2,1

a(n) = first occurrence of n in A001414 (sum of prime divisors of n).

Write n as 2a+3b, then A001414(2^a*3^b) = n, so a(n) <= m.

It appears that this sequence can be computed as follows: if n is prime, a(n) = n. Otherwise, find the k with the smallest value of a(k) such that n - k is prime. Then a(n) = (n-k) * a(k). - Franklin T. Adams-Watters (FrankTAW(AT)Netscape.net), Sep 15 2006

a(n)=3^(n-2-2*floor((n-1)/3))*2^(2-(n-1)mod 3) for n>1. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Nov 11 2007

a(8) = 15 because the smallest number whose prime divisors sum to 8 is 15 (3*5)

a(14)=33, for instance,is the first among the finite set A={33,49,100,120,128,144,162} or {3*11,7^2,2^2*5^2,2^3*3*5,2^7,2^4*3^2,2*3^4} such that sopfr(v)=14, where any v belongs to A.

a = Table[0, {75}]; Do[b = Plus @@ Flatten[ Table[ #1, {#2}] & @@@ FactorInteger[n]]; If[b < 76 && a[[b]] == 0, a[[b]] = n], {n, 2, 1000}]; a (Robert G. Wilson v (rgwv(AT)rgwv.com), May 04 2002)

Cf. A001414, A064502.

First column of array A064364, n>=2.

See A000792 for the maximal numbers whose prime factors sums up to n.

Sequence in context: A091893 A074756 A075162 this_sequence A069968 A086931 A164339

Adjacent sequences: A056237 A056238 A056239 this_sequence A056241 A056242 A056243

nonn,easy

Adam Kertesz (adamkertesz(AT)worldnet.att.net), Aug 19 2000

More terms from James A. Sellers (sellersj(AT)math.psu.edu), Aug 25 2000