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A316706
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Number of solutions to k_1 + 2*k_2 + ... + n*k_n = n, where k_i are from {-1,0,1}, i=1..n.
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14
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1, 1, 1, 2, 5, 12, 27, 69, 178, 457, 1194, 3178, 8538, 23062, 62726, 171804, 473069, 1308397, 3634075, 10133154, 28352421, 79575702, 223981549, 632101856, 1788172541, 5069879063, 14403962756, 41001479103, 116921037003, 333971884899, 955443681814, 2737387314548, 7853533625522, 22560919253095, 64890249175438, 186854616134794
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OFFSET
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0,4
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COMMENTS
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a(n) is the coefficient of both x^n and 1/x^n in Product_{k=1..n} (1/x^k + 1 + x^k), while A007576 gives the constant term in the symmetric product.
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LINKS
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FORMULA
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a(n) = [x^n] Product_{k=1..n} (1/x^k + 1 + x^k).
a(n) = [x^(n*(n-1)/2)] Product_{k=1..n} (1 + x^k + x^(2*k)).
a(n) = [x^(n*(n+3)/2)] Product_{k=1..n} (1 + x^k + x^(2*k)).
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MATHEMATICA
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nmax = 40; p = 1; Flatten[{1, Table[Coefficient[p = Expand[p*(1/x^n + 1 + x^n)], x^n], {n, 1, nmax}]}] (* Vaclav Kotesovec, Jul 11 2018 *)
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PROG
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(PARI) {a(n) = polcoeff( prod(k=1, n, 1/x^k + 1 + x^k) + x*O(x^n), n)}
for(n=0, 40, print1(a(n), ", "))
(Python)
from collections import Counter
c = {0:1}
for k in range(1, n+1):
b = Counter(c)
for j in c:
a = c[j]
b[j+k] += a
b[j-k] += a
c = b
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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