%I
%S 1,3,2,4,12,8,16,48,32,64,192,128,256,768,512,1024,3072,2048,4096,
%T 12288,8192,16384,49152,32768,65536,196608,131072,262144,786432,
%U 524288,1048576,3145728,2097152,4194304,12582912,8388608,16777216,50331648
%N Index at which A137823(n) occurs first in A137822 (gaps in numbers m such that 3  sum( Catalan(k), k=1..2m)).
%C Other characterization of the sequence: concatenate pattern (1,3,2) multiplying it by 4 after each concatenation step. Or: Start with 1,3,2, then iteratively append the whole sequence obtained so far multiplied by 4^(length of the sequence divided by 3)
%C See A137822 and A137823 for more comments and formulas.
%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,4).
%F If n=2 (mod 3) then a(n) = 3*2^[2(n1)/3]; else a(n) = 2^[(2(n1)/3].
%F a(n) = 4*a(n3) for n>3. G.f.: x*(1+x)*(1+2*x)/(14*x^3). [_Colin Barker_, Aug 19 2012]
%o (PARI) A137824(n) = if( n%3==2,3,1)<<(2*(n1)\3)
%o (PARI) A137824(n) = for( i=1,#A137822, A137822[i]==A137823[n] & return(i))
%o (PARI) a=[1,3,2]; for( i=1,5, a=concat( a, 4^(#a/3)*a )); a
%Y Cf. A107755, A122983, A137821A137823.
%K nonn,easy
%O 1,2
%A _M. F. Hasler_, May 15 2008
