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A032801
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Number of unordered sets a, b, c, d of distinct integers from 1..n such that a+b+c+d = 0 (mod n).
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3
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0, 0, 0, 0, 1, 3, 5, 9, 14, 22, 30, 42, 55, 73, 91, 115, 140, 172, 204, 244, 285, 335, 385, 445, 506, 578, 650, 734, 819, 917, 1015, 1127, 1240, 1368, 1496, 1640, 1785, 1947, 2109, 2289, 2470, 2670, 2870, 3090, 3311, 3553, 3795, 4059, 4324, 4612, 4900, 5212, 5525
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OFFSET
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1,6
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COMMENTS
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By reading carefully the proof of Lemma 5.1 (pp. 65-66) in Barnes (1959), we see that he actually proved a general result (even though he does not state it in the lemma). For 1 <= k <= n, let T(n, k) be the number of unordered sets b_1, b_2, ..., b_k of k distinct integers from 1..n such that b_1 + b_2 + ... + b_k = 0 (mod n). The proof of Lemma 5.1 in the paper implies that T(n, k) = (1/n) * Sum_{s | gcd(n, k)} (-1)^(k - (k/s)) * phi(s) * binomial(n/s, k/s).
For fixed k >= 1, the g.f. of the sequence (T(n, k): n >= 1) (with T(n, k) = 0 for 1 <= n < k) is (x^k/k) * Sum_{s|k} phi(s) * (-1)^(k - (k/s)) / (1 - x^s)^(k/s).
For k = 4, we get T(n, k=4) = (1/n) * Sum_{d | gcd(n, 4)} (-1)^(4/s) * phi(d) * binomial(n/d, 4/d), which agrees with Barnes' 3-part formula in Lemma 5.1 and with the formula in N. J. A. Sloane's Maple program below. It also agrees with Colin Barker's formula below.
For k = 4, the g.f. is (x^4/4) * Sum_{s|4} phi(s) * (-1)^(4/s) /(1 - x^s)^(4/s), which agrees with Herbert Kociemba's g.f. below.
Barnes' (1959) formula is a special case of Theorem 4 (p. 66) in Ramanathan (1944). If R(n, k, v) is the number of unordered sets b_1, b_2, ..., b_k of k distinct integers from 1..n such that b_1 + b_2 + ... + b_k = v (mod n), then he proved that R(n, k, v) = (1/n) * Sum_{s | gcd(n,k)} (-1)^(k - (k/s)) * binomial(n/s, k/s) * C_s(v), where C_s(v) = A054533(s, v) is Ramanujan's sum (even though it was discovered first around 1900 by the Austrian mathematician R. D. von Sterneck).
Because C_s(v = 0) = phi(s), we get Barnes' (implicit) result; i.e., R(n, k, v=0) = T(n, k) and a(n) = R(n, k=4, v=0) = T(n, k=4).
For k=2, we have R(n, k=2, v=0) = T(n, k=2) = A004526(n-1) for n >= 1. For k=3, we have R(n, k=3, v=0) = T(n, k=3) = A058212(n) for n >= 1. For k=5, we have R(n, k=5, v=0) = T(n, k=5) = A008646(n-5) for n >= 5.
(End)
Von Sterneck (1902, 1903) dealt with this problem. In his notation, we need to find (n)_i, the number of integer solutions of the congruence n = x_1 + ... + x_i (mod M) such that 0 <= x_1 < x_2 < ... < x_i < M, where (his n) = 0, (his M) = (our n), and (his i) = 4. He gave several formulas for solving this problem for various cases of his M (M = prime, M = product of two primes, M = power of 2, etc.). - Petros Hadjicostas, Aug 20 2019
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REFERENCES
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E. V. McLaughlin, Numbers of factorizations in non-unique factorial domains, Senior Thesis, Allegeny College, Meadville, PA, April 2004.
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LINKS
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FORMULA
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G.f.: x^5*(1+x-x^2+x^3)/((-1+x)^4*(1+x)^2*(1+x^2)). - Herbert Kociemba, Oct 22 2016
a(n) = (-6 * (4 + 2*(-1)^n + (-i)^n + i^n) + (25 + 3*(-1)^n)*n - 12*n^2 + 2*n^3)/48, where i = sqrt(-1). - Colin Barker, Oct 23 2016
a(n) = -A008610(-n), per formulae of Ralf Stephan (A008610) and C. Barker (above). Also, A008610(n) - a(n+4) = (1+(-1)^signum(n mod 4))/2, i.e., (1,0,0,0,1,0,0,0,...) repeating (offset 0). - Gregory Gerard Wojnar, Jul 09 2022
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MAPLE
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f := n-> if n mod 2 <> 0 then (n-1)*(n-2)*(n-3)/24 elif n mod 4 = 0 then (n-4)*(n^2-2*n+6)/24 else (n-2)*(n^2-4*n+6)/24; fi;
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MATHEMATICA
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CoefficientList[Series[(x^3 / 4) (1 / (1 - x)^4 + 1 / (1 - x^2)^2 - 2 / (1 - x^4)), {x, 0, 60}], x] (* Vincenzo Librandi, Jul 13 2019 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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