cp's OEIS Frontend

Since the sum of two distinct integers from 1 to n can be as much as 2n-1, this triangular table cannot show all the possible cases. For larger triangles showing all solutions, see A220691 and A220693. - Antti Karttunen, Feb 18 2013 [based on Robert Israel's mail, May 07 2012]

It is known that a(15) is 31 or 32, a(16)=37 and a(17) is 40, 41 or 42. - N. J. A. Sloane, Feb 11 2013

In other words, the independence number of the (n-1)-triangular grid graph.

Apart from a(3) and a(5) same as A007997(n+4) and A058212(n+2). - Eric W. Weisstein, Jun 14 2017

Also the independence number of the n-triangular honeycomb king graph. - Eric W. Weisstein, Sep 06 2017

Row less the last element is palindrome for n=odd or n=power of 2 where n is the row number (observation-conjecture).

From Petros Hadjicostas, Jul 13 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).

According to the definition of this sequence, for 1 <= k <= n, T(n, k) 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 = 0 (mod n). The proof of Lemma 5.1 in Barnes (1959) implies that T(n, k) = (1/n) * Sum_{s | gcd(n, k)} (-1)^(k - (k/s)) * phi(s) * binomial(n/s, k/s) for 1 <= k <= n.

For fixed k >= 1, the g.f. of the column (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), which generalizes Herbert Kociemba's formula from A032801.

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) = A054535(s, v) = Sum_{d | gcd(s,v)} d * Moebius(s/d) is Ramanujan's sum (even though it was first discovered 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) for 1 <= k <= n.

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=4, we have R(n, k=4, v=0) = A032801(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.

The reason we have T(2*m+1, k) = A037306(2*m+1, k) = A047996(2*m+1, k) for m >= 0 and k >= 1 is the following. When n = 2*m + 1, all divisors s of gcd(n, k) are odd. In such a case, k - (k/s) is even for all k >= 1, and thus (-1)^(k - (k/s)) = 1, and thus T(n = 2*m+1, k) = (1/n) * Sum_{s | gcd(n, k)} phi(s) * binomial(n/s, k/s) = A037306(2*m+1, k) = A047996(2*m+1, k).

By summing the products of the g.f. of column k times y^k from k = 1 to k = infinity, we get the bivariate g.f. for the array: Sum_{n, k >= 1} T(n, k)*x^n*y^k = Sum_{s >= 1} (phi(s)/s) * log((1 - x^s + (-x*y)^s)/(1 - x^s)) = -x/(1 - x) - Sum_{s >= 1} (phi(s)/s) * log(1 - x^s + (-x*y)^s).

Letting y = 1 in the above bivariate g.f., we get the g.f. of the sequence (Sum_{1 <= k <= n} T(n, k): n >= 1) is -x/(1 - x) - Sum_{s >= 1} (phi(s)/s) * log(1 - x^s + (-x)^s) = -x/(1 - x) + Sum_{m >= 0} (phi(2*m + 1)/(2*m + 1)) * log(1 - 2*x^(2*m+1)), which is the g.f. of sequence A082550. Hence, sequence A082550 consists of the row sums.

There is another important interpretation of this array T(n, k) which is related to some of the references for sequence A047996, but because the discussion is too lengthy, we omit the details.

(End)

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