A005721 Central quadrinomial coefficients.

1, 4, 44, 580, 8092, 116304, 1703636, 25288120, 379061020, 5724954544, 86981744944, 1327977811076, 20356299454276, 313095240079600, 4829571309488760, 74683398325804080, 1157402982351003420, 17971185794898859248

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Sum of squares of entries in the n-th row of triangle of quadrinomial coefficients A008287 (Pascal triangle of order 4). - Adi Dani, Jul 03 2011

Central coefficients in triangle A008287 ((1 + x + x^2 + x^3)^n), see link. - Zagros Lalo, Sep 25 2018

References

• L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.
• Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 601, 602.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Robert Israel, Table of n, a(n) for n = 0..597 (first 101 terms from T. D. Noe)
• Hacène Belbachir and Yassine Otmani, Quadrinomial-Like Versions for Wolstenholme, Morley and Glaisher Congruences, Integers (2023) Vol. 23.
• Adi Dani, Restricted compositions of natural numbers-section: Generalized Pascal triangle
• R. K. Guy, editor, Western Number Theory Problems, 1985-12-21 & 23, Typescript, Jul 13 1986, Dept. of Math. and Stat., Univ. Calgary, 11 pages. Annotated scan of pages 1, 3, 7, 9, with permission. See Problem 85:02.
• R. K. Guy, Letter to N. J. A. Sloane, 1987
• Zagros Lalo, Formula for the Central coefficients in triangle A008287 ((1 + x + x^2 + x^3)^n).

Crossrefs

Cf. A002426, A005190, A008287.

Programs

GAP

List([0..20],n->Sum([0..Int(3*n/4)],k->(-1)^k*Binomial(2*n,k)*Binomial(5*n-4*k-1,3*n-4*k))); # Muniru A Asiru, Sep 26 2018

Magma

[(&+[(-1)^k*Binomial(2*n,k)*Binomial(5*n-4*k-1,3*n-4*k): k in [0..Floor(3*n/4)]]): n in [0..30]]; // G. C. Greubel, Oct 06 2018

Maple

F := (t^2-1)*(6*t+t^2+1)^(1/2)/(3*t^3+13*t^2+t-1); G := t/((t+1)^2*(6*t+t^2+1));
Ginv := RootOf(numer(G-x),t); series(eval(F,t=Ginv),x=0,20);
seq(coeff((1+x+x^2+x^3)^(2*n),x,3*n),n=0..50); # Robert Israel, Nov 01 2015

Mathematica

Table[Sum[(-1)^k*Binomial[2*n,k]*Binomial[5*n-4*k-1,3*n-4*k],{k,0,3*n/4}],{n,0,25}] (* Adi Dani, Jul 03 2011 *)
RecurrenceTable[{128*(n-1)*(2*n-3)*(2*n-1)*(5*n-1)*a[n-2] -8*(2*n-1)*(145*n^3-319*n^2+201*n-30)*a[n-1]+3*n*(3*n-2)*(3*n-1)*(5*n-6)*a[n]==0,
a[0]==1,a[1]==4},a,{n,0,5000}] (* Bradley Klee, Jun 25 2018 *)
a[n_] := a[n] = Sum[(2*n)!/((j - n)!*(3*n + i - 2*j)!*(j - 2*i)!*i!), {i, 0, n}, {j, n, 2*n}]; Table[a[n], {n, 0, 20}] (* Zagros Lalo, Sep 25 2018 *)

PARI

a(n)={local(v=Vec((1+x+x^2+x^3)^n));sum(k=1,#v, v[k]^2);}

PARI

a(n)=sum(k=0,3*n/4, (-1)^k*binomial(2*n,k)*binomial(5*n-4*k-1,3*n-4*k));

PARI

vector(30, n, n--; polcoeff((1+x+x^2+x^3)^(2*n), (6*n)>>1)) \\ Altug Alkan, Nov 01 2015

Formula

a(n) = A005190(2*n) = A008287(2*n, 3*n).

G.f.: Let Z(x) be a solution of (-1+16*x)*(32*x-27)^2*Z^6+9*(-9+64*x)*(32*x-27)*Z^4+81*(80*x-27)*Z^2+729 = 0, with Z(0)=1. Compute a Puiseux series for Z(x) at x=0, then Z(x) in C[[x^(1/3)]]. Remove all non-integer powers of x. The result is the generating function for A005721. - Mark van Hoeij, Oct 29 2011

G.f.: F(G^(-1)(x)) where F(t) = (t^2-1)*(6*t+t^2+1)^(1/2)/(3*t^3+13*t^2+t-1) and G(t) = t/((t+1)^2*(6*t+t^2+1)). - Mark van Hoeij, Oct 30 2011

From Bradley Klee, Jun 25 2018: (Start)

128*(n-1)*(2*n-3)*(2*n-1)*(5*n-1)*a(n-2) - 8*(2*n-1)*(145*n^3-319*n^2+201*n-30)*a(n-1) + 3*n*(3*n-2)*(3*n-1)*(5*n-6)*a(n) = 0.

G.f. G(x) satisfies a Picard-Fuchs type differential equation, 0 = Sum_{m=0..5, n=0..3} M_{m,n} x^m*(d^n/dx^n G(x)), with integer matrix:

M={{ 24, -6, 0, 0},

{-768, 1488, -54, 0},

{6144, -16128, 2520, -27},

{ 0, 55296, -29568, 896},

{ 0, 0, 49152, -7936},

{ 0, 0, 0, 8192}}(End)

a(n) = sum_{k=0..floor(3n/4)} (-1)^k binomial(2n,k) * binomial(5n-4k-1,3n-4k). - Muniru A Asiru, Sep 26 2018

a(n) = Sum_{i=0..n} Sum_{j=n..2n}(f); f= ( (2*n)!/((j - n)!*(3*n + i - 2*j)!*(j - 2*i)!*i!) ); f=0 for (3*n + i - 2*j)<0 or (j - 2*i)<0. See also formula in Links section. - Zagros Lalo, Sep 27 2018