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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
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OFFSET
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1,1
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COMMENTS
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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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LINKS
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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. For 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); # Emeric Deutsch, 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] (* Robert G. Wilson v, Sep 30 2009 *)
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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