1,5

Let M = n X n matrix with (i,j)-th entry a(n+1-j, n+1-i), e.g. if n = 3, M = [1 1 1; 3 1 0; 2 0 0]. Given a sequence s = [s(0)..s(n-1)], let b = [b(0)..b(n-1)] be its inverse binomial transform, and let c = [c(0)..c(n-1)] = M^(-1)*transpose(b). Then s(k) = Sum_{i=0..n-1} b(i)*binomial(k,i) = Sum_{i=0..n-1} c(i)*k^i, k=0..n-1. - Gary W. Adamson, Nov 11, 2001.

H. Hasse: Ein Summierungsverfahren fuer die Riemannsche Zeta-Reihe.

A. Riskin and D. Beckwith, Problem 10231, Amer. Math. Monthly, 102 (1995), 175-176.

Author?, Title?

N. J. A. Sloane, Transforms

E.g.f.: -ln(1-y*(exp(x)-1)). - Vladeta Jovovic (vladeta(AT)Eunet.yu), Sep 28 2003

a(n, k) = S2(n, k)*(k-1)! where S2(n, k) is a Stirling number of the second kind (cf. A008277). Also a(n,k) = T(n,k)/k, where T(n, k) = A019538.

Essentially same triangle as triangle [1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, 7, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] where DELTA is Deleham's operator defined in A084938, but the notation is different.

Sum of terms in n-th row = A000629(n) - Gary W. Adamson (qntmpkt(AT)yahoo.com), May 30 2005

The row generating polynomials P(n, t) are given by P(1, t)=t, P(n+1, t)=t(t+1)diff(P(n, t), t) for n>=1 (see the Riskin and Beckwith reference). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 09 2005

Additional comments from Gottfried Helms, Jul 12 2006:

(Start) Delta-matrix as can be read from H.Hasse's proof of a connection

between the Zeta-function and Bernoullinumbers (see link below).

Let P = lower triangular matrix with entries P[row,col] = binom(row,col)

Let J = unit matrix with alternating signs J[r,r]=(-1)^r

Let N(m) = columnmatrix with N(m)(r) = (r+1)^m, N(1)--> natural numbers

Let V = Vandermondematrix with V[r,c] = (r+1)^c

V is then also N(0)||N(1)||N(2)||N(3)... (indices r,c always beginning at 0)

Then Delta = P*J * V and B' = N(-1)' * Delta

where B is the columnmatrix of Bernoullinumbers and ' means transpose,

or for the single k'th bernoullinumber B_k with the appropriate column of Delta

B_k = N(-1)' * Delta[ *,k ] = N(-1)' * P*J * N(k)

Using a single column instead of V and assuming infinite dimension

H. Hasse showed that in x = N(-1) * P*J * N(s),

where s can be any complex number and s*zeta(1-s) = x.

His theorem reads: s*zeta(1-s) = sum_{n=0..inf} ( (n+1)^-1 * delta(n,s) )

where delta(n,s) = sum_{j=0..n} [ (-1)^j * binom(n,j) * (j+1)^s ] (end)

The k-th row (k>=1) contains a(i, k) for i=1 to k, where a(i, k) satisfies Sum_{i=1..n} C(i, 1)^k = 2 * C(n+1, 2) * Sum_{i=1..k} a(i, k) * C(n-1, i-1)/(i+1). E.g. Row 3 contains 1, 3, 2 so Sum_{i=1..n} C(i, 1)^3 = 2 * C(n+1, 2) * [ a(1, 3)/2 +a(2, 3) *C(n-1, 1)/3 +a(3, 3)*C(n-1, 2)/4 ] = [ (n+1)*n ] * [ 1/2 +(3/3)*C(n-1, 1) +(2/4)*C(n-1, 2) ] = ( n^2 +n ) * ( n -1 +[ C(n-1, 2) +1 ]/2 ) = C(n+1, 2)^2. See A000537 for more details ( 1^3 +2^3 +3^3 +4^3 +5^3 +... ). - Andre F. Labossiere (boronali(AT)laposte.net), Sep 22 2003

1; 1,1; 1,3,2; 1,7,12,6; 1,15,50,60,24; ...

Row 5 of triangle is {1,15,50,60,24}, which is {1,15,25,10,1} times {0!,1!,2!,3!,4!}.

a:=(n, k)->add( (-1)^(k-i)*C(k, i)*i^n, i=0..k)/k;

(PARI) T(n, k)=if(k<0|k>n, 0, n!*polcoeff((x/log(1+x+x^2*O(x^n)))^(n+1), n-k))

Dropping the column of 1's gives A053440. See also A008277.

Without the k in the denominator (in the definition), we get A019538. See also the Stirling number triangle A008277.

Cf. A087127, A087107, A087108, A087109, A087110, A087111, A084938 A075263.

Row sums give A000629.

Sequence in context: A122832 A056151 A134436 this_sequence A082038 A090452 A110439

Adjacent sequences: A028243 A028244 A028245 this_sequence A028247 A028248 A028249

nonn,easy,nice,tabl

njas, Doug McKenzie (mckfam4(AT)aol.com)

More terms from James A. Sellers (sellersj(AT)math.psu.edu), Jan 14 2000

Definition corrected by Li Guo, Dec 16 2006