A346510 a(n) is the number of nontrivial divisors of A346507(n) ending with 1.

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

Offset: 1

Author

Stefano Spezia, Jul 21 2021

Keywords

• nonn
• base

Examples

			a(42) = 4 since there are 4 nontrivial divisors of A346507(42) = 2541 ending with 1: 11, 21, 121 and 231.
		

Crossrefs

Cf. A017281, A070824, A346388 (ending with 5), A346389 (ending with 6), A346392, A346507, A346508, A346509.

Programs

Mathematica

b={}; For[n=1, n<=500, n++, For[k=1, kMax[b], AppendTo[b, 10n+1]]]]; (* A346507 *) a={}; For[i =1, i<=Length[b], i++, AppendTo[a, Length[Drop[Select[Divisors[Part[b, i]], (Mod[#, 10]==1&)], -1]]-1]]; a

PARI

f(n) = sumdiv(n, d, (d>1) && (d(f(x)), [1..5000])) \\ Michel Marcus, Jul 28 2021

Python

from sympy import divisors
def f(n): return sum(d%10 == 1 for d in divisors(n)[1:-1])
def A346507upto(lim): return sorted(set(a*b for a in range(11, lim//11+1, 10) for b in range(a, lim//a+1, 10)))
print(list(map(f, A346507upto(5000)))) # Michael S. Branicky, Jul 31 2021

Formula

a(n) = A346392(A346507(n)) - 1.