A201553 Number of arrays of 6 integers in -n..n with sum zero.
1, 141, 1751, 9331, 32661, 88913, 204763, 418503, 782153, 1363573, 2248575, 3543035, 5375005, 7896825, 11287235, 15753487, 21533457, 28897757, 38151847, 49638147, 63738149, 80874529, 101513259, 126165719, 155390809, 189797061, 230044751
Offset: 0
Examples
Some solutions for n=5: ..4....5....4...-2...-4....5...-1...-2...-1...-3...-3....0....2...-4....2...-5 ..1...-4....5....3....4...-4....1....1....1....0....2...-2....1...-2...-1....1 .-2....3...-5....3....1....0...-4....2...-2....3....3....0....4....3....4....3 .-3...-3...-4....2....2...-3....5....4....4....0...-2....2....0....4...-1...-2 ..5....4...-4...-2...-3...-1...-4...-1....1....0...-2....3...-4...-5...-2....4 .-5...-5....4...-4....0....3....3...-4...-3....0....2...-3...-3....4...-2...-1
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..10000 (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]
- Robert Israel, Proof of "empirical" formula.
- Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).
Crossrefs
Cf. A201552.
Programs
-
Mathematica
a[n_] := Coefficient[Expand[Sum[x^k, {k, 0, 2n}]^6, x], x, 6n]; Array[a, 25, 0] (* Amiram Eldar, Dec 14 2018 *) -
PARI
{a(n) = polcoeff((sum(k=0, 2*n, x^k))^6, 6*n, x)} \\ Seiichi Manyama, Dec 14 2018
Formula
Empirical: a(n) = (2*n+1)*(44*n^4+88*n^3+71*n^2+27*n+5)/5.
Empirical formula verified (see link) by Robert Israel, Dec 14 2018.
Empirical: a(n)= integral( (sin((n+1/2)x)/sin(x/2))^6, x=0..Pi)/Pi. - Yalcin Aktar, Dec 03 2011
Conjectures from Colin Barker, May 23 2018: (Start)
G.f.: x*(141 + 905*x + 940*x^2 + 120*x^3 + 7*x^4 - x^5) / (1 - x)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
(End)
a(n) = [x^(6*n)] (Sum_{k=0..2*n} x^k)^6. - Seiichi Manyama, Dec 14 2018
E.g.f.: exp(x)*(5 + 700*x + 3675*x^2 + 3750*x^3 + 1100*x^4 + 88*x^5)/5. - Stefano Spezia, Sep 28 2024
Extensions
a(0)=1 prepended by Seiichi Manyama, Dec 14 2018
Comments