%I #16 Oct 12 2024 14:43:58
%S 1,2,3,4,5,6,7,8,9,111,2112,4224,13131,21112,21312,31113,42624,211112,
%T 234432,1113111,2111112,2114112,2118112,21122112,61111116,111111111,
%U 211121112,211242112,211262112,213141312,2111111112,2112332112,2114114112,2131221312
%N Palindromes which are divisible by the product and by the sum of their digits.
%C Intersection of A082232 and A117057.
%C Are there infinitely many terms that don't contain a 1? - _Derek Orr_, Aug 25 2014
%H Chai Wah Wu, <a href="/A117228/b117228.txt">Table of n, a(n) for n = 1..91</a>
%e 42624 is divisible by 4*2*6*2*4 and by 4+2+6+2+4.
%o (Python)
%o from operator import mul
%o from functools import reduce
%o from gmpy2 import t_mod, mpz
%o A117228 = sorted([mpz(n) for n in (str(x)+str(x)[::-1] for x in range(1, 10**8))
%o if not (n.count('0') or t_mod(mpz(n), sum((mpz(d) for d in n)))
%o or t_mod(mpz(n), reduce(mul, (mpz(d) for d in n))))]+
%o [mpz(n) for n in (str(x)+str(x)[-2::-1] for x in range(10**8))
%o if not (n.count('0') or t_mod(mpz(n), sum((mpz(d) for d in n)))
%o or t_mod(mpz(n), reduce(mul, (mpz(d) for d in n))))])
%o # _Chai Wah Wu_, Aug 25 2014
%o (PARI)
%o rev(n)=r="";d=digits(n);for(i=1,#d,r=concat(Str(d[i]),r));eval(r)
%o for(n=1,10^7,d=digits(n);if(rev(n)==n,p=prod(i=1,#d,d[i]);if(p&&n%p==0&&n%sumdigits(n)==0,print1(n,", ")))) \\ _Derek Orr_, Aug 25 2014
%Y Cf. A002113, A082232, A117057.
%K base,nonn
%O 1,2
%A _Giovanni Resta_, Apr 22 2006
%E More terms from _Chai Wah Wu_, Aug 22 2014