A316744 a(n) is the smallest number having exactly n ways to be represented as sum of at least two consecutive positive integers and expressible as sum of n consecutive positive integers, or 0 if no such number exists.

9, 15, 162, 45, 729, 105, 900, 405, 9765625, 495, 1062882, 9477, 3969, 945, 344373768, 3825, 387420489, 7695, 34650, 413343, 81402749386839761113321, 7245, 202500, 732421875, 38025, 25515, 919973073089479921953602, 58725, 0, 29295, 23619600, 473513931, 60886809, 17325, 300189270593998242

Offset: 2

Author

Jianing Song, Jul 13 2018

Keywords

• nonn

Comments

a(n) is the smallest number such that A069283(n) == n*(n-1)/2 == A142150(n) (mod n). Or equivalently, a(n) is the smallest number of the form 2^t*s, where s is an odd number with exactly n + 1 divisors and divisible by A000265(n), t = 0 for odd n and A007814(n) - 1 for even n.

Let n = 2^e_0*Product_{i=1..m} p_i^e_i where p_i are odd primes; n + 1 = Product_{j=1..s} q_j where q_j are primes, then a(n) != 0 iff there is an injection f from {1,2,..,m} to {1,2,...,s} such that q_f(i) >= e_i + 1 for all 1 <= i <= m, implying s >= m. If such an injection does exist, then the number of k having exactly n ways to be represented as sum of at least two consecutive positive integers and expressible as sum of n consecutive positive integers is finite iff s = m, in which case the number of k is equal to the number of injections such that if i < j and e_i = e_j then q_f(i) <= q_f(j).

If A038547(n) is divisible by A000265(n), then a(n) = 2^t*A038547(n), t defined as above.

If n + 1 is a Fermat prime, then a(n) = (n/2)*3^n. If n = p - 1 = 2^e*q with p, q primes, then a(n) = 2^(e-1)*q^n, which is relatively large.

Examples

			a(2) = 9 = 4 + 5 = 2 + 3 + 4.
a(3) = 15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5.
a(4) = 162 = 53 + 54 + 55 = 39 + 40 + 41 + 42 = 14 + 15 + 16 + ... + 22 = 8 + 9 + 10 + ... + 19.
If a number k has exactly 30 ways to be represented as sum of at least two consecutive positive integers, then it must have exactly 31 odd divisors. On the other hand, the sum of 30 consecutive positive integers is congruent to 15 mod 30, so k must be of the form p^30 where p is an odd prime, which obviously cannot be divisible by 15. So a(30) = 0.
Let n = 225 = 3^2*5^2, n + 1 = 226 = 2*113, so e_1 = 2, e_2 = 2, q_1 = 2, q_2 = 113. An injection from {1,2} to {1,2} such that q_f(1) >= e_1 + 1 and q_f(2) >= e_2 + 1 does not exist, so a(225) = 0.
		

Crossrefs

Cf. A000265, A001222, A005087, A007814, A038547, A069283, A142150.