A265104 a(n) = A265100(n+1) - 6, n >= 1.

8, 26, 35, 80, 89, 107, 116, 242, 251, 269, 278, 323, 332, 350, 359, 728, 737, 755, 764, 809, 818, 836, 845, 971, 980, 998, 1007, 1052, 1061, 1079, 1088, 2186, 2195, 2213, 2222, 2267, 2276, 2294, 2303, 2429, 2438, 2456, 2465, 2510, 2519, 2537

Offset: 1

Author

L. Edson Jeffery, Dec 01 2015

Keywords

• nonn

Comments

In the following, let "gap" and "gap number" be as defined in A265100, and let C(m) denote the m-th Catalan number (A000108).

Conjecture 1: The sequence contains all possible gap numbers.

Conjecture 2: For any gap G, the order |G| of G is the constant |G| = 6.

Conjecture 3: If g is a gap number, then 3*g + 2 is a gap number.

Conjecture 4: If C(m) =!= 0 (mod 3), then C(3*m+1) =!= 0 (mod 3) (=!= means "not congruent") or, what is the same thing, if m lies in a gap, then 3*m + 1 lies in a gap.

Crossrefs

Cf. A265100.

Programs

Mathematica

a005836[1] := 0; a005836[n_] := If[OddQ[n], 3*a005836[Floor[(n + 1)/2]], a005836[n - 1] + 1]; a265100[n_] := 9*a005836[n] + 5; a265104[n_] := a265100[n+1] - 6; Table[a265104[n], {n, 46}]
(* Or: *)
a007814[x_] := IntegerExponent[x, 2]; a003602[x_] := (1 + x/2^a007814[x])/2; a005836[1] := 0; a005836[n_] := If[OddQ[n], 3*a005836[Floor[(n + 1)/2]], a005836[n - 1] + 1]; a265100[n_] := 9*a005836[n] + 5; a265104[n_] := (3^(a007814[n] + 2) - 3)/2 + a265100[2^(a007814[n])*(2*a003602[n] - 1)]; Table[a265104[n], {n, 46}]

Formula

a(n) = (3^(A007814(n) + 2) - 3)/2 + A265100(2^(A007814(n))*(2*A003602(n) - 1)), n >= 1.

Conjecture: a(n) = A265100(n) + A085296(n).