login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A117228
Palindromes which are divisible by the product and by the sum of their digits.
3
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
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
KEYWORD
base,nonn
AUTHOR
Giovanni Resta, Apr 22 2006
EXTENSIONS
More terms from Chai Wah Wu, Aug 22 2014
STATUS
approved