A201552 -id:A201552 - cp's OEIS frontend

A002426 Central trinomial coefficients: largest coefficient of (1 + x + x^2)^n.

1, 1, 3, 7, 19, 51, 141, 393, 1107, 3139, 8953, 25653, 73789, 212941, 616227, 1787607, 5196627, 15134931, 44152809, 128996853, 377379369, 1105350729, 3241135527, 9513228123, 27948336381, 82176836301, 241813226151, 712070156203, 2098240353907, 6186675630819

Offset: 0

N. J. A. Sloane, Simon Plouffe

Number of ordered trees with n + 1 edges, having root of odd degree and nonroot nodes of outdegree at most 2. - Emeric Deutsch, Aug 02 2002
Number of paths of length n with steps U = (1,1), D = (1,-1) and H = (1,0), running from (0,0) to (n,0) (i.e., grand Motzkin paths of length n). For example, a(3) = 7 because we have HHH, HUD, HDU, UDH, DUH, UHD and DHU. - Emeric Deutsch, May 31 2003
Number of lattice paths from (0,0) to (n,n) using steps (2,0), (0,2), (1,1). It appears that 1/sqrt((1 - x)^2 - 4*x^s) is the g.f. for lattice paths from (0,0) to (n,n) using steps (s,0), (0,s), (1,1). - Joerg Arndt, Jul 01 2011
Number of lattice paths from (0,0) to (n,n) using steps (1,0), (1,1), (1,2). - Joerg Arndt, Jul 05 2011
Binomial transform of A000984, with interpolated zeros. - Paul Barry, Jul 01 2003
Number of leaves in all 0-1-2 trees with n edges, n > 0. (A 0-1-2 tree is an ordered tree in which every vertex has at most two children.) - Emeric Deutsch, Nov 30 2003
a(n) is the number of UDU-free paths of n + 1 upsteps (U) and n downsteps (D) that start U. For example, a(2) = 3 counts UUUDD, UUDDU, UDDUU. - David Callan, Aug 18 2004
Diagonal sums of triangle A063007. - Paul Barry, Aug 31 2004
Number of ordered ballots from n voters that result in an equal number of votes for candidates A and B in a three candidate election. Ties are counted even when candidates A and B lose the election. For example, a(3) = 7 because ballots of the form (voter-1 choice, voter-2 choice, voter-3 choice) that result in equal votes for candidates A and B are the following: (A,B,C), (A,C,B), (B,A,C), (B,C,A), (C,A,B), (C,B,A) and (C,C,C). - Dennis P. Walsh, Oct 08 2004
a(n) is the number of weakly increasing sequences (a_1,a_2,...,a_n) with each a_i in [n]={1,2,...,n} and no element of [n] occurring more than twice. For n = 3, the sequences are 112, 113, 122, 123, 133, 223, 233. - David Callan, Oct 24 2004
Note that n divides a(n+1) - a(n). In fact, (a(n+1) - a(n))/n = A007971(n+1). - T. D. Noe, Mar 16 2005
Row sums of triangle A105868. - Paul Barry, Apr 23 2005
Number of paths of length n with steps U = (1,1), D = (1,-1) and H = (1,0), starting at (0,0), staying weakly above the x-axis (i.e., left factors of Motzkin paths) and having no H steps on the x-axis. Example: a(3) = 7 because we have UDU, UHD, UHH, UHU, UUD, UUH and UUU. - Emeric Deutsch, Oct 07 2007
Equals right border of triangle A152227; starting with offset 1, the row sums of triangle A152227. - Gary W. Adamson, Nov 29 2008
Starting with offset 1 = iterates of M * [1,1,1,...] where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 07 2009
Hankel transform is 2^n. - Paul Barry, Aug 05 2009
a(n) is prime for n = 2, 3 and 4, with no others for n <= 10^5 (E. W. Weisstein, Mar 14 2005). It has apparently not been proved that no [other] prime central trinomials exist. - Jonathan Vos Post, Mar 19 2010
a(n) is not divisible by 3 for n whose base-3 representation contains no 2 (A005836).
a(n) = number of (n-1)-lettered words in the alphabet {1,2,3} with as many occurrences of the substring (consecutive subword) [1,2] as those of [2,1]. See the papers by Ekhad-Zeilberger and Zeilberger. - N. J. A. Sloane, Jul 05 2012
a(n) = coefficient of x^n in (1 + x + x^2)^n. - L. Edson Jeffery, Mar 23 2013
a(n) is the number of ordered pairs (A,B) of subsets of {1,2,...,n} such that (i.) A and B are disjoint and (ii.) A and B contain the same number of elements. For example, a(2) = 3 because we have: ({},{}) ; ({1},{2}) ; ({2},{1}). - Geoffrey Critzer, Sep 04 2013
Also central terms of A082601. - Reinhard Zumkeller, Apr 13 2014
a(n) is the number of n-tuples with entries 0, 1, or 2 and with the sum of entries equal to n. For n=3, the seven 3-tuples are (1,1,1), (0,1,2), (0,2,1), (1,0,2), (1,2,0), (2,0,1), and (2,1,0). - Dennis P. Walsh, May 08 2015
The series 2*a(n) + 3*a(n+1) + a(n+2) = 2*A245455(n+3) has Hankel transform of L(2n+1)*2^n, offset n = 1, L being a Lucas number, see A002878 (empirical observation). - Tony Foster III, Sep 05 2016
The series (2*a(n) + 3*a(n+1) + a(n+2))/2 = A245455(n+3) has Hankel transform of L(2n+1), offset n=1, L being a Lucas number, see A002878 (empirical observation). - Tony Foster III, Sep 05 2016
Conjecture: An integer n > 3 is prime if and only if a(n) == 1 (mod n^2). We have verified this for n up to 8*10^5, and proved that a(p) == 1 (mod p^2) for any prime p > 3 (cf. A277640). - Zhi-Wei Sun, Nov 30 2016
This is the analog for Coxeter type B of Motzkin numbers (A001006) for Coxeter type A. - F. Chapoton, Jul 19 2017
a(n) is also the number of solutions to the equation x(1) + x(2) + ... + x(n) = 0, where x(1), ..., x(n) are in the set {-1,0,1}. Indeed, the terms in (1 + x + x^2)^n that produce x^n are of the form x^i(1)*x^i(2)*...*x^i(n) where i(1), i(2), ..., i(n) are in {0,1,2} and i(1) + i(2) + ... + i(n) = n. By setting j(t) = i(t) - 1 we obtain that j(1), ..., j(n) satisfy j(1) + ... + j(n) =0 and j(t) in {-1,0,1} for all t = 1..n. - Lucien Haddad, Mar 10 2018
If n is a prime greater than 3 then a(n)-1 is divisible by n^2. - Ira M. Gessel, Aug 08 2021
Let f(m) = ceiling((q+log(q))/log(9)), where q = -log(log(27)/(2*m^2*Pi)) then f(a(n)) = n, for n > 0. - Miko Labalan, Oct 07 2024
Diagonal of the rational function 1 / (1 - x^2 - y^2 - x*y). - Ilya Gutkovskiy, Apr 23 2025

