A079882 A run of 2^n 1's followed by a run of 2^n 2's, for n=0, 1, 2, ...

1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2

Offset: 0

Author

N. J. A. Sloane, Feb 21 2003

Keywords

• nonn
• easy

Comments

In the sequence of nonnegative integers (cf. A001477) substitute all n by 2^floor(n/2) occurrences of (1 + n mod 2); a(n)=A173920(n+2,3) for n>0. [From Reinhard Zumkeller, Mar 04 2010]

Links

• Robert Israel, Table of n, a(n) for n = 0..10000

Crossrefs

Partial sums give A079945. Equals 1 + A079944. Cf. A080584.

First differences of A080637.

Programs

Maple

f1 := n->[seq(1,i=1..2^n)]; f2 := n->[seq(2,i=1..2^n)]; s := []; for i from 0 to 10 do s := [op(s), op(f1(i)), op(f2(i))]; od: s;

Mathematica

Table[{PadRight[{},2^n,1],PadRight[{},2^n,2]},{n,0,5}]//Flatten (* Harvey P. Dale, Jul 22 2016 *)

Formula

a(n) = floor(log[2](8*(n+2)/3)) - floor(log[2](n+2)). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 22 2003