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Primes of the form (x^2 + y^3)/(x+y), with x,y > 1 two distinct integers.
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%I #7 Apr 22 2019 01:20:51

%S 3,7,13,31,43,67,73,109,139,149,157,179,193,211,229,241,307,317,379,

%T 389,421,457,463,491,499,593,601,647,661,751,757,769,829,839,937,1009,

%U 1021,1033,1123,1171,1213,1231,1283,1319,1381,1459,1481,1483,1549,1621

%N Primes of the form (x^2 + y^3)/(x+y), with x,y > 1 two distinct integers.

%e a(1) = 3 = (1^2 + 2^3)/(1+2).

%e a(2) = 7 = (1^2 + 3^3)/(1+3) or (6^2 + 3^3)/(6+3).

%e a(3) = 13 = (1^2 + 4^3)/(1+4) or (12^2 + 4^3)/ (12+4).

%e a(4) = 31 = (1^2 + 6^3)/(1+6).

%p isA162869 := proc(p) local a,b ; if isprime(p) then for b from 1 to p do for d in numtheory[divisors](b^2*(b+1)) do a := d-b ; if a > 1 and (a^2+b^3)= p*(a+b) then RETURN(true); fi; od: od: RETURN(false) ; else false; fi; end:

%p for n from 1 do p := ithprime(n) ; if isA162869(p) then printf("%d,\n",p) ; fi; od: # _R. J. Mathar_, Sep 22 2009

%t f[a_,b_]:=(a^2+b^3)/(a+b); lst={};Do[Do[If[f[a,b]==IntegerPart[f[a,b]], If[a!=b&&PrimeQ[f[a,b]],AppendTo[lst,f[a,b]]]],{b,4*6!}],{a,4*6!}]; Take[Union[lst],50]

%K nonn

%O 1,1

%A _Vladimir Joseph Stephan Orlovsky_, Jul 15 2009

%E Comment turned into examples by _R. J. Mathar_, Sep 22 2009