|
%I #10 Jun 06 2013 19:08:50
%S 8,9,25,32,81,125,128,169,289,343,512,625,1681,2048,2197,3125,3721,
%T 4913,8192,32761,32768,50653,66049,78125,97969,131072,177241,357911,
%U 524288,707281,1030301,1419857,1442401,1953125,2097152,2476099,3031081,3463321,6355441,7645373
%N Nontrivial prime powers (A025475) which are a sum of a smaller nontrivial prime power and a perfect square.
%e 81 = 32 + 7^2, so 81 is in the sequence.
%e 169 = 25 + 12^2, so 169 is in the sequence.
%p for n from 1 do
%p if isA025475(n) then
%p for j from 1 do
%p pj := n-j^2 ;
%p if pj < 0 then
%p break;
%p elif isA025475(pj) then
%p printf("%d,\n",n);
%p break ;
%p end if;
%p end do:
%p end if:
%p end do: # _R. J. Mathar_, Jun 06 2013
%o (C)
%o #include <stdio.h>
%o #include <stdlib.h>
%o #include <math.h>
%o #define TOP (1ULL<<32)
%o typedef unsigned long long U64;
%o int compare64(const void *p1, const void *p2) {
%o if (*(U64*)p1== *(U64*)p2) return 0;
%o return (*(U64*)p1 < *(U64*)p2) ? -1 : 1;
%o }
%o int main() {
%o U64 i, j, p, t, r, n=0, *pp = (U64*)malloc(TOP/2);
%o unsigned char *c = (unsigned char *)malloc(TOP/8);
%o memset(c, 0, TOP/8);
%o for (pp[n++] = i = 1; i < TOP; i+=2)
%o if ((c[i>>4] & (1<<((i>>1) & 7)))==0) {
%o for (p=i+(i==1), j = p*p; ; j*=p) {
%o pp[n++] = j;
%o double k = ((double)j) * ((double)p);
%o if (k >= ((double)(1ULL<<62)*4.0)) break;
%o }
%o if(i>1) for (j=i*i>>1; j<TOP; j+=i) c[j>>3]|= 1<<(j&7);
%o }
%o qsort(pp, n, 8, compare64);
%o for (i = 0; i < n; i++)
%o for (p=pp[i], j=0; j < i; j++) {
%o t = p - pp[j];
%o r = sqrt(t);
%o if (r*r==t) { printf("%llu, ", p); break; }
%o }
%o return 0;
%o }
%Y Cf. A025475, A226231, A226232.
%K nonn
%O 1,1
%A _Alex Ratushnyak_, May 31 2013
|
