OFFSET
1,1
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
EXAMPLE
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.
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,hard,base
AUTHOR
Charles Ronco, Aug 08 2017
EXTENSIONS
a(6)-a(7) from Giovanni Resta, Aug 09 2017
STATUS
approved