A165356 Primes p such that p + (p^2 - 1)/8 is a perfect square.

3, 19, 211, 1249, 4513, 1445953, 30381331, 286292179, 2959257735801707821729

Offset

1,1

Comments

The primes p = A000040(j) at j= 2, 8, 47, 204, 612, 110340 etc. generating the squares 2^2, 8^2, 76^2, 443^2 etc.

From the ansatz p + (p^2 - 1)/8 = s^2 we conclude p = -4 + sqrt(17 + 8*s^2), so all s are members of A077241.

Links

Table of n, a(n) for n=1..9.

Example

For p=3, p + (p^2-1)/8 = 4 = 2^2. For p=19, p + (p^2-1)/8 = 64 = 8^2. For p=211, p + (p^2-1)/8 = 5776 = 76^2.

Maple

A077241 := proc(n) if n <= 3 then op(n+1, [1, 2, 8, 13]) ; else 6*procname(n-2)-procname(n-4) ; fi; end:
for n from 0 do s := A077241(n) ; p := sqrt(17+8*s^2)-4 ; if isprime(p) then printf("%d, \n", p) ; fi; od: # R. J. Mathar, Sep 21 2009
a := proc (n) if isprime(n) = true and type(sqrt(n+(1/8)*n^2-1/8), integer) = true then n else end if end proc; seq(a(n), n = 1 .. 10000000); # Emeric Deutsch, Sep 21 2009

Mathematica

p = 2; lst = {}; While[p < 10^12, If[ IntegerQ@ Sqrt[p + (p^2 - 1)/8], AppendTo[lst, p]; Print@p]; p = NextPrime@p] (* Robert G. Wilson v, Sep 30 2009 *)

Crossrefs

Cf. A165352, A165353, A165354.

Sequence in context: A345218 A049056 A204262 * A000275 A058165 A074707

Adjacent sequences: A165353 A165354 A165355 * A165357 A165358 A165359

Keyword

nonn,more

Author

Vincenzo Librandi, Sep 16 2009

Extensions

6 more terms from R. J. Mathar, Sep 21 2009

Status

approved