A357299 a(n) is the number of divisors of n whose first digit equals the first digit of n.

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

Offset: 1

Author

Bernard Schott, Sep 23 2022

Keywords

• nonn
• base

Comments

Similar to A330348, but with last digit.

a(n) >= 1 because there is always a divisor that fits: n.

a(n) >= 2 for n>1 in A131835.

Examples

			The divisors of 26 that start in 2 are 2 and 26, so a(26) = 2.
		

Links

• Michel Marcus, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000030, A131835, A330348, A356549, A357300.

Programs

Mathematica

f[n_] := IntegerDigits[n][[1]]; a[n_] := DivisorSum[n, 1 &, f[#] == f[n] &]; Array[a, 100] (* Amiram Eldar, Sep 23 2022 *)

PARI

a(n) = my(fd=digits(n)[1]); sumdiv(n, d, digits(d)[1] == fd); \\ Michel Marcus, Sep 23 2022

Python

from sympy import divisors
def a(n): f = str(n)[0]; return sum(1 for d in divisors(n) if str(d)[0]==f)
print([a(n) for n in range(1, 101)]) # Michael S. Branicky, Sep 23 2022