A286928 -id:A286928 - cp's OEIS frontend

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

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

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.

A318793 Constant term in the expansion of (Sum_{k=0..n} k*(x^k + x^(-k)))^n.

Original entry on oeis.org

1, 0, 10, 84, 12060, 922680, 203474180, 45546045720, 16977056982648, 7385901628225968, 4359210462435545640, 3063111491275816418020, 2669859570203387710219500, 2738752987417403052110951664, 3328615281192062743163487239944

Offset: 0

Seiichi Manyama, Dec 15 2018

nonn

			(2/x^2 + 1/x + 0 + x + 2*x^2)^2 = 4/x^4 + 4/x^3 + 1/x^2 + 4/x + 10 + 4*x + x^2 + 4*x^3 + 4*x^4. So a(2) = 10.

Vaclav Kotesovec, Table of n, a(n) for n = 0..214

Cf. A286928, A318794.

a[n_] := If[n==0, 1, Coefficient[Expand[Sum[k*(x^k + x^(-k)), {k, 0, n}]^n], x, 0]]; Array[a, 15, 0] (* Amiram Eldar, Dec 15 2018 *)

{a(n) = polcoeff((sum(k=0, n, k*(x^k+x^(-k))))^n, 0, x)}

a(n) ~ exp(1) * n^(2*n - 3/2) / sqrt(Pi). - Vaclav Kotesovec, Dec 15 2018

A329076 Constant term in the expansion of ((Sum_{k=-n..n} x^k) * (Sum_{k=-n..n} y^k) - (Sum_{k=-n+1..n-1} x^k) * (Sum_{k=-n+1..n-1} y^k))^n.

Original entry on oeis.org

1, 0, 16, 72, 7008, 162000, 17555520, 1093527120, 140846184640, 16016249944800, 2550757928818680, 419682645514181280, 82389928294166805312, 17418502084657134228768, 4123280170924828458697152, 1054943518137131171386437600, 293933660095874311773617934720, 87968971083026619734709639853632

Offset: 0

Seiichi Manyama, Nov 04 2019

Also number of n-step closed paths (from origin to origin) in 2-dimensional lattice, using steps (t_1,t_2) (|t_1| + |t_2| = 2*n).

Seiichi Manyama, Table of n, a(n) for n = 0..150 (terms 0..53 from Vaclav Kotesovec)
Wikipedia, Taxicab geometry.

Main diagonal of A329074.

Cf. A094061, A286928.

{a(n) = polcoef(polcoef((sum(k=-n, n, x^k)*sum(k=-n, n, y^k)-sum(k=-n+1, n-1, x^k)*sum(k=-n+1, n-1, y^k))^n, 0), 0)}

{a(n) = polcoef(polcoef((sum(k=0, 2*n, (x^k+1/x^k)*(y^(2*n-k)+1/y^(2*n-k)))-x^(2*n)-1/x^(2*n)-y^(2*n)-1/y^(2*n))^n, 0), 0)}

f(n) = (x^(n+1)-1/x^n)/(x-1);
a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n, k)*polcoef(f(n)^k*f(n-1)^(n-k), 0)^2)

Conjecture: a(n) ~ 3 * 2^(3*n - 2) * n^(n-3) / Pi. - Vaclav Kotesovec, Nov 05 2019

A160492 a(n) = number of solutions to an equation x_1 + ... + x_j =0 with 1<=j<=n satisfying -n<=x_i<=n (1<=i<=j).

Original entry on oeis.org

1, 6, 45, 560, 9795, 223524, 6284089, 210208560, 8156750283, 360297117070, 17853149451841, 980844453593160, 59179098916735213, 3890176308574524934, 276750779199166606705, 21185250061147839785120, 1736385140876356212244563, 151719500906542020597450498

Offset: 1

Srikanth K S, May 15 2009

nonn

The number of variables in the equation can be from 1 to n and each variable can have a value of -n to n. See A286928 for the case of exactly n variables. - Andrew Howroyd, May 16 2017

			From _Andrew Howroyd_, May 16 2017 (Start)
Case n=3:
1 variable: {0} is only solution.
2 variables: {-3,3}, {-2,2}, {-1,1}, {0,0}, {1,-1}, {2,-2}, {3,-3}.
3 variables: {-3 0 3}x6, {-3 1 2}x6, {-2 -1 3}x6, {-2 0 2}x6,
             {-2 1 1}x3, {-1 -1 2}x3, {-1 0 1}x6, {0 0 0}x1
In the above, {-3 0 3}x6 means that the values can be expanded to 6 solutions by considering different orderings.
In total there are 1 + 7 + 37 = 45 solutions so a(3)=45.
(End)

Cf. A286928.

zerocompositionswithzero[p_] := Module[{united = {}, i, zerosums = {}, count = 0}, For[i = 1, i <= p, i = i + 1, united = Union[united, Tuples[Table[x, {x, -p, p}], i]] ]; For[i = 1, i <= Length[united], i = i + 1, If[Sum[united[[i, j]], {j, 1, Length[united[[i]]]}] == 0, zerosums = Append[zerosums, united[[i]]]; count = count + 1;]; ]; Return[{count, zerosums}]; ];

\\ nr compositions of r with max value m into exactly k parts.
compositions(r,m,k)=sum(i=0,floor((r-k)/m),(-1)^i*binomial(r-1-i*m, k-1)*binomial(k, i));
a(n)=sum(v=1,n,compositions(v*(n+1),2*n+1,v));  \\ Andrew Howroyd, May 16 2017

from sympy import binomial
def C(r, m, k): return sum([(-1)**i*binomial(r - 1 - i*m, k - 1)*binomial(k, i) for i in range(int((r - k)/m) + 1)])
def a(n): return sum([C(v*(n + 1), 2*n + 1, v) for v in range(1, n + 1)]) # Indranil Ghosh, May 16 2017, after the PARI program by Andrew Howroyd

a(n) = Sum_{k=1..n} Sum_{i=0..floor(k/2)} (-1)^i*binomial(k*(n+1)-i*(2*n+1)-1, k-1)*binomial(k, i). - Andrew Howroyd, May 16 2017

Name clarified and a(6)-a(18) from Andrew Howroyd, May 16 2017

cp's OEIS Frontend

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

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A318793 Constant term in the expansion of (Sum_{k=0..n} k*(x^k + x^(-k)))^n.

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A329076 Constant term in the expansion of ((Sum_{k=-n..n} x^k) * (Sum_{k=-n..n} y^k) - (Sum_{k=-n+1..n-1} x^k) * (Sum_{k=-n+1..n-1} y^k))^n.

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A160492 a(n) = number of solutions to an equation x_1 + ... + x_j =0 with 1<=j<=n satisfying -n<=x_i<=n (1<=i<=j).

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A322514 a(n) = [x^(n^2)] (Sum_{k=0..2*n} (k+1)*x^k)^n.

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A322514 a(n) = [x^(n^2)] (Sum_{k=0..2n} (k+1)x^k)^n.