A003160 -id:A003160 - cp's OEIS frontend

A095774 a(n)=2*A003160(n)-n.

1, 0, -1, 0, 1, 2, 1, 0, 1, 0, -1, 0, -1, -2, -1, 0, 1, 0, -1, 0, 1, 2, 1, 0, 1, 2, 3, 2, 1, 2, 1, 0, 1, 0, -1, 0, 1, 2, 1, 0, 1, 0, -1, 0, -1, -2, -1, 0, 1, 0, -1, 0, -1, -2, -1, -2, -3, -2, -1, 0, -1, -2, -1, 0, 1, 0, -1, 0, 1, 2, 1, 0, 1, 0, -1, 0, -1, -2, -1, 0, 1, 0, -1, 0, 1, 2, 1, 0, 1, 2, 3, 2, 1, 2, 1, 0, 1, 0, -1, 0, 1, 2, 1, 0, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2

Offset: 1

Benoit Cloitre, Jun 05 2004

sign

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

a095774 n = 2 * a003160 n - n  -- Reinhard Zumkeller, Aug 02 2013

m=1106;v=vector(m,j,1);for(n=3,m,g=n-v[v[n-1]]-v[v[n-2]];v[n]=g);a(n)=2*v[n]-n

Offset corrected by Reinhard Zumkeller, Aug 02 2013

A095775 n/2 when 2*A003160(n) = n.

Original entry on oeis.org

1, 2, 4, 5, 6, 8, 9, 10, 12, 16, 17, 18, 20, 21, 22, 24, 25, 26, 30, 32, 33, 34, 36, 37, 38, 40, 41, 42, 44, 48, 49, 50, 52, 60, 64, 65, 66, 68, 69, 70, 72, 73, 74, 76, 80, 81, 82, 84, 85, 86, 88, 89, 90, 94, 96, 97, 98, 100, 101, 102, 104, 105, 106, 110, 118, 120, 121, 122

Offset: 1

Benoit Cloitre, Jun 05 2004

nonn

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Cf. A095774.

a095775 n = a095775_list !! (n-1)
a095775_list = map (`div` 2) $ filter ((== 0) . a095774) [1..]
-- Reinhard Zumkeller, Aug 02 2013

A080426 a(1)=1, a(2)=3; all terms are either 1 or 3; each run of 3's is followed by a run of two 1's; and a(n) is the length of the n-th run of 3's.

Original entry on oeis.org

1, 3, 1, 1, 3, 3, 3, 1, 1, 3, 1, 1, 3, 1, 1, 3, 3, 3, 1, 1, 3, 3, 3, 1, 1, 3, 3, 3, 1, 1, 3, 1, 1, 3, 1, 1, 3, 3, 3, 1, 1, 3, 1, 1, 3, 1, 1, 3, 3, 3, 1, 1, 3, 1, 1, 3, 1, 1, 3, 3, 3, 1, 1, 3, 3, 3, 1, 1, 3, 3, 3, 1, 1, 3, 1, 1, 3, 1, 1, 3, 3, 3, 1, 1, 3, 3, 3, 1, 1, 3, 3, 3, 1, 1, 3, 1, 1, 3, 1, 1, 3, 3, 3, 1, 1

Offset: 1

John W. Layman, Feb 18 2003

nonn

It appears that the sequence can be calculated by any of the following three methods: (1) Start with 1 and repeatedly replace (simultaneously) all 1's with 1,3,1 and all 3's with 1,3,3,3,1. [Equivalently, trajectory of 1 under the morphism 1 -> 1,3,1; 3 -> 1,3,3,3,1. - N. J. A. Sloane, Nov 03 2019] (2) a(n)= A026490(2n). (3) Replace each 2 in A026465 (run lengths in Thue-Morse) with 3.
Length of n-th run of 1's in the Feigenbaum sequence A035263 = 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, .... - Philippe Deléham, Apr 18 2004
Another construction. Let S_0 = 1, and let S_n be obtained by applying the morphism 1 -> 3, 3 -> 113 to S_{n-1}. The sequence is the concatenation S_0, S_1, S_2, ... - D. R. Hofstadter, Oct 23 2014
a(n+1) is the number of times n appears in A003160. - John Keith, Dec 31 2020

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Jean-Paul Allouche, On the morphism 1 -> 121, 2 -> 12221, CNRS France, 2024. See pp. 5-6.
Jean-Paul Allouche, On the morphism 1 -> 121, 2 -> 12221, Preprint, 2024 [Local copy, with permission]
D. R. Hofstadter, Anti-Fibonacci numbers, Oct 23 2014.

Cf. A026465, A026490, A035263, A003156, A328979, A003160.

Arises in the analysis of A075326, A249031 and A249032.

