A338822 a(n) is the number of divisors of n whose last digits equal the number of digits of n.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 0, 2, 0, 1, 0, 1, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 1, 0, 1, 0, 2, 0, 2, 0, 2, 0, 2, 0, 1, 0, 1, 0, 3, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 0, 3, 0, 1, 0, 2, 0, 1, 0

Offset: 1

Author

Bernard Schott, Nov 11 2020

Keywords

• nonn
• base

Comments

Inspired by Project Euler, Problem 474 (see link).

If in the name, one replaces “whose last digits equal” with “whose last digit equals” then, this sequence is finite and the last term should be a(999999999) = 5 because 999999999 has 9 digits and among the 20 divisors of 999999999, five of them (9, 999, 9009009, 12345679 and 999999999) have the last digit 9.

With this name, the next term is a(10^9) = 1 because 10^9 has 10 digits and among the 100 divisors of 10^9, only one of them (10) ends with 10.

For p prime, p >= 7, the number 10^(p - 1) + p - 1 is divisible by p and the number of digits is p, so it has at least one divisor that ends in p. Thus, there is an infinity of terms a(k) > 0. - Marius A. Burtea, Nov 12 2020

Examples

			72 has 2 digits, and among the divisors of 72 (1, 2, 3, 4, 6, 12, 18, 24, 36, 72), three of them (2, 12 and 72) have the last digit 2, hence a(72) = 3.
111 has 3 digits, and among the divisors of 111 (1, 3, 37, 111), only one of them (3) has the last digit 3, hence a(111) = 1.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Project Euler, Problem 474: Last digits of divisors.

Crossrefs

Cf. A000005, A330348.

Programs

Magma

[#[d:d in Divisors(n) | d mod 10^(#Intseq(#Intseq(n))) eq #Intseq(n)]:n in [1..100]]; // Marius A. Burtea, Nov 12 2020

Maple

f:= proc(n) local d, dd;
  d:= ilog10(n)+1;
  dd:= ilog10(d)+1;
  nops(select(t -> t mod 10^dd = d, numtheory:-divisors(n)))
end proc:
map(f, [$1..100]); # Robert Israel, Nov 12 2020

Mathematica

a[n_] := DivisorSum[n, 1 &, Divisible [# - (ndigit = IntegerLength[n]), 10^IntegerLength[ndigit]] &]; Array[a, 100] (* Amiram Eldar, Nov 12 2020 *)

PARI

a(n) = my(nb = #Str(n), nc = #Str(nb)); sumdiv(n, d, if (dMichel Marcus, Nov 16 2020

Formula

For 1-digit numbers: a(n) = 1.

For 2-digit numbers: a(n) = 0 iff n is odd, a(n) >= 1 if n is even.

For 4-digit numbers: a(n) = 0 if n is odd.

For 5-digit numbers, a(n) >= 2 if n ends with 5, a(n) >=1 if n ends with 0, otherwise a(n) = 0.

For 8-digit numbers, a(n) = 0 if n is odd.

Extensions

Name generalized following remark of Marius A. Burtea, Nov 12 2020