A357171 a(n) is the number of divisors of n whose digits are in strictly increasing order (A009993).
1, 2, 2, 3, 2, 4, 2, 4, 3, 3, 1, 6, 2, 4, 4, 5, 2, 6, 2, 4, 3, 2, 2, 8, 3, 4, 4, 6, 2, 6, 1, 5, 2, 4, 4, 9, 2, 4, 4, 5, 1, 6, 1, 3, 6, 4, 2, 10, 3, 4, 3, 5, 1, 7, 2, 8, 4, 4, 2, 8, 1, 2, 4, 5, 3, 4, 2, 6, 4, 6, 1, 11, 1, 3, 5, 5, 2, 8, 2, 6, 4, 2, 1, 9, 3, 2, 3, 4, 2, 9, 3, 5, 2, 3, 3, 10, 1, 5, 3, 5
Offset: 1
Examples
22 has 4 divisors {1, 2, 11, 22} of which two have decimal digits that are not in strictly increasing order: {11, 22}, hence a(22) = 4-2 = 2.
52 has divisors {1, 2, 4, 13, 26, 52} and a(52) = 5 of them have decimal digits that are in strictly increasing order (all except 52 itself).
Crossrefs
Programs
-
Maple
f:= proc(n) local d,L,i,t; t:= 0; for d in numtheory:-divisors(n) do L:= convert(d,base,10); if `and`(seq(L[i]>L[i+1],i=1..nops(L)-1)) then t:= t+1 fi od; t end proc: map(f, [$1..100]); # Robert Israel, Sep 16 2022 -
Mathematica
a[n_] := DivisorSum[n, 1 &, Less @@ IntegerDigits[#] &]; Array[a, 100] (* Amiram Eldar, Sep 16 2022 *)
-
PARI
isok(d) = Set(d=digits(d)) == d; \\ A009993 a(n) = sumdiv(n, d, isok(d)); \\ Michel Marcus, Sep 16 2022
-
Python
from sympy import divisors def c(n): s = str(n); return s == "".join(sorted(set(s))) def a(n): return sum(1 for d in divisors(n, generator=True) if c(d)) print([a(n) for n in range(1, 101)]) # Michael S. Branicky, Sep 16 2022
Formula
G.f.: Sum_{n in A009993} x^n/(1-x^n). - Robert Israel, Sep 16 2022
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{n=2..512} 1/A009993(n) = 4.47614714667538759358... (this is a rational number whose numerator and denominator have 1037 and 1036 digits, respectively). - Amiram Eldar, Jan 06 2024
Comments