-- following Deléham
import Data.List (group)
a080426 n = a080426_list !! n
a080426_list = map length $ filter ((== 1) . head) $ group a035263_list
-- Reinhard Zumkeller, Oct 27 2014

Position[ Nest[ Flatten[# /. {0 -> {0, 2, 1}, 1 -> {0}, 2 -> {0}}]&, {0}, 8], 0] // Flatten // Differences // Prepend[#, 1]& (* Jean-François Alcover, Mar 14 2014, after Philippe Deléham *)
nsteps=7;Flatten[SubstitutionSystem[{1->{3},3->{1,1,3}},{1},nsteps]] (* Paolo Xausa, Aug 12 2022, using D. R. Hofstadter's construction *)

A080426(nmax) = my(a=[1], s=[[1, 3, 1], [], [1, 3, 3, 3, 1]]); while(length(a)A080426(100) \\ Paolo Xausa, Sep 14 2022, using method (1) from comments

def A080426(nmax):
    a, s = "1", "".maketrans({"1":"131", "3":"13331"})
    while len(a) < nmax: a = a.translate(s)
    return list(map(int, a[:nmax]))
print(A080426(100)) # Paolo Xausa, Aug 30 2022, using method (1) from comments

def A080426(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x):
        c, s = x, bin(x)[2:]
        l = len(s)
        for i in range(l&1,l,2):
            c -= int(s[i])+int('0'+s[:i],2)
        return c
    return bisection(lambda x:f(x)+n,n,n)-bisection(lambda x:f(x)+n-1,n-1,n-1)-1 # Chai Wah Wu, Jan 29 2025

a(1) = 1; for n>1, a(n) = A003156(n) - A003156(n-1). - Philippe Deléham, Apr 16 2004

A287422 a(1) = a(2) = 1; a(n) = n - a(a(n-1)) - a(n-a(n-1)) for n > 2.

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 4, 4, 4, 4, 5, 5, 6, 6, 7, 8, 9, 10, 11, 11, 12, 13, 13, 13, 14, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 19, 19, 19, 19, 20, 21, 21, 22, 22, 22, 23, 24, 25, 25, 26, 26, 27, 28, 29, 29, 30, 31, 32, 33, 34, 35, 36, 36, 37, 38, 39, 40, 40, 41, 42, 42, 42, 43, 44, 45, 46, 46

Offset: 1

Altug Alkan, Jun 24 2017

nonn

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Altug Alkan, Plot of a(n)-n/2 for n<=2^20
Altug Alkan, Plot of (a(n-a(n-1))/n, a(a(n-1))/n) for n <= 250000
Rémy Sigrist, Colored scatterplot of (a(n-a(n-1))/n, a(a(n-1))/n) for n <= 100000000 (where the hue is function of n)

Cf. A003160, A004001, A005206, A088462, A189663.

I:=[1,1,1,2]; [n le 4 select I[n] else n - Self(Self(n-1)) - Self(n-Self(n-1)): n in [1..100]]; // Vincenzo Librandi, Aug 08 2018

Nest[Function[{a, n}, Append[a, n - a[[a[[-1]] ]] - a[[-a[[-1]] ]] ] ] @@ {#, Length@ # + 1} &, {1, 1}, 80] (* Michael De Vlieger, Aug 07 2018 *)

q=vector(10000); q[1]=q[2]=1; for(n=3, #q, q[n]=n-q[q[n-1]]-q[n-q[n-1]]); q

A308818 a(n) = a(a(n-1) mod n) + a(a(n-2) mod n) with a(0)=2 and a(1)=3.

Original entry on oeis.org

2, 3, 5, 7, 10, 7, 13, 15, 22, 23, 12, 6, 15, 18, 13, 25, 41, 37, 10, 22, 17, 40, 47, 40, 81, 38, 22, 53, 85, 134, 51, 29, 156, 215, 23, 47, 46, 35, 69, 98, 144, 81, 108, 116, 102, 37, 47, 37, 72, 75, 85, 104, 217, 111, 10, 15, 37, 60, 40, 147, 197, 51, 110

Offset: 0

Arran Ireland, Jun 26 2019

nonn

a(0) and a(1) are chosen to be the smallest starting numbers greater than 1 that are believed to result in a sequence that doesn't cycle.
Empirical observation of the first 10^8 terms suggests that the sequence doesn't enter a cycle.
Conjectures: (i) This sequence doesn't enter a cycle. (ii) There is an integer greater than 1 that can never appear in this sequence.

			a(2) = a(a(2-1) mod 2) + a(a(2-2) mod 2) = a(a(1) mod 2) + a(a(0) mod 2) = a(3 mod 2) + a(2 mod 2) = a(1) + a(0) = 3 + 2 = 5.

Cf. A005185, A046698, A003160.

Cf. A000027 (if a(0)=1 and a(1)=2).

a = [2, 3]
for n in range(2, 10**4 + 3):
    a.append(a[(a[n - 1] % n)] + a[(a[n - 2] % n)])
    print((n - 2), ",", a[n - 2], sep="")

cp's OEIS Frontend

A095774 a(n)=2*A003160(n)-n.

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A095775 n/2 when 2*A003160(n) = n.

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A080426 a(1)=1, a(2)=3; all terms are either 1 or 3; each run of 3's is followed by a run of two 1's; and a(n) is the length of the n-th run of 3's.

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A287422 a(1) = a(2) = 1; a(n) = n - a(a(n-1)) - a(n-a(n-1)) for n > 2.

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A308818 a(n) = a(a(n-1) mod n) + a(a(n-2) mod n) with a(0)=2 and a(1)=3.

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A078474 a(1)=a(2)=a(3)=1, a(n)=n-a(a(n-1))-a(a(n-2))-a(a(n-3)).

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