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%I #13 Mar 13 2024 11:01:39
%S 216,5400,10584,26136,36504,62424,77976,114264,181656,207576,264600,
%T 295704,363096,399384,477144,606744,653400,751896,803736,912600,
%U 969624,1088856,1149984,1151064,1280664,1348056,1488024,1560600,1710936,1788696,1949400,2032344,2203416
%N Prime power perfect numbers: If n = Product p_i^r_i let PPsigma(n) = Product {Sum p_i^s_i, 2<=s_i<=r_i, s_i is prime}; sequence gives numbers k such that PPsigma(k) = 2*k.
%H Amiram Eldar, <a href="/A096290/b096290.txt">Table of n, a(n) for n = 1..1000</a>
%e 5400 is in the sequence because 5400 = 2^3*3^3*5^2 and (2^2+2^3)*(3^2+3^3)*(5^2) = 2*5400.
%p PPsigma := proc(n)
%p option remember;
%p local a, pe, p, e,f,i ;
%p a := 1 ;
%p for pe in ifactors(n)[2] do
%p p := op(1, pe) ;
%p e := op(2, pe) ;
%p f := 0 ;
%p for i from 2 to e do
%p if isprime(i) then
%p f := f+p^i ;
%p end if;
%p end do:
%p a := a*f ;
%p end do;
%p a ;
%p end proc:
%p for n from 1 do
%p if PPsigma(n) = 2*n then
%p print(n) ;
%p end if;
%p end do: # _R. J. Mathar_, Mar 13 2024
%t f[p_, e_] := Plus @@ (p^Select[Range[e], PrimeQ]); s[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[300000], s[#] == 2*# &] (* _Amiram Eldar_, Sep 19 2022 *)
%Y Cf. A100509.
%K nonn
%O 1,1
%A _Yasutoshi Kohmoto_, Jun 24 2004
%E Corrected and extended by _Farideh Firoozbakht_, Nov 17 2004
%E More terms from _Amiram Eldar_, Sep 19 2022
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