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 A226152 Numbers n such that n^2 is an average of 4 consecutive primes. 0
 3, 9, 12, 21, 24, 35, 72, 126, 129, 179, 189, 194, 198, 214, 243, 253, 255, 279, 304, 322, 432, 443, 480, 487, 511, 523, 663, 681, 696, 699, 711, 717, 721, 734, 738, 796, 802, 838, 910, 975, 1008, 1034, 1070, 1144, 1215, 1230, 1237, 1265, 1276, 1370, 1375, 1386, 1469 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Integers of the form sqrt(A102655(k)) for any k. - R. J. Mathar, Jun 06 2013 LINKS FORMULA a(n) = A051395(n)/2. MAPLE A034963 := proc(n)     add(ithprime(i), i=n..n+3) ; end proc: for n from 1 to 90000 do     s := A034963(n)/4 ;     if type(s, 'integer') then     if issqr(s) then         printf("%d, ", sqrt(s)) ;     end if;     end if; end do: # R. J. Mathar, Jun 06 2013 PROG (C) #include #include #include #define TOP (1ULL<<30) int main() {   unsigned long long i, j, p1, p2, p3, r, s;   unsigned char *c = (unsigned char *)malloc(TOP/8);   memset(c, 0, TOP/8);   for (i=3; i < TOP; i+=2)     if ((c[i>>4] & (1<<((i>>1) & 7)))==0 /*&& i<(1ULL<<32)*/)         for (j=i*i>>1; j>3] |= 1 << (j&7);   for (p3=2, p2=3, p1=5, i=7; i < TOP; i+=2)     if ((c[i>>4] & (1<<((i>>1) & 7)))==0) {       s = p3 + p2 + p1 + i;       if (s%4==0) {         s/=4;         r = sqrt(s);         if (r*r==s) printf("%llu, ", r);       }       p3 = p2, p2 = p1, p1 = i;     }   return 0; } CROSSREFS Cf. A102655, A051395, A206280. Sequence in context: A272027 A310323 A212059 * A155504 A270672 A211217 Adjacent sequences:  A226149 A226150 A226151 * A226153 A226154 A226155 KEYWORD nonn AUTHOR Alex Ratushnyak, May 28 2013 STATUS approved

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Last modified September 25 10:58 EDT 2021. Contains 347654 sequences. (Running on oeis4.)