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 A212861 Numbers n such that the sum of prime factors of n (counted with repetition) equals three times the largest prime divisor. 3
 8, 24, 27, 125, 150, 160, 180, 343, 490, 588, 700, 840, 896, 945, 1008, 1134, 1331, 2197, 3388, 3718, 4840, 4913, 5445, 5808, 6292, 6534, 6859, 8085, 8624, 9464, 9625, 9702, 10647, 11550, 12167, 12274, 12320, 12675, 13520, 13750, 13860, 14784, 15015, 15028 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS The numbers prime(n)^3 are in the sequence. LINKS Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 FORMULA sopfr(n) = 3*gpf(n), where gpf = A006530. - Charles R Greathouse IV, May 29 2012 EXAMPLE 150 is in the sequence because 150 = 2*3*5^2 => sum of prime divisors = 2+3 + 5*2 = 15 = 3*5 where 5 is the greatest prime divisor. MAPLE with(numtheory):A:= proc(n) local e, j; e := ifactors(n)[2]: add (e[j][1]*e[j][2], j=1..nops(e)) end: for m from 2 to 20000 do: x:=factorset(m):n1:=nops(x):if A(m)=3*x[n1] then printf(`%d, `, m):else fi:od: MATHEMATICA spfQ[n_]:=Module[{f=FactorInteger[n]}, Total[Flatten[Table[#[[1]], #[[2]]]&/@ f]]==3*f[[-1, 1]]]; Select[Range[16000], spfQ] (* Harvey P. Dale, Jul 26 2016 *) PROG (PARI) is(n)=my(f=factor(n), k=#f[, 1]); k && sum(i=1, k, f[i, 1]*f[i, 2]) == 3*f[k, 1] \\ Charles R Greathouse IV, May 29 2012 CROSSREFS Cf. A001414, A163836. Sequence in context: A175496 A048109 A068781 * A333961 A038524 A261394 Adjacent sequences:  A212858 A212859 A212860 * A212862 A212863 A212864 KEYWORD nonn AUTHOR Michel Lagneau, May 29 2012 STATUS approved

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Last modified June 21 19:59 EDT 2021. Contains 345365 sequences. (Running on oeis4.)