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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.
18
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
1,3
COMMENTS
Equivalently, the number of compositions of n*(k + 1) into n parts with maximum part size 2*k+1. - Andrew Howroyd, Oct 14 2017
LINKS
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.
FORMULA
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
EXAMPLE
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,...
MAPLE
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
MATHEMATICA
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 *)
PROG
(PARI)
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
KEYWORD
nonn,tabl,easy
AUTHOR
R. H. Hardin, Dec 02 2011
STATUS
approved