

A215692


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.


10



0, 127, 127343, 1275122197, 127125100019683, 127125100012167148877, 1271251000106481038233442951, 127125100010648103823100000014348907, 127125100010648103823100000010077696108531333, 1271251000106481038231000000100776961005446251939096223
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OFFSET

1,2


COMMENTS

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


LINKS



FORMULA

a(n) ~ 10^(n(n+1)/2)*0.1271251000106481038231000000100776961... (conjectured)  M. F. Hasler, Dec 31 2020


EXAMPLE

a(1) = 0 because no 1digit cube {0,1,8} is prime.
a(2) = 127 because 127 is prime and is the concatenation of 1=1^3 and 27 = 3^3.


PROG

(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


CROSSREFS

Cf. A340115 (largest prime of the given form), A000217 (triangular numbers: length of nth term).


KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



