A383729 Numbers k such that omega(k) = 5 and the largest prime factor of k equals the sum of its remaining distinct prime factors, where omega(k) = A001221(k).
3570, 7140, 8970, 10626, 10710, 14280, 16530, 17850, 17940, 20706, 21252, 21420, 24738, 24882, 24990, 26910, 28560, 31878, 32130, 33060, 35700, 35880, 36890, 38130, 41412, 42504, 42840, 44330, 44850, 49476, 49590, 49764, 49938, 49980, 52170, 53550, 53820, 54834, 55986, 57120
Offset: 1
Keywords
Examples
10710 is a term because it has 5 distinct prime factors (2, 3, 5, 7 and 17) and the largest one is the sum of the others (2 + 3 + 5 + 7 = 17).
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
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Maple
N:= 10^5: # for terms <= N P:= select(isprime,[2,seq(i,i=3..N/(2*3*5*7),2)]): V:= NULL: i:= 1: for j from i+1 while P[i]*P[j]^3*(P[i]+3*P[j]) < N do for k from j+1 while P[i]*P[j]*P[k]^2*(P[i]+P[j]+2*P[k]) < N do for l from k+1 while P[i]*P[j]*P[k]*P[l] * (P[i]+P[j]+P[k]+P[l]) <= N do p5:= P[i]+P[j]+P[k]+P[l]; if not isprime(p5) then next fi; for d1 from 1 while P[i]^d1 * P[j] * P[k] * P[l] * p5 <= N do for d2 from 1 while P[i]^d1 * P[j]^d2 * P[k] * P[l] * p5 <= N do for d3 from 1 while P[i]^d1 * P[j]^d2 * P[k]^d3 * P[l] * p5 <= N do for d4 from 1 while P[i]^d1 * P[j]^d2 * P[k]^d3 * P[l]^d4 * p5 <= N do for d5 from 1 while P[i]^d1 * P[j]^d2 * P[k]^d3 * P[l]^d4 * p5^d5 <= N do V:= V,P[i]^d1 * P[j]^d2 * P[k]^d3 * P[l]^d4 * p5^d5 od od od od od od od od: sort([V]); # Robert Israel, Jun 09 2025
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Mathematica
A383729Q[k_] := Length[#] == 5 && Total[Most[#]] == Last[#] & [FactorInteger[k][[All, 1]]]; Select[Range[10^5], A383729Q]