			For n = 2, (x^2 + x + 1)^2 = x^4 + 2*x^3 + 3*x^2 + 2*x + 1, so a(2) = 3. - _Michael B. Porter_, Sep 06 2016

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Index entries for sequences of k-nomial coefficients
Index entries for "core" sequences

INVERT transform is A007971. Partial sums are A097893. Squares are A168597.
Main column of A027907. Column k=2 of A305161. Column k=0 of A328347. Column 1 of A201552(?).
Cf. A001006, A002878, A005043, A005717, A082758 (bisection), A273055 (bisection), A102445, A113302, A113303, A113304, A113305 (divisibility of central trinomial coefficients), A152227, A277640.

a002426 n = a027907 n n  -- Reinhard Zumkeller, Jan 22 2013

P:=PolynomialRing(Integers()); [Max(Coefficients((1+x+x^2)^n)): n in [0..26]]; // Bruno Berselli, Jul 05 2011

A002426 := proc(n) local k;
    sum(binomial(n, k)*binomial(n-k, k), k=0..floor(n/2));
end proc: # Detlef Pauly (dettodet(AT)yahoo.de), Nov 09 2001
# Alternatively:
a := n -> simplify(GegenbauerC(n,-n,-1/2)):
seq(a(n), n=0..29); # Peter Luschny, May 07 2016

Table[ CoefficientList[ Series[(1 + x + x^2)^n, {x, 0, n}], x][[ -1]], {n, 0, 27}] (* Robert G. Wilson v *)
a=b=1; Join[{a,b}, Table[c=((2n-1)b + 3(n-1)a)/n; a=b; b=c; c, {n,2,100}]]; Table[Sqrt[-3]^n LegendreP[n,1/Sqrt[-3]],{n,0,26}] (* Wouter Meeussen, Feb 16 2013 *)
a[ n_] := If[ n < 0, 0, 3^n Hypergeometric2F1[ 1/2, -n, 1, 4/3]]; (* Michael Somos, Jul 08 2014 *)
Table[4^n *JacobiP[n,-n-1/2,-n-1/2,-1/2], {n,0,29}] (* Peter Luschny, May 13 2016 *)
a[n_] := a[n] = Sum[n!/((n - 2*i)!*(i!)^2), {i, 0, n/2}]; Table[a[n], {n, 0, 29}] (* Shara Lalo and Zagros Lalo, Oct 03 2018 *)

trinomial(n,k):=coeff(expand((1+x+x^2)^n),x,k);
makelist(trinomial(n,n),n,0,12); /* Emanuele Munarini, Mar 15 2011 */

makelist(ultraspherical(n,-n,-1/2),n,0,12); /* Emanuele Munarini, Dec 20 2016 */

{a(n) = if( n<0, 0, polcoeff( (1 + x + x^2)^n, n))};

/* as lattice paths: same as in A092566 but use */
steps=[[2, 0], [0, 2], [1, 1]];
/* Joerg Arndt, Jul 01 2011 */

a(n)=polcoeff(sum(m=0, n, (2*m)!/m!^2 * x^(2*m) / (1-x+x*O(x^n))^(2*m+1)), n) \\ Paul D. Hanna, Sep 21 2013

from math import comb
def A002426(n): return sum(comb(n,k)*comb(k,n-k) for k in range(n+1)) # Chai Wah Wu, Nov 15 2022

A002426 = lambda n: hypergeometric([-n/2, (1-n)/2], [1], 4)
[simplify(A002426(n)) for n in (0..29)]
# Peter Luschny, Sep 17 2014

def A():
    a, b, n = 1, 1, 1
    yield a
    while True:
        yield b
        n += 1
        a, b = b, ((3 * (n - 1)) * a + (2 * n - 1) * b) // n
A002426 = A()
print([next(A002426) for  in range(30)])  # _Peter Luschny, May 16 2016

