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a(n) = binomial(n, 2^floor(log_2(n))).
1

%I #21 Dec 15 2018 23:45:21

%S 1,1,3,1,5,15,35,1,9,45,165,495,1287,3003,6435,1,17,153,969,4845,

%T 20349,74613,245157,735471,2042975,5311735,13037895,30421755,67863915,

%U 145422675,300540195,1,33,561,6545,58905,435897,2760681,15380937,76904685,350343565

%N a(n) = binomial(n, 2^floor(log_2(n))).

%C All terms are odd. It follows from the definition and Kummer theorem on 2-adic order of binomial coefficients.

%C From _Robert Israel_, Aug 30 2017: (Start)

%C a(n) = 1 if n is a power of 2.

%C a(2^k+m) = a(2^k+m-1)*(1 + 2^k/m) if 1 <= m < 2^k. (End)

%H Robert Israel, <a href="/A291665/b291665.txt">Table of n, a(n) for n = 1..3421</a>

%H Vladimir Shevelev, <a href="https://arxiv.org/abs/1708.08096">On a Luschny question</a>, arXiv:1708.08096 [math.NT], 2017.

%F a(n) = binomial(n, A053644(n)). - _Michel Marcus_, Dec 15 2018

%e For n=11, k=8. So a(n) = binomial(11,8) = 165.

%p seq(binomial(n,2^ilog2(n)),n=1..100); # _Robert Israel_, Aug 30 2017

%t Table[Binomial[n, 2^Floor@ Log2@ n], {n, 41}] (* _Michael De Vlieger_, Aug 29 2017 *)

%o (PARI) a(n) = binomial(n, 2^logint(n, 2)); \\ _Michel Marcus_, Dec 15 2018

%Y Cf. A053644.

%K nonn,look

%O 1,3

%A _Vladimir Shevelev_, Aug 29 2017

%E More precise definition from _Michael De Vlieger_, Aug 30 2017