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%I #24 May 03 2023 09:13:38
%S 0,1,4,21,64,321,1092,5461,16384,81921,278532,1392661,4210752,
%T 21053761,71582788,357913941,1073741824,5368709121,18253611012,
%U 91268055061,275951648832,1379758244161,4691178030148,23455890150741
%N a(n) = (A001317(2n)-1)/4.
%C a(n) = (A038192(n) - 3)/12 = (A038183(n) - 1)/4, reflecting the fact that, if you look at each row of the Pascal triangle's parity as a binary number (cf. A001317), then the numbers in odd rows are thrice the numbers in even rows.
%C Conjecture: a(2^k) = 2^(2^(k+1)-2). [This conjecture is true. - _Vladimir Shevelev_, Nov 28 2010]
%C Conjectures: lim(n->inf, a(2n+1)/a(2n)) = 5, lim(n->inf, a(4n+2)/a(4n+1)) = 17/5, lim(n->inf, a(8n+4)/a(8n+3)) = 257/85 etc. [This follows from the formula, for n>=0, t>=1: ( 4*a(2^t*n+2^(t-1))+1 )/( 4*a(2^t*n+2^(t-1)-1)+1 ) = 3*F_t/(F_t-2), where F_t= A000215(t) - _Vladimir Shevelev_, Nov 28 2010]
%H Vladimir Shevelev, <a href="http://arxiv.org/abs/1011.6083">On Stephan's conjectures concerning Pascal triangle modulo 2</a>, arXiv:1011.6083 [math.NT], 2010-2012.
%t a1317[n_] := Sum[2^k Mod[Binomial[n, k], 2] , {k, 0, n}];
%t a[n_] := (a1317[2n] - 1)/4;
%t Table[a[n], {n, 0, 23}] (* _Jean-François Alcover_, Jan 18 2019 *)
%o (PARI) a(n)=(sum(k=0,2*n+1,(binomial(2*n+1,k)%2)*2^k)-3)/12
%o (Python)
%o def A089893(n): return sum((bool(~(m:=n<<1)&m-k)^1)<<k for k in range((n<<1)+1))-1>>2 # _Chai Wah Wu_, May 02 2023
%Y Cf. A000215, A001317, A038183, A038192.
%K nonn
%O 0,3
%A _Ralf Stephan_, Jan 10 2004
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