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a(n) = Sum_{k=0..n} ({binomial(n+2k,2k)*binomial(n,k)} mod 2).
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%I #26 May 29 2017 12:59:21

%S 1,2,2,2,2,4,2,2,2,4,4,4,2,4,2,2,2,4,4,4,4,8,4,4,2,4,4,4,2,4,2,2,2,4,

%T 4,4,4,8,4,4,4,8,8,8,4,8,4,4,2,4,4,4,4,8,4,4,2,4,4,4,2,4,2,2,2,4,4,4,

%U 4,8,4,4,4,8,8,8,4,8,4,4,4,8,8,8,8,16,8

%N a(n) = Sum_{k=0..n} ({binomial(n+2k,2k)*binomial(n,k)} mod 2).

%C Equals the run length transform of A040000: 1,2,2,2,2,2,...

%H Chai Wah Wu, <a href="/A277561/b277561.txt">Table of n, a(n) for n = 0..10000</a>

%H Chai Wah Wu, <a href="https://arxiv.org/abs/1610.06166">Sums of products of binomial coefficients mod 2 and run length transforms of sequences</a>, arXiv:1610.06166 [math.CO], 2016.

%H <a href="/index/Ru#rlt">Index entries for sequences computed with run length transform</a>

%F a(n) = 2^A069010(n). a(2n) = a(n), a(4n+1) = 2a(n), a(4n+3) = a(2n+1). - _Chai Wah Wu_, Nov 04 2016

%F a(n) = A034444(A005940(1+n)). - _Antti Karttunen_, May 29 2017

%t Table[Sum[Mod[Binomial[n + 2 k, 2 k] Binomial[n, k], 2], {k, 0, n}], {n, 0, 86}] (* _Michael De Vlieger_, Oct 21 2016 *)

%o (Python)

%o def A277561(n):

%o return sum(int(not (~(n+2*k) & 2*k) | (~n & k)) for k in range(n+1))

%o (PARI) a(n) = sum(k=0, n, binomial(n+2*k, 2*k)*binomial(n,k) % 2); \\ _Michel Marcus_, Oct 21 2016

%Y Cf. A005940, A034444, A040000, A069010, A106737.

%K nonn

%O 0,2

%A _Chai Wah Wu_, Oct 19 2016