cp's OEIS Frontend

A228018 Prime powers p (A025475) such that the sum of the proper divisors of p is also a prime power.

%I A228018 #21 Sep 18 2024 08:43:35
%S A228018 9,49,243,961,16129,67092481,17179607041,274876858369,
%T A228018 4611686014132420609
%N A228018 Prime powers p (A025475) such that the sum of the proper divisors of p is also a prime power.
%C A228018 Numbers k such that both k and A001065(k) are in A025475.
%C A228018 Eight of the first nine terms are squares of Mersenne primes (A133049).
%C A228018 From _Pontus von Brömssen_, Sep 17 2024: (Start)
%C A228018 All squares of Mersenne primes are terms.
%C A228018 10^20 < a(10) <= A133049(9) = (2^61-1)^2.
%C A228018 All terms are odd. Otherwise, 2^k would be a term for some positive k >= 2 and then A001065(2^k) = 2^k-1 would be a prime power in A025475, which is impossible by Catalan's conjecture (Mihăilescu's theorem).
%C A228018 (End)
%e A228018 Proper divisors of 243 are 1, 3, 9, 27, 81, their sum is 121 = 11^2, so 243 is in the sequence.
%o A228018 (C)
%o A228018 #include <stdio.h>
%o A228018 #include <stdlib.h>
%o A228018 #define TOP (1ULL<<32)
%o A228018 typedef unsigned long long U64;
%o A228018 int compare64(const void *p1, const void *p2) {
%o A228018   if (*(U64*)p1== *(U64*)p2) return 0;
%o A228018   return (*(U64*)p1 < *(U64*)p2) ? -1 : 1;
%o A228018 }
%o A228018 U64 findElement(U64 *a, U64 start, U64 end, U64 element) {
%o A228018   if (start+1==end)  return (a[start]==element);
%o A228018   U64 mid = (start+end)/2;
%o A228018   if (a[mid] > element)
%o A228018     return findElement(a, start, mid, element);
%o A228018   return findElement(a, mid, end, element);
%o A228018 }
%o A228018 int main() {
%o A228018   U64 i, j, p, n=0, *pp = (U64*)malloc(TOP/2), sum;
%o A228018   unsigned char *c = (unsigned char *)malloc(TOP/16);
%o A228018   if (!c || !pp) exit(1);
%o A228018   memset(c, 0, TOP/16);
%o A228018   pp[n++] = 1;
%o A228018   for (i=1; i < TOP; i+=2) {
%o A228018     if ((c[i>>4] & (1<<((i>>1) & 7)))==0) {
%o A228018       for (p=i+(i==1), j = p*p; ; j*=p) {
%o A228018         pp[n++] = j;
%o A228018         double k = ((double)j) * ((double)p);
%o A228018         if (k >= ((double)(1ULL<<60)*16.0)) break;
%o A228018       }
%o A228018       if (i>1)
%o A228018         for (j=i*i>>1; j<TOP/2; j+=i)  c[j>>3] |= 1<<(j&7);
%o A228018     }
%o A228018     if ((i&(i-2))==1)  printf("%llu ", i);
%o A228018   }
%o A228018   printf("// %llu\n\n", n);
%o A228018   qsort(pp, n, 8, compare64);
%o A228018   for (i=1; i < TOP; i+=2)
%o A228018     if ((c[i>>4] & (1<<((i>>1) & 7)))==0)
%o A228018       for (p=i+(i==1), sum=1+p, j = p*p; ; j*=p) {
%o A228018         if (findElement(pp, 0, n, sum)) printf("%llu, ", j);
%o A228018         sum += j;
%o A228018         double k = ((double)j) * ((double)p);
%o A228018         if (k >= ((double)(1ULL<<60)*16.0)) break;
%o A228018       }
%o A228018   return 0;
%o A228018 }
%o A228018 (PARI) for(n=1,10^6,if(!isprime(n),v=factor(n);if(matsize(v)[1]==1,s=sumdiv(n,d,d)-n;if(!isprime(s),vv=factor(s);if(matsize(vv)[1]==1,print(n)))))) /* _Ralf Stephan_, Aug 05 2013 */
%Y A228018 Cf. A025475, A001065, A133049.
%K A228018 nonn,more
%O A228018 1,1
%A A228018 _Alex Ratushnyak_, Aug 02 2013