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A235743
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Primes p(k) such that p(k) + p(k+3) = p(k+1) + p(k+2) + 4.
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1
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17, 41, 79, 131, 149, 173, 227, 233, 239, 347, 349, 379, 439, 463, 521, 599, 641, 673, 677, 983, 1013, 1091, 1231, 1277, 1427, 1429, 1453, 1487, 1549, 1607, 1811, 1949, 2099, 2203, 2309, 2579, 2609, 2687, 2689, 2833, 2857, 2903, 2909, 2917, 3083, 3167, 3299
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OFFSET
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1,1
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COMMENTS
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If p(k) is in the sequence, then the four consecutive primes p(k), p(k+1), p(k+2), p(k+3) possess a property of quadruplet of consecutive squares: n^2 + (n+3)^2 = (n+1)^2 + (n+2)^2 + 4.
Cf. A022885, where such quadruplets possess a linear property: n + (n+3) = (n+1) + (n+2).
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LINKS
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EXAMPLE
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17 is in the sequence since 17, 19, 23, and 29 are four consecutive primes and it holds 17 + 29 = 19 + 23 + 4.
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MATHEMATICA
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f[{a_, b_, c_, d_}]:= a-b-c+d; First /@ Select[Partition[Prime@ Range@ 500, 4, 1], f@# == 4 &] (* Giovanni Resta, Jan 16 2014 *)
Transpose[Select[Partition[Prime[Range[5000]], 4, 1], First[#] + Last[#]==#[[2]] + #[[3]] + 4&]][[1]] (* Vincenzo Librandi, Feb 02 2014 *)
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PROG
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(PARI) isok(p) = { my(k = primepi(p)); (p == prime(k)) && ((prime(k) + prime(k+3)) == (prime(k+1) + prime(k+2) + 4)); } \\ Michel Marcus, Jan 15 2014
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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