G.f.: 1/sqrt(1 - 2*x - 3*x^2).
E.g.f.: exp(x)*I_0(2x), where I_0 is a Bessel function. - Michael Somos, Sep 09 2002
a(n) = 2*A027914(n) - 3^n. - Benoit Cloitre, Sep 28 2002
a(n) is asymptotic to d*3^n/sqrt(n) with d around 0.5.. - Benoit Cloitre, Nov 02 2002, d = sqrt(3/Pi)/2 = 0.4886025119... - Alec Mihailovs (alec(AT)mihailovs.com), Feb 24 2005
D-finite with recurrence: a(n) = ((2*n - 1)*a(n-1) + 3*(n - 1)*a(n-2))/n; a(0) = a(1) = 1; see paper by Barcucci, Pinzani and Sprugnoli.
Inverse binomial transform of A000984. - Vladeta Jovovic, Apr 28 2003
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(k, k/2)*(1 + (-1)^k)/2; a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*binomial(2*k, k). - Paul Barry, Jul 01 2003
a(n) = Sum_{k>=0} binomial(n, 2*k)*binomial(2*k, k). - Philippe Deléham, Dec 31 2003
a(n) = Sum_{i+j=n, 0<=j<=i<=n} binomial(n, i)*binomial(i, j). - Benoit Cloitre, Jun 06 2004
a(n) = 3*a(n-1) - 2*A005043(n). - Joost Vermeij (joost_vermeij(AT)hotmail.com), Feb 10 2005
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(k, n-k). - Paul Barry, Apr 23 2005
a(n) = (-1/4)^n*Sum_{k=0..n} binomial(2*k, k)*binomial(2*n-2*k, n-k)*(-3)^k. - Philippe Deléham, Aug 17 2005
a(n) = A111808(n,n). - Reinhard Zumkeller, Aug 17 2005
a(n) = Sum_{k=0..n} (((1 + (-1)^k)/2)*Sum_{i=0..floor((n-k)/2)} binomial(n, i)*binomial(n-i, i+k)*((k + 1)/(i + k + 1))). - Paul Barry, Sep 23 2005
a(n) = 3^n*Sum_{j=0..n} (-1/3)^j*C(n, j)*C(2*j, j); follows from (a) in A027907. - Loic Turban (turban(AT)lpm.u-nancy.fr), Aug 31 2006
a(n) = (1/2)^n*Sum_{j=0..n} 3^j*binomial(n, j)*binomial(2*n-2*j, n) = (3/2)^n*Sum_{j=0..n} (1/3)^j*binomial(n, j)*binomial(2*j, n); follows from (c) in A027907. - Loic Turban (turban(AT)lpm.u-nancy.fr), Aug 31 2006
a(n) = (1/Pi)*Integral_{x=-1..3} x^n/sqrt((3 - x)*(1 + x)) is moment representation. - Paul Barry, Sep 10 2007
G.f.: 1/(1 - x - 2x^2/(1 - x - x^2/(1 - x - x^2/(1 - ... (continued fraction). - Paul Barry, Aug 05 2009
a(n) = sqrt(-1/3)*(-1)^n*hypergeometric([1/2, n+1], [1], 4/3). - Mark van Hoeij, Nov 12 2009
a(n) = (1/Pi)*Integral_{x=-1..1} (1 + 2*x)^n/sqrt(1 - x^2) = (1/Pi)*Integral_{t=0..Pi} (1 + 2*cos(t))^n. - Eli Wolfhagen, Feb 01 2011
In general, g.f.: 1/sqrt(1 - 2*a*x + x^2*(a^2 - 4*b)) = 1/(1 - a*x)*(1 - 2*x^2*b/(G(0)*(a*x - 1) + 2*x^2*b)); G(k) = 1 - a*x - x^2*b/G(k+1); for g.f.: 1/sqrt(1 - 2*x - 3*x^2) = 1/(1 - x)*(1 - 2*x^2/(G(0)*(x - 1) + 2*x^2)); G(k) = 1 - x - x^2/G(k+1), a = 1, b = 1; (continued fraction). - Sergei N. Gladkovskii, Dec 08 2011
a(n) = Sum_{k=0..floor(n/3)} (-1)^k*binomial(2*n-3*k-1, n-3*k)*binomial(n, k). - Gopinath A. R., Feb 10 2012
G.f.: A(x) = x*B'(x)/B(x) where B(x) satisfies B(x) = x*(1 + B(x) + B(x)^2). - Vladimir Kruchinin, Feb 03 2013 (B(x) = x*A001006(x) - Michael Somos, Jul 08 2014)
G.f.: G(0), where G(k) = 1 + x*(2 + 3*x)*(4*k + 1)/(4*k + 2 - x*(2 + 3*x)*(4*k + 2)*(4*k + 3)/(x*(2 + 3*x)*(4*k + 3) + 4*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 29 2013
E.g.f.: exp(x) * Sum_{k>=0} (x^k/k!)^2. - Geoffrey Critzer, Sep 04 2013
G.f.: Sum_{n>=0} (2*n)!/n!^2*(x^(2*n)/(1 - x)^(2*n+1)). - Paul D. Hanna, Sep 21 2013
0 = a(n)*(9*a(n+1) + 9*a(n+2) - 6*a(n+3)) + a(n+1)*(3*a(n+1) + 4*a(n+2) - 3*a(n+3)) + a(n+2)*(-a(n+2) + a(n+3)) for all n in Z. - Michael Somos, Jul 08 2014
a(n) = hypergeometric([-n/2, (1-n)/2], [1], 4). - Peter Luschny, Sep 17 2014
a(n) = A132885(n,0), that is, a(n) = A132885(A002620(n+1)). - Altug Alkan, Nov 29 2015
a(n) = GegenbauerC(n,-n,-1/2). - Peter Luschny, May 07 2016
a(n) = 4^n*JacobiP[n,-n-1/2,-n-1/2,-1/2]. - Peter Luschny, May 13 2016
From Alexander Burstein, Oct 03 2017: (Start)
G.f.: A(4*x) = B(-x)*B(3*x), where B(x) is the g.f. of A000984.
G.f.: A(2*x)*A(-2*x) = B(x^2)*B(9*x^2).
G.f.: A(x) = 1 + x*M'(x)/M(x), where M(x) is the g.f. of A001006. (End)
a(n) = Sum_{i=0..n/2} n!/((n - 2*i)!*(i!)^2). [Cf. Lalo and Lalo link. It is Luschny's terminating hypergeometric sum.] - Shara Lalo and Zagros Lalo, Oct 03 2018
From Peter Bala, Feb 07 2022: (Start)
a(n)^2 = Sum_{k = 0..n} (-3)^(n-k)*binomial(2*k,k)^2*binomial(n+k,n-k) and has g.f. Sum_{n >= 0} binomial(2*n,n)^2*x^n/(1 + 3*x)^(2*n+1). Compare with the g.f. for a(n) given above by Hanna.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all prime p and positive integers n and k.
Conjecture: The stronger congruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for all prime p >= 5 and positive integers n and k. (End)
a(n) = A005043(n) + A005717(n) for n >= 1. - Amiram Eldar, May 17 2024
For even n, a(n) = (n-1)!!* 2^{n/2}/ (n/2)!* 2F1(-n/2,-n/2;1/2;1/4). For odd n, a(n) = n!! *2^(n/2-1/2) / (n/2-1/2)! * 2F1(1/2-n/2,1/2-n/2;3/2;1/4). - R. J. Mathar, Mar 19 2025

