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A262706
Triangle: Newton expansion of C(n,m)^5, read by rows.
2
1, 0, 1, 0, 30, 1, 0, 150, 240, 1, 0, 240, 6810, 1020, 1, 0, 120, 63540, 94890, 3120, 1, 0, 0, 271170, 2615340, 740640, 7770, 1, 0, 0, 604800, 32186070, 47271840, 4029690, 16800, 1, 0, 0, 730800, 214628400, 1281612570, 518276640, 17075940, 32760, 1, 0, 0, 453600, 859992000, 18459063000, 26947757970, 4027831080, 60171300, 59040, 1
OFFSET
0,5
COMMENTS
Triangle here T_5(n,m) is such that C(n,m)^5 = Sum_{j=0..n} C(n,j)*T_5(j,m).
Equivalently, lower triangular matrix T_5 such that
|| C(n,m)^5 || = P * T_5 = A007318 * T_5.
T_5(n,m) = 0 for n < m and for 5*m < n.
Refer to comment to A262704.
Example:
C(x,2)^5 = x^5*(x-1)^5/32 = 1*C(x,2) + 240*C(x,3) + 6810*C(x,4) + 63540*C(x,5) + 271170*C(x,6) + 604800*C(x,7) + 730800*C(x,8) + 453600*C(x,9) + 113400*C(x,10);
C(5,2)^5 = C(5,3)^5 = 100000 = 1*C(5,2) + 240*C(5,3) + 6810*C(5,4) + 63540*C(5,5) = 1*C(5,3) + 1020*C(5,4) + 94890*C(5,5).
LINKS
P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, arXiv:quant-ph/0402027, 2004.
FORMULA
T_5(n,m) = Sum_{j=0..n} (-1)^(n-j)*C(n,j)*C(j,m)^5.
Also, let S(r,s)(n,m) denote the Generalized Stirling2 numbers as defined in the link above, then T_5(n,m) = n! / (m!)^5 * S(m,m)(5,n).
EXAMPLE
Triangle starts:
[1];
[0, 1];
[0, 30, 1];
[0, 150, 240, 1];
[0, 240, 6810, 1020, 1];
[0, 120, 63540, 94890, 3120, 1];
[0, 0, 271170, 2615340, 740640, 7770, 1];
MATHEMATICA
T5[n_, m_] := Sum[(-1)^(n - j) * Binomial[n, j] * Binomial[j, m]^5, {j, 0, n}]; Table[T5[n, m], {n, 0, 9}, {m, 0, n}] // Flatten (* Jean-François Alcover, Oct 01 2015 *)
PROG
(MuPAD)
// as a function
T_5:=(n, m)->_plus((-1)^(n-j)*binomial(n, j)*binomial(j, m)^5 $ j=0..n):
// as a matrix h x h
_P:=h->matrix([[binomial(n, m) $m=0..h]$n=0..h]):
_P_5:=h->matrix([[binomial(n, m)^5 $m=0..h]$n=0..h]):
_T_5:=h->_P(h)^-1*_P_5(h):
(Magma) [&+[(-1)^(n-j)*Binomial(n, j)*Binomial(j, m)^5: j in [0..n]]: m in [0..n], n in [0..10]]; // Bruno Berselli, Oct 01 2015
(PARI) T_5(nmax) = {for(n=0, nmax, for(m=0, n, print1(sum(j=0, n, (-1)^(n-j)*binomial(n, j)*binomial(j, m)^5), ", ")); print())} \\ Colin Barker, Oct 01 2015
CROSSREFS
Second diagonal (T_5(n+1,n)) is A061167(n+1).
Column T_5(n,2) is A122193(5,n).
Cf. A109983 (transpose of), A262704, A262705.
Sequence in context: A030128 A137348 A137737 * A062513 A040906 A040907
KEYWORD
nonn,tabl,easy
AUTHOR
Giuliano Cabrele, Sep 30 2015
STATUS
approved