A136412 -id:A136412 - cp's OEIS frontend

A162910 Denominators of Bird tree fractions.

1, 2, 1, 3, 3, 1, 2, 5, 4, 4, 5, 2, 1, 3, 3, 8, 7, 5, 7, 7, 5, 7, 8, 3, 3, 1, 2, 5, 4, 4, 5, 13, 11, 9, 12, 9, 6, 10, 11, 11, 10, 6, 9, 12, 9, 11, 13, 5, 4, 4, 5, 2, 1, 3, 3, 8, 7, 5, 7, 7, 5, 7, 8, 21, 18, 14, 19, 16, 11, 17, 19, 14, 13, 7, 11, 17, 13, 15, 18, 18, 15, 13, 17, 11, 7, 13, 14, 19

Offset: 1

Ralf Hinze (ralf.hinze(AT)comlab.ox.ac.uk), Aug 05 2009

The Bird tree is an infinite binary tree labeled with rational numbers. The root is labeled with 1. The tree enjoys the following fractal property: it can be transformed into its left subtree by first incrementing and then reciprocalizing the elements; for the right subtree interchange the order of the two steps: the elements are first reciprocalized and then incremented. Like the Stern-Brocot tree, the Bird tree enumerates all the positive rationals (A162909(n)/A162910(n)).
From Yosu Yurramendi, Jul 11 2014: (Start)
If the terms (n>0) are written as an array (left-aligned fashion) with rows of length 2^m, m = 0,1,2,3,...
   1,
   2, 1,
   3, 3,1, 2,
   5, 4,4, 5,2,1, 3, 3,
   8, 7,5, 7,7,5, 7, 8, 3, 3,1,2, 5,4, 4, 5,
   13,11,9,12,9,6,10,11,11,10,6,9,12,9,11,13,5,4,4,5,2,1,3,3,8,7,5,7,7,5,7,8,
then the sum of the m-th row is 3^m (m = 0,1,2,), each column k is a Fibonacci sequence.
If the rows are written in a right-aligned fashion:
                                                                        1,
                                                                      2,1,
                                                                  3,3,1,2,
                                                          5,4,4,5,2,1,3,3,
                                          8,7,5,7,7,5,7,8,3,3,1,2,5,4,4,5,
13,11,9,12,9,6,10,11,11,10,6,9,12,9,11,13,5,4,4,5,2,1,3,3,8,7,5,7,7,5,7,8,
then each column k also is a Fibonacci sequence.
The Fibonacci sequences of both triangles are equal except the first terms of second triangle.
If the sequence is considered by blocks of length 2^m, m = 0,1,2,..., the blocks of this sequence are the reverses of blocks of A162909 ( a(2^m+k) = A162909(2^(m+1)-1-k), m = 0,1,2,..., k = 0,1,2,...,2^m-1).
(End)

			The first four levels of the Bird tree: [1/1] [1/2, 2/1] [2/3, 1/3, 3/1, 3/2], [3/5, 3/4, 1/4, 2/5, 5/2, 4/1, 4/3, 5/3].

Ralf Hinze, Functional pearls: the bird tree, J. Funct. Programming 19 (2009), no. 5, 491-508.
Index entries for fraction trees

This sequence is the composition of A162912 and A059893: a(n) = A162912(A059893(n)). This sequence is a permutation of A002487(n+2).

import Ratio; bird :: [Rational]; bird = branch (recip . succ) (succ . recip) 1; branch f g a = a : branch f g (f a) \/ branch f g (g a); (a : as) \/ bs = a : (bs \/ as); a162909 = map numerator bird; a162910 = map denominator bird

blocklevel <- 6 # arbitrary
a <- 1
for(m in 1:blocklevel) for(k in 0:(2^(m-1)-1)){
a[2^m+k]         = a[2^m-k-1] + a[2^(m-1)+k]
a[2^m+2^(m-1)+k] = a[2^m-k-1]
}
a
# Yosu Yurramendi, Jul 11 2014

a(2^m+k) = a(2^m-k-1) + a(2^(m-1)+k), a(2^m+2^(m-1)+k) = a(2^m-k-1), a(1) = 1, m=0,1,2,3,..., k=0,1,...,2^(m-1)-1. - Yosu Yurramendi, Jul 11 2014
If k is odd a(A080675(n)*2^m+k) = A268087(2^m+k), if k is even a(A136412(2^m+k+1)*2^m+k) = A268087(2^m+k), m >= 0, 0 <= k < 2^m, n > 0. a(A081254(n)) = 1, n > 0. - Yosu Yurramendi, Feb 21 2017
a(n) = A002487(1+A258996(A059893(n))) = A002487(1+A059893(A258746(n))), n > 0. - Yosu Yurramendi, Jul 14 2021

A007302 Optimal cost function between two processors at distance n.

