A054898 a(n) = Sum_{k>0} floor(n/9^k).

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11

Offset: 0

Author

Henry Bottomley, May 23 2000

Keywords

• nonn

Comments

Different from the highest power of 9 dividing n!, A090618.

Examples

			a(100)=12.
a(10^3)=124.
a(10^4)=1248.
a(10^5)=12498.
a(10^6)=124996.
a(10^7)=1249997.
a(10^8)=12499996.
a(10^9)=124999997.
		

Links

• Hieronymus Fischer, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A011371 and A054861 for analogs involving powers of 2 and 3.

Cf. A054899, A067080, A098844, A132033.

Programs

Mathematica

Table[t = 0; p = 9; While[s = Floor[n/p]; t = t + s; s > 0, p *= 9]; t, {n, 0, 100} ]
Table[Sum[Floor[n/9^k],{k,n}],{n,0,100}] (* Harvey P. Dale, Jul 10 2024 *)

Formula

a(n) = floor(n/9) + floor(n/81) + floor(n/729) + floor(n/6561) + ....

a(n) = (n-A053830(n))/8.

From Hieronymus Fischer, Aug 14 2007: (Start)

Recurrence:

a(n) = floor(n/9) + a(floor(n/9));

a(9*n) = n + a(n);

a(n*9^m) = n*(9^m-1)/8 + a(n).

a(k*9^m) = k*(9^m-1)/8, for 0<=k<9, m>=0.

Asymptotic behavior:

a(n) = n/8 + O(log(n)),

a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.

a(n) <= (n-1)/8; equality holds for powers of 9.

a(n) >= (n-8)/8 - floor(log_9(n)); equality holds for n=9^m-1, m>0.

lim inf (n/8 - a(n)) =1/8, for n-->oo.

lim sup (n/8 - log_9(n) - a(n)) = 0, for n-->oo.

lim sup (a(n+1) - a(n) - log_9(n)) = 0, for n-->oo.

G.f.: g(x) = sum{k>0, x^(9^k)/(1-x^(9^k))}/(1-x). (End)

Extensions

Examples added by Hieronymus Fischer, Jun 06 2012