cp's OEIS Frontend

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

%I A117057 #19 May 17 2021 04:54:22
%S A117057 1,2,3,4,5,6,7,8,9,11,111,212,1111,2112,4224,11111,11711,13131,21112,
%T A117057 21312,31113,42624,111111,211112,234432,1111111,1113111,2111112,
%U A117057 2112112,2114112,2118112,11111111,21111112,21122112,61111116,111111111
%N A117057 Palindromes which are divisible by the product of their digits.
%C A117057 Are there infinitely many terms that do not contain a 1? - _Derek Orr_, Aug 26 2014
%H A117057 Chai Wah Wu, <a href="/A117057/b117057.txt">Table of n, a(n) for n = 1..221</a>
%e A117057 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.
%t A117057 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_ *)
%o A117057 (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
%o A117057 (Python)
%o A117057 from operator import mul
%o A117057 from functools import reduce
%o A117057 from gmpy2 import t_mod, mpz
%o A117057 A117057 = sorted([mpz(n) for n in (str(x)+str(x)[::-1] for x in range(1,10**6))
%o A117057         if not (n.count('0') or t_mod(mpz(n), reduce(mul,(mpz(d) for d in n))))]+
%o A117057         [mpz(n) for n in (str(x)+str(x)[-2::-1] for x in range(10**6))
%o A117057         if not (n.count('0') or t_mod(mpz(n), reduce(mul,(mpz(d) for d in n))))])
%o A117057 # _Chai Wah Wu_, Aug 26 2014
%Y A117057 Cf. A002113.
%K A117057 base,nonn
%O A117057 1,2
%A A117057 Luc Stevens (lms022(AT)yahoo.com), Apr 16 2006
%E A117057 a(23) to a(36) from _Klaus Brockhaus_, Apr 17 2006