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

Original entry on oeis.org

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

David W. Wilson

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

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

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

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

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

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

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

A025014 Central "nonomial" coefficient: largest coefficient of (1+x+...+x^8)^n.

Original entry on oeis.org

1, 1, 9, 61, 489, 3951, 32661, 273127, 2306025, 19610233, 167729959, 1441383219, 12434998005, 107632809909, 934263293679, 8129320828911, 70886845397481, 619288973447049, 5419332253680705, 47494787636620701, 416800775902696839

Offset: 0

David W. Wilson

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

T. D. Noe, Table of n, a(n) for n=0..200
Michelle Rudolph-Lilith and Lyle E. Muller, On a link between Dirichlet kernels and central multinomial coefficients, Discrete Mathematics 338.9 (2015): 1567-1572.
Index entries for sequences of k-nomial coefficients

Column 4 of A201552. Row 9 of A077042.

seq(add((-1)^k * binomial(n, k)*binomial(5*n-9*k-1, n-1), k = 0..floor(4*n/9)), n = 0..20) ; # Peter Bala, Oct 16 2024

Flatten[{1,Table[Coefficient[Expand[Sum[x^j,{j,0,8}]^n],x^(4*n)],{n,1,20}]}] (* Vaclav Kotesovec, Aug 09 2013 *)

The Almkvist-Zeilberger algorithm in EKHAD establishes the following recurrence:
-6561*(4*n+17)*(4*n+13)*(5*n+24)*(5*n+19)*(5*n+14)*(5*n+23)*(n+4)*(n+3)*(n+2)*(n+1)*a(n)+1458*(5*n+24)*(5*n+19)*(4*n+17)*(5*n+9)*(4*n+9)*(5*n+18)*(2*n+9)*(n+4)*(
n+3)*(n+2)*a(n+1)+162*(5*n+24)*(5*n+14)*(4*n+13)*(5*n+23)*(n+4)*(n+3)*(1020*n^4+12291*n^3+53378*n^2+98617*n+65610)*a(n+2)-18*(4*n+17)*(4*n+9)*(5*n+19)*(2*n+9)*(5
*n+9)*(5*n+18)*(n+4)*(385*n^3+4158*n^2+14551*n+16610)*a(n+3)-(5*n+23)*(4*n+13)*(4*n+9)*(5*n+24)*(5*n+14)*(5*n+9)*(2101*n^4+33616*n^3+201391*n^2+535416*n+532980)*
a(n+4)+8*(4*n+19)*(5*n+19)*(5*n+14)*(5*n+9)*(2*n+9)*(4*n+17)*(4*n+13)*(4*n+9)*(5*n+18)*(n+5)*a(n+5) = 0. - Doron Zeilberger, Apr 02 2013.
a(n) ~ 9^n * sqrt(3/(40*Pi*n)). - Vaclav Kotesovec, Aug 09 2013
a(n) = Sum_{k = 0..floor(4*n/9)} (-1)^k * binomial(n, k)*binomial(5*n-9*k-1, n-1). - Peter Bala, Oct 16 2024

A077042 Square array read by falling antidiagonals of central polynomial coefficients: largest coefficient in expansion of (1 + x + x^2 + ... + x^(n-1))^k = ((1-x^n)/(1-x))^k, i.e., the coefficient of x^floor(k(n-1)/2) and of x^ceiling(k(n-1)/2); also number of compositions of floor(k*(n+1)/2) into exactly k positive integers each no more than n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 3, 3, 1, 1, 0, 1, 6, 7, 4, 1, 1, 0, 1, 10, 19, 12, 5, 1, 1, 0, 1, 20, 51, 44, 19, 6, 1, 1, 0, 1, 35, 141, 155, 85, 27, 7, 1, 1, 0, 1, 70, 393, 580, 381, 146, 37, 8, 1, 1, 0, 1, 126, 1107, 2128, 1751, 780, 231, 48, 9, 1, 1, 0, 1, 252, 3139

Offset: 0

Henry Bottomley, Oct 22 2002

From Michel Marcus, Dec 01 2012: (Start)
A pair of numbers written in base n are said to be comparable if all digits of the first number are at least as big as the corresponding digit of the second number, or vice versa. Otherwise, this pair will be defined as uncomparable. A set of pairwise uncomparable integers will be called anti-hierarchic.
T(n,k) is the size of the maximal anti-hierarchic set of integers written with k digits in base n.
For example, for base n=2 and k=4 digits:
- 0 (0000) and 15 (1111) are comparable, while 6 (0110) and 9 (1001) are uncomparable,
- the maximal antihierarchic set is {3 (0011), 5 (0101), 6 (0110), 9 (1001), 10 (1010), 12 (1100)} with 6 elements that are all pairwise uncomparable. (End)

