

A100571


Cubes m^3 such that m^3 is the sum of m1 consecutive primes plus a larger prime.


0



8, 27, 64, 125, 216, 343, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 27000, 29791, 32768, 35937, 39304, 42875, 46656, 50653, 54872, 59319, 64000
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OFFSET

1,1


COMMENTS

Or, triangular cubic numbers with prime indices. [Comment is not clear to me!  N. J. A. Sloane, Feb 23 2021]
Conjecture: sequence consists of all the cubes > 1 except 8^3=512.  Giovanni Teofilatto, Apr 23 2015


LINKS



EXAMPLE

a(2)=27 because 3^3=3+5+19 and p is 19;
a(3)=64 because 4^3=5+7+11+41 and p is 41;
a(4)=125 because 5^3=5+7+11+13+89 and p is 89.


MAPLE

N:= 100; # to get all terms <= N^3
pmax:= ithprime(N+numtheory:pi((N+1)^2)):
kmax:= (pmax1)/2:
Primes:= select(isprime, [2, seq(2*k+1, k=1..kmax)]):
C:= ListTools:PartialSums(Primes):
A:= NULL:
for m from 1 to N1 do
for t from 0 do
if t = 0 then q:= (m+1)^3  C[m]
else q:= (m+1)^3  C[t+m] + C[t]
fi;
if q <= Primes[t+m] then break fi;
if isprime(q) then A:= A, (m+1)^3; break fi;
od
od:


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



