then

%C p[n,k](1) = -1,-3,2,-9,14,-6,-27,74,-72,24,-81,350,-582,432,-120,...

%C (-1)^n*(-2)^(n-k)*p[n,k](-1/2))=1,3,2,9,16,6,27,98,90,24,81,544,924,576,120,..

%C (-1)^n*(-2)^(n-k)*p[n,k](-3/2))=1,1,2,1,8,6,1,26,54,24,1,80,348,384,120,... (End)

%C Variant of A199400.

%H Vincenzo Librandi, <a href="/A162508/b162508.txt">Rows n = 1..50, flattened</a>

%F From _Peter Bala_, Jul 21 2014: (Start)

%F T(n,k) = (-1)^k*k!*( Stirling2(n+1,k+1) - Stirling2(n,k+1) ), 1 <= k <= n.

%F T(n,k) = (-1)^k*(k + 1)*A038719(n+1,k+1).

%F E.g.f.: - B(-x,z)^2, where B(x,z) = 1/((1 + x)*exp(-z) - x) = 1 + (1 + x)*z + (1 + 3*x + 2*x^2)*z^2/2! + ... is an e.g.f. for A028246 (with an offset of 0).

%F Recurrence: T(n,k) = (k + 1)*T(n-1,k) - k*T(n-1,k-1).

%F The unsigned version of the triangle equals the matrix product A007318*A019538.

%F Assuming this triangle is a signed version of A199400 then the n-th row polynomial R(n,x) = 1/(1 - x)*( sum {k = 1..inf} k*(k + 1)^(n-1)*(x/(x - 1))^k ), valid for x in the open interval (-inf, 1/2). (End)

%e For n >= 1, k >= 1:

%e ..................... -1

%e ................... -2, 2

%e ................. -4, 10, -6

%e .............. -8, 38, -54, 24

%e ......... -16, 130, -330, 336, -120

%e ..... -32, 422, -1710, 3000, -2400, 720

%e -64, 1330, -8106, 21840, -29400, 19440, -5040

%p T := proc(n,k) local v; if n=0 and k=0 then 1 else

%p add((-1)^v*v*binomial(k,v)*(v+1)^(n-1),v=0..k) fi end:

%p # _Peter Bala_'s e.g.f. assuming offset 0:

%p egf := (x, z) -> -((1-x)/exp(z) + x)^(-2):

%p ser := series(egf(x, z), z, 10): coz := n -> n!*coeff(ser, z, n):

%p row := n -> seq(coeff(coz(n), x, k), k = 0..n):

%p seq(print(row(n)), n = 0..9); # _Peter Luschny_, Jan 28 2021

%t t[n_, k_] := Sum[(-1)^v*v*Binomial[k, v]*(v + 1)^(n - 1), {v, 0, k}]; Table[t[n, k], {n, 1, 8}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Jul 26 2013, after Maple *)

%o (Sage)

%o def A162508(n, k):

%o if n==0 and k==0: return 1

%o return add((-1)^v*v*binomial(k, v)*(v+1)^(n-1) for v in (0..k))

%o for n in (1..8): [A162508(n, k) for k in (1..n)] # _Peter Luschny_, Jul 21 2014

%Y Cf. A162509, A073146, A051716, A051717, A199400. A019538, A028246, A038719.

%K sign,tabl

%O 1,2

%A _Peter Luschny_, Jul 05 2009

%E More terms from _Philippe Deléham_, Nov 05 2011