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%I #22 Mar 19 2021 06:59:25
%S 3,7,19,31,79,97,139,199,211,229,271,307,331,337,367,379,439,499,547,
%T 577,601,607,619,691,727,811,829,937,967,1009,1069,1171,1279,1297,
%U 1399,1429,1459,1531,1609,1627,1759,1867,2011,2029,2089,2131,2179,2221,2281
%N Odd primes p such that 2p-1 is prime and no p is equal to 2q-1 with q in the sequence.
%C At each step, the smallest possible p is chosen.
%C These are the primes described in lemma 2 of the paper by Holt. - _T. D. Noe_, Sep 28 2012
%C This sequence was used by Holt (2003) to prove that there are at least two solutions k to phi(n+k) = phi(k) for all even n <= 1.38*10^26595411. - _Amiram Eldar_, Mar 19 2021
%H Michel Marcus, <a href="/A217199/b217199.txt">Table of n, a(n) for n = 1..1000</a>
%H Jeffery J. Holt, <a href="http://dx.doi.org/10.1090/S0025-5718-03-01509-6">The minimal number of solutions to phi(n)=phi(n+k)</a>, Math. Comp., 72 (2003), 2059-2061.
%H A. Schinzel and Andrzej Wakulicz, <a href="http://matwbn.icm.edu.pl/ksiazki/aa/aa4/aa431.pdf">Sur l'équation phi(x+k)=phi(x), I.</a>, Acta Arith. 4 (1958), 181-184.
%t t = {}; p = 2; Do[p = NextPrime[p]; If[PrimeQ[2*p - 1] && ! MemberQ[2*t - 1, p], AppendTo[t, p]], {PrimePi[2281]}]; t
%o (PARI) intab(val, tab) = {for (ii=1, length(tab),if (tab[ii] == val, return (1);););return(0);}
%o lista(nn) = {tab = []; for (i=1, nn, len = length(tab); if (len == 0, p = 3, p = nextprime(tab[len]+1)); while (! isprime(2*p-1) || intab((p+1)/2, tab) , p = nextprime(p+1);); tab = concat(tab, p); print1(p, ", "););}
%Y Cf. A007015, A110581, A217198, A342701.
%K nonn
%O 1,1
%A _Michel Marcus_, Sep 27 2012
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