A330988 a(1)=2, a(2)=3; then a(n+1) = smallest k such that S(k) = S(a(n)) + S(a(n-1)), (n>=2), where S is sopfr (A001414).

2, 3, 5, 15, 13, 38, 93, 106, 89, 695, 233, 1492, 1821, 3932, 1597, 12895, 16708, 13526, 76573, 70828, 28657, 787967, 1125255, 4005507, 6087997, 10487301, 514229, 30784111, 68658699, 150301527, 38770237, 290846217, 525964251, 164233751, 193262488, 1368085495, 1075181473, 8903068701, 10762707995, 4554542743, 433494437

Offset: 1

Author

David James Sycamore, Jan 05 2020

Keywords

• nonn

Comments

The subsequence of prime terms is A005478; a term is prime if and only if it is a Fibonacci prime (proved by Giovanni Resta).

Examples

			a(3)=5 since 5 is the smallest number whose sum of prime divisors is soprf(2) + sopfr(3) = 2 + 3 = 5. a(4)=15 since sopfr(3)+sopfr(5)=8, and 15 is the smallest number whose sum of prime divisors is 8.
		

Links

• Giovanni Resta, Table of n, a(n) for n = 1..51

Crossrefs

Cf. A000045, A001414, A005478, A056240, A330821.

Programs

Mathematica

sopfr[n_] := Plus @@ Times @@@ FactorInteger[n]; a[1] = 2; a[2] = 3; a[n_] := a[n] = Block[{t = sopfr@ a[n-1] + sopfr@ a[n-2], k=3}, While[ sopfr[k] != t, k++]; k]; Array[a, 21] (* Giovanni Resta, Jan 07 2020 *)

Formula

a(n+1) = A056240(A001414(a(n)) + A001414(a(n-1))).

a(n) = A056240(A000045(n+2)). - Giovanni Resta, Jan 07 2020