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%I #6 Dec 09 2017 03:20:21
%S 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,
%T 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,
%U 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
%N Maximal power of 2 dividing the binomial coefficient binomial(m, 2^k) where m >= 1 and 1 <= 2^k <= m.
%C Provided m and k are given, the sequence index n is n=A001855(m)+k+1. Ordered by m as rows and k as columns the sequence forms a sort of a logarithmically distorted triangle.
%C a(n) is the maximal power of 2 dividing A130067(n).
%C Equivalent propositions: (1) a(n)=0; (2) A130067(n) is odd; (3) the k-th digit of m is 1; (4) A030308(n)=1.
%F 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(m). 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.
%e 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).
%e 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).
%Y Cf. A130067, A065040, A001855, A030308.
%K nonn,tabl
%O 1,6
%A _Hieronymus Fischer_, May 05 2007, Sep 10 2007
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