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URL: https://oeis.org/A063527

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A063527
Numbers that are divisible by all of their 1 and 2 digit substrings.
5
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22, 24, 33, 36, 44, 48, 55, 66, 77, 88, 99, 1111, 1155, 1248, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999, 27216, 31248, 111111, 116688, 121212, 142128, 212184, 222222, 242424, 313131, 321216, 333333, 363636, 368424, 444444
OFFSET
1,2
COMMENTS
Subsequence of A034838. - Michel Marcus, Sep 19 2014
EXAMPLE
1155 is divisible by 1, 1, 5, 5, 11, 15 and 55.
MATHEMATICA
d12Q[n_]:=Module[{idn=IntegerDigits[n], idn2}, idn2=FromDigits/@Partition[ idn, 2, 1]; FreeQ[idn, 0]&&And@@Divisible[n, idn]&&And@@Divisible[n, idn2]]; Select[Range[400000], d12Q] (* Harvey P. Dale, Aug 11 2015 *)
PROG
(Python)
from itertools import product
A063527_list = []
for g in range(1, 7):
for n in product('123456789', repeat=g):
s = ''.join(n)
m = int(s)
if not any([m % int(d) for d in s]):
for i in range(len(s)-1):
if m % int(s[i:i+2]):
break
else:
A063527_list.append(m) # Chai Wah Wu, Sep 18 2014
(PARI) is(n) = {my(d = digits(n), t = 0); s = Set(d); if(s[1] == 0, return(0)); for(i = 1, 2, for(j = 1, #d - i + 1, t++; fr = fromdigits(vector(i, k, d[j+k-1])); if(n % fr != 0, return(0)); ) ); 1 } \\ David A. Corneth, Sep 17 2019
CROSSREFS
Cf. A034838 (integers divisible by all their digits).
Sequence in context: A178158 A337184 A034838 * A209933 A182183 A308472
KEYWORD
base,easy,nonn
AUTHOR
Erich Friedman, Aug 01 2001
EXTENSIONS
More terms from David A. Corneth, Sep 17 2019
STATUS
approved