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A271441
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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.
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1
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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
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OFFSET
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1,1
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COMMENTS
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Choosing the initial value a(1) = 2 seems to produce the most irregular sequence.
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LINKS
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EXAMPLE
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Since a(11) = 28, gpf(28) = 7 and a(7) = 12, then a(12) = 28 + 12 = 40.
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MATHEMATICA
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Nest[Append[#, (x = #[[-1]]) + If[(p = FactorInteger[x][[-1, 1]]) <= Length@#, #[[p]], 1]] &, {2}, 49] (* Ivan Neretin, Jan 27 2017 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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