A052379 Number of integers from 1 to 10^(n+1)-1 that lack 0 and 1 as a digit.
8, 72, 584, 4680, 37448, 299592, 2396744, 19173960, 153391688, 1227133512, 9817068104, 78536544840, 628292358728, 5026338869832, 40210710958664, 321685687669320, 2573485501354568, 20587884010836552, 164703072086692424, 1317624576693539400, 10540996613548315208
Offset: 0
Examples
For n=1, the numbers from 1 to 99 which have 0 or 1 as a digit are the numbers 1 and 10, 20, 30, ..., 90 and 11, 12, ..., 18, 19 and 21, 31, ..., 91. So a(1) = 99 - 27 = 72.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..400
- Index entries for linear recurrences with constant coefficients, signature (9,-8).
Programs
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Magma
[(8^(n+2)-1)/7-1: n in [0..20]]; // Vincenzo Librandi, Jul 04 2011
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Maple
A052379:=n->(8^(n+2)-1)/7-1: seq(A052379(n), n=0..20); # Wesley Ivan Hurt, Sep 26 2014
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Mathematica
(8^(Range[0,20]+2)-1)/7-1 (* or *) LinearRecurrence[{9,-8},{8,72},20] (* Harvey P. Dale, Sep 22 2013 *) -
PARI
a(n)=8^(n+2)\7 - 1 \\ Charles R Greathouse IV, Aug 25 2014
Formula
a(n) = (8^(n+2) - 1)/7 - 1.
G.f.: 8/((1-x)*(1-8*x)). - R. J. Mathar, Nov 19 2007
a(n) = 8*a(n-1) + 8.
a(n) = Sum_{k=1..n} 8^k. - corrected by Michel Marcus, Sep 25 2014
Conjecture: a(n) = A023001(n+2)-1. - R. J. Mathar, May 18 2007. Comment from Vim Wenders, Mar 26 2008: The conjecture is true: the g.f. leads to the closed form a(n) = -(8/7)*(1^n) + (64/7)*(8^n) = (-8 + 64*8^n)/7 = (-8 + 8^(n+2))/7 = (8^(n+2) - 1)/7 - 1 = A023001(n+2) - 1.
a(n) = 9*a(n-1) - 8*a(n-2); a(0)=8, a(1)=72. - Harvey P. Dale, Sep 22 2013
a(n) = 8*A023001(n+1). - Alois P. Heinz, Feb 15 2023
Extensions
More terms and revised description from James Sellers, Mar 13 2000