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%I #7 Aug 02 2015 14:38:18
%S 3,19,211,1249,4513,1445953,30381331,286292179,2959257735801707821729
%N Primes p such that p + (p^2 - 1)/8 is a perfect square.
%C 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.
%C 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.
%e 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.
%p A077241 := proc(n) if n <= 3 then op(n+1,[1,2,8,13]) ; else 6*procname(n-2)-procname(n-4) ; fi; end:
%p 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
%p 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
%t 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 *)
%Y Cf. A165352, A165353, A165354.
%K nonn,more
%O 1,1
%A _Vincenzo Librandi_, Sep 16 2009
%E 6 more terms from _R. J. Mathar_, Sep 21 2009
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