%I #12 Nov 15 2024 09:06:17

%S 15,490,21637,1121505,3913864295,1131238503938606,78801255302666615,

%T 5589233202595404488,29349508915133986374841,

%U 2163909235608484556362424,913865816485680423486405066750,191623400974625892978847721669762887224010

%N Number of partitions of cuban primes.

%C Number of partitions of prime numbers that are the difference of two consecutive cubes.

%C Number of partitions of primes p such that p=(3*k^2 + 1)/4 for some integer k (A121259).

%H Robert Israel, <a href="/A377045/b377045.txt">Table of n, a(n) for n = 1..153</a>

%F a(n) = A000041(A002407(n)).

%F a(n) = A000041((3*A121259(n)^2 + 1)/4).

%p R:= NULL: count:= 0:

%p for i from 1 while count < 30 do

%p p:= (i+1)^3 - i^3;

%p if isprime(p) then count:= count+1; v:= combinat:-numbpart(p); R:= R,v; fi

%p od:

%p R; # _Robert Israel_, Nov 14 2024

%t PartitionsP[Select[Table[(3k^2 + 1)/4,{k,50}],PrimeQ]]

%Y Cf. A000041, A002407, A121259.

%K nonn

%O 1,1

%A _Paul F. Marrero Romero_, Oct 14 2024