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

Author

David James Sycamore, Jan 14 2022

Keywords

• nonn

Comments

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.

Examples

			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.
		

Links

• 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.

Crossrefs

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

Programs

Mathematica

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 *)

PARI

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

Extensions

More terms from Michael De Vlieger, Jan 14 2022