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

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

Offset: 0

Author

Giuliano Cabrele, Sep 27 2015

Keywords

• nonn
• tabl
• easy

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.

Examples

			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]  ...
		

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.

Crossrefs

Row sums are A172634, 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.

Programs

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

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 *)

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

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

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).