A346953 a(n) is the number of divisors of A346950(n) ending with 3.

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

Offset

1,2

Links

Table of n, a(n) for n=1..88.

Example

a(17) = 4 since there are 4 divisors of A346950(17) = 429 ending with 3: 3, 13, 33 and 143.

Mathematica

b={}; For[n=0, n<=450, n++, For[k=0, k<=n, k++, If[Mod[10*n+9, 10*k+3]==0 && Mod[(10*n+9)/(10*k+3), 10]==3 && 10*n+9>Max[b], AppendTo[b, 10*n+9]]]]; (* A346950 *) a={}; For[i =1, i<=Length[b], i++, AppendTo[a, Length[Drop[Select[Divisors[Part[b, i]], (Mod[#, 10]==3&)]]]]]; a

Prog

(Python)
from sympy import divisors
def f(n): return sum(d%10 == 3 for d in divisors(n)[1:-1])
def A346950upto(lim): return sorted(set(a*b for a in range(3, lim//3+1, 10) for b in range(a, lim//a+1, 10)))
print(list(map(f, A346950upto(2129)))) # Michael S. Branicky, Aug 11 2021

Crossrefs

Cf. A000005, A017377, A346388 (ending with 5), A346389 (ending with 6), A346950, A346951, A346952.

Sequence in context: A087010 A098220 A297032 * A235508 A271778 A326702

Adjacent sequences: A346950 A346951 A346952 * A346954 A346955 A346956

Keyword

nonn,base

Author

Stefano Spezia, Aug 08 2021

Status

approved