A226230
Nontrivial prime powers (A025475) which are a sum of a smaller nontrivial prime power and a perfect square.
Original entry on oeis.org
8, 9, 25, 32, 81, 125, 128, 169, 289, 343, 512, 625, 1681, 2048, 2197, 3125, 3721, 4913, 8192, 32761, 32768, 50653, 66049, 78125, 97969, 131072, 177241, 357911, 524288, 707281, 1030301, 1419857, 1442401, 1953125, 2097152, 2476099, 3031081, 3463321, 6355441, 7645373
Offset: 1
81 = 32 + 7^2, so 81 is in the sequence.
169 = 25 + 12^2, so 169 is in the sequence.
-
#include
#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;
}
int main() {
U64 i, j, p, t, r, n=0, *pp = (U64*)malloc(TOP/2);
unsigned char *c = (unsigned char *)malloc(TOP/8);
memset(c, 0, TOP/8);
for (pp[n++] = 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<<62)*4.0)) break;
}
if(i>1) for (j=i*i>>1; j>3]|= 1<<(j&7);
}
qsort(pp, n, 8, compare64);
for (i = 0; i < n; i++)
for (p=pp[i], j=0; j < i; j++) {
t = p - pp[j];
r = sqrt(t);
if (r*r==t) { printf("%llu, ", p); break; }
}
return 0;
}
-
for n from 1 do
if isA025475(n) then
for j from 1 do
pj := n-j^2 ;
if pj < 0 then
break;
elif isA025475(pj) then
printf("%d,\n",n);
break ;
end if;
end do:
end if:
end do: # R. J. Mathar, Jun 06 2013
A226231
Nontrivial prime powers (A025475) which are a sum of a smaller nontrivial prime power and a perfect square in more than 1 way.
Original entry on oeis.org
25, 125, 512, 3125, 4913, 50653, 78125, 7645373, 48828125, 69343957, 310288733, 410338673, 1220703125, 4103684801, 24820429213, 164296466333, 296709280757, 762939453125, 1772000266301, 10368641602001, 11392143817501, 19073486328125, 29724740184833, 72079590632113
Offset: 1
125 = 4 + 11^2 = 25 + 10^2 = 121 + 4^2. Because there are three representations, 125 is in the sequence. Note that 100 + 5^2 is not listed, because 100 is not a prime power.
-
#include
#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;
}
int main() {
U64 i, j, k, p, t, r, n=0, *pp = (U64*)malloc(TOP/2);
unsigned char *c = (unsigned char *)malloc(TOP/8);
memset(c, 0, TOP/8);
for (pp[n++] = 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<<62)*4.0)) break;
}
if(i>1) for (j=i*i>>1; j>3]|= 1<<(j&7);
}
qsort(pp, n, 8, compare64);
for (i = 0; i < n; i++)
for (p=pp[i], j = k = 0; j < i; j++) {
t = p - pp[j];
r = sqrt(t);
if (r*r==t && ++k==2) { printf("%llu, ", p); break; }
}
return 0;
}
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