A205596 -id:A205596 - cp's OEIS frontend

A205592 a(2) = 1, a(3k) = a(3k+1) = a(2k), a(3k+2) = 2a(2k+1) for k >= 1.

1, 1, 1, 2, 1, 1, 4, 1, 1, 2, 4, 4, 2, 1, 1, 4, 4, 4, 8, 2, 2, 2, 1, 1, 8, 4, 4, 8, 8, 8, 4, 2, 2, 4, 1, 1, 2, 8, 8, 8, 4, 4, 16, 8, 8, 16, 4, 4, 4, 2, 2, 8, 1, 1, 2, 2, 2, 16, 8, 8, 16, 4, 4, 8, 16, 16, 16, 8, 8, 32, 4, 4, 8, 4, 4, 4, 2, 2, 16, 1, 1, 2, 2, 2, 4

Offset: 2

Joseph Myers, Jan 29 2012

Joseph Myers, Table of n, a(n) for n = 2..1000
2011/12 British Mathematical Olympiad Round 2, Problem 2.
Index to sequences related to Olympiads and other Mathematical competitions

Cf. A205591, A205593, A205594, A205595, A205596.

Cf. A335933.

a(n) = A205591(n) - A205591(n-1).

A205593 a(2) = 0, a(3k) = a(3k+1) = a(2k), a(3k+2) = a(2k+1) + 1 for k >= 1.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 2, 0, 0, 1, 2, 2, 1, 0, 0, 2, 2, 2, 3, 1, 1, 1, 0, 0, 3, 2, 2, 3, 3, 3, 2, 1, 1, 2, 0, 0, 1, 3, 3, 3, 2, 2, 4, 3, 3, 4, 2, 2, 2, 1, 1, 3, 0, 0, 1, 1, 1, 4, 3, 3, 4, 2, 2, 3, 4, 4, 4, 3, 3, 5, 2, 2, 3, 2, 2, 2, 1, 1, 4, 0, 0, 1, 1, 1, 2, 1, 1, 5

Offset: 2

Joseph Myers, Jan 29 2012

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

Cf. A205591, A205592, A205594, A205595, A205596.

a(n) = log_2(A205592(n)).

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

Original entry on oeis.org

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

Joseph Myers, Jan 29 2012

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

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

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

A205594 n such that A205592(n) > n.

Original entry on oeis.org

242, 818, 1841, 2762, 6215, 9323, 13985, 14003, 14012, 20978, 21005, 26243, 29888, 31467, 31468, 31481, 31508, 31529, 31589, 39365, 47201, 47202, 47203, 47222, 47262, 47263, 47294, 47384, 59048, 59156, 59183, 61235, 70802, 70805, 70895, 88572, 88573, 88775, 91853

Offset: 1

Joseph Myers, Jan 29 2012

nonn

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

Cf. A205591, A205592, A205593, A205595, A205596.

A205595 A205592(A205594(n)).

Original entry on oeis.org

256, 1024, 2048, 4096, 8192, 16384, 32768, 16384, 16384, 65536, 32768, 32768, 32768, 65536, 65536, 32768, 65536, 32768, 32768, 65536, 131072, 65536, 65536, 65536, 65536, 65536, 65536, 65536, 131072, 65536, 65536, 65536, 262144, 131072, 131072, 131072, 131072, 131072

Offset: 1

Joseph Myers, Jan 29 2012

nonn

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

Cf. A205591, A205592, A205593, A205594, A205596.

cp's OEIS Frontend

A205592 a(2) = 1, a(3k) = a(3k+1) = a(2k), a(3k+2) = 2a(2k+1) for k >= 1.

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A205593 a(2) = 0, a(3k) = a(3k+1) = a(2k), a(3k+2) = a(2k+1) + 1 for k >= 1.

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Formula

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

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A205594 n such that A205592(n) > n.

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A205595 A205592(A205594(n)).

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