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Quintan primes: p = (x^5 + y^5)/(x + y).
(Formerly M4792 N2044)
2

%I M4792 N2044 #29 Jan 11 2018 01:09:03

%S 11,61,181,421,461,521,991,1621,1871,3001,4441,4621,6871,9091,9931,

%T 12391,13421,14821,19141,25951,35281,35401,55201,58321,61681,62071,

%U 72931,74731,91331,92921,95881,108421,117911,117991,131041,132661,141961

%N Quintan primes: p = (x^5 + y^5)/(x + y).

%C (x^5 + y^5)/(x + y) = x^4 - y*x^3 + y^2*x^2 - y^3*x + y^4. - _Jens Kruse Andersen_, Jul 14 2014

%D A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929; see Vol. 2, p. 201.

%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H Jens Kruse Andersen, <a href="/A002650/b002650.txt">Table of n, a(n) for n = 1..10000</a>

%H A. J. C. Cunningham, <a href="/A001912/a001912.pdf">Binomial Factorisations</a>, Vols. 1-9, Hodgson, London, 1923-1929. [Annotated scans of a few pages from Volumes 1 and 2]

%e (3^5 + 1^5)/(3 + 1) = 61. This is prime and therefore in the sequence. - _Jens Kruse Andersen_, Jul 14 2014

%t Take[Select[Union[(#[[1]]^5+#[[2]]^5)/Total[#]&/@Tuples[Range[200],2]], #>0&& PrimeQ[#]&],50] (* _Harvey P. Dale_, May 21 2012 *)

%o (PARI) m=10^6; v=[]; for(x=1, (2*m)^(1/4), for(y=1, x, n=(x^5+y^5)/(x+y); if(n<=m && isprime(n), v=concat(v,n)))); vecsort(v) \\ _Jens Kruse Andersen_, Jul 14 2014

%Y Cf. A002649.

%K nonn

%O 1,1

%A _N. J. A. Sloane_

%E More terms from _Sean A. Irvine_, May 08 2014