%I #20 Jan 04 2024 01:43:09

%S 1,1,1,1,1,1,1,1,1,2,2,1,1,1,1,1,1,1,1,2,1,2,1,1,1,1,1,1,1,2,1,1,2,1,

%T 1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,1,2,1,1,1,1,2,1,1,1,1,1,2,1,1,

%U 1,2,1,1,1,1,1,1,2,1,1,2,1,1,1,1,1,1,1,2,1,2,1,1,1,1,1,1,1,1,2,3,2,1,1,1,1

%N Number of divisors of n with largest digit = 1 (base 10).

%H Robert Israel, <a href= "/A083888/b083888.txt" >Table of n, a(n) for n = 1..10000</a>

%F a(n) = A000005(n) - A083889(n) - A083890(n) - A083891(n) - A083892(n) - A083893(n) - A083894(n) - A083895(n) - A083896(n).

%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=1} 1/A007088(k) = 1.23840561530559480971.... - _Amiram Eldar_, Jan 04 2024

%e n=110, 4 of the divisors of 110 {1,2,5,10,11,22,55,110} have largest digit =1: {1,10,11,110}, therefore a(110)=4.

%p f:= proc(n) nops(select(t -> max(convert(t,base,10))=d, numtheory:-divisors(n))) end proc:

%p d:= 1:

%p map(f, [$1..200]); # _Robert Israel_, Oct 06 2019

%t With[{k = 1}, Array[DivisorSum[#, 1 &, And[#[[k]] > 0, Total@ #[[k + 1;; 9]] == 0] &@ DigitCount[#] &] &, 105]] (* _Michael De Vlieger_, Oct 06 2019 *)

%o (Magma) [#[d:d in Divisors(n) | Max(Intseq(d)) eq 1]: n in [1..110]]; // _Marius A. Burtea_, Oct 06 2019

%Y Cf. A054055, A000005, A007088, A083889, A083890, A083891, A083892, A083893, A083894, A083895, A083896.

%K nonn,base

%O 1,10

%A _Reinhard Zumkeller_, May 08 2003