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Search: id:A165356
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| A165356 |
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Primes p such that p+(p^2-1)/8 is a perfect square. |
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+0 1
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| 3, 19, 211, 1249, 4513, 1445953, 30381331, 286292179, 2959257735801707821729
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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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.
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EXAMPLE
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For p=3, p+(p^2-1)/8=4=2^2. For p=19, p+(p^2-1)/8=64=8^2. p=211, p+(p^2-1)/8=5776=76^2.
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MAPLE
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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); [From Emeric Deutsch (deutsch(AT)duke.poly.edu), Sep 21 2009]
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MATHEMATICA
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p = 2; lst = {}; While[p < 10^12, If[ IntegerQ@ Sqrt[p + (p^2 - 1)/8], AppendTo[lst, p]; Print@p]; p = NextPrime@p] [From Robert G. Wilson v (rgwv(AT)rgwv.com), Sep 30 2009]
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CROSSREFS
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Cf. A165352, A165353, A165354
Sequence in context: A052886 A079144 A049056 this_sequence A000275 A058165 A074707
Adjacent sequences: A165353 A165354 A165355 this_sequence A165357 A165358 A165359
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KEYWORD
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nonn,more
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AUTHOR
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Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Sep 16 2009
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EXTENSIONS
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6 more terms from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 21 2009
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