A052379 - OEIS

%I #50 Feb 15 2023 11:29:57

%S 8,72,584,4680,37448,299592,2396744,19173960,153391688,1227133512,

%T 9817068104,78536544840,628292358728,5026338869832,40210710958664,

%U 321685687669320,2573485501354568,20587884010836552,164703072086692424,1317624576693539400,10540996613548315208

%N Number of integers from 1 to 10^(n+1)-1 that lack 0 and 1 as a digit.

%H Vincenzo Librandi, <a href="/A052379/b052379.txt">Table of n, a(n) for n = 0..400</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (9,-8).

%F a(n) = (8^(n+2) - 1)/7 - 1.

%F G.f.: 8/((1-x)*(1-8*x)). - _R. J. Mathar_, Nov 19 2007

%F a(n) = 8*a(n-1) + 8.

%F a(n) = Sum_{k=1..n} 8^k. - corrected by _Michel Marcus_, Sep 25 2014

%F 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.

%F a(n) = 9*a(n-1) - 8*a(n-2); a(0)=8, a(1)=72. - _Harvey P. Dale_, Sep 22 2013

%F a(n) = 8*A023001(n+1). - _Alois P. Heinz_, Feb 15 2023

%e 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.

%p A052379:=n->(8^(n+2)-1)/7-1: seq(A052379(n), n=0..20); # _Wesley Ivan Hurt_, Sep 26 2014

%t (8^(Range[0,20]+2)-1)/7-1 (* or *) LinearRecurrence[{9,-8},{8,72},20] (* _Harvey P. Dale_, Sep 22 2013 *)

%o (Magma) [(8^(n+2)-1)/7-1: n in [0..20]]; // _Vincenzo Librandi_, Jul 04 2011

%o (PARI) a(n)=8^(n+2)\7 - 1 \\ _Charles R Greathouse IV_, Aug 25 2014

%Y Cf. A023001, A024101, A052386.

%K easy,nonn,base

%O 0,1

%A _Odimar Fabeny_, Mar 12 2000

%E More terms and revised description from _James A. Sellers_, Mar 13 2000