%I
%S 0,127,127343,1275122197,127125100019683,127125100012167148877,
%T 1271251000106481038233442951,127125100010648103823100000014348907,
%U 127125100010648103823100000010077696108531333,1271251000106481038231000000100776961005446251939096223
%N Smallest prime whose decimal expansion consists of the concatenation of a 1digit cube, a 2digit cube, a 3digit cube, ..., and an ndigit cube, or 0 if there is no such prime.
%C This is to cubes A000578 as A215689 is to squares A000290.
%C The nth term has A000217(n) = n(n+1)/2 digits. 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 smallest k digit cube, cf. formula. The number of such primes between a(n) and A340115(n) (the largest of this form) is (0, 2, 2, 9, 177, 6909, 570166, ...). (In particular, for n = 2 and 3, a(n) and A340115(n) are the only two primes of this form.) 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^((k1)/3), k=1..n) and the density of primes in that range according to the PNT.  _M. F. Hasler_, Dec 31 2020
%H M. F. Hasler, <a href="/A215692/b215692.txt">Table of n, a(n) for n = 1..44</a> (all terms < 10^1000), Dec 31 2020.
%F a(n) ~ 10^(n(n+1)/2)*0.1271251000106481038231000000100776961... (conjectured)  _M. F. Hasler_, Dec 31 2020
%e a(1) = 0 because no 1digit cube {0,1,8} is prime.
%e a(2) = 127 because 127 is prime and is the concatenation of 1=1^3 and 27 = 3^3.
%o (PARI) apply( {A215692(n)=forvec(v=vector(n,k,[ceil(10^((k1)/3)),sqrtnint(10^k1,3)]),ispseudoprime(n=eval(concat([Str(k^3)k<v])))&&return(n))}, [1..12]) \\ _M. F. Hasler_, Dec 31 2020
%Y Cf. A000040 (primes), A000578 (cubes), A215689, A215641, A215647 (analog for squares, primes, semiprimes).
%Y Cf. A340115 (largest prime of the given form), A000217 (triangular numbers: length of nth term).
%K nonn,base
%O 1,2
%A _Jonathan Vos Post_, Aug 20 2012
%E More terms (up to a(10)) from _Alois P. Heinz_, Aug 21 2012
