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A337184
Numbers divisible by their first digit and their last digit.
1
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22, 24, 33, 36, 44, 48, 55, 66, 77, 88, 99, 101, 102, 104, 105, 111, 112, 115, 121, 122, 123, 124, 125, 126, 128, 131, 132, 135, 141, 142, 144, 145, 147, 151, 152, 153, 155, 156, 161, 162, 164, 165, 168, 171, 172, 175, 181, 182
OFFSET
1,2
COMMENTS
The first 23 terms are the same first 23 terms of A034838 then a(24) = 101 while A034838(24) = 111.
Terms of A034709 beginning with 1 and terms of A034837 ending with 1 are terms.
All positive repdigits (A010785) are terms.
There are infinitely many terms m for any of the 53 pairs (first digit, last digit) of m described below: when m begins with {1, 3, 7, 9} then m ends with any digit from 1 to 9; when m begins with {2, 4, 6, 8}, then m must also end with {2, 4, 6, 8}; to finish, when m begins with 5, m must only end with 5. - Metin Sariyar, Jan 29 2021
MATHEMATICA
Select[Range[175], Mod[#, 10] > 0 && And @@ Divisible[#, IntegerDigits[#][[{1, -1}]]] &] (* Amiram Eldar, Jan 29 2021 *)
PROG
(Python)
def ok(n): s = str(n); return s[-1] != '0' and n%int(s[0])+n%int(s[-1]) == 0
print([m for m in range(180) if ok(m)]) # Michael S. Branicky, Jan 29 2021
(PARI) is(n) = n%10>0 && n%(n%10)==0 && n % (n\10^logint(n, 10)) == 0 \\ David A. Corneth, Jan 29 2021
CROSSREFS
Intersection of A034709 and A034837.
Subsequences: A010785\{0}, A034838, A043037, A043040, A208259.
Cf. A139138.
Sequence in context: A084434 A034709 A178158 * A034838 A063527 A209933
KEYWORD
nonn,base
AUTHOR
Bernard Schott, Jan 29 2021
STATUS
approved