|
| |
|
|
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, 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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,3
|
|
|
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
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
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.
|
|
|
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 *)
|
|
|
PROG
|
(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
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
