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A338610 Integers m such that there exist one prime p and one positive integer k, for which the expression k^3 + k^2*p is a perfect cube m^3.

%I A338610 #34 Nov 13 2020 14:20:13
%S A338610 2,12,36,80,252,810,1100,1452,2366,2940,5202,12696,14400,16250,20412,
%T A338610 22736,27900,33792,40460,52022,56316,70602,75852,93150,112896,120050,
%U A338610 143312,169400,198476,242172,254016,291852,305252,410700,518400,538002,643452,689216,737100
%N A338610 Integers m such that there exist one prime p and one positive integer k, for which the expression k^3 + k^2*p is a perfect cube m^3.
%C A338610 This concerns Problem 131 of Project Euler (see link).
%C A338610 For each such term m with this property, the values of k and of p are unique.
%C A338610 The solution to the Diophantine equation is: (q^3)^3 + (q^3)^2 * ((q+1)^3 - q^3) = ((q+1) * q^2)^3, where
%C A338610 - the prime p is the cuban prime (q+1)^3 - q^3 = A002407(n),
%C A338610 - corresponding to q = A111251(n),
%C A338610 - the positive integer k = q^3, and,
%C A338610 - the resulting m = (q+1)*q^2 = (A111251(n)+1)*(A111251(n))^2.
%H A338610 Robert Israel, <a href="/A338610/b338610.txt">Table of n, a(n) for n = 1..10000</a>
%H A338610 Project Euler, <a href="https://projecteuler.net/problem=131">Problem 131: Prime cube partnership</a>.
%F A338610 a(n) = (A111251(n) + 1)*(A111251(n))^2.
%F A338610 a(n) = A011379(A111251(n)).
%e A338610 For n=1, q=A111251(1)=1 and 1^3 + 1^2*(2^3 - 1^3) = 1+1*7 = 2^3, hence, k=1^3, cuban prime=7, and a(1)=m=2.
%e A338610 For n=3, q=A111251(3)=3 and (3^3)^3 + (3^3)^2*(4^3 - 3^3) = 27^3 + 27^2*37 = 46656 = 36^3, hence, k=3^3, cuban prime=37, and a(3)=m=36.
%p A338610 for q from 1 to 90 do
%p A338610 p:=3*q^2+3*q+1;
%p A338610 if isprime(p) then print((q+1)*q^2); else fi; od:
%t A338610 f[n_] := n^2*(n+1); f /@ Select[Range[100], PrimeQ[3*#^2 + 3*# + 1] &] (* _Amiram Eldar_, Nov 05 2020 *)
%o A338610 (PARI) lista(nn) =  apply(x->x^2*(x+1), select(x->isprime(3*x^2 + 3*x + 1), [1..nn])); \\ _Michel Marcus_, Nov 05 2020
%Y A338610 Cf. A002407, A111251.
%Y A338610 Subsequence of A011379.
%K A338610 nonn
%O A338610 1,1
%A A338610 _Bernard Schott_, Nov 03 2020