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).
1, 6, 45, 560, 9795, 223524, 6284089, 210208560, 8156750283, 360297117070, 17853149451841, 980844453593160, 59179098916735213, 3890176308574524934, 276750779199166606705, 21185250061147839785120, 1736385140876356212244563, 151719500906542020597450498
Offset: 1
Keywords
Examples
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)
Crossrefs
Cf. A286928.
Programs
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Mathematica
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}]; ];
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PARI
\\ 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
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Python
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
Formula
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
Extensions
Name clarified and a(6)-a(18) from Andrew Howroyd, May 16 2017
Comments