

A039004


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


13



0, 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60, 63, 66, 72, 75, 78, 90, 96, 99, 102, 105, 108, 111, 114, 120, 123, 126, 129, 132, 135, 141, 144, 147, 150, 153, 156, 159, 165, 177, 180, 183, 189, 192, 195, 198, 201, 204, 207, 210, 216, 219
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OFFSET

1,2


COMMENTS

Numbers such that sum (1)^k*b(k) = 0 where b(k)=kth binary digit of n.  Benoit Cloitre, Nov 18 2003
Conjecture: a(C(2n,n)1) = 4^n  1. (A000984 is C(2n,n).  Gerald McGarvey, Nov 18 2007
From Russell Jay Hendel, Jun 23 2015: (Start)
We prove the McGarvey conjecture (A) a(e(n,n)1) = 4^n1, 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) = #{ndigit, base4 numbers with nk more 1digits than 2digits}. 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 ndigit, base4 numbers with nk more 1digits than 2digits.
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 1digits and 2digits. The biggest ndigit, base4 number is 333...3 (n copies of 3). Since 333...33 has zero 1digits and zero 2digits 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^n1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n1 is the (e(n,n)1)st member of A039004, proving the McGarvey conjecture. (End)


LINKS

Alois P. Heinz, Table of n, a(n) for n = 1..10000


FORMULA

Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n.  Benoit Cloitre, Nov 24 2002
That conjecture is false. The number of members of the sequence from 0 to 4^d1 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^d1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction.  Robert Israel, Jun 24 2015


MAPLE

N:= 1000: # to get all terms up to N, which should be divisible by 4
B:= Array(0..N1):
d:= ceil(log[4](N));
S:= Array(0..N1, [seq(op([0, 1, 1, 0]), i=1..N/4)]):
for i from 1 to d do
B:= B + S;
S:= Array(0..N1, i> S[floor(i/4)]);
od:
select(t > B[t]=0, [$0..N1]); # Robert Israel, Jun 24 2015


PROG

(PARI) for(n=0, 219, if(sum(i=1, length(binary(n)), (1)^i*component(binary(n), i))==0, print1(n, ", ")))
See link in A139351 for Fortran program.


CROSSREFS

Cf. A139370, A139371, A139372, A139373.
Cf. A139351, A139352, A139353, A139354, A139355.
Sequence in context: A194226 A193803 A284601 * A070021 A083354 A156242
Adjacent sequences: A039001 A039002 A039003 * A039005 A039006 A039007


KEYWORD

nonn,base,easy


AUTHOR

Olivier Gérard


STATUS

approved



