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%I #50 May 19 2023 06:19:13
%S 0,1,2,3,4,5,6,9,10,11,12,13,18,19,20,21,22,25,26,27,36,37,38,41,42,
%T 43,44,45,50,51,52,53,54,73,74,75,76,77,82,83,84,85,86,89,90,91,100,
%U 101,102,105,106,107,108,109,146,147,148,149,150,153,154,155,164,165,166
%N Numbers without 3 consecutive equal binary digits.
%C Complement of A136037; intersection of A003796 and A003726. - _Reinhard Zumkeller_, Dec 11 2007
%C From _Emeric Deutsch_, Jan 27 2018: (Start)
%C Also 0 together with the indices of the compositions that have no parts larger than 2. For the definition of the index of a composition see A298644.
%C For example, 105 is in the sequence since its binary form is 1101001 and the composition [2,1,1,2,1] has no parts larger than 2.
%C On the other hand, 132 is not in the sequence since its binary form is 10000100 and the composition [1,4,1,2] has a part larger than 2.
%C The command c(n) from the Maple program yields the composition having index n. (End)
%C The sequence contains A000045(n+1) positive terms with binary length n. - _Rémy Sigrist_, Sep 30 2022
%H Reinhard Zumkeller, <a href="/A063037/b063037.txt">Table of n, a(n) for n = 1..10000</a>
%F It appears (but has not yet been proved) that the terms of {a(n)} can be computed recursively as follows. Let {c(i)} be defined for i >= 4 by c(i) = 2c(i-1) + 1, if i is a multiple of 3, else c(i) = 2c(i-1) - 1, with c(4) = 1. I.e., {c(i)} = {1,1,3,5,9,19,37,73,147,...}, for i=4,5,6,... . Let a(1)=1, a(2)=2, a(3)=3. For n > 3, choose k so that F(k)-2 < n <= F(k+1)-2, where F(k) denotes the k-th Fibonacci number (A000045). Then a(n) = c(k) + 2a(F(k)-2) - a(2F(k)-n-3). This has been verified for n up to 1100. - _John W. Layman_, May 26 2009
%e The binary representation of 9 (1001) has no 3 consecutive equal digits.
%p isA063037 := proc(n)
%p local bdgs,rep,d,i ;
%p if n = 0 then
%p return true;
%p end if;
%p bdgs := convert(n,base,2) ;
%p rep := 1;
%p d := op(1,bdgs) ;
%p for i from 2 to nops(bdgs) do
%p if op(i,bdgs) = op(i-1,bdgs) then
%p rep := rep+1 ;
%p else
%p rep :=1 ;
%p end if ;
%p if rep > 2 then
%p return false;
%p end if;
%p end do:
%p return true ;
%p end proc:
%p for n from 0 to 50 do
%p if isA063037(n) then
%p printf("%d,",n) ;
%p end if;
%p end do: # _R. J. Mathar_, Dec 18 2013
%p # Second Maple program:
%p Runs := proc (L) local j, r, i, k: j := 1: r[j] := L[1]: for i from 2 to nops(L) do if L[i] = L[i-1] then r[j] := r[j], L[i] else j := j+1: r[j] := L[i] end if end do: [seq([r[k]], k = 1 .. j)] end proc: RunLengths := proc (L) map(nops, Runs(L)) end proc: c := proc (n) ListTools:-Reverse(convert(n, base, 2)): RunLengths(%) end proc: A := {0}: for n to 250 do if `subset`(convert(c(n), set), {1, 2}) then A := `union`(A, {n}) else end if end do: A; # most of this Maple program is due to _W. Edwin Clark_. - _Emeric Deutsch_, Jan 27 2018
%t Select[Range[0, 168], AllTrue[Length /@ Split@ IntegerDigits[#, 2], # < 3 &] &] (* _Michael De Vlieger_, Aug 20 2017 *)
%o (PARI) { n=0; for (m=0, 10^9, x=m; t=1; b=2; while (x>0, d=x-2*(x\2); x\=2; if (d==b, c++; if (c==3, t=x=0), b=d; c=1)); if (t, write("b063037.txt", n++, " ", m); if (n==1000, break)) ) } \\ _Harry J. Smith_, Aug 16 2009
%o (PARI) a(n) = { n--; if (n<=1, return (n), my (s=1); for (i=1, oo, my (f=fibonacci(i+1)); if (n<s+f, return (2^i-1-a(2*s-n)), s+=f))) } \\ _Rémy Sigrist_, Sep 30 2022
%o (Python)
%o from itertools import count, islice
%o def A063037_gen(startvalue=0): # generator of terms >= startvalue
%o return filter(lambda n: not ('000' in (s:=bin(n)[2:]) or '111' in s), count(max(0,startvalue)))
%o A063037_list = list(islice(A063037_gen(),20)) # _Chai Wah Wu_, Oct 04 2022
%Y Cf. A000045, A000975, A166535, A286262.
%K easy,nonn
%O 1,3
%A _Lior Manor_, Jul 05 2001
%E Missing "less than" sign supplied in the conjectured recurrence (thanks to _Franklin T. Adams-Watters_ for pointing this out) by _John W. Layman_, Nov 09 2009
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