A201551 -id:A201551 - cp's OEIS frontend

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

N. J. A. Sloane

nonn

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

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).

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

Cf. A035343, A349936 (bisection).

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

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

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

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 *)

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

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

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

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 */

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

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

A201552 Square array read by diagonals: T(n,k) = number of arrays of n integers in -k..k with sum equal to 0.

Original entry on oeis.org

1, 1, 3, 1, 5, 7, 1, 7, 19, 19, 1, 9, 37, 85, 51, 1, 11, 61, 231, 381, 141, 1, 13, 91, 489, 1451, 1751, 393, 1, 15, 127, 891, 3951, 9331, 8135, 1107, 1, 17, 169, 1469, 8801, 32661, 60691, 38165, 3139, 1, 19, 217, 2255, 17151, 88913, 273127, 398567, 180325, 8953, 1

Offset: 1

R. H. Hardin, Dec 02 2011

Equivalently, the number of compositions of n*(k + 1) into n parts with maximum part size 2*k+1. - Andrew Howroyd, Oct 14 2017

			Some solutions for n=7, k=3:
..1...-2....1...-1....1...-3....0....0....1....2....3...-3....0....2....1....0
.-1....2...-2....2....2....2...-1....0....2....2...-2...-1...-2...-1....2...-1
.-3...-1....1...-3....2....1....0....1....3....0....2....0...-1....2...-2...-1
..0....3....3....3...-2...-2....3....3...-3...-3....0...-1...-1...-1....0....3
..2...-1...-1...-1...-3....0...-3...-2....1...-1...-1....1....1....0....3...-1
..2...-1...-3....0....2....3....0....1...-2....1....1....1....3...-2...-3...-3
.-1....0....1....0...-2...-1....1...-3...-2...-1...-3....3....0....0...-1....3
Table starts:
.   1,      1,       1,        1,        1,         1,...
.   3,      5,       7,        9,       11,        13,...
.   7,     19,      37,       61,       91,       127,...
.  19,     85,     231,      489,      891,      1469,...
.  51,    381,    1451,     3951,     8801,     17151,...
. 141,   1751,    9331,    32661,    88913,    204763,...
. 393,   8135,   60691,   273127,   908755,   2473325,...
.1107,  38165,  398567,  2306025,  9377467,  30162301,...
.3139, 180325, 2636263, 19610233, 97464799, 370487485,...

R. H. Hardin, Table of n, a(n) for n = 1..9999
Michelle Rudolph-Lilith and Lyle E. Muller, On a link between Dirichlet kernels and central multinomial coefficients, Discrete Mathematics 338.9 (2015): 1567-1572.
Wikipedia, Dirichlet kernel.

Columns 1-10: A002426, A005191, A025012, A025014, A201549, A201550, A201551, A322538, A322539, A322540.
Rows 3-10: A003215, A063496(n+1), A083669, A201553, A201554, A322535, A322536, A322537.
Cf. A286928.

seq(print(seq(add((-1)^i*binomial(n, i)*binomial((k+1)*n-(2*k+1)*i-1, n-1), i = 0..floor((1/2)*n)), k = 1..10)), n = 1..10); # Peter Bala, Oct 16 2024

comps[r_, m_, k_] := Sum[(-1)^i*Binomial[r - 1 - i*m, k - 1]*Binomial[k, i], {i, 0, Floor[(r - k)/m]}];  T[n_, k_] := comps[n*(k + 1), 2*k + 1, n]; Table[T[n - k + 1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Oct 31 2017, after Andrew Howroyd *)

comps(r, m, k)=sum(i=0, floor((r-k)/m), (-1)^i*binomial(r-1-i*m, k-1)*binomial(k, i));
T(n,k) = comps(n*(k+1), 2*k+1, n); \\ Andrew Howroyd, Oct 14 2017

Empirical: T(n,k) = Sum_{i=0..floor(k*n/(2*k+1))} (-1)^i*binomial(n,i)* binomial((k+1)*n-(2*k+1)*i-1,n-1).
The above empirical formula is true and can be derived from the formula for the number of compositions with given number of parts and maximum part size. - Andrew Howroyd, Oct 14 2017
Empirical for rows:
T(1,k) = 1
T(2,k) = 2*k + 1
T(3,k) = 3*k^2 + 3*k + 1
T(4,k) = (16/3)*k^3 + 8*k^2 + (14/3)*k + 1
T(5,k) = (115/12)*k^4 + (115/6)*k^3 + (185/12)*k^2 + (35/6)*k + 1
T(6,k) = (88/5)*k^5 + 44*k^4 + 46*k^3 + 25*k^2 + (37/5)*k + 1
T(7,k) = (5887/180)*k^6 + (5887/60)*k^5 + (2275/18)*k^4 + (357/4)*k^3 + (6643/180)*k^2 + (259/30)*k + 1
T(m,k) = (1/Pi)*integral_{x=0..Pi} (sin((k+1/2)x)/sin(x/2))^m dx; for the proof see Dirichlet Kernel link; so f(m,n) = (1/Pi)*integral_{x=0..Pi} (Sum_{k=-n..n} exp(I*k*x))^m dx = sum(integral(exp(I(k_1+...+k_m).x),x=0..Pi)/Pi,{k_1,...,k_m=-n..n}) = sum(delta_0(k1+...+k_m),{k_1,...,k_m=-n..n}) = number of arrays of m integers in -n..n with sum zero. - Yalcin Aktar, Dec 03 2011
T(n, k) = the constant term in the expansion of (x^(-k) + ... + x^(-1) + 1 + x + ... + x^k)^n = the coefficient of x^(k*n) (i.e., the central coefficient) in the expansion of (1 + x + ... + x^(2*k))^n = the coefficient of x^(k*n) in the expansion of ( (1 - x^(2*k+1))/(1 - x) )^n. Expanding the binomials and collecting terms gives the empirical formula above. - Peter Bala, Oct 16 2024

A208596 Number of n-bead necklaces labeled with numbers -7..7 not allowing reversal, with sum zero.

Original entry on oeis.org

1, 8, 57, 568, 6077, 69784, 833253, 10259448, 129245091, 1658145128, 21589248803, 284548542120, 3789094334455, 50900085245304, 688944374917247, 9386664978851448, 128633790260673263, 1771859642698543096, 24518513933529549357, 340679786167936420216

Offset: 1

R. H. Hardin, Feb 29 2012

nonn

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

Andrew Howroyd, Table of n, a(n) for n = 1..100

Column 7 of A208597.

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_Integer, k_] := DivisorSum[n, EulerPhi[n/#] comps[#*(k + 1), 2 k + 1, #] &]/n; a[n_] = a[n, 7]; Array[a, 20] (* Jean-François Alcover, Nov 01 2017, after Andrew Howroyd *)

a(n) = (1/n) * Sum_{d | n} totient(n/d) * A201551(d). - Andrew Howroyd, Mar 02 2017

a(14)-a(20) from Andrew Howroyd, Mar 02 2017

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

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A201552 Square array read by diagonals: T(n,k) = number of arrays of n integers in -k..k with sum equal to 0.

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A208596 Number of n-bead necklaces labeled with numbers -7..7 not allowing reversal, with sum zero.

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