Original entry on oeis.org

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

Offset: 0

Simon Plouffe

Also the number of nonzero digits in the symmetric signed digit expansion of n with q=2 (i.e., the representation of n in the (-1,0,1)2 number system). - _Ralf Stephan, Jun 30 2003
Volger (1985) proves that a(n) <= ceiling(log_2(3n/2) / 2) and uses a(n) to derive an upper bound on the length of the minimum addition-subtraction chain for n. - Steven G. Johnson (stevenj(AT)math.mit.edu), May 01 2007
Starting from 0, the smallest number of steps to reach n, where each step involves moving a power of 2 in either direction. - Dmitry Kamenetsky, Jul 04 2023

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Jean-Paul Allouche and Jeffrey Shallit, The Ring of k-regular Sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
Alma D'Aniello, Aldo de Luca, and Alessandro De Luca, On Christoffel and standard words and their derivatives, arXiv:1602.03231 [cs.DM], 2016 (mentions this sequence).
Clemens Heuberger and Helmut Prodinger, On minimal expansions in redundant number systems: Algorithms and quantitative analysis, Computing 66(2001), 377-393.
Sara Kropf and Stephan Wagner, q-Quasiadditive functions, arXiv:1605.03654 [math.CO], 2016. See section "The nonadjacent form and its Hamming weight".
Gurmeet Singh Manku and Joe Sawada, A loopless Gray code for minimal signed-binary representations, 13th Annual European Symposium on Algorithms (ESA), LNCS 3669 (2005), 438-447.
Nathan Mihm, Optimal Addition-Subtraction Chains, Bachelor Honors Thesis, Univ. Minnesota-Twin Cities (2025). See p. 4.
Aline Parreau, Michel Rigo, Eric Rowland, and Elise Vandomme, A new approach to the 2-regularity of the l-abelian complexity of 2-automatic sequences, arXiv preprint arXiv:1405.3532 [cs.FL], 2014.
Yaniv Sadeh, Ori Rottenstreich, and Haim Kaplan, Codes for Load Balancing in TCAMs: Size Analysis, arXiv:2212.13256 [cs.NI], 2022.
Joe Sawada, A simple Gray code to list all minimal signed binary representations, SIAM Journal on Discrete Mathematics, Vol. 21 No. 1 (2007), 16-25.
Hugo Volger, Some results on addition/subtraction chains, Information Processing Letters, vol. 20, p. 155-160 (1985).
A. Weitzman, Transformation of parallel programs guided by micro-analysis, pp. 155-159 of Algorithms Seminars 1992-1993, ed. B. Salvy, Report #2130, INRIA, Rocquencourt, Dec. 1993.
Index to sequences related to the complexity of n

Cf. A005578, A057526.
Subtracting 1 gives A280737.
Cf. A007583 (indices of record highs).

import Data.Bits (xor)
a007302 n = a000120 $ xor n (3 * n) :: Integer
-- Reinhard Zumkeller, Jun 17 2012

a[n_] := Count[ BitXor[ b1 = IntegerDigits[n, 2]; b3 = IntegerDigits[3*n, 2]; PadLeft[b1, Length[b3]], b3], 1]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, May 20 2014, after Ramasamy Chandramouli *)

ep(r,n)=local(t=n/2^(r+2));floor(t+5/6)-floor(t+4/6)-floor(t+2/6)+floor(t+1/6)
a(n)=sum(r=0,log(3*n)\log(2)-1,!!ep(r,n))
for(n=1,100,print1(a(n)", "))
/* corrected by Charles R Greathouse IV, Jun 16 2012 */

a(n)=hammingweight(bitxor(n,3*n)) \\ Charles R Greathouse IV, Jan 03 2017

a(0) = 0; a(n) = 1 if n is a power of 2; a(n) = 1 + min { a(n-2^k), a(2^(k+1)-n) } if 2^k < n < 2^(k+1).
a(n) = 0 if n = 0, = 1 if n = 1, = a(n/2) if n > 1 and n even and = min(a(n-1), a(n+1))+1 if n > 1 and n odd. - David W. Wilson, Dec 28 2005
a(n) = hammingweight( XOR(n, 3*n) ). - Ramasamy Chandramouli, Aug 20 2010
A007302(n) = A000120(n) - sum (A213629(n,A136412(k))). - Reinhard Zumkeller, Jun 17 2012
a(0) = 0; a(2n) = a(n); a(4n-1) = a(n) + 1; a(4n+1) = a(n) + 1. - Nathan Fox, Mar 12 2013

A213629 In binary representation: T(n,k) = number of (possibly overlapping) occurrences of k in n, triangle read by rows, 1<=k<=n.

