OFFSET
1,2
COMMENTS
If a(n) exists it has A000217(n) = n*(n+1)/2 digits.
The similar smallest primes are in A215692.
We can conjecture that a(n) > 0 for all n > 1 and the terms converge to the concatenation of (c(1), c(2), c(3), ...) where c(k) is the largest k digit cube. The number of such primes between A215692(n) and a(n) is (0, 2, 2, 9, 177, 6909, 570166, ...). This is very close to what we expect given the number of concatenations of cubes of the respective length (product of 10^(k/3)-10^((k-1)/3), k=1..n) and the density of primes in that range according to the PNT. - M. F. Hasler, Dec 31 2020
LINKS
M. F. Hasler, Table of n, a(n) for n = 1..44 (all terms < 10^1000), Dec 31 2020
EXAMPLE
a(1) = 0 because no 1-digit cube {0, 1, 8} is prime.
a(2) = 827 because 827 is prime and is the concatenation of 8 = 2^3 and 27 = 3^3.
a(3) = 164729 because 827343, 827729, 864343 and 864729 are not primes and 164729, concatenation of 1 = 1^3, 64 = 4^3 and 729 = 9^3 is prime.
PROG
(Python)
from sympy import isprime
from itertools import product
def a(n):
cubes = [str(k**3) for k in range(1, int((10**n)**(1/3))+2)]
revcbs = [[k3 for k3 in cubes if len(k3)==i+1][::-1] for i in range(n)]
for t in product(*revcbs):
intt = int("".join(t))
if isprime(intt): return intt
return 0
print([a(n) for n in range(1, 11)]) # Michael S. Branicky, Dec 28 2020
(PARI) A340115(n)=forvec(v=vector(n, k, -[sqrtnint(10^k-1, 3), ceil(10^((k-1)/3))]), ispseudoprime(n=eval(concat([Str(-k^3)|k<-v])))&&return(n)) \\ M. F. Hasler, Dec 31 2020
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Bernard Schott, Dec 28 2020
EXTENSIONS
a(4)-a(10) from Michael S. Branicky, Dec 28 2020
STATUS
approved