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%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
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