# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/

Search: id:a169689
Showing 1-1 of 1

%I A169689 #9 Feb 24 2021 02:48:19
%S A169689 0,1,6,4,24,4,20,12,84,4,20,12,76,12,60,36,276,4,20,12,76,12,60,36,
%T A169689 260,12,60,36,228,36,180,108,876,4,20,12,76,12,60,36,260,12,60,36,228,
%U A169689 36,180,108,844,12,60,36,228,36,180,108,780,36,180,108,684,108,540,324,2724,4
%N A169689 (A169648(4n+4) - A147582(4n+5))/4.
%C A169689 A169648 and A147582 agree except at these terms.
%H A169689 David Applegate, Omar E. Pol and N. J. A. Sloane, <a href="/A000695/a000695_1.pdf">The Toothpick Sequence and Other Sequences from Cellular Automata</a>, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
%H A169689 N. J. A. Sloane, <a href="/wiki/Catalog_of_Toothpick_and_CA_Sequences_in_OEIS">Catalog of Toothpick and Cellular Automata Sequences in the OEIS</a>
%F A169689 a(-1)=0, a(0)=1, a(1)=6. For n >= 2, let n = 2^k+j with 0 <= j < 2^k, and write j+1 = 2^m*(2t+1). Then a(n) = 4*(3^(m+1)-2^(m+1))*3^wt(t), except if j=2^k-1 we must add 2^(k+1) to the result (here wt(t) = A000120(t)).
%F A169689 Recurrence: a(-1)=0, a(0)=1, a(1)=6. For n>=2, write n = 2^k + j, with 0 <= j < 2^k. If j+1 is a power of 2, say j+1 = 2^r, set f=j+1 if r<k, f=3(j+1) if r=k, and otherwise set f=0. Then a(n) = 3*a(j) + f. (The explicit formula in the previous line is better.)
%F A169689 Since there is a simple explicit formula for A147582(n), this provides a simple way to generate A169648.
%e A169689 Can be written in the form of a triangle:
%e A169689 0,
%e A169689 1,
%e A169689 6,
%e A169689 4,24,
%e A169689 4,20,12,84,
%e A169689 4,20,12,76,12,60,36,276,
%e A169689 4,20,12,76,12,60,36,260,12,60,36,228,36,180,108,876,
%e A169689 4,20,12,76,12,60,36,260,12,60,36,228,36,180,108,844,12,60,36,228,36,180,108,780,36,180,108,684,108,540,324,2724,
%e A169689 ...
%p A169689 a:=proc(n) option remember; local f,j,k,t1;
%p A169689 if n=-1 then RETURN(0); elif n=0 then RETURN(1); elif n=1 then RETURN(6);
%p A169689 else k:=floor(log(n)/log(2)); j:=n-2^k; t1 := 2^floor(log(j+1)/log(2));
%p A169689 if t1=j+1 and j < 2^k-1 then f := j+1 elif t1=j+1 then f := 3*(j+1) else f := 0; fi;
%p A169689 RETURN(3*a(j)+f);
%p A169689 fi;
%p A169689 end;
%p A169689 [seq(a(n),n=-1..200)];
%Y A169689 Equals A169688/4. Cf. A169697 (limit of rows).
%K A169689 nonn,tabf
%O A169689 -1,3
%A A169689 _N. J. A. Sloane_, Apr 14 2010

# Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE