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A363739 a(n) is the length of the n-th run of identical terms of A349509.

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%I A363739 #34 Aug 08 2023 09:13:08
%S A363739 2,1,5,1,17,1,2,1,5,1,17,1,26,1,2,1,23,1,8,1,44,1,80,1,2,1,5,1,17,1,
%T A363739 53,1,26,1,134,1,80,1,29,1,23,1,107,1,2,1,5,1,71,1,161,1,2,1,77,1,35,
%U A363739 1,65,1,59,1,164,1,26,1,50,1,8,1,233,1,194,1,290,1,2
%N A363739 a(n) is the length of the n-th run of identical terms of A349509.
%H A363739 Kevin Ryde, <a href="/A363739/b363739.txt">Table of n, a(n) for n = 1..10000</a>
%F A363739 Conjectures: (Start)
%F A363739 a(2*n) = 1.
%F A363739 a(2*n+1) = A349929(n) - A349929(n-1) - 1, with A349929(0) = 0. (End)
%t A363739 A349509[n_]:=Denominator[Binomial[n^3+6n^2-6n+2, n^3-1]/n^3]; a={}; imax=2200; For[i=1, i<=imax, i++, For[j=1, A349509[j+i-1]==A349509[j+i], j++]; If[i>1 && A349509[j+i-1]!=A349509[j+i], AppendTo[a,1]]; i+=j; AppendTo[a, j]]; a
%Y A363739 Cf. A349509, A349929.
%K A363739 nonn
%O A363739 1,1
%A A363739 _Stefano Spezia_, Jun 19 2023