A330397 Composite numbers k such that gpf(k)^2 + gpf(k) == 0 (mod k), where gpf(k) = A006530(k) is the greatest prime dividing k.
6, 10, 12, 14, 15, 22, 26, 28, 30, 33, 34, 38, 44, 46, 51, 56, 58, 62, 66, 69, 74, 76, 82, 86, 87, 91, 92, 94, 95, 102, 106, 118, 122, 123, 124, 132, 134, 138, 141, 142, 145, 146, 153, 158, 159, 166, 172, 174, 177, 178, 182, 184, 188, 190, 194, 202, 206, 213, 214, 218, 226, 236
Offset: 1
Keywords
Links
- Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Magma
[k: k in [4..240] | -Maximum(PrimeDivisors(k))^2 mod k eq Maximum(PrimeDivisors(k))];
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Mathematica
Select[Range[250], (g = FactorInteger[#][[-1, 1]]) < # && Divisible[g^2 + g, #] &] (* Amiram Eldar, Feb 25 2020 *)
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PARI
gpf(n,f=factor(n))=f=f[,1]; if(#f==0,1,f[#f]); is(n)=my(g=gpf(n)); g
Charles R Greathouse IV, Mar 26 2020 -
PARI
list(lim)=my(v=List()); forfactored(N=6,lim\1, my(n=N[1],f=N[2][,1],i=#f); if(i>1 && (f[i]^2+f[i])%n==0, listput(v,n))); Vec(v); \\ Charles R Greathouse IV, Mar 27 2020
Comments