proposed

approved

editing

proposed

Maximal power of 2 dividing the binomial coefficient binomial(m, 2^k) where m >= 1 and 1 <= 2^k <= m.

approved

editing

Hieronymus Fischer (_Hieronymus. Fischer(AT)gmx.de), _, May 05 2007, Sep 10 2007

Provided m and k are given, the sequence index n is n=A001855(m)+k+1. Ordered by m as lines rows and k as rows columns the sequence forms a sort of a logarithmically distorted triangle. a(n) is the maxmal power of 2 dividing A130067(n). Equivalent propositions: (1) a(n)=0; (2) A130067(n) is odd; (3) the k-th digit of m is 1; (4) A030308(n)=1.

a(n) is the maximal power of 2 dividing A130067(n).

Equivalent propositions: (1) a(n)=0; (2) A130067(n) is odd; (3) the k-th digit of m is 1; (4) A030308(n)=1.

a(n)=g(m)-g(m-2^k) where g(x)=sum(floor(x/2^i), k<i<=floor(log_2(x))), m=max(j | A001855(j)<n) and k=n-1-A001855(jm). Also true: a(n)=sum(product(1-b(i), k<=i<j), k<j<=floor(log_2(m))) where b(i) is the i-th digit of the binary representation of m. Example: n=35 gives m=12 and k=1, so that m=1100 and a(n)=1-b(1)+(1-b(1))(1-b(2))=1+0=1.

nonn,tabl,new

Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), May 05 2007, Sep 10 2007

Maximal power of 2 dividing the binomial coefficient binomial(m,2^k) where m>=1 and 1<=2^k<=m.

0, 1, 0, 0, 0, 2, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 3, 2, 1, 0, 0, 2, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 2, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 4, 3, 2, 1, 0, 0, 3, 2, 1, 0, 1, 0, 2, 1, 0, 0, 0, 2, 1, 0, 2, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 3, 2, 1, 0, 0, 0, 2, 1, 0, 0, 1, 0, 1, 0, 0, 0

1,6

Provided m and k are given, the sequence index n is n=A001855(m)+k+1. Ordered by m as lines and k as rows the sequence forms a sort of logarithmically distorted triangle. a(n) is the maxmal power of 2 dividing A130067(n). Equivalent propositions: (1) a(n)=0; (2) A130067(n) is odd; (3) the k-th digit of m is 1; (4) A030308(n)=1.

a(n)=g(m)-g(m-2^k) where g(x)=sum(floor(x/2^i), k<i<=floor(log_2(x))), m=max(j | A001855(j)<n) and k=n-1-A001855(j). Also true: a(n)=sum(product(1-b(i), k<=i<j), k<j<=floor(log_2(m))) where b(i) is the i-th digit of the binary representation of m. Example: n=35 gives m=12 and k=1, so that m=1100 and a(n)=1-b(1)+(1-b(1))(1-b(2))=1+0=1.

a(6)=2 since 2^2 divides binomial(4,2^0)=4 and 2^3 is not a factor (here n=6 gives m=4, k=0).

a(20)=1 since 2^1 divides binomial(8,2^2)=70 and 2^2 is not a factor (here n=20 gives m=8, k=2).

Cf. A130067, A065040, A001855, A030308.

nonn,tabl,new

Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), May 05 2007

approved