A205592 -id:A205592 - cp's OEIS frontend

A205594 n such that A205592(n) > n.

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.

A205596 Least k such that A205592(k) = 2^n.

Original entry on oeis.org

2, 5, 8, 20, 44, 71, 107, 161, 242, 545, 818, 1841, 2762, 6215, 9323, 13985, 20978, 47201, 70802, 159305, 238958, 358910, 807548, 1814615, 2721923, 4082885, 6124328, 13779557, 20669336, 40222412, 87267041

Offset: 0

Joseph Myers, Jan 29 2012

nonn

2011/12 British Mathematical Olympiad Round 2, Problem 2.

Cf. A205591, A205592, A205593, A205594, A205595.

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.

A343856 Irregular table read by rows; the first row is [1]; to obtain the next row, replace each odd-indexed term u with (u, u), and each even-indexed term v with (2*v).

Original entry on oeis.org

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

Offset: 1

Rémy Sigrist, May 01 2021

Sequence A061419 and A343857 gives row lengths and partial sums, respectively.
The n-th row sums to 2^(n-1).
This sequence has fractal features.
As with Jim Conant's iterative dissection of a square (A328078), at each iteration, we split in two odd-indexed elements.
This sequence has similarities with A205592: in A205592:
- we start with A205592(1) = 1,
- for k = 1, 2, ...:
    if k is odd: append two copies of A205592(k),
    if k is even: append 2*A205592(k).

			Table begins:
1:  [1]
2:  [1, 1]
3:  [1, 1, 2]
4:  [1, 1, 2, 2, 2]
5:  [1, 1, 2, 2, 2, 4, 2, 2]
6:  [1, 1, 2, 2, 2, 4, 2, 2, 8, 2, 2, 4]
7:  [1, 1, 2, 2, 2, 4, 2, 2, 8, 2, 2, 4, 8, 8, 4, 2, 2, 8]

Cf. A061419, A205592, A328078, A343857.

{ a = r = [1]; for (n=1, 8, i = 0; a=concat(a, r = concat(apply (v -> if (i++%2, [v,v], [2*v]), r)))); print (a) }

def auptorow(rows):
  alst, row, newrow = [1], [1], []
  for r in range(2, rows+1):
    for i, v in enumerate(row, start=1): newrow += [v, v] if i%2 else [2*v]
    alst, row, newrow = alst + newrow, newrow, []
  return alst
print(auptorow(9)) # Michael S. Branicky, May 04 2021

cp's OEIS Frontend

A205594 n such that A205592(n) > n.

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A205595 A205592(A205594(n)).

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A205596 Least k such that A205592(k) = 2^n.

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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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A343856 Irregular table read by rows; the first row is [1]; to obtain the next row, replace each odd-indexed term u with (u, u), and each even-indexed term v with (2*v).

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PARI

Python