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 A039004 Numbers n such that representation in base 4 has same number of 1's and 2's. 13

%I

%S 0,3,6,9,12,15,18,24,27,30,33,36,39,45,48,51,54,57,60,63,66,72,75,78,

%T 90,96,99,102,105,108,111,114,120,123,126,129,132,135,141,144,147,150,

%U 153,156,159,165,177,180,183,189,192,195,198,201,204,207,210,216,219

%N Numbers n such that representation in base 4 has same number of 1's and 2's.

%C Numbers such that sum (-1)^k*b(k) = 0 where b(k)=k-th binary digit of n. - _Benoit Cloitre_, Nov 18 2003

%C Conjecture: a(C(2n,n)-1) = 4^n - 1. (A000984 is C(2n,n). - _Gerald McGarvey_, Nov 18 2007

%C From _Russell Jay Hendel_, Jun 23 2015: (Start)

%C We prove the McGarvey conjecture (A) a(e(n,n)-1) = 4^n-1, with e(n,m) = a034870(n,m) = binomial(2n,m), the even rows of Pascal's triangle. By the comment from Hendel in A034870, we have the function s(n,k) = #{n-digit, base-4 numbers with n-k more 1-digits than 2-digits}. As shown in A034870, (B) #s(n,k)= e(n,k) with # indicating cardinality, that is, e(n,k) = binomial(2n,k) gives the number of n-digit, base-4 numbers with n-k more 1-digits than 2-digits.

%C We now show that (B) implies (A). By definition, s(n,n) contains the e(n,n) = binomial(2n,n) numbers with an equal number of 1-digits and 2-digits. The biggest n-digit, base-4 number is 333...3 (n copies of 3). Since 333...33 has zero 1-digits and zero 2-digits it follows that 333...333 is a member of s(n,n) and hence it is the biggest member of s(n,n). But 333...333 (n copies of 3) in base 4 has value 4^n-1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n-1 is the (e(n,n)-1)st member of A039004, proving the McGarvey conjecture. (End)

%H Alois P. Heinz, <a href="/A039004/b039004.txt">Table of n, a(n) for n = 1..10000</a>

%F Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n. - _Benoit Cloitre_, Nov 24 2002

%F That conjecture is false. The number of members of the sequence from 0 to 4^d-1 is binomial(2d,d) which by Stirling's formula is asymptotic to 4^d/sqrt(Pi*d). If Cloitre's conjecture were true we would have 4^d-1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction. - _Robert Israel_, Jun 24 2015

%p N:= 1000: # to get all terms up to N, which should be divisible by 4

%p B:= Array(0..N-1):

%p d:= ceil(log[4](N));

%p S:= Array(0..N-1,[seq(op([0,1,-1,0]),i=1..N/4)]):

%p for i from 1 to d do

%p B:= B + S;

%p S:= Array(0..N-1,i-> S[floor(i/4)]);

%p od:

%p select(t -> B[t]=0, [\$0..N-1]); # _Robert Israel_, Jun 24 2015

%o (PARI) for(n=0,219,if(sum(i=1,length(binary(n)),(-1)^i*component(binary(n),i))==0,print1(n,",")))

%o See link in A139351 for Fortran program.

%Y Cf. A139370, A139371, A139372, A139373.

%Y Cf. A139351, A139352, A139353, A139354, A139355.

%K nonn,base,easy

%O 1,2