A061286 -id:A061286 - cp's OEIS frontend

A350767 a(1)=1. Thereafter, a(n+1) is the least unused number k such that either d(j(n)) properly divides d(k) or d(k) properly divides d(j(n)), where j(n) = a(n)+1 and d is the divisor counting function A000005.

1, 6, 8, 12, 10, 14, 2, 15, 48, 18, 20, 3, 28, 21, 5, 7, 11, 4, 22, 24, 32, 13, 17, 9, 19, 23, 26, 29, 27, 25, 30, 33, 31, 37, 40, 34, 41, 35, 49, 43, 47, 16, 38, 42, 39, 46, 44, 53, 51, 59, 45, 54, 56, 60, 50, 61, 66, 52, 55, 57, 67, 71, 58, 62, 72, 63, 192, 65

Offset: 1

David James Sycamore, Jan 14 2022

nonn

If d(j(n)) is prime p then d(a(n+1)) must be properly divisible by p. In practice the proper divisor for computation of a(n+1) toggles between d(j(n)) and d(k).
Conjecture: This is a permutation of the positive integers. Numbers with the same number (tau) of divisors appear in their natural orders (e.g., primes, semiprimes, squares).
The plot, after the first few terms, resolves itself into points tightly packed on and around a straight line of slope 1, with exceptional points appearing as significant upward or downward "spikes".
When d(j(n)) is prime p appearing for the first time in the sequence J = {d(j(a(n)), n>=1}, then a(n+1) is the smallest number with 2p divisors, which produces a significantly large upward spike above the straight line (6, 12, 48, 192, 3072, 12288, ...).
When d(j(a(n)) is 2p, seen for the first time in J, then a(n+1) is the smallest number with p divisors, which produces a large downward spike, below the straight line (2, 4, 16, 64, 1024, 4096, ...).
The sequence of fixed points starts: 1, 46, 69, 74, 110, 140, 142, 152, 154, 178, ... apparently becoming denser as n increases.

			a(1)=1, so j(1)=2, d(j(1))=2, a prime, so we need the smallest unused k such that d(k) is properly divisible by 2, hence a(2)=6.
a(2)=6, j(2)=4, d(j(2))=3, a prime so we need the smallest unused k such that d(k) is properly divisible by 3, hence a(3)=8.

Michael De Vlieger, Table of n, a(n) for n = 1..10001
Michael De Vlieger, Scatterplot of a(n), n = 1..256, color coded to show primes in red, evens in blue, 2 in magenta, odd composites in black. A gold highlight indicates terms such that (a(n)+1) | a(n+1). Outlying points are annotated.
Michael De Vlieger, Log-log scatterplot of a(n), n = 1..2^16. Terms in A3680(p) such that p is prime appear annotated in red, and those in A061286 appear in blue.

Cf. A000040, A001248, A030514, A054753, A030516, A030626, A085986, A178739, A030629, A030630, A030631, A030632, A030633, A030634, A005179, A061286, A003680.

Block[{c, d, j, k, u, nn}, nn = 120; j = u = 2; c[] = 0; c[1] = 1; Do[d[i] = DivisorSigma[0, i], {i, 2^(Ceiling@ Log2[nn] + 3)}]; {1}~Join~Reap[Do[k = u; While[Nand[Or[Divisible[d[j], d[k]], Divisible[d[k], d[j]]], d[j] != d[k], c[k] == 0], k++]; Sow[k]; c[k] = i; j = k + 1; If[k == u, While[c[u] != 0, u++]], {i, nn}] ][[-1, -1]] ] (* _Michael De Vlieger, Jan 14 2022 *)

isok(k, last, set) = if (!setsearch(set, k), my(ndk=numdiv(k), ndl=numdiv(last+1)); (ndl != ndk) && ((!(ndk % ndl)) || (!(ndl % ndk))));
lista(nn) = {my(last = 1, list = List(last), set = Set(list)); for (n=2, nn, my(k=1); while (!isok(k, last, set), k++); listput(list, k); set = Set(list); last = k;); Vec(list);} \\ Michel Marcus, Jan 15 2022

More terms from Michael De Vlieger, Jan 14 2022

A172442 A sixth of the smallest integer for which the number of divisors is the n-th composite.

Original entry on oeis.org

1, 2, 4, 6, 8, 10, 32, 24, 20, 30, 40, 96, 512, 60, 216, 2048, 150, 160, 120, 140, 1536, 32768, 864, 210, 131072, 6144, 280, 480, 2560, 600, 2097152, 420, 7776, 1080, 98304, 10240, 1050, 13824, 1120, 393216, 134217728, 840, 536870912, 2400, 1260, 55296, 7680, 163840, 6291456

Offset: 1

Juri-Stepan Gerasimov, Mar 12 2011

			The first composite is A002808(1)=4. The smallest integer with 4 divisors is A005179(4) = 6, which divided by 6 yields a(1)=1.

Cf. A002808, A005179, A061286, A187678.

nd[n_]:=Module[{k=6},While[DivisorSigma[0,k]!=n,k+=6];k]; With[{cmps=Select[Range[45],CompositeQ]},(Table[nd[c],{c,cmps}])/6] (* The program generates the first 30 terms of the sequence. *) (* Harvey P. Dale, Jun 15 2025 *)

a(n) = A005179(A002808(n))/6.

A379332 Numbers k such that k^sigma(k) == k (mod sigma(k)).

Original entry on oeis.org

1, 2, 4, 9, 10, 16, 25, 33, 36, 64, 81, 121, 136, 145, 261, 289, 528, 529, 624, 625, 729, 841, 865, 897, 900, 1024, 1089, 1441, 1681, 1989, 2016, 2209, 2241, 2304, 2353, 2401, 2809, 3481, 3808, 3984, 4069, 4096, 4320, 5041, 5185, 6400, 6889, 7201, 7921, 9801

Offset: 1

Juri-Stepan Gerasimov, Dec 21 2024

nonn

			2 is in the sequence because 2^sigma(2) == 2 (mod sigma(2)) or 2^3 == 2 (mod 3),
3 not is in the sequence because not 3^sigma(3) == 3 (mod sigma(3)) or 3^4 == 1 (mod 4),
4 is in the sequence because 4^sigma(4) == 2 (mod sigma(4)) or 4^7 == 2 (mod 7).

Supersequence of A061286.

Cf. A000203, A002476.

q[k_] := Module[{s = DivisorSigma[1, k]}, PowerMod[k, s, s] == k]; q[1] = True; Select[Range[10^4], q] (* Amiram Eldar, Dec 21 2024 *)

isok(k) = my(s=sigma(k)); Mod(k, s)^s == k; \\ Michel Marcus, Dec 21 2024

cp's OEIS Frontend

A350767 a(1)=1. Thereafter, a(n+1) is the least unused number k such that either d(j(n)) properly divides d(k) or d(k) properly divides d(j(n)), where j(n) = a(n)+1 and d is the divisor counting function A000005.

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A172442 A sixth of the smallest integer for which the number of divisors is the n-th composite.

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A379332 Numbers k such that k^sigma(k) == k (mod sigma(k)).

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Mathematica

PARI