A290675 For a given n, the nonzero digits of n are the prime indices for the factorization of n.

14, 154, 1196, 66079, 279174, 302768895, 2249805789

Offset: 1

Author

Charles Ronco, Aug 08 2017

Keywords

• nonn
• hard
• base

Comments

a(8) > 10^35, if it exists. - Giovanni Resta, Aug 09 2017

a(n) != 1 mod 10. a(8) > 10^44, if it exists. - Chai Wah Wu, Aug 10 2017

Fixed points of A113581. - Alois P. Heinz, May 11 2023

Examples

			14 = 2*7, which are the 1st and 4th primes. 154 = 2*11*7 which are the 1st, 5th, and 4th primes, respectively. So use the digits of n (excluding zero) to find the corresponding primes, and the product of those primes equals n.
		

Crossrefs

Supersequence of A097227.

Cf. A113581.

Programs

Mathematica

x = 10^7; (* this number is the upper end of the search *) Do[If[n == Times @@ Prime /@ DeleteCases[RealDigits[n][[1]], 0], Print[n]], {n, x}] (* or *)
up = 3*^9; ric[n_, e_, k_] := Block[{m=n, j=0}, If[k == 10, If[Most@ DigitCount[n] == e, Print@n; Sow@n], While[m < up, ric[m, Append[e, j], k+1]; j++; m *= Prime[k] ]]]; Sort@ Reap[ric[1, {}, 1]][[2, 1]] (* faster, Giovanni Resta, Aug 09 2017 *)

PARI

is(n) = my(d=digits(n), prd=1); for(k=1, #d, if(d[k]!=0, prd=prd*prime(d[k]))); prd==n \\ Felix Fröhlich, Aug 09 2017

Python

from functools import reduce
from operator import mul
from itertools import combinations_with_replacement
A290675_list, lmax, ptuple = [], 12, (2,3,5,7,11,13,17,19,23)
for l in range(1,lmax+1):
    for d in combinations_with_replacement(range(1,10),l):
        n = reduce(mul,(ptuple[i-1] for i in d))
        if n < 10**lmax and tuple(sorted((int(x) for x in str(n) if x != '0'))) == d:
            A290675_list.append(n) # Chai Wah Wu, Aug 10 2017

Extensions

a(6)-a(7) from Giovanni Resta, Aug 09 2017