cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A340115 Largest prime whose decimal expansion consists of the concatenation of a 1-digit cube, a 2-digit cube, a 3-digit cube, ..., and an n-digit cube, or 0 if there is no such prime.

Original entry on oeis.org

0, 827, 164729, 8642164913, 864729685979507, 864729926197336531441, 8647299261973369702994826809, 864729926197336970299980034443986977, 864729926197336970299993837599897344909853209, 8647299261973369702999938375998973449970029998036054027
Offset: 1

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Author

Bernard Schott, Dec 28 2020

Keywords

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

Examples

			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.
		

Crossrefs

Cf. A338968 (with concatenated primes), A339978 (with concatenated squares).

Programs

  • 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
  • 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
    

Extensions

a(4)-a(10) from Michael S. Branicky, Dec 28 2020