A117057 Palindromes which are divisible by the product of their digits.

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 111, 212, 1111, 2112, 4224, 11111, 11711, 13131, 21112, 21312, 31113, 42624, 111111, 211112, 234432, 1111111, 1113111, 2111112, 2112112, 2114112, 2118112, 11111111, 21111112, 21122112, 61111116, 111111111

Offset

1,2

Comments

Are there infinitely many terms that do not contain a 1? - Derek Orr, Aug 26 2014

Links

Chai Wah Wu, Table of n, a(n) for n = 1..221

Example

4224 is in the sequence because (1) it is a palindrome, (2) the product of its digits is 4*2*2*4=64 and 4224 is divisible by 64.

Mathematica

fQ[n_] := Block[{id = IntegerDigits@n}, Reverse@id == id && Count[id, 0] == 0 && Mod[n, Times @@ id] == 0]; Do[ If[ fQ@n, Print@n], {n, 10^7}] (* Robert G. Wilson v *)

Prog

(PARI) {m=120000000; for(n=1, m, k=n; rev=0; while(k>0, d=divrem(k, 10); k=d[1]; rev=10*rev+d[2]); if(n==rev, p=1; h=n; while(h>0, d=divrem(h, 10); h=d[1]; p=p*d[2]); if(p>0&&n%p==0, print1(n, ", "))))} \\ Klaus Brockhaus, Apr 17 2006
(Python)
from operator import mul
from functools import reduce
from gmpy2 import t_mod, mpz
A117057 = sorted([mpz(n) for n in (str(x)+str(x)[::-1] for x in range(1, 10**6))
        if not (n.count('0') or t_mod(mpz(n), reduce(mul, (mpz(d) for d in n))))]+
        [mpz(n) for n in (str(x)+str(x)[-2::-1] for x in range(10**6))
        if not (n.count('0') or t_mod(mpz(n), reduce(mul, (mpz(d) for d in n))))])
# Chai Wah Wu, Aug 26 2014

Crossrefs

Cf. A002113.

Sequence in context: A369127 A229623 A277856 * A249516 A239090 A248889

Adjacent sequences: A117054 A117055 A117056 * A117058 A117059 A117060

Keyword

base,nonn

Author

Luc Stevens (lms022(AT)yahoo.com), Apr 16 2006

Extensions

a(23) to a(36) from Klaus Brockhaus, Apr 17 2006

Status

approved