A337157 a(n) is the smallest m such that A054024(m) = prime(n), where A054024(m) is A000203(m) mod m, or -1 if there is no such m.

20, 4, -1, 8, 21, 27, 39, 36, 57, 115, 32, 155, 63, 50, 129, 235, 265, 371, 305, 201, 98, 365, 237, 171, 245, 291, 485, 309, 325, 327, 128, 189, 279, 917, 1507, 1529, 242, 785, 489, 835, 865, 1211, 385, 605, 579, 324, 338, 2321, 669, 1115, 687, 1165, 399, 1936

Offset: 1

Author

Michel Marcus, Jan 28 2021

Keywords

• sign

Comments

From Bernard Schott, Jan 28 2021: (Start)

a(n) > 0 when n <> 3.

Proof: When n = 1, 2, 4 then a(n) = 20, 4, 8; then, following Goldbach conjecture when p = prime(n) >= 11 with p-1 = p1 + p2, then sigma(p1*p2) = 1+p1+p2+p1*p2 == 1+p1+p2 = p (mod p1*p2); hence, for each n>= 5, a(n) exists and a(n) <= p1*p2 (see examples).

Now, when n = 3, no k is known such that sigma(k) == 5 (mod k)? (End)

Examples

			For prime(5) = 11, 11-1=3+7 with 3*7 = 21, so a(5) <= 21 and a(5) = 21.
For prime(6) = 13, 13-1=5+7 with 5*7 = 35, so a(6) <= 35 but sigma(27) = 40 == 13 (mod 27), hence a(6)=27.
		

Links

• Michel Marcus, Table of n, a(n) for n = 1..4000

Crossrefs

Cf. A000203, A054024, A308150.

Programs

PARI

f(n) = sigma(n) % n; \\ A054024
a(n) = if (n==3, return (-1)); my(k=1, p=prime(n)); while (f(k) != p, k++); k;