A262704 -id:A262704 - cp's OEIS frontend

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

1, 0, 1, 0, 14, 1, 0, 36, 78, 1, 0, 24, 978, 252, 1, 0, 0, 4320, 8730, 620, 1, 0, 0, 8460, 103820, 46890, 1290, 1, 0, 0, 7560, 581700, 1159340, 185430, 2394, 1, 0, 0, 2520, 1767360, 13387570, 8314880, 595476, 4088, 1, 0, 0, 0, 3087000, 85806000, 170429490, 44341584, 1642788, 6552, 1

Offset: 0

Giuliano Cabrele, Sep 30 2015

Triangle here T_4(n,m) is such that C(n,m)^4 = Sum_{j=0..n} C(n,j)*T_4(j,m).
Equivalently, lower triangular matrix T_4 such that
|| C(n,m)^4 || = A202750 =  P * T_4 = A007318 * T_4.
T_4(n,m) = 0 for n < m and for 4*m < n.
Refer to comment to A262704.
Example:
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(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).

			Triangle starts:
[1];
[0,  1];
[0, 14,    1];
[0, 36,   78,      1];
[0, 24,  978,    252,     1];
[0,  0, 4320,   8730,   620,    1];
[0,  0, 8460, 103820, 46890, 1290, 1];

P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, arXiv:quant-ph/0402027, 2004.

Row sums are, by definition, the inverse binomial transform of A005260.
Second diagonal (T_4(n+1,n)) is A058895(n+1).
Column T_4(n,2) is A122193(4,n).
Cf. A109983 (transpose of), A262704, A262706.
Cf. A078739, A078741.

[&+[(-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

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 *)

// as a function
T_4:=(n,m)->_plus((-1)^(n-j)*binomial(n,j)*binomial(j,m)^4 $ j=0..n):
// as a matrix h x h
_P:=h->matrix([[binomial(n,m) $m=0..h]$n=0..h]):
_P_4:=h->matrix([[binomial(n,m)^4 $m=0..h]$n=0..h]):
_T_4:=h->_P(h)^-1*_P_4(h):

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

T_4(n,m) = Sum_{j=0..n} (-1)^(n-j)*C(n,j)*C(j,m)^4.

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).

A262706 Triangle: Newton expansion of C(n,m)^5, read by rows.

Original entry on oeis.org

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

Giuliano Cabrele, Sep 30 2015

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).

			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];

P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, arXiv:quant-ph/0402027, 2004.

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.
Cf. A078739, A078741.

[&+[(-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

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 *)

// 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):

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

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).

A274786 Diagonal of the rational function 1/(1 - (wxz + wy + wz + xy + xz + y + z)).

Original entry on oeis.org

1, 6, 114, 2940, 87570, 2835756, 96982116, 3446781624, 126047377170, 4712189770860, 179275447715364, 6918537571788024, 270178056420497316, 10656693484898995800, 423937118582497715400, 16989669600664370275440, 685277433339552643145490, 27797911234749454227812460, 1133299570662800455270517700

Offset: 0

Gheorghe Coserea, Jul 14 2016

nonn

Gheorghe Coserea, Table of n, a(n) for n = 0..200
A. Bostan, S. Boukraa, J.-M. Maillard and J.-A. Weil, Diagonals of rational functions and selected differential Galois groups, arXiv preprint arXiv:1507.03227 [math-ph], 2015.
Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
Jacques-Arthur Weil, Supplementary Material for the Paper "Diagonals of rational functions and selected differential Galois groups"

Cf. A000984, A005258, A262704, A268545 - A268555.

a[n_] := Sum[(-1)^j Binomial[2n, j] Binomial[j, n]^3, {j, n, 2n}];
(* or much faster *)
a[0] = 1; a[1] = 6; a[n_] := a[n] = (2*(2*n - 1)*(11*n^2 - 11*n + 3)*a[n - 1] + 4*(n - 1)*(2*n - 3)*(2*n - 1)*a[n - 2])/n^3;
Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Dec 01 2017, after Vaclav Kotesovec *)

a(n) = sum(j=n, 2*n, (-1)^(j)*binomial(2*n, 2*n - j)*binomial(j, n)^3);

my(x='x, y='y, z='z, w='w);
R = 1/(1-(w*x*z+w*y+w*z+x*y+x*z+y+z));
diag(n, expr, var) = {
  my(a = vector(n));
  for (i = 1, #var, expr = taylor(expr, var[#var - i + 1], n));
  for (k = 1, n, a[k] = expr;
       for (i = 1, #var, a[k] = polcoeff(a[k], k-1)));
  return(a);
};
diag(18, R, [x,y,z,w])

a(n) = Sum_{j=0..2*n} (-1)^j * binomial(2*n,j) * binomial(j,n)^3.
a(n) = T(2*n,n), where triangle T(n,k) is defined by A262704.
0 = (-x^2+44*x^3+16*x^4)*y''' + (-3*x+198*x^2+96*x^3)*y'' + (-1+144*x+108*x^2)*y' + (6+12*x)*y, where y is the g.f.
Recurrence: n^3*a(n) = 2*(2*n - 1)*(11*n^2 - 11*n + 3)*a(n-1) + 4*(n-1)*(2*n - 3)*(2*n - 1)*a(n-2). - Vaclav Kotesovec, Dec 01 2017
a(n) ~ 2^(2*n - 1) * phi^(5*n + 5/2) / (5^(1/4) * (Pi*n)^(3/2)), where phi = A001622 = (1 + sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Dec 01 2017
Conjecture: a(n) = [x^n] (1 + x)^(2*n) * P(n,(1 + x)/(1 - x))^2, where P(n,x) denotes the n-th Legendre polynomial. Cf. A005260(n) = [x^n] (1 - x)^(2*n) * P(n,(1 + x)/(1 - x))^2, due to Carlitz. - Peter Bala, Sep 21 2021
a(n) = A000984(n) * A005258(n). - Peter Bala, Oct 12 2024

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

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A262706 Triangle: Newton expansion of C(n,m)^5, read by rows.

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A274786 Diagonal of the rational function 1/(1 - (wxz + wy + wz + xy + xz + y + z)).

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