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A225927
Least number of prime powers greater than 1 needed to sum up to n, or 0 if n cannot be represented as a sum of prime powers.
1
0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 2, 2, 0, 0, 1, 2, 2, 0, 2, 3, 3, 0, 2, 1, 3, 1, 3, 2, 4, 2, 1, 2, 2, 2, 2, 3, 3, 3, 2, 2, 3, 2, 3, 3, 4, 3, 2, 1, 2, 3, 2, 2, 2, 4, 3, 2, 2, 2, 3, 3, 3, 3, 1, 2, 3, 3, 2, 3, 3, 4, 2, 2, 2, 3, 2, 3, 3, 3, 2, 1, 3, 3, 3, 2, 3, 4, 3, 2, 2
OFFSET
1,12
COMMENTS
Nontrivial prime powers: A025475 except 1.
EXAMPLE
26 = 9 + 9 + 8, three summands, so a(26) = 3.
PROG
(C)
#include <stdio.h> // GCC -O3
#define TOP (1ULL<<15) // ~140 seconds // (1ULL<<17) is ok
#define TOP2 (TOP*TOP)
typedef unsigned long long U64;
int compare64(const void *p1, const void *p2) {
if (*(U64*)p1 < *(U64*)p2) return -1;
return (*(U64*)p1 == *(U64*)p2) ? 0 : 1;
}
int main() {
U64 i, j, k, p, pp=1, pfp=0, *primes, *pwFlat = (U64*)malloc(TOP*2);
primes = (U64*)malloc(TOP2);
char *c = (char*)pwFlat, *f = (char*)primes, *ff;
memset(c, 0, TOP);
for (primes[0]=2, i=3; i<TOP; i+=2) if (c[i>>1]==0)
for (primes[pp++]=i, j=i*i>>1; j<TOP/2; j+=i) c[j]=1;
for (i = 0; i < pp; ++i)
for (j=primes[i]*primes[i]; j<TOP2; j*=primes[i]) pwFlat[pfp++]=j;
qsort(pwFlat, pfp, 8, compare64);
memset(f, 100, TOP2);
for (pwFlat[pfp]=TOP2, i=0; (p=pwFlat[i])<TOP2; printf("."), ++i)
for(f[p]=1, j=0; j<=i && (pp=p+pwFlat[j])<TOP2; ++j)
for(f[pp]=2, k=0; k<=j && (pfp=pp+pwFlat[k])<TOP2; ++k) f[pfp]=3;
for (k=1; k < TOP2; k++) {
if (f[k]==100) {
for (j=0, ff=&f[k], pp=99; (p=pwFlat[j]) < k; j++)
if (ff[-p] < pp) { pp = ff[-p]; if (pp<=3) break; }
f[k] = pp+1;
}
if (k<200) printf("%llu, ", f[k] > 99 ? 0 : f[k]);
else if (f[k]>4) printf("\n%llu at %llu ", f[k], k);
}
return 0;
}
CROSSREFS
Sequence in context: A328802 A355245 A103822 * A029392 A035379 A280452
KEYWORD
nonn
AUTHOR
Alex Ratushnyak, May 21 2013
STATUS
approved