Original entry on oeis.org

1, 1, 1, 2, 0, 1, 1, 1, 0, 1, 2, 1, 0, 0, 1, 2, 1, 1, 0, 0, 1, 3, 0, 2, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 2, 1, 0, 1, 0, 0, 0, 0, 1, 2, 2, 0, 0, 1, 0, 0, 0, 0, 1, 3, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 2, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 3, 1, 1, 0, 1, 1, 0, 0

Offset: 1

Reinhard Zumkeller, Jun 17 2012

The definition is based on the definition of pattern functions in the paper of Allouche and Shallit;
sum of n-th row = A029931(n);
T(n,1) = A000120(n);
T(n,2) = A033264(n) for n > 1;
T(n,3) = A014081(n) for n > 2;
T(n,4) = A056978(n) for n > 3;
T(n,5) = A056979(n) for n > 4;
T(n,6) = A056980(n) for n > 5;
T(n,7) = A014082(n) for n > 6;
T(n,k) = 0 for k with floor(n/2) < k < n;
T(n,n) = 1;
A122953(n) = Sum_{k=1..n} A057427(T(n,k));
A005811(n) = T(n,1) + T(n,2) - T(n,3);
A007302(n) = A000120(n) - sum (A213629(n,A136412(k))).

			The triangle begins:
.   1:                        1
.   2:                      1   1
.   3:                    2   0   1
.   4:                  1   1   0   1
.   5:                2   1   0   0   1
.   6:              2   1   1   0   0   1
.   7:            3   0   2   0   0   0   1
.   8:          1   1   0   1   0   0   0   1
.   9:        2   1   0   1   0   0   0   0   1
.  10:      2   2   0   0   1   0   0   0   0   1
.  11:    3   1   1   0   1   0   0   0   0   0   1
.  12:  2   1   1   1   0   1   0   0   0   0   0   1.

Reinhard Zumkeller, Rows n = 1..150 of triangle, flattened
J.-P. Allouche, J. Shallit, The Ring of k-regular Sequences II, Example 4, p. 12
Index entries for sequences related to binary expansion of n

Cf. A030308, A007088.

import Data.List (inits, tails, isPrefixOf)
a213629 n k = a213629_tabl !! (n-1) !! (k-1)
a213629_row n = a213629_tabl !! (n-1)
a213629_tabl = map f $ tail $ inits $ tail $ map reverse a030308_tabf where
   f xss = map (\xs ->
           sum $ map (fromEnum . (xs `isPrefixOf`)) $ tails $ last xss) xss

t[n_, k_] := (idn = IntegerDigits[n, 2]; idk = IntegerDigits[k, 2]; ln = Length[idn]; lk = Length[idk]; For[cnt = 0; i = 1, i <= ln - lk + 1, i++, If[idn[[i ;; i + lk - 1]] == idk, cnt++]]; cnt); Table[t[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, Oct 22 2012 *)

A191669 Dispersion of A004767 (4k+3, k>=0), by antidiagonals.

Original entry on oeis.org

1, 3, 2, 11, 7, 4, 43, 27, 15, 5, 171, 107, 59, 19, 6, 683, 427, 235, 75, 23, 8, 2731, 1707, 939, 299, 91, 31, 9, 10923, 6827, 3755, 1195, 363, 123, 35, 10, 43691, 27307, 15019, 4779, 1451, 491, 139, 39, 12, 174763, 109227, 60075, 19115, 5803, 1963, 555, 155

