A032801 Number of unordered sets a, b, c, d of distinct integers from 1..n such that a+b+c+d = 0 (mod n).

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

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

From Petros Hadjicostas, Jul 12 2019: (Start)

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

References

• E. V. McLaughlin, Numbers of factorizations in non-unique factorial domains, Senior Thesis, Allegeny College, Meadville, PA, April 2004.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Eric Stephen Barnes, The construction of perfect and extreme forms I, Acta Arith., 5 (1959); see pp. 65-66.
• David Broadhurst and Xavier Roulleau, Number of partitions of modular integers, arXiv:2502.19523 [math.NT], 2025. See pp. 4, 14, 19.
• K. G. Ramanathan, Some applications of Ramanujan's trigonometrical sum C_m(n), Proc. Indian Acad. Sci., Sect. A 20 (1944), 62-69; see p. 66.
• R. D. von Sterneck, Ein Analogon zur additiven Zahlentheorie, Sitzungsber. Akad. Wiss. Sapientiae Math.-Naturwiss. Kl. 111 (1902), 1567-1601 (Abt. IIa). [Accessible only in the USA through the HathiTrust Digital Library.]
• R. D. von Sterneck, Über ein Analogon zur additiven Zahlentheorie, Jahresbericht der Deutschen Mathematiker-Vereinigung 12 (1903), 110-113. [This is a summary of the 1902 paper.]
• Index entries for linear recurrences with constant coefficients, signature (2,0,-2,2,-2,0,2,-1).

Crossrefs

Cf. A001399, A001400, A004526, A008646, A054533, A058212, A008610.

Column k=4 of A267632.

Programs

Maple

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;

Mathematica

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 *)

Formula

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

Extensions

Offset changed by David A. Corneth, Oct 23 2016