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

0,3

Links

Amiram Eldar, Table of n, a(n) for n = 0..10000

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

Example

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.

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 *)

Prog

(PARI) a(n)=n--; 2^(log(n>>valuation(n, 2)+.5)\log(2)) \\ Charles R Greathouse IV, May 06 2013

Crossrefs

Cf. A000005, A000079, A000265, A006519, A007318, A082907.

Sequence in context: A153279 A278425 A309019 * A086449 A321088 A070556

Adjacent sequences: A082905 A082906 A082907 * A082909 A082910 A082911

Keyword

nonn

Author

Labos Elemer, Apr 23 2003

Status

approved