Offset: 1

Clark Kimberling, Jun 11 2011

For a background discussion of dispersions, see A191426.
...
Each of the sequences (4n, n>2), (4n+1, n>0), (3n+2, n>=0), generates a dispersion. Each complement (beginning with its first term >1) also generates a dispersion. The six sequences and dispersions are listed here:
...
A191452=dispersion of A008586 (4k, k>=1)
A191667=dispersion of A016813 (4k+1, k>=1)
A191668=dispersion of A016825 (4k+2, k>=0)
A191669=dispersion of A004767 (4k+3, k>=0)
A191670=dispersion of A042968 (1 or 2 or 3 mod 4 and >=2)
A191671=dispersion of A004772 (0 or 1 or 3 mod 4 and >=2)
A191672=dispersion of A004773 (0 or 1 or 2 mod 4 and >=2)
A191673=dispersion of A004773 (0 or 2 or 3 mod 4 and >=2)
...
EXCEPT for at most 2 initial terms (so that column 1 always starts with 1):
A191452 has 1st col A042968, all else A008486
A191667 has 1st col A004772, all else A016813
A191668 has 1st col A042965, all else A016825
A191669 has 1st col A004773, all else A004767
A191670 has 1st col A008486, all else A042968
A191671 has 1st col A016813, all else A004772
A191672 has 1st col A016825, all else A042965
A191673 has 1st col A004767, all else A004773
...
Regarding the dispersions A191670-A191673, there is a formula for sequences of the type "(a or b or c mod m)", (as in the Mathematica program below):
   If f(n)=(n mod 3), then (a,b,c,a,b,c,a,b,c,...) is given by
   a*f(n+2)+b*f(n+1)+c*f(n), so that "(a or b or c mod m)" is given by
   a*f(n+2)+b*f(n+1)+c*f(n)+m*floor((n-1)/3)), for n>=1.

			Northwest corner:
1...3....11....43....171
2...7....27....107...427
4...15...59....235...939
5...19...75....299...1195
6...23...91....363...1451

Ivan Neretin, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals, flattened)

Row 1: A007583, Row 2: A136412.

Cf. A004773, A004767, A191673, A191426.

(* Program generates the dispersion array T of the increasing sequence f[n] *)
r = 40; r1 = 12; c = 40; c1 = 12;
f[n_] := 4*n-1
Table[f[n], {n, 1, 30}] (* A004767 *)
mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]
rows = {NestList[f, 1, c]};
Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}];
t[i_, j_] := rows[[i, j]];
TableForm[Table[t[i, j], {i, 1, 10}, {j, 1, 10}]] (* A191669 *)
Flatten[Table[t[k, n - k + 1], {n, 1, c1}, {k, 1, n}]] (* A191669 *)

A199210 a(n) = (11*4^n + 1)/3.

Original entry on oeis.org

4, 15, 59, 235, 939, 3755, 15019, 60075, 240299, 961195, 3844779, 15379115, 61516459, 246065835, 984263339, 3937053355, 15748213419, 62992853675, 251971414699, 1007885658795, 4031542635179, 16126170540715, 64504682162859

Offset: 0

Vincenzo Librandi, Nov 04 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Amya Luo, Pattern Avoidance in Nonnesting Permutations, Undergraduate Thesis, Dartmouth College (2024). See p. 16.
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Sequences of the form (m*4^n + 1)/3: A007583 (m=2), A136412 (m=5), this sequence (m=11), A199210 (m=11), A206373 (m=14).

Magma
```
[(11*4^n+1)/3: n in [0..30]];
```

LinearRecurrence[{5,-4}, {4,15}, 31] (* G. C. Greubel, Jan 19 2023 *)

[(11*4^n+1)/3 for n in range(31)] # G. C. Greubel, Jan 19 2023

a(n) = 4*a(n-1) - 1.
a(n) = 5*a(n-1) - 4*a(n-2).
G.f.: (4-5*x)/((1-x)*(1-4*x)). - Bruno Berselli, Nov 04 2011
E.g.f.: (1/3)*(11*exp(4*x) + exp(x)). - G. C. Greubel, Jan 19 2023

A255138 a(n) = (1 + 2^n(3 + 2(-1)^n))/3.

Original entry on oeis.org

2, 1, 7, 3, 27, 11, 107, 43, 427, 171, 1707, 683, 6827, 2731, 27307, 10923, 109227, 43691, 436907, 174763, 1747627, 699051, 6990507, 2796203, 27962027, 11184811, 111848107, 44739243, 447392427, 178956971

Offset: 0

L. Edson Jeffery, May 04 2015

Let N_1 be the set of odd natural numbers and v(y) the 2-adic valuation of y. Define the map F : N_1 -> N_1 by F(x) = (3*x+1)/2^v(3*x+1) (see A075677). Let F^(k)(x) denote k-fold iteration of F, with recurrence F^(k)(x) = F(F^(k-1)(x)), k > 0, and initial condition F^(0)(x) = x. Then, for n>0, a(n) is the least m such that F^(n)(4*m-3) == 1 (mod 4). Cf. A257499.

