A178545 Primes p such that q = p^2 + p + 1 is an emirp.
3, 5, 41, 59, 839, 857, 1811, 1931, 3011, 3221, 3407, 3671, 8387, 8543, 8627, 9719, 9743, 9803, 10781, 11549, 12647, 13469, 13487, 13499, 13613, 13931, 14087, 17477, 17573, 17837, 18089, 18269, 19319, 19403, 19661, 19991, 27191, 27947, 31223, 33311, 34313
Offset: 1
Examples
3^2 + 3 + 1 = 13 = prime(6), R(13) = prime(11), 3 is first term. 5^2 + 5 + 1 = 31 = prime(11), R(31) = prime(6), 5 is 2nd term. q = 1811^2 + 1811 + 1 = 3281533 = prime(235691), R(q) = prime(240351), first case that p = 1811 = prime(280) = emirp(87) is itself an emirp.
References
- M. Gardner: Die magischen Zahlen des Dr. Matrix, Krueger Verlag, Frankfurt am Main, 1987
- R. Guy: Unsolved Problems in Number Theory,3rd edition, Springer, New York, 2004
- G. H. Hardy, E. M. Wright: Einfuehrung in die Zahlentheorie, R. Oldenburg, Muenchen, 1958
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
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Maple
filter:= proc(p) local q,qr; if not isprime(p) then return false fi; q:= p^2+p+1; if not isprime(q) then return false fi; qr:= revdigs(q); qr <> q and isprime(qr); end proc: select(filter, [3,seq(i,i=5..50000,6)]); # Robert Israel, Dec 04 2016
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Mathematica
EmirpQ[n_] := If[ PrimeQ@n, Block[{id = IntegerDigits@n}, rid = Reverse@ id; rid != id && PrimeQ@ FromDigits@ rid]]; Select[ Prime@ Range@ 3700, EmirpQ[ #^2 + # + 1] &] (* Robert G. Wilson v, Jul 26 2010 *) p2emrpQ[p_]:=With[{q=p^2+p+1},!PalindromeQ[q]&&AllTrue[{q,IntegerReverse[q]},PrimeQ]]; Select[Prime[Range[3700]],p2emrpQ] (* Harvey P. Dale, Mar 10 2025 *)
Extensions
More terms from Robert G. Wilson v, Jul 26 2010
Comments