OFFSET
1,3
COMMENTS
Inspired by A360179, but uses a simpler rule for non-novel terms.
It is an obvious conjecture that every number eventually appears, but is there a proof?
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..40000
Michael De Vlieger, Scatterplot of a(n), n = 1..2^16, showing records in red, smallest missing numbers in blue (small until they enter sequence, then large), terms deriving from novel predecessors in gold, otherwise green.
EXAMPLE
The initial terms (in the third column, N = novel term, D = non-novel term):
.n.a(n).....t
.1,..1,.N,
.2,..1,.D,..1
.3,..2,.N,
.4,..2,.D,..2
.5,..3,.N,
.6,..2,.D,..3
.7,..4,.N,
.8,..3,.D,..4
.9,..5,.N,
10,..2,.D,..5
11,..4,.D,..6
12,..6,.N,
13,..4,.D,..7
14,..7,.N,
15,..2,.D,..8
16,..4,.D,..9
17,..6,.D,.10
18,..8,.N,
19,..4,.D,.11
20,..7,.D,.12
21,.11,.N,
22,..2,.D,.13
...
If n=8, for example, a(8) = 3 is a non-novel term, the 4th such, so a(9) = a(8) + d(a(4)) = 3 + d(2) = 5.
Comment from Michael De Vlieger, Apr 08 2023 (Start)
Can be read as an irregular triangle of increasing subsequences:
1;
1, 2;
2, 3;
2, 4;
3, 5;
2, 4, 6;
4, 7;
2, 4, 6, 8;
4, 7, 11;
2, 5, 7, 9;
3, 6, 10;
4, 8, 11, 13;
2, 4, 6, 8, 10, 13, 15;
4, 8, 12;
6, 9, 13, 15, 17;
2, 4, 7, 11, 15, 19;
etc.
(End)
MATHEMATICA
nn = 120; c[_] = False; f[n_] := DivisorSigma[0, n]; a[1] = m = 1; Do[(If[c[#], a[n] = # + f[a[m]] ; m++, a[n] = f[#] ]; c[#] = True) &[a[n - 1]], {n, 2, nn}]; Array[a, nn] (* Michael De Vlieger, Apr 08 2023 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
N. J. A. Sloane, Apr 08 2023
STATUS
approved