A137824 Index at which A137823(n) occurs first in A137822 (gaps in numbers m such that 3 | sum( Catalan(k), k=1..2m)).

1, 3, 2, 4, 12, 8, 16, 48, 32, 64, 192, 128, 256, 768, 512, 1024, 3072, 2048, 4096, 12288, 8192, 16384, 49152, 32768, 65536, 196608, 131072, 262144, 786432, 524288, 1048576, 3145728, 2097152, 4194304, 12582912, 8388608, 16777216, 50331648, 33554432, 67108864, 201326592

Offset

1,2

Comments

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)

See A137822 and A137823 for more comments and formulas.

Links

Paolo Xausa, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (0,0,4).

Formula

If n==2 (mod 3) then a(n) = 3*2^[2*(n-1)/3]; else a(n) = 2^[2*(n-1)/3].

From Colin Barker, Aug 19 2012: (Start)

a(n) = 4*a(n-3) for n>3.

G.f.: x*(1+x)*(1+2*x)/(1-4*x^3). (End)

Mathematica

A137824[n_] := 2^Quotient[2*(n-1), 3]*If[Mod[n, 3] == 2, 3, 1];
Array[A137824, 40] (* Paolo Xausa, May 25 2026 *)
(* Alternative: *)
LinearRecurrence[{0, 0, 4}, {1, 3, 2}, 40] (* Paolo Xausa, May 25 2026 *)

Prog

(PARI) A137824(n) = if( n%3==2, 3, 1)<<(2*(n-1)\3)
(PARI) A137824(n) = for( i=1, #A137822, A137822[i]==A137823[n] & return(i))
(PARI) a=[1, 3, 2]; for( i=1, 5, a=concat( a, 4^(#a/3)*a )); a

Crossrefs

Cf. A107755, A122983, A137821, A137822, A137823.

Sequence in context: A332647 A290333 A386841 * A019321 A336435 A279261

Adjacent sequences: A137821 A137822 A137823 * A137825 A137826 A137827

Keyword

nonn,easy,changed

Author

M. F. Hasler, May 15 2008

Extensions

More terms from Paolo Xausa, May 25 2026

Status

approved