Let k == 1 mod 4, and k(r) be the r-th iteration at which k appears in a Collatz sequence. When n >= 2 and k(r) == [2^(n+1) - a(n)] mod 2^(n+1), then n is the number of halving steps following k(r+1). For instance, since a(5) = 11, there are 5 halving steps following k(r+1) when k(r) == 53 mod 64, because 2^(5+1) = 64 and 64-11 = 53; e.g., k(r) = 117: 117 -> 352 -> 176 -> 88 -> 44 -> 22 -> 11. - Bob Selcoe, Feb 09 2017

Index entries for linear recurrences with constant coefficients, signature (1,4,-4).

Cf. A007583, A075677, A096773, A136412, A257499.

[(1 + 2^n*(3 + 2*(-1)^n))/3: n in [0..50]]; // Wesley Ivan Hurt, Nov 05 2015

A255138:=n->(1 + 2^n*(3 + 2*(-1)^n))/3: seq(A255138(n), n=0..50); # Wesley Ivan Hurt, Nov 05 2015

a[n_] := (1 + 2^n*(3 + 2*(-1)^n))/3; Table[a[n], {n, 0, 29}]
LinearRecurrence[{1,4,-4},{2,1,7},30] (* Harvey P. Dale, Aug 03 2024 *)

vector(30, n, n--; (1 + 2^n*(3 + 2*(-1)^n))/3) \\ Altug Alkan, Nov 05 2015

a(2*n) = A136412(n); a(2*n+1) = A007583(n).
G.f.: (2-x-2*x^2)/((x-1)*(2*x-1)*(2*x+1)). - R. J. Mathar, Jul 25 2015
a(n) = a(n-1) + 4*a(n-2) - 4*a(n-3) for n > 2. - Wesley Ivan Hurt, Nov 05 2015
a(n) = 4*a(n-2) - 1. - Bob Selcoe, Feb 09 2017
a(n) = 2^(n+1) - A096773(n+1). - Ruud H.G. van Tol, Sep 04 2023

A199115 a(n) = 5*4^n+1.

Original entry on oeis.org

6, 21, 81, 321, 1281, 5121, 20481, 81921, 327681, 1310721, 5242881, 20971521, 83886081, 335544321, 1342177281, 5368709121, 21474836481, 85899345921, 343597383681, 1374389534721, 5497558138881, 21990232555521, 87960930222081

Offset: 0

Vincenzo Librandi, Nov 04 2011

An Engel expansion of 4/5 to the base 4 as defined in A181565, with the associated series expansion 4/5 = 4/6 + 4^2/(6*21) + 4^3/(6*21*81) + 4^4/(6*21*81*321) + ... . Cf. A136412 and A140660. - Peter Bala, Oct 29 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A136412, A140660, A199115.

Magma
```
[5*4^n+1: n in [0..30]];
```

5*4^Range[0,30]+1 (* or  *) LinearRecurrence[{5,-4},{6,21},30] (* Harvey P. Dale, Oct 19 2024 *)

a(n) = 3*A136412(n).
a(n) = 4*a(n-1)-3.
a(n) = 5*a(n-1)-4*a(n-2).
G.f.: 3*(2-3*x)/((1-x)*(1-4*x)). - Bruno Berselli, Nov 04 2011

A206373 a(n) = (14*4^n + 1)/3.

Original entry on oeis.org

5, 19, 75, 299, 1195, 4779, 19115, 76459, 305835, 1223339, 4893355, 19573419, 78293675, 313174699, 1252698795, 5010795179, 20043180715, 80172722859, 320690891435, 1282763565739, 5131054262955, 20524217051819, 82096868207275, 328387472829099, 1313549891316395

Offset: 0

Brad Clardy, Feb 07 2012

A generalized Engel expansion of 2/7 to the base b := 4/3 as defined in A181565 with associated series expansion 2/7 = b/5 + b^2/(5*19) + b^3/(5*19*75) + b^4/(5*19*75*299) + .... - Peter Bala, Oct 30 2013

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Sequences of the form (m*4^n + 1)/3: A007583 (m=2), A136412 (m=5), A199210 (m=11), A199210 (m=11), this sequence (m=14).

Cf. A181565.

