A330348 a(n) is the number of divisors of n whose last digit equals the last digit of n.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 2, 2, 2, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 2, 2, 1, 1, 1, 3, 2, 2, 1, 2, 3, 1, 1, 2, 1, 2, 2, 2, 1, 1, 2, 1, 1, 1, 1, 4, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 3, 1, 1, 4, 1, 2, 1, 1, 4, 2, 2, 1, 3, 2, 1, 1, 2, 1

Offset

1,11

Comments

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

a(n) >= 1.

When n > 10 ends with 0, 1, 2 or 5, then a(n) >= 2.

The first 19 terms are the same as A038769, but a(20) = 2 and A038769(20) = 1.

From Robert Israel, Jun 04 2020: (Start)

a(10*n) = A000005(n).

If n is odd, then a(2*n) = a(n) and a(5*n) = A000005(n). (End)

Integers all of whose divisors end with the same last digit (which is necessarily 1) are in A004615. - Bernard Schott, May 07 2021

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Project Euler, Problem 474: Last digits of divisors

Example

The divisors of 12 that end in 2 are 2 and 12, so a(12) = 2.

Maple

f:= proc(n) local t;
t:= n mod 10;
nops(select(k -> k mod 10 = t, numtheory:-divisors(n)))
end proc:
map(f, [$1..100]); # Robert Israel, Jun 04 2020

Mathematica

a[n_] := DivisorSum[n, 1 &, Mod[# - n, 10] == 0 &]; Array[a, 100] (* Amiram Eldar, Jun 04 2020 *)

Prog

(PARI) a(n) = my(u=n%10); sumdiv(n, d, (d%10) == u); \\ Michel Marcus, Jun 04 2020
(Python)
from sympy import divisors
def a(n): return sum((n-d)%10 == 0 for d in divisors(n, generator=True))
print([a(n) for n in range(1, 90)]) # Michael S. Branicky, Aug 15 2022

Crossrefs

Cf. A000005, A004615, A038769.

Sequence in context: A327818 A255481 A241418 * A117229 A264119 A225518

Adjacent sequences: A330345 A330346 A330347 * A330349 A330350 A330351

Keyword

nonn,base

Author

Bernard Schott, Jun 04 2020

Status

approved