			Rows of square array start:
  1,    0,    0,    0,    0,    0,    0, ...
  1,    1,    1,    1,    1,    1,    1, ...
  1,    1,    2,    3,    6,   10,   20, ...
  1,    1,    3,    7,   19,   51,  141, ...
  1,    1,    4,   12,   44,  155,  580, ...
  1,    1,    5,   19,   85,  381, 1751, ...
  ...
Read by antidiagonals:
  1;
  0, 1;
  0, 1, 1;
  0, 1, 1, 1;
  0, 1, 2, 1, 1;
  0, 1, 3, 3, 1, 1;
  0, 1, 6, 7, 4, 1, 1;
  ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Denis Bouyssou, Thierry Marchant, Marc Pirlot, The size of the largest antichains in products of linear orders, arXiv:1903.07569 [math.CO], 2019.
J. W. Sander, On maximal antihierarchic sets of integers, Discrete Mathematics, Volume 113, Issues 1-3, 5 April 1993, Pages 179-189.
Index entries for sequences related to compositions

Rows include A000007, A000012, A001405, A002426, A005190, A005191, A018901, A025012, A025013, A025014, A025015, A201549, A225779, A201550. Columns include A000012, A000012, A001477, A077043, A005900, A077044, A071816, A133458.
Central diagonal is A077045, with A077046 and A077047 either side. Cf. A067059, A270918, A201552.
Cf. A273975.

t[n_, k_] := Max[ CoefficientList[ Series[ ((1-x^n) / (1-x))^k, {x, 0, k*(n-1)}], x]]; t[0, 0] = 1; t[0, ] = 0; Flatten[ Table[ t[n-k, k], {n, 0, 12}, {k, n, 0, -1}]] (* _Jean-François Alcover, Nov 04 2011 *)

T(n,k)=if(n<1 || k<1,k==0,vecmax(Vec(((1-x^n)/(1-x))^k)))

By the central limit theorem, T(n,k) is roughly n^(k-1)*sqrt(6/(Pi*k)).

T(n,k) = Sum{j=0,h/n} (-1)^j*binomial(k,j)*binomial(k-1+h-n*j,k-1) with h=floor(k*(n-1)/2), k>0. - Michel Marcus, Dec 01 2012

A208597 T(n,k) = number of n-bead necklaces labeled with numbers -k..k not allowing reversal, with sum zero.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 7, 6, 1, 5, 13, 23, 11, 1, 6, 21, 60, 77, 26, 1, 7, 31, 125, 291, 297, 57, 1, 8, 43, 226, 791, 1564, 1163, 142, 1, 9, 57, 371, 1761, 5457, 8671, 4783, 351, 1, 10, 73, 568, 3431, 14838, 39019, 49852, 20041, 902, 1, 11, 91, 825, 6077, 34153, 129823

Offset: 1

R. H. Hardin, Feb 29 2012

			Table starts
...1....1.....1......1.......1.......1........1........1........1.........1
...2....3.....4......5.......6.......7........8........9.......10........11
...3....7....13.....21......31......43.......57.......73.......91.......111
...6...23....60....125.....226.....371......568......825.....1150......1551
..11...77...291....791....1761....3431.....6077....10021....15631.....23321
..26..297..1564...5457...14838...34153....69784...130401...227314....374825
..57.1163..8671..39019..129823..353333...833253..1764925..3438877...6267735
.142.4783.49852.288317.1172298.3770475.10259448.24627705.53630854.108036775

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 185 terms from R. H. Hardin)

Columns 1-7 are A208602, A208591, A208592, A208593, A208594, A208595, A208596.
Rows 3-7 are A002061(n+1), A208598, A208599, A208600, A208601.
Main diagonal is A208590.
Cf. A201552, A286928, A208825.

comps[r_, m_, k_] := Sum[(-1)^i*Binomial[r-1-i*m, k-1]*Binomial[k, i], {i, 0, Floor[(r-k)/m]}]; a[n_, k_] := DivisorSum[n, EulerPhi[n/#] comps[#*(k + 1), 2k+1, #]&]/n; Table[a[n-k+1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Oct 07 2017, after Andrew Howroyd's PARI code *)

comps(r,m,k)=sum(i=0,floor((r-k)/m),(-1)^i*binomial(r-1-i*m, k-1)*binomial(k, i));
a(n,k)=sumdiv(n,d,eulerphi(n/d)*comps(d*(k+1), 2*k+1, d))/n;
for(n=1,8,for(k=1,10,print1(a(n,k),", ")); print()); \\ Andrew Howroyd, May 16 2017

from sympy import binomial, divisors, totient, floor
def comps(r, m, k): return sum([(-1)**i*binomial(r - 1 - i*m, k - 1)*binomial(k, i) for i in range(floor((r - k)/m) + 1)])
def a(n, k): return sum([totient(n//d)*comps(d*(k + 1), 2*k + 1, d) for d in divisors(n)])//n
for n in range(1, 12): print([a(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Nov 07 2017, after PARI code

require(numbers)
comps <- function(r, m, k) {
  S <- numeric()
  for (i in 0:floor((r-k)/m)) S <- c(S, (-1)^i*choose(r-1-i*m, k-1)*choose(k, i))
  return(sum(S))
}
a <- function(n, k) {
  S <- numeric()
  for (d in divisors(n)) S <- c(S, eulersPhi(n/d)*comps(d*(k+1), 2*k+1, d))
  return(sum(S)/n)
}
for (n in 1:11) {
  for (k in 1:n) {
    print(a(k,n-k+1))
  }
} # Indranil Ghosh, Nov 07 2017, after PARI code

T(n,k) = Sum_{d|n} phi(n/d) * A201552(d, k). - Andrew Howroyd, Oct 14 2017
Empirical for row n:
n=1: a(k) = 1.
n=2: a(k) = k + 1.
n=3: a(k) = k^2 + k + 1.
n=4: a(k) = (4/3)*k^3 + 2*k^2 + (5/3)*k + 1.
n=5: a(k) = (23/12)*k^4 + (23/6)*k^3 + (37/12)*k^2 + (7/6)*k + 1.
n=6: a(k) = (44/15)*k^5 + (22/3)*k^4 + (23/3)*k^3 + (14/3)*k^2 + (12/5)*k + 1.
n=7: a(k) = (841/180)*k^6 + (841/60)*k^5 + (325/18)*k^4 + (51/4)*k^3 + (949/180)*k^2 + (37/30)*k + 1.

