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%I #28 Apr 21 2021 04:34:26
%S 0,1,8,625,13402696,19720133460129649,
%T 126747521841153485025455279433135688,
%U 15141471069096667541622192498608408980462133134430650704600552060872705905
%N Positions of breadth-first-wise encodings (A002542) of the complete binary trees (A084107) in A014486.
%H Alexander Adamchuk, Nov 10 2007, <a href="/A083942/b083942.txt">Table of n, a(n) for n = 0..11</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CatalanNumber.html">Catalan Number</a>.
%F a(n) = A057118(A084108(n)).
%F a(n) = A080300(A002542(n)) [provided that 2^((2^n)-1)*((2^((2^n)-1))-1) is indeed the formula for A002542].
%F Conjecture: a(n) = A014138(2^n-2) for n>0. - _Alexander Adamchuk_, Nov 10 2007
%F Conjecture: a(n) = Sum_{k=1..2^n-1} A000108(k). - _Alexander Adamchuk_, Nov 10 2007
%F Let h(n) = -((C(2*n,n)*hypergeom([1,1/2+n],[2+n],4))/(1+n)+I*sqrt(3)/2+1/2). Assuming Adamchuk's conjecture a(n) = h(2^n) and A014138(n) = h(n+1). - _Peter Luschny_, Mar 09 2015
%Y Cf. A014138 (partial sums of Catalan numbers), A000108 (Catalan Numbers).
%K nonn
%O 0,3
%A _Antti Karttunen_, May 13 2003