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%I #9 Jun 06 2013 08:09:53
%S 3,9,12,21,24,35,72,126,129,179,189,194,198,214,243,253,255,279,304,
%T 322,432,443,480,487,511,523,663,681,696,699,711,717,721,734,738,796,
%U 802,838,910,975,1008,1034,1070,1144,1215,1230,1237,1265,1276,1370,1375,1386,1469
%N Numbers n such that n^2 is an average of 4 consecutive primes.
%C Integers of the form sqrt(A102655(k)) for any k. - _R. J. Mathar_, Jun 06 2013
%F a(n) = A051395(n)/2.
%p A034963 := proc(n)
%p add(ithprime(i), i=n..n+3) ;
%p end proc:
%p for n from 1 to 90000 do
%p s := A034963(n)/4 ;
%p if type(s,'integer') then
%p if issqr(s) then
%p printf("%d, ", sqrt(s)) ;
%p end if;
%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<<30)
%o int main() {
%o unsigned long long i, j, p1, p2, p3, r, s;
%o unsigned char *c = (unsigned char *)malloc(TOP/8);
%o memset(c, 0, TOP/8);
%o for (i=3; i < TOP; i+=2)
%o if ((c[i>>4] & (1<<((i>>1) & 7)))==0 /*&& i<(1ULL<<32)*/)
%o for (j=i*i>>1; j<TOP; j+=i) c[j>>3] |= 1 << (j&7);
%o for (p3=2, p2=3, p1=5, i=7; i < TOP; i+=2)
%o if ((c[i>>4] & (1<<((i>>1) & 7)))==0) {
%o s = p3 + p2 + p1 + i;
%o if (s%4==0) {
%o s/=4;
%o r = sqrt(s);
%o if (r*r==s) printf("%llu, ", r);
%o }
%o p3 = p2, p2 = p1, p1 = i;
%o }
%o return 0;
%o }
%Y Cf. A102655, A051395, A206280.
%K nonn
%O 1,1
%A _Alex Ratushnyak_, May 28 2013
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