A273975 Three-dimensional array written by antidiagonals in k,n: T(k,n,h) with k >= 1, n >= 0, 0 <= h <= n*(k-1) is the coefficient of x^h in the polynomial (1 + x + ... + x^(k-1))^n = ((x^k-1)/(x-1))^n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 3, 2, 1, 1, 3, 6, 7, 6, 3, 1, 1, 4, 6, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 4, 3, 2, 1, 1, 3, 6, 10, 12, 12, 10, 6, 3, 1, 1, 4, 10

Offset: 1

Andrey Zabolotskiy, Nov 10 2016

Equivalently, T(k,n,h) is the number of ordered sets of n nonnegative integers < k with the sum equal to h.
From Juan Pablo Herrera P., Nov 21 2016: (Start)
T(k,n,h) is the number of possible ways of randomly selecting h cards from k-1 sets, each with n different playing cards. It is also the number of lattice paths from (0,0) to (n,h) using steps (1,0), (1,1), (1,2), ..., (1,k-1).
Shallow diagonal sums of each triangle with fixed k give the k-bonacci numbers. (End)
T(k,n,h) is the number of n-dimensional grid points of a k X k X ... X k grid, which are lying in the (n-1)-dimensional hyperplane which is at an L1 distance of h from one of the grid's corners, and normal to the corresponding main diagonal of the grid. - Eitan Y. Levine, Apr 23 2023

			For first few k and for first few n, the rows with h = 0..n*(k-1) are given:
k=1:  1;  1;  1;  1;  1; ...
k=2:  1;  1, 1;  1, 2, 1;  1, 3, 3, 1;  1, 4, 6, 4, 1; ...
k=3:  1;  1, 1, 1;  1, 2, 3, 2, 1;  1, 3, 6, 7, 6, 3, 1; ...
k=4:  1;  1, 1, 1, 1;  1, 2, 3, 4, 3, 2, 1; ...
For example, (1 + x + x^2)^3 = 1 + 3*x + 6*x^2 + 7*x^3 + 6*x^4 + 3*x^5 + x^6, hence T(3,3,2) = T(3,3,4) = 6.
From _Eitan Y. Levine_, Apr 23 2023: (Start)
Example for the repeated cumulative sum formula, for (k,n)=(3,3) (each line is the cumulative sum of the previous line, and the first line is the padded, alternating 3rd row from Pascal's triangle):
  1  0  0 -3  0  0  3  0  0 -1
  1  1  1 -2 -2 -2  1  1  1
  1  2  3  1 -1 -3 -2 -1
  1  3  6  7  6  3  1
which is T(3,3,h). (End)

Andrey Zabolotskiy, slices with k+n = 1..20, flattened
Florentin Smarandache, K-Nomial Coefficients, arXiv:math/0612062 [math.GM], 2006 (originally published in French in: F. Smarandache, Généralisations et Généralités, Ed. Nouvelle, 1984, pp. 24-26).

k-nomial arrays for fixed k=1..10: A000012, A007318, A027907, A008287, A035343, A063260, A063265, A171890, A213652, A213651.
Arrays for fixed n=0..6: A000012, A000012, A004737, A109439, A277949, A277950, A277951.
Central n-nomial coefficients for n=1..9, i.e., sequences with h=floor(n*(k-1)/2) and fixed n: A000012, A000984 (A001405), A002426, A005721 (A005190), A005191, A063419 (A018901), A025012, (A025013), A025014, A174061 (A025015), A201549, (A225779), A201550. Arrays: A201552, A077042, see also cfs. therein.
Triangle n=k-1: A181567. Triangle n=k: A163181.

a = Table[CoefficientList[Sum[x^(h-1),{h,k}]^n,x],{k,10},{n,0,9}];
Flatten@Table[a[[s-n,n+1]],{s,10},{n,0,s-1}]
(* alternate program *)
row[k_, n_] := Nest[Accumulate,Upsample[Table[((-1)^j)*Binomial[n,j],{j,0,n}],k],n][[;;n*(k-1)+1]] (* Eitan Y. Levine, Apr 23 2023 *)

T(k,n,h) = Sum_{i = 0..floor(h/k)} (-1)^i*binomial(n,i)*binomial(n+h-1-k*i,n-1). [Corrected by Eitan Y. Levine, Apr 23 2023]
From Eitan Y. Levine, Apr 23 2023: (Start)
(T(k,n,h))_{h=0..n*(k-1)} = f(f(...f(g(P))...)), where:
(x_i)_{i=0..m} denotes a tuple (in particular, the LHS contains the values for 0 <= h <= n*(k-1)),
f repeats n times,
f((x_i){i=0..m}) = (Sum{j=0..i} x_j)_{i=0..m} is the cumulative sum function,
g((x_i){i=0..m}) = (x(i/k) if k|i, otherwise 0)_{i=0..m*k} is adding k-1 zeros between adjacent elements,
and P=((-1)^i*binomial(n,i))_{i=0..n} is the n-th row of Pascal's triangle, with alternating signs. (End)
From Eitan Y. Levine, Jul 27 2023: (Start)
Recurrence relations, the first follows from the sequence's defining polynomial as mentioned in the Smarandache link:
T(k,n+1,h) = Sum_{i = 0..s-1} T(k,n,h-i)
T(k+1,n,h) = Sum_{i = 0..n} binomial(n,i)*T(k,n-i,h-i*k) (End)

A083669 Number of ordered quintuples (a,b,c,d,e), -n <= a,b,c,d,e <= n, such that a+b+c+d+e = 0.

