A205591 a(1) = 1, a(n) = a(floor((2n-1)/3)) + a(floor(2n/3)) for n > 1.

1, 2, 3, 4, 6, 7, 8, 12, 13, 14, 16, 20, 24, 26, 27, 28, 32, 36, 40, 48, 50, 52, 54, 55, 56, 64, 68, 72, 80, 88, 96, 100, 102, 104, 108, 109, 110, 112, 120, 128, 136, 140, 144, 160, 168, 176, 192, 196, 200, 204, 206, 208, 216, 217, 218, 220, 222, 224, 240, 248, 256

Offset: 1

Author

Joseph Myers, Jan 29 2012

Keywords

• easy
• nonn

Comments

In other words, a(1)=1 and then any term is a sum of two earliest possible previous terms (not necessarily distinct), given that each term must be used in summation no more than three times. So a(2)=1+1 (thus 1 gets used twice), a(3)=1+2 (thus 1 gets used for the third and final time, then 2 steps in), and so on. - Ivan Neretin, Jul 09 2015

Links

• Joseph Myers, Table of n, a(n) for n = 1..1000
• 2011/12 British Mathematical Olympiad Round 2, Problem 2.

Crossrefs

Cf. A205592, A205593, A205594, A205595, A205596.