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a(n+1) = least k with sum of prime factors (with repetition) = a(n)+1 with a(0) = 2.
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%I #20 Apr 29 2019 09:24:03

%S 2,3,4,5,8,14,26,92,356,1412,5636,185559,556671,21152738,42305474,

%T 2919075981,14595379885,102167659153,3882371047054,361060507372953,

%U 16969843846526629,1561225633880447476,6244902535521789892

%N a(n+1) = least k with sum of prime factors (with repetition) = a(n)+1 with a(0) = 2.

%H Jason Earls, <a href="https://pdfs.semanticscholar.org/4559/ac50797ddeda688576630c4d92229440a0a3.pdf#page=274">A note on the Smarandache divisors of divisors sequence and two similar sequences</a>, in Smarandache Notions Journal (2004), Vol. 14.1, page 275. [broken link]

%e 92 is a term because it is the smallest number whose sum of prime factors is equal to the previous term + 1; 92 = 2^2*23 and 2+2+23 = 26+1.

%t Nest[Append[#, Block[{k = #[[-1]] + 1, m}, m = k; While[Total@ Flatten[ConstantArray[#1, #2] & @@@ FactorInteger@ k] != m, k++]; k]] &, {2}, 12] (* _Michael De Vlieger_, Mar 28 2018 *)

%o (PARI) v = vector(200); count = 0; m = 2; print1("2 3 4 5 8 "); n = 8; while (count < 199, f = factor(m); s = sum(i = 1, matsize(f)[1], f[i, 1]*f[i, 2]); if (s <= 200 && v[s] == 0, count++; v[s] = m); m++); for (i = 1, 20, p = precprime(n + 1); if (p == n + 1, n++; print1(n, " "), b = v[n + 1 - p]; c = p; while (b > n + 1 - p, p = precprime(p - 1); m = v[n + 1 - p]; if (m < b, b = m; c = p)); n = b*c; print1(n, " "))); \\ _David Wasserman_, Jan 23 2005

%K nonn

%O 0,1

%A _Jason Earls_, Oct 03 2002

%E More terms from _David Wasserman_, Jan 23 2005