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A190105 a(n) = (3*A002145(n) - 1)/4.

%I A190105 #12 Nov 08 2018 08:49:18
%S A190105 2,5,8,14,17,23,32,35,44,50,53,59,62,77,80,95,98,104,113,122,125,134,
%T A190105 143,149,158,167,170,179,188,197,203,212,230,233,248,260,269,275,284,
%U A190105 287,314,323,329,332,347,350,359,365,368,374,377,392,410,422,428,440
%N A190105 a(n) = (3*A002145(n) - 1)/4.
%C A190105 For primes p of the form 4n+3, in the order of A002145, let us seek solutions for prime p|(a^x + b^y) or p|(a^y + b^x) subject to the conditions p = a+b = x+y and 0 < a,b,x,y < p. The larger of the two exponents x and y is inserted into the sequence.
%C A190105 If either of (a,b) is a primitive root of p, there is a unique solution, either p|(a^x + b^y) or p|(a^y + b^x). If neither is a primitive root (see A060749), there are multiple solutions and p|(a^x + b^y) or p|(a^y + b^x) will always be one of them for some of the given exponents (x,y) contributing to the sequence.
%H A190105 G. C. Greubel, <a href="/A190105/b190105.txt">Table of n, a(n) for n = 1..10000</a>
%e A190105 For p=43=A002145(7), (x,y)=(11,32) because 43-(43+1)/4=32; hence x=43-32.  With (a,b)=(12,31) the unique solution is 43|(12^11 + 31^32) because 12 is a primitive root of 43. The larger of 11 and 32 is a(7)=32 in the sequence. For 43 multiple solutions occur when neither of the pairs (a,b) is a primitive root of 43: p divides each of 11^4 + 32^39, 11^18 + 32^25, 11^32 + 32^11; note that the exponents (11,32) occur in the last entry.
%p A190105 for n from 1 to 200 do p:=4*n-1: if(isprime(p))then printf("%d, ", (3*p-1)/4); fi: od: # _Nathaniel Johnston_, May 18 2011
%t A190105 A002145 := Select[4 Range[300] - 1, PrimeQ]; Table[(3*A002145[[n]] - 1)/4, {n, 1, 60}] (* _G. C. Greubel_, Nov 07 2018 *)
%Y A190105 Cf. A005099 is the list of x in (x,y).
%K A190105 nonn,easy
%O A190105 1,1
%A A190105 _J. M. Bergot_, May 04 2011