A337184 Numbers divisible by their first digit and their last digit.

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

Author

Bernard Schott, Jan 29 2021

Keywords

• nonn
• base
• easy

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

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000
• Index entries for 10-automatic sequences.

Crossrefs

Intersection of A034709 and A034837.

Subsequences: A010785\{0}, A034838, A043037, A043040, A208259, A066622.

Cf. A139138.

Programs

Mathematica

Select[Range[175], Mod[#, 10] > 0 && And @@ Divisible[#, IntegerDigits[#][[{1, -1}]]] &] (* Amiram Eldar, 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

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

Formula

(10n-9)/9 <= a(n) < 45n. (I believe the liminf of a(n)/n is 3.18... and the limsup is 6.18....) - Charles R Greathouse IV, Nov 26 2024

Conjecture: 3n < a(n) < 7n for n > 75. - Charles R Greathouse IV, Dec 02 2024