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

9, 49, 243, 961, 16129, 67092481, 17179607041, 274876858369, 4611686014132420609

Offset: 1

Author

Alex Ratushnyak, Aug 02 2013

Keywords

• nonn
• more

Comments

Numbers k such that both k and A001065(k) are in A025475.

Eight of the first nine terms are squares of Mersenne primes (A133049).

From Pontus von Brömssen, Sep 17 2024: (Start)

All squares of Mersenne primes are terms.

10^20 < a(10) <= A133049(9) = (2^61-1)^2.

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).

(End)

Examples

			Proper divisors of 243 are 1, 3, 9, 27, 81, their sum is 121 = 11^2, so 243 is in the sequence.
		

Crossrefs

Cf. A025475, A001065, A133049.

Programs

C

#include 
#include 
#define TOP (1ULL<<32)
typedef unsigned long long U64;
int compare64(const void *p1, const void *p2) {
  if (*(U64*)p1== *(U64*)p2) return 0;
  return (*(U64*)p1 < *(U64*)p2) ? -1 : 1;
}
U64 findElement(U64 *a, U64 start, U64 end, U64 element) {
  if (start+1==end)  return (a[start]==element);
  U64 mid = (start+end)/2;
  if (a[mid] > element)
    return findElement(a, start, mid, element);
  return findElement(a, mid, end, element);
}
int main() {
  U64 i, j, p, n=0, *pp = (U64*)malloc(TOP/2), sum;
  unsigned char *c = (unsigned char *)malloc(TOP/16);
  if (!c || !pp) exit(1);
  memset(c, 0, TOP/16);
  pp[n++] = 1;
  for (i=1; i < TOP; i+=2) {
    if ((c[i>>4] & (1<<((i>>1) & 7)))==0) {
      for (p=i+(i==1), j = p*p; ; j*=p) {
        pp[n++] = j;
        double k = ((double)j) * ((double)p);
        if (k >= ((double)(1ULL<<60)*16.0)) break;
      }
      if (i>1)
        for (j=i*i>>1; j>3] |= 1<<(j&7);
    }
    if ((i&(i-2))==1)  printf("%llu ", i);
  }
  printf("// %llu\n\n", n);
  qsort(pp, n, 8, compare64);
  for (i=1; i < TOP; i+=2)
    if ((c[i>>4] & (1<<((i>>1) & 7)))==0)
      for (p=i+(i==1), sum=1+p, j = p*p; ; j*=p) {
        if (findElement(pp, 0, n, sum)) printf("%llu, ", j);
        sum += j;
        double k = ((double)j) * ((double)p);
        if (k >= ((double)(1ULL<<60)*16.0)) break;
      }
  return 0;
}

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 */