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 A331121 a(n) is the smallest positive integer k for which tau(k) does not divide sigma(n). 0
 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 (list; graph; refs; listen; history; text; internal format)
 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 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 A122977 Adjacent sequences:  A331118 A331119 A331120 * A331122 A331123 A331124 KEYWORD nonn AUTHOR Lechoslaw Ratajczak, Jan 10 2020 STATUS approved

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Last modified June 24 12:30 EDT 2021. Contains 345416 sequences. (Running on oeis4.)