A040163 a(n) is the absolute value of (the first digit of n minus the last digit of n).

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 0, 1, 2, 3, 4, 5

Offset: 1

Author

Felice Russo

Keywords

• nonn
• base

Examples

			a(371) = abs(3 - 1) = 2.
a(567) = abs(5 - 7) = 2.
		

Links

• David F. Marrs, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000030 (first digit of n), A010879 (last digit of n).

Cf. A037904, A040114, A040115, A040997.

Programs

Mathematica

Array[Abs[First@ # - Last@ #] &@ IntegerDigits@ # &, 106] (* Michael De Vlieger, Oct 15 2018 *)

PARI

a(n) = my(digs = digits(n)); abs(digs[1] - digs[#digs]); \\ Michel Marcus, Sep 27 2013

PARI

apply( {A040163(n)=abs(n\10^logint(n+!n,10)-n%10)}, [0..111]) \\ M. F. Hasler, Apr 22 2024

Python

for n in range(1,51): print(abs(int(str(n)[0])-int(str(n)[-1]))) # David F. Marrs, Oct 14 2018

Formula

a(n) = abs(floor(n / 10 ^ floor(log_10(n))) - (n - floor(n / 10) * 10)) - David F. Marrs, Oct 14 2018