A331121 a(n) is the smallest positive integer k for which tau(k) does not divide sigma(n).

2, 2, 4, 2, 6, 16, 4, 2, 2, 6, 16, 4, 4, 16, 16, 2, 6, 2, 4, 6, 4, 16, 16, 24, 2, 6, 4, 4, 6, 16, 4, 2, 16, 6, 16, 2, 4, 24, 4, 6, 6, 16, 4, 16, 6, 16, 16, 4, 2, 2, 16, 4, 6, 36, 16, 36, 4, 6, 24, 16, 4, 16, 4, 2, 16, 16

Offset: 1

Author

Lechoslaw Ratajczak, Jan 10 2020

Keywords

• nonn

Comments

Consecutive t satisfying the equation a(t) = 2 are consecutive elements of A028982 (squares and twice squares).

Conjecture: consecutive u satisfying the equation a(u) = 4 are consecutive elements of a sequence defined as follows: (A024614 \ A088535) \ A074384. The conjecture was checked for 10^6 consecutive integers.

Examples

			a(10) = 6 because sigma(10) = 18 is divisible by (tau(1) = 1), (tau(2) = 2), (tau(3) = 2), (tau(4) = 3), (tau(5) = 2), and is not divisible by (tau(6) = 4).
		

Crossrefs

Cf. A000005, A000203, A024614, A028982, A074384, A088535.

Programs

Mathematica

Array[Block[{k = 1}, While[Mod[DivisorSigma[1, #], DivisorSigma[0, k]] == 0, k++]; k] &, 66] (* Michael De Vlieger, Jan 31 2020 *)

Maxima

a(n):=(for k:1 while mod(divsum(n), length(divisors(k))) = 0 do z:k, z+1) $ makelist(a(n), n, 1, 100, 1);

PARI

a(n) = my(k=1, sn=sigma(n)); while ((sn % numdiv(k)) == 0, k++); k; \\ Michel Marcus, Jan 10 2020