A343327 Start with a(1)=1; thereafter a(n+1) = Sum_{k=1..n such that d(a(k)) = d(a(n))} d(a(k)) where d = A000005.

1, 1, 2, 2, 4, 3, 6, 4, 6, 8, 12, 6, 16, 5, 8, 20, 12, 18, 24, 8, 24, 16, 10, 28, 30, 24, 32, 36, 9, 9, 12, 42, 40, 48, 10, 32, 48, 20, 54, 56, 64, 7, 10, 36, 18, 60, 12, 66, 72, 24, 80, 30, 88, 96, 36, 27, 40, 104, 112, 40, 120, 16, 15, 44, 72, 48, 50, 78, 128, 136, 144, 15

Offset: 1

Author

David James Sycamore, Apr 12 2021

Keywords

• nonn

Comments

Every nonzero number occurs in the sequence since all terms in A000005 are included. The multiplicity of any term m (> 2) is d(m)-1 (since m > 2 cannot be created by adding m 1s). Prime 2 occurs twice, consecutively (see example), and thereafter no two adjacent terms can be prime because the term prior to a prime q is the first occurrence of 2^(q-1) (composite for q > 2), and since d(q)=2, the following term is a multiple (> 1) of 2. Thus each prime > 2 is flanked by composite even numbers, and occurs once only. Conjecture: The primes, after the first appearance of 2, appear in their natural order.

The primes appear in order at a(n) for n in {3, 4, 6, 14, 42, 507, 1939, 22454, 90322, ...}, with a(3) = a(4) = 2. No further primes appear for n <= 2^20. - Michael De Vlieger, May 01 2021

Examples

			Given a(1)=1, a(2) is also 1, since a(1) has no prior term. a(3)=1+1=2 and a(4)=2 because a(3) has 2 divisors and there is no prior term with 2 divisors. a(5)=2+2=4.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Michael De Vlieger, Scatterplot of a(n) for n = 1..2^20 showing trajectories of tau(a(n)).
• Michael De Vlieger, Annotated scatterplot of a(n) for n = 1..256, noting and illustrating tau(a(n)) via color function.

Crossrefs

Cf. A000005, A009087, A005179, A061286.

Programs

Mathematica

Block[{a = {1}, c}, Do[(If[! IntegerQ[c[#]], Set[c[#], 1], c[#]++]; AppendTo[a, # c[#]]) &@ DivisorSigma[0, a[[-1]]], 71]; a] (* Michael De Vlieger, Apr 12 2021 *)

PARI

lista(nn) = {my(va = vector(nn)); va[1] = 1; for (n=2, nn, va[n] = sum(k=1, n-1, if (numdiv(va[k]) == numdiv(va[n-1]), numdiv(va[k])));); va;} \\ Michel Marcus, Apr 28 2021