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

%I A262705 #24 Sep 08 2022 08:46:14
%S A262705 1,0,1,0,14,1,0,36,78,1,0,24,978,252,1,0,0,4320,8730,620,1,0,0,8460,
%T A262705 103820,46890,1290,1,0,0,7560,581700,1159340,185430,2394,1,0,0,2520,
%U A262705 1767360,13387570,8314880,595476,4088,1,0,0,0,3087000,85806000,170429490,44341584,1642788,6552,1
%N A262705 Triangle: Newton expansion of C(n,m)^4, read by rows.
%C A262705 Triangle here T_4(n,m) is such that C(n,m)^4 = Sum_{j=0..n} C(n,j)*T_4(j,m).
%C A262705 Equivalently, lower triangular matrix T_4 such that
%C A262705 || C(n,m)^4 || = A202750 =  P * T_4 = A007318 * T_4.
%C A262705 T_4(n,m) = 0 for n < m and for 4*m < n.
%C A262705 Refer to comment to A262704.
%C A262705 Example:
%C A262705 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 A262705 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).
%H A262705 P. Blasiak, K. A. Penson and A. I. Solomon, <a href="http://www.arxiv.org/abs/quant-ph/0402027">The general boson normal ordering problem</a>, arXiv:quant-ph/0402027, 2004.
%F A262705 T_4(n,m) = Sum_{j=0..n} (-1)^(n-j)*C(n,j)*C(j,m)^4.
%F A262705 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).
%e A262705 Triangle starts:
%e A262705 [1];
%e A262705 [0,  1];
%e A262705 [0, 14,    1];
%e A262705 [0, 36,   78,      1];
%e A262705 [0, 24,  978,    252,     1];
%e A262705 [0,  0, 4320,   8730,   620,    1];
%e A262705 [0,  0, 8460, 103820, 46890, 1290, 1];
%t A262705 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 *)
%o A262705 (MuPAD)
%o A262705 // as a function
%o A262705 T_4:=(n,m)->_plus((-1)^(n-j)*binomial(n,j)*binomial(j,m)^4 $ j=0..n):
%o A262705 // as a matrix h x h
%o A262705 _P:=h->matrix([[binomial(n,m) $m=0..h]$n=0..h]):
%o A262705 _P_4:=h->matrix([[binomial(n,m)^4 $m=0..h]$n=0..h]):
%o A262705 _T_4:=h->_P(h)^-1*_P_4(h):
%o A262705 (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
%o A262705 (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
%Y A262705 Row sums are, by definition, the inverse binomial transform of A005260.
%Y A262705 Second diagonal (T_4(n+1,n)) is A058895(n+1).
%Y A262705 Column T_4(n,2) is A122193(4,n).
%Y A262705 Cf. A109983 (transpose of), A262704, A262706.
%Y A262705 Cf. A078739, A078741.
%K A262705 nonn,tabl,easy
%O A262705 0,5
%A A262705 _Giuliano Cabrele_, Sep 30 2015