cp's OEIS Frontend

A262704 Triangle: Newton expansion of C(n,m)^3, read by rows.

%I A262704 #44 Sep 08 2022 08:46:14
%S A262704 1,0,1,0,6,1,0,6,24,1,0,0,114,60,1,0,0,180,690,120,1,0,0,90,2940,2640,
%T A262704 210,1,0,0,0,5670,21840,7770,336,1,0,0,0,5040,87570,107520,19236,504,
%U A262704 1,0,0,0,1680,189000,735210,407400,42084,720,1,0,0,0,0,224700,2835756,4280850,1284360,83880,990,1
%N A262704 Triangle: Newton expansion of C(n,m)^3, read by rows.
%C A262704 Triangle here T_3(n,m) is such that C(n,m)^3 = Sum_{j=0..n} C(n,j)*T_3(j,m).
%C A262704 Equivalently, lower triangular matrix T_3 such that
%C A262704 || C(n,m)^3 || = A181583 =  P * T_3 = A007318 * T_3.
%C A262704 T_3(n,m) = 0 for n < m and for 3*m < n. In fact:
%C A262704 C(x,m)^q and C(x,m), with m nonnegative and q positive integer, are polynomial in x of degree m*q and m respectively, and C(x,m) is a divisor of C(x,m)^q.
%C A262704 Therefore the Newton series will give C(x,m)^q = T_q(m,m)*C(x,m) + T_q(m+1,m)*C(x,m+1) + ... + T_q(q*m,m)*C(x,q*m), where T_q(n,m) is the n-th forward finite difference of C(x,m)^q at x = 0.
%C A262704 Example:
%C A262704 C(x,2)^3 = x^3*(x-1)^3 / 8 = 1*C(x,2) + 24*C(x,3) + 114*C(x,4) + 180*C(x,5) + 90*C(x,6);
%C A262704 C(5,2)^3 = C(5,3)^3 = 1000 = 1*C(5,2) + 24*C(5,3) + 114*C(5,4) + 180*C(5,5) = 1*C(5,3) + 60*C(5,4) + 690*C(5,5).
%C A262704 So we get the expansion of the 3rd power of the binomial coefficient in terms of the binomial coefficients on the same row.
%C A262704 T_1 is the unitary matrix,
%C A262704 T_2 is the transpose of A109983,
%C A262704 T_3 is this sequence,
%C A262704 T_4, T_5 are A262705, A262706.
%H A262704 Gheorghe Coserea, <a href="/A262704/b262704.txt">Rows n = 0..200, flattened</a>
%H A262704 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 A262704 T_3(n,m) = Sum_{j=0..n} (-1)^(n-j)*C(n,j)*C(j,m)^3.
%F A262704 Also, let S(r,s)(n,m) denote the Generalized Stirling2 numbers as defined in the link above,then T_3(n,m) = n! / (m!)^3 * S(m,m)(3,n).
%e A262704 Triangle starts:
%e A262704 n\m  [0]     [1]     [2]     [3]     [4]     [5]     [6]     [7]     [8]
%e A262704 [0]  1;
%e A262704 [1]  0,      1;
%e A262704 [2]  0,      6,      1;
%e A262704 [3]  0,      6,      24,     1;
%e A262704 [4]  0,      0,      114,    60,     1;
%e A262704 [5]  0,      0,      180,    690,    120,    1;
%e A262704 [6]  0,      0,      90,     2940,   2640,   210,    1;
%e A262704 [7]  0,      0,      0,      5670,   21840,  7770,   336,    1;
%e A262704 [8]  0,      0,      0,      5040,   87570,  107520, 19236,  504,    1;
%e A262704 [9]  ...
%t A262704 T3[n_, m_] := Sum[(-1)^(n - j) * Binomial[n, j] * Binomial[j, m]^3, {j, 0, n}]; Table[T3[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* _Jean-François Alcover_, Oct 01 2015 *)
%o A262704 (MuPAD)
%o A262704 // as a function
%o A262704 T_3:=(n,m)->_plus((-1)^(n-j)*binomial(n,j)*binomial(j,m)^3 $ j=0..n):
%o A262704 // as a matrix h x h
%o A262704 _P:=h->matrix([[binomial(n,m) $m=0..h]$n=0..h]):
%o A262704 _P_3:=h->matrix([[binomial(n,m)^3 $m=0..h]$n=0..h]):
%o A262704 _T_3:=h->_P(h)^-1*_P_3(h):
%o A262704 (PARI) T_3(nmax) = {for(n=0, nmax, for(m=0, n, print1(sum(j=0, n, (-1)^(n-j)*binomial(n,j)*binomial(j,m)^3), ", ")); print())} \\ _Colin Barker_, Oct 01 2015
%o A262704 (Magma) [&+[(-1)^(n-j)*Binomial(n,j)*Binomial(j,m)^3: j in [0..n]]: m in [0..n], n in [0..10]]; // _Bruno Berselli_, Oct 01 2015
%o A262704 (PARI) t3(n,m) = sum(j=0, n,  (-1)^((n-j)%2)* binomial(n,j)*binomial(j,m)^3);
%o A262704 concat(vector(11, n, vector(n, k, t3(n-1,k-1)))) \\ _Gheorghe Coserea_, Jul 14 2016
%Y A262704 Row sums are A172634, the inverse binomial transform of the Franel numbers (A000172).
%Y A262704 Column sums are the A126086, per the comment given thereto by _Brendan McKay_.
%Y A262704 Second diagonal (T_3(n+1,n)) is A007531 (n+2).
%Y A262704 Column T_3(n,2) is A122193(3,n).
%Y A262704 Cf. A109983 (transpose of), A262705, A262706.
%Y A262704 Cf. A078739, A078741, A274786.
%K A262704 nonn,tabl,easy
%O A262704 0,5
%A A262704 _Giuliano Cabrele_, Sep 27 2015