A083890 Number of divisors of n with largest digit = 3 (base 10).

0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 2, 1, 1, 2, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 2, 0, 1, 1, 1, 1, 2, 0, 0, 2, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 2, 1, 1, 2, 0, 0, 2, 0, 0, 2, 0, 0, 1, 1, 1, 1

Offset

1,30

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Formula

a(n) = A000005(n) - A083888(n) - A083889(n) - A083891(n) - A083892(n) - A083893(n) - A083894(n) - A083895(n) - A083896(n) = A083898(n) - A083897(n).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=1} 1/A277965(k) = 0.84217457724798904648... . - Amiram Eldar, Jan 04 2024

Example

n=132, 3 of the 12 divisors of 132 have largest digit =3: {3,33,132}, therefore a(132)=3.

Maple

f:= proc(n) nops(select(t -> max(convert(t, base, 10))=d, numtheory:-divisors(n))) end proc:
d:= 3:
map(f, [$1..200]); # Robert Israel, Oct 06 2019

Mathematica

With[{k = 3}, Array[DivisorSum[#, 1 &, And[#[[k]] > 0, Total@ #[[k + 1 ;; 9]] == 0] &@ DigitCount[#] &] &, 105]] (* Michael De Vlieger, Oct 06 2019 *)
Table[Count[Divisors[n], _?(Max[IntegerDigits[#]]==3&)], {n, 120}] (* Harvey P. Dale, Sep 05 2020 *)

Prog

(Magma) [#[d:d in Divisors(n) | Max(Intseq(d)) eq 3]: n in [1..130]]; // Marius A. Burtea, Oct 06 2019

Crossrefs

Cf. A054055, A000005, A083898, A083888, A083889, A083891, A083892, A083893, A083894, A083895, A083896, A277965.

Sequence in context: A329334 A062979 A114781 * A000989 A132401 A104273

Adjacent sequences: A083887 A083888 A083889 * A083891 A083892 A083893

Keyword

nonn,base

Author

Reinhard Zumkeller, May 08 2003

Status

approved