A061217 Number of zeros in the concatenation n(n-1)(n-2)(n-3)...321.
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, 9, 9, 9, 9, 9, 9, 9, 9, 9, 11, 12, 13, 14
Offset: 1
Examples
a(30) = 3 since number of zeros in 302928272625242322212019181716151413121110987654321 is 3. (This example implies offset = 1.)
Links
- Hieronymus Fischer, Table of n, a(n) for n = 1..10000
- Tanya Khovanova and Gregory Marton, Archive Labeling Sequences, arXiv:2305.10357 [math.HO], 2023.
Programs
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Haskell
a061217 n = a061217_list !! (n-1) a061217_list = scanl1 (+) $ map a055641 [1..] -- Reinhard Zumkeller, Oct 27 2013
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Mathematica
Table[Count[Flatten[IntegerDigits/@Table[x-n,{n,0,x-1}]],0],{x,110}] (* Harvey P. Dale, Aug 10 2011 *) -
PARI
a(n) = my(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)))) \\ adapted from formula by Hieronymus Fischer \\ David A. Corneth, Jan 23 2019
Formula
From Hieronymus Fischer, Jun 12 2012: (Start)
a(n) = (m+1)*(n+1) - (10^(m+1)-1)/9 + (1/2)*Sum_{j=1..m+1} (floor(n/10^j)*(2*n + 2 - (1 + floor(n/10^j))*10^j) - floor(n/10^j + 9/10)*(2*n + 2 + (4/5 - floor(n/10^j + 9/10))*10^j)), where m=floor(log_10(n)).
a(n) = A117804(n+1) - (n+1)*A054640(n) + (1/2)*Sum_{j=1..m+1} ((floor(n/10^j + 9/10)^2 - floor(n/10^j)^2)*10^j - (4/5*floor(n/10^j + 9/10) + floor(n/10^j))*10^j), where m=floor(log_10(n)).
a(10^m-1) = m*10^(m-1) - (10^m-1)/9.
(This is the total number of zeros occurring in all the numbers 1..10^m-1 or numbers with <= m places excluding zero.)
G.f.: g(x) = (1/(1-x)^2)*Sum_{j>=0} (1-x^10^j)*x^10^(j+1)/(1-x^10^(j+1)). (End)
Extensions
Corrected and extended by Patrick De Geest, Jun 05 2001
Offset changed to 1 by Hieronymus Fischer, Jun 12 2012
Comments