Original entry on oeis.org

1, 51, 381, 1451, 3951, 8801, 17151, 30381, 50101, 78151, 116601, 167751, 234131, 318501, 423851, 553401, 710601, 899131, 1122901, 1386051, 1692951, 2048201, 2456631, 2923301, 3453501, 4052751, 4726801, 5481631, 6323451, 7258701

Offset: 0

Benoit Cloitre, Jun 14 2003

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Row 5 of A201552.

Cf. A003215, A063496.

[1+5*n*(n+1)*(23*n^2+23*n+14)/12: n in [0..30]]; // Vincenzo Librandi, Dec 15 2018

LinearRecurrence[{5, -10, 10, -5, 1}, {1, 51, 381, 1451, 3951}, 30] (* Vincenzo Librandi, Dec 15 2018 *)

a(n)=115/12*n^4+115/6*n^3+185/12*n^2+35/6*n+1

{a(n) = polcoeff((sum(k=0, 2*n, x^k))^5, 5*n, x)} \\ Seiichi Manyama, Dec 14 2018

a(n) = 1 + 5*n*(n+1)*(23*n^2 + 23*n + 14)/12.
a(n) = (1/Pi)*Integral_{x=0..Pi} (sin((n+1/2)*x)/sin(x/2))^5. - Yalcin Aktar, Dec 03 2011
G.f.: ( -1 - 46*x - 136*x^2 - 46*x^3 - x^4 ) / (x-1)^5. - R. J. Mathar, Dec 17 2011
a(n) = [x^(5*n)] (Sum_{k=0..2*n} x^k)^5. - Seiichi Manyama, Dec 14 2018

A201549 Number of arrays of n integers in -5..5 with sum zero.

Original entry on oeis.org

1, 1, 11, 91, 891, 8801, 88913, 908755, 9377467, 97464799, 1018872811, 10701243741, 112835748609, 1193692544825, 12663809507129, 134678108144591, 1435345208419771, 15326122342137035, 163920458145421109

Offset: 0

R. H. Hardin, Dec 02 2011

nonn

Also largest coefficient of (1+x+...+x^10)^n. - Vaclav Kotesovec, Aug 09 2013

			Some solutions for n=6
.-5...-5...-1...-2...-1....3....4....1....1...-1....3....4....5....0...-5....5
.-2....4...-1....1....0...-4....2....1....1...-5...-4...-4....1...-3....5...-5
..0...-3....1....3...-1...-4....0...-1....2...-4...-4...-4...-5...-3...-2....5
..4....3....3...-3....4....5....1....2....2....5....3....5...-3....2....2...-5
..5...-4...-1...-4...-4....1...-2...-4...-5....2....0....4....1....4...-4....3
.-2....5...-1....5....2...-1...-5....1...-1....3....2...-5....1....0....4...-3

Seiichi Manyama, Table of n, a(n) for n = 0..500 (terms 1..210 from R. H. Hardin) [It was suggested that the initial terms of this b-file were wrong, but in fact they are correct. - _N. J. A. Sloane_, Jan 19 2019]
Index entries for sequences of k-nomial coefficients

Column 5 of A201552. Row 11 of A077042.

seq(add((-1)^k * binomial(n, k)*binomial(6*n-11*k-1, n-1), k = 0..floor(n/2)), n = 0..20); # Peter Bala, Oct 16 2024

Table[Coefficient[Expand[Sum[x^j,{j,0,10}]^n],x^(5*n)],{n,1,20}] (* Vaclav Kotesovec, Aug 09 2013 *)

{a(n) = polcoeff((sum(k=0, 10, x^k))^n, 5*n, x)} \\ Seiichi Manyama, Dec 14 2018

From Vaclav Kotesovec, Aug 09 2013: (Start)
Recurrence: 726*(n-2) * (n-1) * (2*n-1) * (3*n-7) * (3*n-1) * (5*n-19) * (5*n-9) * (5*n-8) * (6*n-25) * (6*n-13) * (6*n-1) * (2280*n^4  -  18164*n^3  +  49523*n^2  -  53013*n  +  18900)*a(n-3)  -  3*(2*n-1) * (3*n-7) * (3*n-1) * (5*n-19) * (5*n-14) * (5*n-9) * (5*n-8) * (6*n-25) * (6*n-19) * (6*n-13) * (6*n-1) * (4651*n^4  -  18604*n^3  +  27451*n^2  -  17694*n  +  4200)*a(n-1)  +  3993*(n-3) * (n-2) * (n-1) * (3*n-4) * (5*n-14) * (5*n-4) * (5*n-3) * (6*n-19) * (6*n-7) * (6*n-1) * (5310*n^5  -  65313*n^4  +  295326*n^3  -  594091*n^2  +  499480*n  -  112320)*a(n-4)  -  33*(n-1) * (3*n-4) * (5*n-19) * (5*n-14) * (5*n-4) * (5*n-3) * (6*n-25) * (6*n-19) * (6*n-7) * (45306*n^6  -  385101*n^5  +  1267841*n^4  -  2002349*n^3  +  1504595*n^2  -  451668*n  +  42120)*a(n-2)  -  161051*(n-5) * (n-4) * (n-3) * (n-2) * (n-1) * (3*n-4) * (3*n-1) * (5*n-14) * (5*n-9) * (5*n-4) * (5*n-3) * (6*n-19) * (6*n-13) * (6*n-7) * (6*n-1)*a(n-6)  -  43923*(n-4) * (n-3) * (n-2) * (n-1) * (2*n-1) * (3*n-7) * (3*n-1) * (5*n-19) * (5*n-9) * (5*n-8) * (5*n-4) * (6*n-25) * (6*n-13) * (6*n-7) * (6*n-1)*a(n-5)  +  5*n*(3*n-7) * (3*n-4) * (5*n-19) * (5*n-14) * (5*n-9) * (5*n-8) * (5*n-4) * (5*n-3) * (5*n-2) * (5*n-1) * (6*n-25) * (6*n-19) * (6*n-13) * (6*n-7)*a(n) = 0.
a(n) ~ 11^n / sqrt(20*Pi*n). (End)
a(n) = Sum_{k = 0..floor(n/2)} (-1)^k * binomial(n, k)*binomial(6*n-11*k-1, n-1). - Peter Bala, Oct 16 2024

a(0)=1 prepended by Seiichi Manyama, Dec 14 2018

A201550 Number of arrays of n integers in -6..6 with sum zero.

