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A261395
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Primes p such that (prime(p)-1)^2 = (prime(q)-1)*(prime(r)-1) for some pair of distinct primes q and r.
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2
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13, 47, 137, 191, 193, 223, 227, 313, 701, 857, 907, 947, 991, 1009, 1069, 1291, 1531, 1889, 2281, 2411, 2447, 2647, 3181, 3389, 3539, 3593, 4093, 4099, 4409, 4481, 4603, 4721, 5557, 5647, 6581, 6793, 6869, 6961, 7211, 7349, 7523, 7723, 7753, 8461, 8537, 8543, 8807, 9137, 9241, 9281
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OFFSET
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1,1
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COMMENTS
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Conjecture: Let d be any nonzero integer. Then there are infinitely many prime triples (p,q,r) with p,q,r distinct such that (prime(p)+d)^2 = (prime(q)+d)*(prime(r)+d). In other words, the set {prime(p)+d: p is prime} contains infinitely many nontrivial three-term geometric progressions.
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REFERENCES
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Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.
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LINKS
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EXAMPLE
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a(1) = 13 since (prime(13)-1)^2 = (41-1)^2 = 1600 = (5-1)*(401-1) = (prime(3)-1)*(prime(79)-1) with 13, 3, 79 all prime.
a(2) = 47 since (prime(47)-1)^2 = 210^2 = 44100 = 30*1470 = (prime(11)-1)*(prime(233)-1) with 47, 11, 233 all prime.
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MATHEMATICA
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Dv[n_]:=Divisors[n]
L[n_]:=Length[Dv[n]]
f[n_]:=Prime[Prime[n]]-1
PQ[p_]:=PrimeQ[p]&&PrimeQ[PrimePi[p]]
n=0; Do[Do[If[PQ[Part[Dv[f[k]^2], i]+1]&&PQ[Part[Dv[f[k]^2], L[f[k]^2]-i+1]+1], n=n+1; Print[n, " ", Prime[k]]; Goto[aa]]; Continue, {i, 1, (L[f[k]^2]-1)/2}];
Label[aa]; Continue, {k, 1, 1150}]
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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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