A025012 Central heptanomial coefficients: largest coefficient of (1+x+...+x^6)^n.

1, 1, 7, 37, 231, 1451, 9331, 60691, 398567, 2636263, 17538157, 117224317, 786588243, 5295520699, 35751527189, 241958082737, 1641010879207, 11150608945863, 75894449584849, 517331384963959, 3531097638576781, 24131083600660801, 165090433568378523

Offset: 0

Author

David W. Wilson

Keywords

• easy
• nonn

Comments

Apparently: number of n-step 1-D walks ending at the origin with steps of size 0, 1, 2 or 3. - David Scambler, Apr 09 2012

Generally, largest coefficient of (1+x+...+x^k)^n is asymptotic to (k+1)^n * sqrt(6/(k*(k+2)*Pi*n)). - Vaclav Kotesovec, Aug 09 2013

References

• Rudolph-Lilith, Michelle, and Lyle E. Muller. "On a link between Dirichlet kernels and central multinomial coefficients." Discrete Mathematics 338.9 (2015): 1567-1572.

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000 (first 201 terms from T. D. Noe)
• 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.
• Doron Zeilberger, EKHAD; Local copy.
• Index entries for sequences of k-nomial coefficients

Crossrefs

Cf. column 3 of A201552. Row 7 of A077042.

Programs

Maple

g := RootOf((2*g+1)*(g^3+5*g^2+6*g+1)*x^2+g*(g^3+3+11*g+7*g^2)*x-g, g);
ogf := sqrt((-4*x*(7*x+6)*(1+2*x)*g^3-4*x*(77*x^2+117*x+43)*g^2-4*x*(119*x^2+188*x+72)*g-307*x^2-217*x^3+27-67*x)/((7*x-1)*(49*x^2-14*x-27)))/(1+x);
series(ogf,x=0,30); # Mark van Hoeij, May 06 2013

Mathematica

With[{poly=Plus@@(x^Range[0,6])},Table[Max[CoefficientList[Expand[poly^i],x]], {i,0,20}]] (* Harvey P. Dale, Mar 09 2011 *)

Formula

D-finite with recurrence: 3*n*(3*n-1)*(3*n-2)*a(n) +(41*n^3-723*n^2+1132*n-456)*a(n-1) +7*(-383*n^3+2607*n^2-5992*n+4608)*a(n-2) +49*(-83*n^3+1317*n^2-6706*n +10512)*a(n-3) +343*(199*n^3-2487*n^2+10394*n-14496)*a(n-4) +2401*(n-4)*(43*n^2-437*n+1116)*a(n-5) -16807*(n-4)*(n-5)*(5*n-24)*a(n-6) -117649*(n-5)*(n-6)*(n-4)*a(n-7) = 0. - R. J. Mathar, Feb 21 2010

I checked that Mathar's conjectured formula fits the 200 terms of the b-file. Note that the coefficients are powers of 7: 49=7^2, 343=7^3, 2401=7^4, 16807=7^5, 117649=7^6. - T. D. Noe, Nov 15 2010

Comment from Doron Zeilberger, Apr 12 2011: The recurrence can be rigorously proved (and also discovered at the same time, no need for a guessing program) automatically via the Almkvist-Zeilberger algorithm (procedure AZd in the EKHAD link). In fact the program finds a 4th-order recurrence, not a 7th-order one: 343/3*(4*n+15)*(n+3)*(n+2)*(n+1)/(3*n+7)/(4*n+7)/(11+3*n)/(n+4)*a(n)+98/3*(4*n+15)*(n+3)*(n+2)*(2*n+7) /(11+3*n)/(3*n+10)/(4*n+11)/(n+4)*a(n+1) -14/3*(n+3)*(92*n^3+713*n^2+1747*n+1372)/(3*n+7)/(4*n+7)/(11+3*n)/(n+4)*a(n+2) -2/3*(4*n+15)*(2*n+7)*(37*n^2 +222*n+332)/(11+3*n)/(3*n+10)/(4*n+1)/(n+4)*a(n+3) +a(n+4) = 0.

G.f.: sqrt((-4*x*(7*x+6)*(1+2*x)*g^3-4*x*(77*x^2+117*x+43)*g^2-4*x*(119*x^2+188*x+72)*g-307*x^2-217*x^3+27-67*x)/((7*x-1)*(49*x^2-14*x-27)))/(1+x) where (2*g+1)*(g^3+5*g^2+6*g+1)*x^2+g*(g^3+3+11*g+7*g^2)*x-g = 0. - Mark van Hoeij, May 06 2013

a(n) ~ 7^n / sqrt(8*Pi*n). - Vaclav Kotesovec, Aug 09 2013