A330807 - OEIS

%I #34 Sep 08 2022 08:46:24

%S 2,2,4,4,8,6,5,10,7,14,9,12,21,10,28,11,22,13,26,15,8,18,16,24,18,32,

%T 20,27,36,30,40,33,14,45,44,30,50,12,35,24,54,55,16,40,66,32,60,36,70,

%U 28,77,18,48,88,17,34,19,38,21,80,39,48,99,51,20,63,52,68,42,48,110,36,90,65,54,121,22,78,72,60,72,84,42,96

%N a(1) = 2; a(n+1) = Sum_{k=1..n} {S(a(k)): S(a(k)) = S(a(n))}, where S is sopfr (A001414).

%C a(n+1) is k(n)*sopfr(a(n)), where k(n) is the number of times (up to and including a(n)) that a term having the same sopfr as a(n) has occurred in the sequence so far.

%C The smallest possible initial term is a(1)=2, since 2 is the smallest number having a prime divisor. Not every integer appears; for example, 3 cannot occur unless chosen as a(1) (in which case 2 cannot appear).

%C The primes do not appear in their natural order (e.g., 31 precedes 29). If the lesser of twin primes p is a(k) and the greater twin has not already occurred, then a(k+2) is the greater twin. Whereas if a composite number appears, it may appear more than once, a prime > 2, if it appears, can appear once only. Open question: Do all primes (except for 3) eventually appear?

%H Amiram Eldar, <a href="/A330807/b330807.txt">Table of n, a(n) for n = 1..10000</a>

%e a(2)=2 because it is the sopfr of a(1) and there are no prior terms which could contribute to the sum.

%e a(3)=2+2=4 because 2 has occurred twice already as sopf of prior terms.

%e a(7)=5 because a(6)=6 and has sopfr 5 which has not been seen before.

%t s[1] = 0; s[n_] := Plus @@ Times @@@ FactorInteger[n]; a[1] = 2; a[n_] := a[n] = (s1 = s[a[n - 1]])*(1 + Sum[Boole[s[a[k]] == s1], {k, 1, n - 2}]); Array[a, 100] (* _Amiram Eldar_, Jan 01 2020 *)

%o (Magma) sopfr:=func<n|&+[j[1]*j[2]: j in Factorization(n)]>; a:=[2]; for n in [2..90] do  Append(~a,&+[sopfr(a[k]):k in [1..n-1]|sopfr(a[k]) eq  sopfr(a[n-1])]); end for; a; // _Marius A. Burtea_, Jan 01 2020

%Y Cf. A001414, A279818.

%K nonn,look

%O 1,1

%A _David James Sycamore_, Jan 01 2020