In the following pattern, zeroes mark odds, other integers count factors of two, arranged in triangular array to show binary structure. To obtain non-zero values on each line, copy sequence P from line above then append 1 followed by P+1. 0, 1, 0, 1, 1, 2, 0, 1, 1, 2, 1, 2, 2, 3, 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 0, and the next line is: 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 0. This should all be provable from the recurrence relation: a(n) = 2*(2*n+1)*a(n-1)/(n+1) when you take the denominator as (n+1)=2^k, but I didn't bother to work out any more details, sorry. This type of structure is reminiscent of Cellular Automaton Rule 90, see also: https://demonstrations.wolfram.com/Code686BuildsTheChairTiling/ This reminds me that the parity pattern of pascal's triangle is also code 90, so you could probably prove it that way. --Brad On Tue, Jan 26, 2021 at 6:12 PM Dan Asimov <asimov@msri.org> wrote:
The OEIS sequence for
binomial(2n+1, n+1) = (2n+1)! / (n! (n+1)!)
(<https://oeis.org/A001700>) begins with
1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052,
(0 ≤ n ≤ 25), for which
binomial(2n+1, n+1) = is odd if and only if n is of the form 2^K - 1.
Is this true for all n ≥ 0 ???
—Dan
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