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%I #16 Nov 29 2013 21:20:27
%S 0,0,0,0,1,0,0,0,0,0,0,0,1,2,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,
%T 0,0,1,1,1,2,3,2,1,1,1,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,2,
%U 1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1
%N Occurrences of '11' in base 3 expansion of n.
%C Occurrences of '11' in A007089(n). This is to base 3 and A007089 as A014081 is to base 2 A007088.
%C First occurrence of k>0 = 4, 13, 40, 121, 364, ..., = A003462(k). (* _Robert G. Wilson v_, Sep 27 2011 *)
%H T. D. Noe, <a href="/A196096/b196096.txt">Table of n, a(n) for n = 0..1000</a>
%e a(4) = 1 because 4 in base 3 is "11" which has one instance of "11".
%e a(13) = 2 because the number 13 in base 3 is "111" which has two substrings of "11".
%p A196096 := proc(n)
%p local a,dgs3 ;
%p a := 0 ;
%p dgs3 := convert(n,base,3) ;
%p for i from 1 to nops(dgs3)-1 do
%p if op(i,dgs3)=1 and op(i+1,dgs3)=1 then
%p a := a+1 ;
%p end if;
%p end do;
%p a ;
%p end proc:
%p seq(A196096(n),n=0..80) ; # _R. J. Mathar_, Sep 28 2011
%t f[n_] := Count[ Partition[ IntegerDigits[ n, 3], 2, 1], {1, 1}]; Array[f, 100, 0] (* _Robert G. Wilson v_, Sep 27 2011 *)
%Y Cf. A007089, A014081.
%K easy,nonn
%O 0,14
%A _Jonathan Vos Post_, Sep 27 2011
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