

A205591


a(1) = 1, a(n) = a(floor((2n1)/3)) + a(floor(2n/3)) for n > 1.


6



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
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OFFSET

1,2


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.
Sequence in context: A198034 A016027 A265347 * A191282 A191281 A032900
Adjacent sequences: A205588 A205589 A205590 * A205592 A205593 A205594


KEYWORD

easy,nonn


AUTHOR

Joseph Myers, Jan 29 2012


STATUS

approved



