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 A262704 Triangle: Newton expansion of C(n,m)^3, read by rows. 4
 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, 210, 1, 0, 0, 0, 5670, 21840, 7770, 336, 1, 0, 0, 0, 5040, 87570, 107520, 19236, 504, 1, 0, 0, 0, 1680, 189000, 735210, 407400, 42084, 720, 1, 0, 0, 0, 0, 224700, 2835756, 4280850, 1284360, 83880, 990, 1 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS Triangle here T_3(n,m) is such that C(n,m)^3 = Sum_{j=0..n} C(n,j)*T_3(j,m). Equivalently, lower triangular matrix T_3 such that || C(n,m)^3 || = A181583 =  P * T_3 = A007318 * T_3. T_3(n,m) = 0 for n < m and for 3*m < n. In fact: 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. 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. Example: 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(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). So we get the expansion of the 3rd power of the binomial coefficient in terms of the binomial coefficients on the same row. T_1 is the unitary matrix, T_2 is the transpose of A109983, T_3 is this sequence, T_4, T_5 are A262705, A262706. LINKS Gheorghe Coserea, Rows n = 0..200, flattened P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, arXiv:quant-ph/0402027, 2004. FORMULA T_3(n,m) = Sum_{j=0..n} (-1)^(n-j)*C(n,j)*C(j,m)^3. 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). EXAMPLE Triangle starts: n\m  [0]     [1]     [2]     [3]     [4]     [5]     [6]     [7]     [8] [0]  1; [1]  0,      1; [2]  0,      6,      1; [3]  0,      6,      24,     1; [4]  0,      0,      114,    60,     1; [5]  0,      0,      180,    690,    120,    1; [6]  0,      0,      90,     2940,   2640,   210,    1; [7]  0,      0,      0,      5670,   21840,  7770,   336,    1; [8]  0,      0,      0,      5040,   87570,  107520, 19236,  504,    1; [9]  ... MATHEMATICA 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 *) PROG (MuPAD) // as a function T_3:=(n, m)->_plus((-1)^(n-j)*binomial(n, j)*binomial(j, m)^3 \$ j=0..n): // as a matrix h x h _P:=h->matrix([[binomial(n, m) \$m=0..h]\$n=0..h]): _P_3:=h->matrix([[binomial(n, m)^3 \$m=0..h]\$n=0..h]): _T_3:=h->_P(h)^-1*_P_3(h): (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 (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 (PARI) t3(n, m) = sum(j=0, n,  (-1)^((n-j)%2)* binomial(n, j)*binomial(j, m)^3); concat(vector(11, n, vector(n, k, t3(n-1, k-1)))) \\ Gheorghe Coserea, Jul 14 2016 CROSSREFS Row sums are, by definition, the A208446, the inverse binomial transform of the Franel numbers (A000172). Column sums are the A126086, per the comment given thereto by Brendan McKay. Second diagonal (T_3(n+1,n)) is A007531 (n+2). Column T_3(n,2) is A122193(3,n). Cf. A109983 (transpose of), A262705, A262706. Cf. A078739, A078741, A274786. Sequence in context: A090203 A120113 A074395 * A195402 A176402 A204013 Adjacent sequences:  A262701 A262702 A262703 * A262705 A262706 A262707 KEYWORD nonn,tabl,easy AUTHOR Giuliano Cabrele, Sep 27 2015 STATUS approved

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Last modified August 17 22:32 EDT 2018. Contains 313817 sequences. (Running on oeis4.)