A082908 Largest value of gcd(2^n, binomial(n,j)) with j=0..n-1; maximum value of largest power of 2 dividing binomial(n,j) in the n-th row of Pascal's triangle.
1, 1, 2, 1, 4, 2, 4, 1, 8, 4, 8, 2, 8, 4, 8, 1, 16, 8, 16, 4, 16, 8, 16, 2, 16, 8, 16, 4, 16, 8, 16, 1, 32, 16, 32, 8, 32, 16, 32, 4, 32, 16, 32, 8, 32, 16, 32, 2, 32, 16, 32, 8, 32, 16, 32, 4, 32, 16, 32, 8, 32, 16, 32, 1, 64, 32, 64, 16, 64, 32, 64, 8, 64, 32, 64, 16, 64, 32, 64, 4, 64, 32
Offset: 0
Keywords
Examples
For n = 10: the 10th row = {1,10,45,120,210,252,210,120,45,10,1}, the largest powers of 2 dividing the entries: {1,2,1,8,2,4,2,8,1,2,1}; maximum 2^k-divisor is a(10) = 8.
Links
- Amiram Eldar, Table of n, a(n) for n = 0..10000
Programs
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Mathematica
Table[Max[Table[GCD[2^n, Binomial[n, j]], {j, 0, n}]], {n, 0, 128}] a[n_] := 2^Floor[Log2[(n+1) / 2^IntegerExponent[n+1, 2]]]; Array[a, 82, 0] (* Amiram Eldar, Mar 15 2025 *) -
PARI
a(n)=n--; 2^(log(n>>valuation(n, 2)+.5)\log(2)) \\ Charles R Greathouse IV, May 06 2013
Formula
a(n) = Max_{gcd(2^n, binomial(n, j)), j=0..n}.
a(n-1) = 2^floor(log_2(A000265(n))). - Brad Clardy, May 06 2013