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%I #30 Feb 23 2021 12:37:22
%S 8,27,64,125,216,343,729,1000,1331,1728,2197,2744,3375,4096,4913,5832,
%T 6859,8000,9261,10648,12167,13824,15625,17576,19683,21952,24389,27000,
%U 29791,32768,35937,39304,42875,46656,50653,54872,59319,64000
%N Cubes m^3 such that m^3 is the sum of m-1 consecutive primes plus a larger prime.
%C Or, triangular cubic numbers with prime indices. [Comment is not clear to me! - _N. J. A. Sloane_, Feb 23 2021]
%C Conjecture: sequence consists of all the cubes > 1 except 8^3=512. - _Giovanni Teofilatto_, Apr 23 2015
%e a(2)=27 because 3^3=3+5+19 and p is 19;
%e a(3)=64 because 4^3=5+7+11+41 and p is 41;
%e a(4)=125 because 5^3=5+7+11+13+89 and p is 89.
%p N:= 100; # to get all terms <= N^3
%p pmax:= ithprime(N+numtheory:-pi((N+1)^2)):
%p kmax:= (pmax-1)/2:
%p Primes:= select(isprime,[2,seq(2*k+1,k=1..kmax)]):
%p C:= ListTools:-PartialSums(Primes):
%p A:= NULL:
%p for m from 1 to N-1 do
%p for t from 0 do
%p if t = 0 then q:= (m+1)^3 - C[m]
%p else q:= (m+1)^3 - C[t+m] + C[t]
%p fi;
%p if q <= Primes[t+m] then break fi;
%p if isprime(q) then A:= A,(m+1)^3; break fi;
%p od
%p od:
%p A; # _Robert Israel_, Apr 24 2015
%Y Subsequence of A000578.
%K nonn
%O 1,1
%A _Giovanni Teofilatto_, Nov 29 2004
%E Definition corrected by _Robert Israel_, Apr 24 2015
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