A182660 -id:A182660 - cp's OEIS frontend

A135416 a(n) = A036987(n)*(n+1)/2.

1, 0, 2, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 16, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 32, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

N. J. A. Sloane, based on a message from Guy Steele and Don Knuth, Mar 01 2008

nonn

Guy Steele defines a family of 36 integer sequences, denoted here by GS(i,j) for 1 <= i, j <= 6, as follows. a[1]=1; a[2n] = i-th term of {0,1,a[n],a[n]+1,2a[n],2a[n]+1}; a[2n+1] = j-th term of {0,1,a[n],a[n]+1,2a[n],2a[n]+1}. The present sequence is GS(1,5).
The full list of 36 sequences:
GS(1,1) = A000007
GS(1,2) = A000035
GS(1,3) = A036987
GS(1,4) = A007814
GS(1,5) = A135416 (the present sequence)
GS(1,6) = A135481
GS(2,1) = A135528
GS(2,2) = A000012
GS(2,3) = A000012
GS(2,4) = A091090
GS(2,5) = A135517
GS(2,6) = A135521
GS(3,1) = A036987
GS(3,2) = A000012
GS(3,3) = A000012
GS(3,4) = A000120
GS(3,5) = A048896
GS(3,6) = A038573
GS(4,1) = A135523
GS(4,2) = A001511
GS(4,3) = A008687
GS(4,4) = A070939
GS(4,5) = A135529
GS(4,6) = A135533
GS(5,1) = A048298
GS(5,2) = A006519
GS(5,3) = A080100
GS(5,4) = A087808
GS(5,5) = A053644
GS(5,6) = A000027
GS(6,1) = A135534
GS(6,2) = A038712
GS(6,3) = A135540
GS(6,4) = A135542
GS(6,5) = A054429
GS(6,6) = A003817
(with a(0)=1): Moebius transform of A038712.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences related to binary expansion of n

Equals A048298(n+1)/2. Cf. A036987, A182660.

GS:=proc(i,j,M) local a,n; a:=array(1..2*M+1); a[1]:=1;
for n from 1 to M do
a[2*n] :=[0,1,a[n],a[n]+1,2*a[n],2*a[n]+1][i];
a[2*n+1]:=[0,1,a[n],a[n]+1,2*a[n],2*a[n]+1][j];
od: a:=convert(a,list); RETURN(a); end;
GS(1,5,200):

i = 1; j = 5; Clear[a]; a[1] = 1; a[n_?EvenQ] := a[n] = {0, 1, a[n/2], a[n/2]+1, 2*a[n/2], 2*a[n/2]+1}[[i]]; a[n_?OddQ] := a[n] = {0, 1, a[(n-1)/2], a[(n-1)/2]+1, 2*a[(n-1)/2], 2*a[(n-1)/2]+1}[[j]]; Array[a, 105] (* Jean-François Alcover, Sep 12 2013 *)

A048298(n) = if(!n,0,if(!bitand(n,n-1),n,0));
A135416(n) = (A048298(n+1)/2); \\ Antti Karttunen, Jul 22 2018

def A135416(n): return int(not(n&(n+1)))*(n+1)>>1 # Chai Wah Wu, Jul 06 2022

G.f.: sum{k>=1, 2^(k-1)*x^(2^k-1) }.

Recurrence: a(2n+1) = 2a(n), a(2n) = 0, starting a(1) = 1.

Formulae and comments by Ralf Stephan, Jun 20 2014

A082691 a(1)=1, a(2)=2, then if the first 32^k-1 terms are a(1), a(2), ..., a(32^k - 1), the first 32^(k+1)-1 terms are a(1), a(2), ..., a(32^k - 1), a(1), a(2), ..., a(32^k - 1), a(32^k-1) + 1.

Original entry on oeis.org

1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 6, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 6, 7, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3

Offset: 1

Benoit Cloitre, Apr 12 2003

nonn

Consider the subsequence b(k) such that a(b(k))=1. Then 3k - b(k) = A063787(k+1) and b(k) = 1 + A004134(k-1).
From Sam Alexander, Nov 27 2010: (Start)
A naive way to try and guess whether a sequence is periodic, based on its first k terms (n1, ..., nk), is to look at all sequences which have period less than k, and guess "periodic" if any of them extend (n1, ..., nk), "nonperiodic" otherwise.
a(1)=1, a(2)=2. Suppose a(1), ..., a(n) have been defined, n > 1.
1. If the above guessing method guesses that (a(1), ..., a(n)) is an initial segment of a periodic sequence, then let a(n+1) be the least nonzero number not appearing in (a(1), ..., a(n)).
2. Otherwise, let (a(n+1), ..., a(2n)) be a copy of (a(1), ..., a(n)).
This sequence thwarts the guessing attempt, tricking the guesser into changing his mind infinitely many times as n->infinity. (End)
As n increases, the average value of the first n terms approaches 7/3 = 2.333... - Maxim Skorohodov, Dec 15 2022

