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

1,1

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.

Links

Table of n, a(n) for n=1..66.

Example

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).

Mathematica

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

Prog

(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

Crossrefs

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

Sequence in context: A275331 A308828 A325683 * A057767 A207329 A348219

Adjacent sequences: A331118 A331119 A331120 * A331122 A331123 A331124

Keyword

nonn

Author

Lechoslaw Ratajczak, Jan 10 2020

Status

approved