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A117057
Palindromes which are divisible by the product of their digits.
3
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
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
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