A055562 -id:A055562 - cp's OEIS frontend

A022441 a(n) = c(n) + c(n-1) where c (A055562) is the sequence of numbers not in a.

1, 5, 7, 10, 14, 17, 20, 23, 25, 28, 31, 34, 37, 40, 43, 46, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 97, 100, 103, 106, 109, 112, 115, 118, 121, 124, 127, 130, 133, 136, 139, 142, 145, 148, 151, 154, 157, 160, 163, 166, 169, 172, 175, 178

Offset: 0

Clark Kimberling

Ivan Neretin, Table of n, a(n) for n = 0..10000
J-P. Bode, H. Harborth, C. Kimberling, Complementary Fibonacci sequences, Fibonacci Quarterly 45 (2007), 254-264.
R. Stephan, Some divide-and-conquer sequences ...
R. Stephan, Table of generating functions

Cf. A055562 (complement), A022424.

[1] cat [3*n + 2 - (Floor((Log(n)/Log(2))) mod 2): n in [0..10]]; // G. C. Greubel, Mar 08 2018

A022441 := n-> `if`(n=0, 1, 3*n + 2 - (ilog2(n) mod 2)):
seq(A022441(n), n= 0..59);

Fold[Append[#1, Plus @@ Complement[Range[Max@#1 + 3], #1][[{#2 + 1, #2 + 2}]]] &, {1, 5}, Range[58]] (* Ivan Neretin, Mar 30 2017 *)
Table[If[n==0,1, 3*n+2 - Mod[Floor[Log[n]/Log[2]], 2]], {n,0,30}] (* G. C. Greubel, Mar 08 2018 *)

for(n=0,30, print1(if(n==0,1, 3*n+2 - (floor(log(n)/log(2))%2)), ", ")) \\ G. C. Greubel, Mar 08 2018

a(n) + a(n-1) = 3n + 2 - (floor(log_2 n) mod 2) for n >= 1. - Jeffrey Shallit, Jun 08 2000
For n>0, a(n) = b(n) with b(0)=0, b(2n) = -b(n)+9n+3, b(2n+1) = -b(n)+9n+6-[n==0]. - Ralf Stephan, Oct 24 2003
a(n) = A210770(2*n+1). - Reinhard Zumkeller, Mar 25 2012

More terms from Winston C. Yang (winston(AT)cs.wisc.edu), Aug 23 2000
Term a(16)=50 fixed by Ivan Neretin, Mar 30 2017
Updated by Clark Kimberling, Feb 19 2018

A022424 Solution a( ) of the complementary equation a(n) = b(n-1) + b(n-2), where a(0) = 1, a(1) = 2; see Comments.

Original entry on oeis.org

1, 2, 7, 9, 11, 14, 18, 22, 25, 28, 31, 33, 36, 39, 41, 44, 47, 50, 53, 56, 59, 62, 66, 69, 72, 75, 78, 82, 85, 88, 91, 94, 97, 100, 103, 106, 109, 112, 115, 118, 121, 124, 127, 129, 132, 135, 138, 141, 144, 147, 150, 153, 156, 159, 161, 164, 167, 170

Offset: 0

Clark Kimberling

nonn

From the Bode-Harborth-Kimberling link:
a(n) = b(n-1) + b(n-2) for n > 2;
b(0) = least positive integer not in {a(0),a(1)};
b(n) = least positive integer not in {a(0),...,a(n),b(0),...,b(n-1)} for n > 1.
Note that (b(n)) is strictly increasing and is the complement of (a(n)).
***
In the following guide to solutions a( ) and b( ) of a(n) = b(n-1) + b(n-2), an asterisk (*) indicates that a( ) differs from the indicated A-sequence in one or two initial terms:
(a(n))     (b(n))    a(0)   a(1)
A022424    A055563     1      2
A022425    A299407     1      4
A022441*   A055562     1      5
A022426    A299411     2      3
A022442*   A099467*    2      4
A299416    A299417     3      4
A299418    A299419     3      5
A299420    A299421     4      5
A022441*   A055562     1      1
***
Guide to solutions a( ) and b( ) of a(n) = b(n-1) + b(n-2) + b(n-3) for various initial values:
(a(n))     (b(n))    a(0)   a(1)   a(2)
A299486    A299487     1      2      3
A299488    A299489     1      2      4
A299490    A299491     1      3      5
A299492    A299492     2      4      5
A299494    A299493     2      4      6
A299496    A299494     3      4      5
***
Guide to other complementary equations:
A022427-A022440:  a(n) = b(n-1) + b(n-3)
A299531-A299532:  a(n) = 2*b(n-1) + b(n-2), a(0) = 1, a(1) = 2
A296220, A299534: a(n) = b(n-1) + 2*b(n-2), a(0) = 1, a(1) = 2
A022437, A299536: a(n) = b(n-1) + b(n-3), a(0) = 1, a(1) = 2, a(2) = 3
A022437, A299538: a(n) = b(n-1) + b(n-3), a(0) = 2, a(1) = 3, a(2) = 4
A022438-A299540:  a(n) = b(n-1) + b(n-3), a(0) = 2, a(1) = 3, a(2) = 5
A299541-A299542:  a(n) = b(n-1) + b(n-3), a(0) = 2, a(1) = 4, a(2) = 6
A299543-A299544:  a(n) = 2*b(n-1) + b(n-2) - b(n-3), a(0) = 1, a(1) = 2, a(2) = 3
A299545-A299546:  a(n) = b(n-1) + 2*b(n-2) - b(n-3), a(0) = 1, a(1) = 2, a(2) = 3
A299547: a(n) = b(n-1) + b(n-2) + ... + b(0), a(0) = 1, a(1) = 2, a(2) = 3

Ivan Neretin, Table of n, a(n) for n = 0..10000
J-P. Bode, H. Harborth, C. Kimberling, Complementary Fibonacci sequences, Fibonacci Quarterly 45 (2007), 254-264.

