A262705 Triangle: Newton expansion of C(n,m)^4, read by rows.

1, 0, 1, 0, 14, 1, 0, 36, 78, 1, 0, 24, 978, 252, 1, 0, 0, 4320, 8730, 620, 1, 0, 0, 8460, 103820, 46890, 1290, 1, 0, 0, 7560, 581700, 1159340, 185430, 2394, 1, 0, 0, 2520, 1767360, 13387570, 8314880, 595476, 4088, 1, 0, 0, 0, 3087000, 85806000, 170429490, 44341584, 1642788, 6552, 1

Offset: 0

Author

Giuliano Cabrele, Sep 30 2015

Keywords

• nonn
• tabl
• easy

Comments

Triangle here T_4(n,m) is such that C(n,m)^4 = Sum_{j=0..n} C(n,j)*T_4(j,m).

Equivalently, lower triangular matrix T_4 such that

|| C(n,m)^4 || = A202750 = P * T_4 = A007318 * T_4.

T_4(n,m) = 0 for n < m and for 4*m < n.

Refer to comment to A262704.

Example:

C(x,2)^4 = x^4*(x-1)^4 /16 = 1*C(x,2) + 78*C(x,3) + 978*C(x,4) + 4320*C(x,5) + 8460*C(x,6) + 7560*C(x,7) + 2520*C(x,8);

C(5,2)^4 = C(5,3)^4 = 10000 = 1*C(5,2) + 78*C(5,3) + 978*C(5,4) + 4320*C(5,5) = 1*C(5,3) + 252*C(5,4) + 8730*C(5,5).

Examples

			Triangle starts:
[1];
[0,  1];
[0, 14,    1];
[0, 36,   78,      1];
[0, 24,  978,    252,     1];
[0,  0, 4320,   8730,   620,    1];
[0,  0, 8460, 103820, 46890, 1290, 1];
		

Links

• P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, arXiv:quant-ph/0402027, 2004.

Crossrefs

Row sums are, by definition, the inverse binomial transform of A005260.

Second diagonal (T_4(n+1,n)) is A058895(n+1).

Column T_4(n,2) is A122193(4,n).

Cf. A109983 (transpose of), A262704, A262706.

Cf. A078739, A078741.

Programs

Magma

[&+[(-1)^(n-j)*Binomial(n,j)*Binomial(j,m)^4: j in [0..n]]: m in [0..n], n in [0..10]]; // Bruno Berselli, Oct 01 2015

Mathematica

T4[n_, m_] := Sum[(-1)^(n - j) * Binomial[n, j] * Binomial[j, m]^4, {j, 0, n}]; Table[T4[n, m], {n, 0, 9}, {m, 0, n}] // Flatten (* Jean-François Alcover, Oct 01 2015 *)

MuPAD

// as a function
T_4:=(n,m)->_plus((-1)^(n-j)*binomial(n,j)*binomial(j,m)^4 $ j=0..n):
// as a matrix h x h
_P:=h->matrix([[binomial(n,m) $m=0..h]$n=0..h]):
_P_4:=h->matrix([[binomial(n,m)^4 $m=0..h]$n=0..h]):
_T_4:=h->_P(h)^-1*_P_4(h):

PARI

T_4(nmax) = {for(n=0, nmax, for(m=0, n, print1(sum(j=0, n, (-1)^(n-j)*binomial(n,j)*binomial(j,m)^4), ", ")); print())} \\ Colin Barker, Oct 01 2015

Formula

T_4(n,m) = Sum_{j=0..n} (-1)^(n-j)*C(n,j)*C(j,m)^4.

Also, let S(r,s)(n,m) denote the Generalized Stirling2 numbers as defined in the link above, then T_4(n,m) = n! / (m!)^4 * S(m,m)(4,n).