Original entry on oeis.org

1, 1, 13, 127, 1469, 17151, 204763, 2473325, 30162301, 370487485, 4577127763, 56813989827, 707972099627, 8851373201919, 110976634957761, 1394804756117877, 17567994350713469, 221690794842728445, 2802194053806820153

Offset: 0

R. H. Hardin, Dec 02 2011

Also largest coefficient of (1+x+...+x^12)^n. - Vaclav Kotesovec, Aug 09 2013

			Some solutions for n=5
.-2...-5...-2...-1....3...-6....0...-3....1....6...-6...-2....5....0...-4...-3
..2...-3...-2....3....0...-1....6...-4....6....1....5....2...-1....3....2....3
..0...-4....4...-6...-4....1...-3....0...-4...-5....0...-6...-3....0....4...-4
..0....6....3....5...-5....6....0....4....3...-4....4....0...-5...-3....3...-1
..0....6...-3...-1....6....0...-3....3...-6....2...-3....6....4....0...-5....5

Seiichi Manyama, Table of n, a(n) for n = 0..450 (terms 1..210 from R. H. Hardin) [It was suggested that the initial terms of this b-file were wrong, but in fact they are correct. - _N. J. A. Sloane_, Jan 19 2019]
Vaclav Kotesovec, Recurrence
Index entries for sequences of k-nomial coefficients

Column 6 of A201552. Row 13 of A077042.

seq(add((-1)^k * binomial(n, k)*binomial(7*n-13*k-1, n-1), k = 0..floor(n/2)), n = 0..20); # Peter Bala, Oct 16 2024

Table[Coefficient[Expand[Sum[x^j,{j,0,12}]^n],x^(6*n)],{n,1,20}] (* Vaclav Kotesovec, Aug 09 2013 *)

{a(n) = polcoeff((sum(k=0, 12, x^k))^n, 6*n, x)} \\ Seiichi Manyama, Dec 14 2018

a(n) ~ 13^n / sqrt(28*Pi*n). - Vaclav Kotesovec, Aug 09 2013

a(n) = Sum_{k = 0..floor(n/2)} (-1)^k * binomial(n, k)*binomial(7*n-13*k-1, n-1). - Peter Bala, Oct 16 2024

a(0)=1 prepended by Seiichi Manyama, Dec 14 2018

A201551 Number of arrays of n integers in -7..7 with sum zero.

Original entry on oeis.org

1, 1, 15, 169, 2255, 30381, 418503, 5832765, 82073295, 1163205475, 16581420835, 237481736823, 3414582082055, 49258226347903, 712601187601395, 10334165623697259, 150186639579545295, 2186774434431445455, 31893473567409732813, 465851764737061437765

Offset: 0

R. H. Hardin, Dec 02 2011

nonn

			Some solutions for n=5
.-6....6....2...-1...-4...-5...-2...-2....4...-6....2....7...-5...-5....3....5
.-2....0....7...-3....2....6...-3....7...-4...-2...-7...-3....6....2...-3...-7
..0...-6...-5...-2....6....5....5...-5...-6....5....5....1...-4....4....1...-4
..2...-2...-4....0...-7...-5....6...-7....6....6...-5....1....4....5...-3....7
..6....2....0....6....3...-1...-6....7....0...-3....5...-6...-1...-6....2...-1

Seiichi Manyama, Table of n, a(n) for n = 0..400 (terms 1..210 from R. H. Hardin) [It was suggested that the initial terms of this b-file were wrong, but in fact they are correct. - _N. J. A. Sloane_, Jan 19 2019]

Column 7 of A201552.

seq(add((-1)^k*binomial(n, k)*binomial(8*n-15*k-1, n-1), k = 0..floor(n/2)), n = 0..20); # Peter Bala, Oct 19 2024

a[n_] := If[n==0, 1, Coefficient[Expand[Sum[x^k, {k, 0, 14}]^n], x^(7n)]]; Array[a, 25, 0] (* Amiram Eldar, Dec 14 2018 *)

{a(n) = polcoeff((sum(k=0, 14, x^k))^n, 7*n, x)} \\ Seiichi Manyama, Dec 14 2018

a(n) ~ sqrt(3) * 15^n / (4*sqrt(7*Pi*n)). - Vaclav Kotesovec, Dec 15 2018

a(0)=1 prepended by Seiichi Manyama, Dec 14 2018

cp's OEIS Frontend

A002426 Central trinomial coefficients: largest coefficient of (1 + x + x^2)^n.

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

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A025014 Central "nonomial" coefficient: largest coefficient of (1+x+...+x^8)^n.

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A208597 T(n,k) = number of n-bead necklaces labeled with numbers -k..k not allowing reversal, with sum zero.

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A273975 Three-dimensional array written by antidiagonals in k,n: T(k,n,h) with k >= 1, n >= 0, 0 <= h <= n*(k-1) is the coefficient of x^h in the polynomial (1 + x + ... + x^(k-1))^n = ((x^k-1)/(x-1))^n.

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