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A274876
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The number of ways (2n)^2 is expressible as (p+1)(q+1) where p and q are distinct primes.
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2
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0, 0, 1, 0, 0, 3, 0, 1, 2, 0, 0, 4, 0, 1, 1, 1, 0, 4, 0, 2, 3, 1, 0, 4, 0, 1, 1, 2, 0, 5, 0, 1, 2, 1, 1, 4, 0, 1, 3, 1, 0, 7, 0, 2, 4, 0, 0, 4, 0, 2, 3, 1, 0, 2, 0, 2, 0, 0, 0, 9, 0, 0, 2, 0, 0, 5, 0, 3, 0, 2, 0, 8, 0, 0, 2, 2, 2, 6, 0, 2, 2, 1, 0, 6, 0, 1, 1, 2, 0, 8, 1, 1, 1, 0, 0, 2, 0, 1, 0, 4, 0, 4, 0, 2, 5
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OFFSET
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1,6
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COMMENTS
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No odd number squared is expressible as (p+1)(q+1) where p and q are distinct primes, since q must be odd and therefore (q+1) is even.
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LINKS
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EXAMPLE
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a(1) = 0 since 2 is not expressible as (p+1)(q+1); same for a(2); a(3) = 1 since 6^2 = 36 = (2+1)(11+1); a(6) = 3 since 12^2 = 144 = (2+1)(47+1) = (5+1)(23+1) = (7+1)(17+1); a(9) = 2 since 18^2 = 324 = (2+1)(107+1) = (5+1)(53+1); etc.
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MATHEMATICA
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f[n_] := Block[{c = 0, p = 2}, While[p < 2n -1, If[ PrimeQ[(2n)^2/(p +1) -1], c++]; p = NextPrime@ p]; c]; Array[f, 105]
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PROG
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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