|
%I #21 Jan 04 2024 03:24:04
%S 0,0,0,0,1,0,0,0,0,1,0,0,0,0,2,0,0,0,0,1,0,0,0,0,2,0,0,0,0,2,0,0,0,0,
%T 2,0,0,0,0,1,0,0,0,0,3,0,0,0,0,3,1,1,1,1,2,0,0,0,0,2,0,0,0,0,1,0,0,0,
%U 0,2,0,0,0,0,3,0,0,0,0,1,0,0,0,0,1,0,0,0,0,3,0,0,0,0,1,0,0,0,0,3,0,1,0,1,4
%N Number of divisors of n with largest digit = 5 (base 10).
%H Robert Israel, <a href="/A083892/b083892.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A000005(n) - A083888(n) - A083889(n) - A083890(n) - A083891(n) - A083893(n) - A083894(n) - A083895(n) - A083896(n) = A083900(n) - A083899(n).
%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=1} 1/A283608(k) = 1.32926350368137107677... . - _Amiram Eldar_, Jan 04 2024
%e n=125, 3 of the 4 divisors of 125 have largest digit =5: {5,25,125}, therefore a(125)=3.
%p f:= proc(n) nops(select(t -> max(convert(t, base, 10))=d, numtheory:-divisors(n))) end proc:
%p d:= 5:
%p map(f, [$1..200]); # _Robert Israel_, Oct 06 2019
%t Table[Count[Divisors[n],_?(Max[IntegerDigits[#]]==5&)],{n,110}] (* _Harvey P. Dale_, Aug 08 2015 *)
%o [#[d:d in Divisors(n) | Max(Intseq(d)) eq 5]: n in [1..150]]; // _Marius A. Burtea_, Oct 06 2019
%Y Cf. A054055, A000005, A083900, A083888, A083889, A083890, A083891, A083893, A083894, A083895, A083896, A283608.
%K nonn,base
%O 1,15
%A _Reinhard Zumkeller_, May 08 2003
|
