A155881 a(n) is the number of zeros needed to write the integers 1 through Fibonacci(n).
0, 0, 0, 0, 0, 0, 1, 2, 3, 5, 8, 24, 43, 67, 121, 188, 409, 708, 1228, 1946, 4131, 6241, 10525, 17866, 29428, 58369, 87881, 156261, 255242, 412545, 767846, 1280460, 2059307, 3343656, 5510186, 9861418, 16472261, 26422596, 43917688, 73697381, 125281166, 206655249
Offset: 1
Examples
F(9)=34, so writing the numbers F(1)..F(9) requires 3 zeros (one each at 10, 20, and 30), thus a(9)=3.
Links
- David A. Corneth, Table of n, a(n) for n = 1..4770
Programs
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Maple
A055641 := proc(n) option remember ; local a,d; if n = 0 then RETURN(a) ; fi; a := 0 ; for d in convert(n,base,10) do if d = 0 then a := a+1 ; fi; od: a ; end: A155881 := proc(n) add(A055641(i),i=1..combinat[fibonacci](n)) ; end: for n from 1 do printf("%d,\n",A155881(n)) ; od; # R. J. Mathar, Feb 19 2009
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Mathematica
Block[{n = 32, s}, s = DigitCount[Range@ Fibonacci@ n, 10, 0]; Array[Total@ Take[s, Fibonacci@ #] &, n]] (* Michael De Vlieger, Jan 23 2019 *)
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PARI
nb(n) = #Set(select(x->(x==0), digits(n))); \\ A055641 a(n) = sum(k=1, fibonacci(n), nb(k)); \\ Michel Marcus, Jan 23 2019
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PARI
a(n) = my(n = fibonacci(n), m=logint(n, 10)); (m+1)*(n+1) - (10^(m+1)-1)/9 + (1/2) * sum(j=1, m+1, (n\10^j * (2*n+2 - (1 + n\10^j) * 10 ^ j) - floor(n/10^j+9/10) * (2*n+2 + ((4/5 - floor(n / 10^j + 9 / 10))*10^j)))) \\ David A. Corneth, Jan 23 2019
Formula
Extensions
8 more terms from R. J. Mathar, Feb 19 2009
9 more terms from Sean A. Irvine, Dec 10 2009
Edited by Jon E. Schoenfield, Jan 22 2019
More terms from David A. Corneth, Jan 23 2019
Comments