			To construct the sequence: start with (1, 2); concatenating those 2 terms gives (1,2,1,2). Appending 3 gives the first 5 terms: (1,2,1,2,3). Concatenating those 5 terms gives (1,2,1,2,3,1,2,1,2,3). Appending 4 gives the first 11 terms: (1,2,1,2,3,1,2,1,2,3,4), etc.

Maxim Skorohodov, Table of n, a(n) for n = 1..10000
Samuel Alexander, On Guessing Whether A Sequence Has A Certain Property, arxiv:1011.6626 [math.LO], 2010-2012; J. Int. Seq. 14 (2011) # 11.4.4

Cf. A082692 (partial sums), A182659, A182660 (other sequences engineered to spite naive guessers).

A182659 A canonical permutation designed to thwart a certain naive attempt to guess whether sequences are permutations.

Original entry on oeis.org

0, 2, 3, 1, 5, 6, 7, 8, 9, 4, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 10, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 22, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68

Offset: 0

Sam Alexander, Nov 26 2010

A naive way to guess whether a function f:N->N is a permutation, based on just an initial subsequence (f(0),...,f(n)), is to guess "no" if (f(0),...,f(n)) contains a repeated entry or if there is some i in {0,...,n} such that i is not in {f(0),...,f(n)} and 2 i<=n; and guess "yes" otherwise. a(n) thwarts that method, causing it to change its mind infinitely often as n->infinity.
a(0)=0. Suppose a(0),...,a(n) have been defined.
1. If the above method guesses that (a(0),...,a(n)) is NOT an initial subsequence of a permutation, then unmark any "marked" numbers.
2. If the above method guesses that (a(0),...,a(n)) IS an initial subsequence of a permutation, then "mark" the smallest number not in {a(0),...,a(n)}.
3. Let a(n+1) be the least unmarked number not in {a(0),...,a(n)}.
A030301 can be derived by a similar method, where instead of trying to guess whether sequences are permutations, the naive victim is trying to guess whether sequences contain infinitely many 0s.

S. Alexander, On Guessing Whether A Sequence Has A Certain Property, arXiv:1011.6626 [math.LO], 2010-2012, J. Int. Seq. 14 (2011) # 11.4.4.

Cf. A030301, A082691, A182660, A068156, A033484.

a(0) = 0; if n = A068156(k+1) = 6*2^k - 3 for some k >= 0 then a(n) = A033484(k) = (n-1)/2; otherwise, a(n) = n+1. - Andrey Zabolotskiy, Feb 27 2025

a(22) corrected and further terms added by Andrey Zabolotskiy, Feb 27 2025

cp's OEIS Frontend

A135416 a(n) = A036987(n)*(n+1)/2.

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A082691 a(1)=1, a(2)=2, then if the first 32^k-1 terms are a(1), a(2), ..., a(32^k - 1), the first 32^(k+1)-1 terms are a(1), a(2), ..., a(32^k - 1), a(1), a(2), ..., a(32^k - 1), a(32^k-1) + 1.

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A182659 A canonical permutation designed to thwart a certain naive attempt to guess whether sequences are permutations.

Views

Author

Keywords

Comments

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Crossrefs

Formula

Extensions

cp's OEIS Frontend

A135416 a(n) = A036987(n)*(n+1)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

Extensions

A082691 a(1)=1, a(2)=2, then if the first 3*2^k-1 terms are a(1), a(2), ..., a(3*2^k - 1), the first 3*2^(k+1)-1 terms are a(1), a(2), ..., a(3*2^k - 1), a(1), a(2), ..., a(3*2^k - 1), a(3*2^k-1) + 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A182659 A canonical permutation designed to thwart a certain naive attempt to guess whether sequences are permutations.

Views

Author

Keywords

Comments

Links

Crossrefs

Formula

Extensions

A082691 a(1)=1, a(2)=2, then if the first 32^k-1 terms are a(1), a(2), ..., a(32^k - 1), the first 32^(k+1)-1 terms are a(1), a(2), ..., a(32^k - 1), a(1), a(2), ..., a(32^k - 1), a(32^k-1) + 1.