A005191 Central pentanomial coefficients: largest coefficient of (1 + x + ... + x^4)^n.

1, 1, 5, 19, 85, 381, 1751, 8135, 38165, 180325, 856945, 4091495, 19611175, 94309099, 454805755, 2198649549, 10651488789, 51698642405, 251345549849, 1223798004815, 5966636799745, 29125608152345, 142330448514875, 696235630761115, 3408895901222375

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn

Comments

Coefficient of x^n in ((1-x^10)/((1-x^5)(1-x^2)(1-x)))^n. - Michael Somos, Sep 24 2003

Note that n divides a(n+1) - a(n). - T. D. Noe, Mar 16 2005

Terms that are not a multiple of 5 have zero density, namely, there are fewer than n^(log(4)/log(5)) such terms among A005191(1..n). In particular, A005191(5k+2) and A005191(5k+4) are multiples of 5 for every k. - Max Alekseyev, Apr 25 2005

Number of n-step 1-D walks ending at the origin with steps of size 0, 1 or 2. - David Scambler, Apr 09 2012

Number of compositions of 2n into exactly n nonnegative parts <= four. a(2) = 5: [4,0], [3,1], [2,2], [1,3], [0,4]. - Alois P. Heinz, Sep 13 2018

Let f(m) = ceiling((q+log(q))/log(25)), where q = -log(log(5)/(2*m^2*Pi)) then f(a(n)) = n, for n > 0. - Miko Labalan, Oct 07 2024

References

• Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 603-604.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000 (first 201 terms from T. D. Noe)
• Armen G. Bagdasaryan and Ovidiu Bagdasar, On some results concerning generalized arithmetic triangles, Electronic Notes in Discrete Mathematics (2018) Vol. 67, 71-77.
• V. E. Hoggatt, Jr. and M. Bicknell, Diagonal sums of generalized Pascal triangles, Fib. Quart., 7 (1969), 341-358, 393.
• Zagros Lalo, Formula for Central terms in triangle A035343 ((1 + x + x^2 + x^3 + x^4)^n).
• Lyle E. Muller and Michelle Rudolph-Lilith, On a link between Dirichlet kernels and central multinomial coefficients, Discrete Mathematics, Volume 338, Issue 9, 6 September 2015, Pages 1567-1572.
• Project Euler, Quintinomial coefficients, Problem 588
• M. Rudolph-Lilith and L. E. Muller, On an explicit representation of central (2k+1)-nomial coefficients, arXiv preprint arXiv:1403.5942 [math.CO], 2014.
• Index entries for sequences of k-nomial coefficients

Crossrefs

Cf. A035343, A349936 (bisection).

Row 5 of A077042. Column 2 of A201551 (?)

Programs

GAP

List([0..25],n->Sum([0..Int(2*n/5)],k->Binomial(n,k)*Binomial(-n,2*n-5*k))); # Muniru A Asiru, Sep 26 2018

Maple

seq(coeff(series(((1-x^10)/((1-x^5)*(1-x^2)*(1-x)))^n,x,n+1), x, n), n = 0 .. 25); # Muniru A Asiru, Sep 26 2018

Mathematica

Flatten[{1,Table[Coefficient[Expand[Sum[x^j,{j,0,4}]^n],x^(2*n)],{n,1,20}]}] (* Vaclav Kotesovec, Aug 09 2013 *)
a[n_] := a[n] = Sum[n!/((q - n)!*(j - 2*q + 2*n)!*(i - 2*j + q)!*(j - 2*i)!*i!), {i, 0, n/2}, {j, 0, n}, {q, n, 2*n}]; Table[a[n], {n, 0, 29}] (* Zagros Lalo, Sep 25 2018 *)
CoefficientList[Series[Sqrt[(-5x+2+2Sqrt[5x^2-6x+1])/(25x^3-10x^2-19x+4)],{x,0,30}],x] (* Harvey P. Dale, Aug 04 2021 *)

PARI

a(n)=if(n<0,0,polcoeff(((1-x^5)/(1-x)+x*O(x^(2*n)))^n,2*n))

PARI

a(n)=if(n<0,0,polcoeff(((1-x^10)/((1-x^5)*(1-x^2)*(1-x))+x*O(x^n))^n,n))

PARI

a(n) = sum(k=0,(2*n)\5,binomial(n,k)*binomial(-n,2*n-5*k)) /* Max Alekseyev */

PARI

a(n) = round((5^n+sum(j=1,2*n-1,(sin(5*Pi*j/2/n)/sin(Pi*j/2/n))^n))/2/n)-2 /* Max Alekseyev */

PARI

a(n) = vecmax(Vec(Pol(vector(5,k,1))^n)); \\ Michel Marcus, Jan 29 2017

Formula

a(n) = Sum_{k=0..floor(2n/5)} binomial(n,k)*binomial(-n, 2n-5k); a(n) = (5^n + Sum_{j=1..2n-1} (sin(5j*Pi/(2n))/sin(j*Pi/(2n)))^n)/(2n) - 2. - Max Alekseyev, Mar 04 2005

D-finite with recurrence: 2*n*(2*n-1)*(3*n-4)*a(n) - (3*n-1)*(19*n^2-38*n+18)*a(n-1) - 5*(n-1)*(3*n-4)*(2*n-1)*a(n-2) + 25*(n-1)*(n-2)*(3*n-1)*a(n-3) = 0. - R. J. Mathar, Feb 21 2010 [Proved using the Almkvist-Zeilberger algorithm in EKHAD. - Doron Zeilberger, Apr 02 2013]

G.f.: sqrt((-5*x+2+2*sqrt(5*x^2-6*x+1))/(25*x^3-10*x^2-19*x+4)). - Mark van Hoeij, May 06 2013

a(n) ~ 5^n/(2*sqrt(Pi*n)). - Vaclav Kotesovec, Aug 09 2013

a(n) = Sum_{i=0..n/2} Sum_{j=0..n} Sum_{q=n..2*n}(f); f=( n!/((q - n)!*(j - 2*q + 2*n)!*(i - 2*j + q)!*(j - 2*i)!*i!) ); f=0 for (j - 2*q + 2*n)<0 or (i - 2*j + q)<0 or (j - 2*i)<0. Also see formula in Links section. - Zagros Lalo, Sep 25 2018