A117228 Palindromes which are divisible by the product and by the sum of their digits.

1, 2, 3, 4, 5, 6, 7, 8, 9, 111, 2112, 4224, 13131, 21112, 21312, 31113, 42624, 211112, 234432, 1113111, 2111112, 2114112, 2118112, 21122112, 61111116, 111111111, 211121112, 211242112, 211262112, 213141312, 2111111112, 2112332112, 2114114112, 2131221312

Offset

1,2

Comments

Intersection of A082232 and A117057.

Are there infinitely many terms that don't contain a 1? - Derek Orr, Aug 25 2014

Links

Chai Wah Wu, Table of n, a(n) for n = 1..91

Example

42624 is divisible by 4*2*6*2*4 and by 4+2+6+2+4.

Prog

(Python)
from operator import mul
from functools import reduce
from gmpy2 import t_mod, mpz
A117228 = sorted([mpz(n) for n in (str(x)+str(x)[::-1] for x in range(1, 10**8))
..........if not (n.count('0') or t_mod(mpz(n), sum((mpz(d) for d in n)))
..........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**8))
..........if not (n.count('0') or t_mod(mpz(n), sum((mpz(d) for d in n)))
..........or t_mod(mpz(n), reduce(mul, (mpz(d) for d in n))))])
# Chai Wah Wu, Aug 25 2014
(PARI)
rev(n)=r=""; d=digits(n); for(i=1, #d, r=concat(Str(d[i]), r)); eval(r)
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

Crossrefs

Cf. A002113, A082232, A117057.

Sequence in context: A283868 A087995 A082232 * A032567 A134853 A193407

Adjacent sequences: A117225 A117226 A117227 * A117229 A117230 A117231

Keyword

base,nonn

Author

Giovanni Resta, Apr 22 2006

Extensions

More terms from Chai Wah Wu, Aug 22 2014

Status

approved