Magma
```
[(14*4^n+1)/3 : n in [0..30]];
```

(14*4^Range[0,30]+1)/3 (* or *) LinearRecurrence[{5,-4},{5,19},30] (* Harvey P. Dale, Jan 13 2023 *)

a(n)=(14*4^n + 1)/3 \\ Charles R Greathouse IV, Jun 01 2015

[(7*2^(2*n+1)+1)/3 for n in range(31)] # G. C. Greubel, Jan 19 2023

a(n) = (14*4^n + 1)/3.
From Peter Bala, Oct 30 2013: (Start)
a(n+1) = 4*a(n) - 1 with a(0) = 5.
a(n) = 5*a(n-1) - 4*a(n-2) with a(0) = 5 and a(1) = 19.
O.g.f. (5 - 6*x)/((1 - x)*(1 - 4*x)). (End)
E.g.f.: (1/3)*(14*exp(4*x) + exp(x)). - G. C. Greubel, Jan 19 2023

A345290 a(n) is obtained by replacing 2^k in binary expansion of n with Fibonacci(-k-2).

Original entry on oeis.org

0, -1, 2, 1, -3, -4, -1, -2, 5, 4, 7, 6, 2, 1, 4, 3, -8, -9, -6, -7, -11, -12, -9, -10, -3, -4, -1, -2, -6, -7, -4, -5, 13, 12, 15, 14, 10, 9, 12, 11, 18, 17, 20, 19, 15, 14, 17, 16, 5, 4, 7, 6, 2, 1, 4, 3, 10, 9, 12, 11, 7, 6, 9, 8, -21, -22, -19, -20, -24

Offset: 0

Rémy Sigrist, Jun 13 2021

This sequence is a variant of A022290; here we consider Fibonacci numbers with negative indices (A039834), there Fibonacci numbers with positive indices (A000045).

After the initial 0, the sequence alternates runs of positive terms and runs of negative terms, the k-th run having 2^(k-1) terms.

			For n = 3:
- 3 = 2^1 + 2^0,
- so a(3) = A039834(2+1) + A039834(2+0) = 2 - 1 = 1.

Rémy Sigrist, Table of n, a(n) for n = 0..8191

Cf. A000045, A000695, A020989, A022290, A039834, A062880, A063694, A063695, A072197, A080675, A136412, A345291, A345292.

a(n) = { my (v=0, e); while (n, n-=2^e=valuation(n, 2); v+=fibonacci(-2-e)); v }

a(n) = A022290(A063695(n)) - A022290(A063694(n)).
a(n) = A022290(n) iff n belongs to A062880.
a(n) = -A022290(n) iff n belongs to A000695.
a(n) = 0 iff n = 0.
a(n) = 1 iff n belongs to A072197.
a(n) = 2 iff n belongs to A080675.
a(n) = -1 iff n belongs to A020989.
a(n) = -2 iff n belongs to A136412.

A171231 a(n) = (10*2^n + 3 - (-1)^n)/6.

Original entry on oeis.org

2, 4, 7, 14, 27, 54, 107, 214, 427, 854, 1707, 3414, 6827, 13654, 27307, 54614, 109227, 218454, 436907, 873814, 1747627, 3495254, 6990507, 13981014, 27962027, 55924054, 111848107, 223696214, 447392427, 894784854, 1789569707

Offset: 0

Paul Curtz, Dec 05 2009

From 14, the last 2 digits are of period 4: repeat [14, 27, 54, 07]. - Paul Curtz, Nov 22 2024

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

Cf. A000035, A000975, A048573, A136412 (1st bisection), 2*A136412 (2nd bisection).

Cf. A010712, A160113.

[( 10*2^n+3-(-1)^n )/6: n in [0..40]]; // Vincenzo Librandi, Aug 05 2011

LinearRecurrence[{2,1,-2},{2,4,7},40] (* Harvey P. Dale, Feb 11 2015 *)

a(n)=(10<Charles R Greathouse IV, Jul 07 2011

a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3), n >= 3.
a(n+1) - a(n) = A048573(n-1).
a(n) = 2*A000975(n+1) - 3*A000975(n-1).
a(n) - a(n-2) = 5*2^n.
a(n+1) - 2*a(n) = ((-1)^n-1)/2 = -A000035(n).
G.f.: ( 2-3*x^2 ) / ( (x-1)*(2*x-1)*(1+x) ). - R. J. Mathar, Jul 07 2011
a(n) = ceiling( (5/3)*(2^n) ). - Wesley Ivan Hurt, Jun 28 2013

Definition replaced by the Lava formula of 2009. Contents converted to formulas. - R. J. Mathar, Jul 07 2011

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