A102684 Number of times the digit 9 appears in the decimal representations of all integers from 0 to n.
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20
Offset: 0
Links
- Hieronymus Fischer, Table of n, a(n) for n = 0..10000
Crossrefs
Programs
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Maple
p:=proc(n) local b,ct,j: b:=convert(n,base,10): ct:=0: for j from 1 to nops(b) do if b[j]>=9 then ct:=ct+1 else ct:=ct fi od: ct: end: seq(add(p(i),i=0..n), n=0..105); # Emeric Deutsch, Feb 23 2005
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Mathematica
Accumulate[DigitCount[Range[0,100],10,9]] (* Harvey P. Dale, Mar 30 2018 *)
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PARI
a(n) = sum(k=0, n, #select(x->(x==9), digits(k))); \\ Michel Marcus, Oct 03 2023
Formula
From Hieronymus Fischer, Jun 10 2012: (Start)
a(n) = (1/2)*Sum_{j=1..m+1} (floor(n/10^j + 1/10)*(2n + 2 - (4/5 + floor(n/10^j + 1/10))*10^j) - floor(n/10^j)*(2n + 2 - (1+floor(n/10^j)) * 10^j)), where m = floor(log_10(n)).
a(n) = (n+1)*A102683(n) + (1/2)*Sum_{j=1..m+1} ((-4/5*floor(n/10^j + 1/10) + floor(n/10^j))*10^j - (floor(n/10^j + 1/10)^2 - floor(n/10^j)^2)*10^j), where m = floor(log_10(n)).
a(10^m-1) = m*10^(m-1).
(this is total number of digits = 9 occurring in all the numbers with <= m places).
G.f.: g(x) = (1/(1-x)^2)*Sum_{j>=0} (x^(9*10^j) - x^(10*10^j))/(1-x^10^(j+1)). (End)
Extensions
More terms from Emeric Deutsch, Feb 23 2005
Definition revised by N. J. A. Sloane, Mar 30 2018
Comments