Cf. A055563 (complement), A022425, A299407, A299486-A299494.

Another pair is given in A324142, A324143.

Fold[Append[#1, Plus @@ Complement[Range[Max@#1 + 3], #1][[{#2, #2 + 1}]]] &, {1, 2}, Range[56]] (* Ivan Neretin, Mar 28 2017 *)

Edited by Clark Kimberling, Feb 16 2018

A210770 a(1) = 1, a(2) = 2; for n > 1, a(2n+2) = smallest number not yet seen, a(2n+1) = a(2n) + a(2n+2).

Original entry on oeis.org

1, 2, 5, 3, 7, 4, 10, 6, 14, 8, 17, 9, 20, 11, 23, 12, 25, 13, 28, 15, 31, 16, 34, 18, 37, 19, 40, 21, 43, 22, 46, 24, 50, 26, 53, 27, 56, 29, 59, 30, 62, 32, 65, 33, 68, 35, 71, 36, 74, 38, 77, 39, 80, 41, 83, 42, 86, 44, 89, 45, 92, 47, 95, 48, 97, 49, 100

Offset: 1

Reinhard Zumkeller, Mar 25 2012

nonn

Permutation of natural numbers with inverse A210771.
From Jeffrey Shallit, Jun 18 2021: (Start)
This sequence is "2-sychronized"; there is a 23-state finite automaton that recognizes the base-2 representations of n and a(n), in parallel.
It obeys the identities
a(4n+3) = a(2n+1) - a(4n) + 2 a(4n+2)
a(8n) = 2a(4n)
a(8n+1) = a(2n+1) + 3a(4n)
a(8n+2) = a(2n+1) + 2 a(4n) - a(4n+1) + a(4n+2)
a(8n+4) = a(2n+1) + a(4n+2)
a(8n+5) = 3a(2n+1) - a(4n) +2a(4n+2)
a(8n+6) = 2a(2n+1) - a(4n) + a(4n+2). (End)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for sequences that are permutations of the natural numbers

Cf. A064736.

import Data.List (delete)
a210770 n = a210770_list !! (n-1)
a210770_list = 1 : 2 : f 1 2 [3..] where
   f u v (w:ws) = u' : w : f u' w (delete u' ws) where u' = v + w

def aupton(terms):
    alst, seen = [1, 2], {1, 2}
    for n in range(2, terms, 2):
        anp1 = alst[-1] + 1
        while anp1 in seen: anp1 += 1
        an = alst[n-1] + anp1
        alst, seen = alst + [an, anp1], seen | {an, anp1}
    return alst[:terms]
print(aupton(67)) # Michael S. Branicky, Jun 18 2021

a(2*n-1) = A022441(n-1); a(2*n) = A055562(n-1).

Definition corrected by Jeffrey Shallit, Jun 18 2021

A055563 a(n) = least number greater than a(n-1) not the sum of an earlier pair of consecutive terms, a(0) = 3.

Original entry on oeis.org

3, 4, 5, 6, 8, 10, 12, 13, 15, 16, 17, 19, 20, 21, 23, 24, 26, 27, 29, 30, 32, 34, 35, 37, 38, 40, 42, 43, 45, 46, 48, 49, 51, 52, 54, 55, 57, 58, 60, 61, 63, 64, 65, 67, 68, 70, 71, 73, 74, 76, 77, 79, 80, 81, 83, 84, 86, 87, 89, 90, 92, 93, 95, 96, 98, 99, 101, 102, 104, 105

Offset: 0

Henry Bottomley, May 26 2000

nonn

			a(3) = 6 because a(2) = 5, 6 <> a(0) + a(1) and 6 <> a(1) + a(2);
a(4) = 8 because a(3) = 6, 7 = a(0) + a(1), 8 <> a(0) + a(1), 8 <> a(1) + a(2) and 8 <> a(2) + a(3).

Ivan Neretin, Table of n, a(n) for n = 0..10000

Complement of A022424. See A001651 for a(0) = 1 and A055562 for a(0) = 2.

a(n) = A022424(n) - a(n-1) for n > 0.

cp's OEIS Frontend

A022441 a(n) = c(n) + c(n-1) where c (A055562) is the sequence of numbers not in a.

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A022424 Solution a( ) of the complementary equation a(n) = b(n-1) + b(n-2), where a(0) = 1, a(1) = 2; see Comments.

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A210770 a(1) = 1, a(2) = 2; for n > 1, a(2*n+2) = smallest number not yet seen, a(2*n+1) = a(2*n) + a(2*n+2).

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A055563 a(n) = least number greater than a(n-1) not the sum of an earlier pair of consecutive terms, a(0) = 3.

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A210770 a(1) = 1, a(2) = 2; for n > 1, a(2n+2) = smallest number not yet seen, a(2n+1) = a(2n) + a(2n+2).