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A293621
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Numbers k such that (2*k)^2 + 1 and (2*k+2)^2 + 1 are both primes.
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1
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1, 2, 7, 12, 27, 62, 102, 192, 232, 317, 322, 357, 547, 572, 587, 622, 637, 657, 687, 782, 807, 837, 842, 982, 1027, 1042, 1047, 1202, 1227, 1267, 1332, 1417, 1462, 1567, 1652, 1767, 1877, 1887, 2012, 2077, 2087, 2182, 2302, 2307, 2367, 2392, 2397, 2477, 2507
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OFFSET
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1,2
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COMMENTS
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Sierpiński proved that under Schinzel's hypothesis H this sequence is infinite. He gives the 24 terms below 10^3.
Sierpiński noted that the only triple of consecutive primes of the form (2n)^2 + 1 are for n = 1 (i.e., 1 and 2 are the only consecutive terms in this sequence), since every triple of consecutive terms contains at least one term which is divisible by 5.
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LINKS
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FORMULA
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EXAMPLE
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1 is in the sequence since (2*1)^2 + 1 = 5 and (2*1+2)^2 + 1 = 17 are both primes.
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MATHEMATICA
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Select[Range[10^4], AllTrue[{(2#)^2+1, (2#+2)^2+1}, PrimeQ] &]
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PROG
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(PARI) isok(n) = isprime((2*n)^2 + 1) && isprime((2*n+2)^2 + 1); \\ Michel Marcus, Oct 13 2017
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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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