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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.

%I A350767 #23 Jul 01 2025 23:01:39
%S A350767 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,
%T A350767 26,29,27,25,30,33,31,37,40,34,41,35,49,43,47,16,38,42,39,46,44,53,51,
%U A350767 59,45,54,56,60,50,61,66,52,55,57,67,71,58,62,72,63,192,65
%N 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.
%C A350767 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).
%C A350767 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).
%C A350767 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".
%C A350767 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, ...).
%C A350767 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, ...).
%C A350767 The sequence of fixed points starts: 1, 46, 69, 74, 110, 140, 142, 152, 154, 178, ... apparently becoming denser as n increases.
%H A350767 Michael De Vlieger, <a href="/A350767/b350767.txt">Table of n, a(n) for n = 1..10001</a>
%H A350767 Michael De Vlieger, <a href="/A350767/a350767.png">Scatterplot of a(n)</a>, 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.
%H A350767 Michael De Vlieger, <a href="/A350767/a350767_1.png">Log-log scatterplot of a(n)</a>, n = 1..2^16. Terms in A3680(p) such that p is prime appear annotated in red, and those in A061286 appear in blue.
%e A350767 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.
%e A350767 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.
%t A350767 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 *)
%o A350767 (PARI) isok(k, last, set) = if (!setsearch(set, k), my(ndk=numdiv(k), ndl=numdiv(last+1)); (ndl != ndk) && ((!(ndk % ndl)) || (!(ndl % ndk))));
%o A350767 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
%Y A350767 Cf. A000040, A001248, A030514, A054753, A030516, A030626, A085986, A178739, A030629, A030630, A030631, A030632, A030633, A030634, A005179, A061286, A003680.
%K A350767 nonn
%O A350767 1,2
%A A350767 _David James Sycamore_, Jan 14 2022
%E A350767 More terms from _Michael De Vlieger_, Jan 14 2022