A271441 a(1) = 2; if gpf(a(n-1)) <= n-1 then a(n) = a(n-1) + a(gpf(a(n-1))), else a(n) = a(n-1) + 1, where gpf(m) is the greatest prime factor of m.

2, 3, 4, 7, 8, 11, 12, 16, 19, 20, 28, 40, 48, 52, 100, 108, 112, 124, 125, 133, 258, 259, 260, 308, 336, 348, 349, 350, 362, 363, 391, 651, 1042, 1043, 1044, 1406, 1407, 1408, 1436, 1437, 1438, 1439, 1440, 1448, 1449, 1709, 1710, 1835, 1836, 1948

Offset: 1

Author

Cody M. Haderlie, Apr 07 2016

Keywords

• nonn
• hear

Comments

Choosing the initial value a(1) = 2 seems to produce the most irregular sequence.

Examples

			Since a(11) = 28, gpf(28) = 7 and a(7) = 12, then a(12) = 28 + 12 = 40.
		

Links

• Ivan Neretin, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A006530 (gpf).

Programs

Mathematica

Nest[Append[#, (x = #[[-1]]) + If[(p = FactorInteger[x][[-1, 1]]) <= Length@#, #[[p]], 1]] &, {2}, 49] (* Ivan Neretin, Jan 27 2017 *)

PARI

gpf(n) = if (n==1, 1, vecmax(factor(n)[,1]));
lista(nn) = {va = vector(nn); print1(va[1] = 2, ", "); for (n=2, nn, if (gpf(va[n-1]) <= n-1, va[n] = va[n-1] + va[gpf(va[n-1])], va[n] = va[n-1]+1); print1(va[n], ", "););} \\ Michel Marcus, Apr 09 2016