A112387 -id:A112387 - cp's OEIS frontend

A374098 a(n) = A112387(n)^2.

1, 1, 4, 1, 16, 9, 64, 25, 256, 121, 1024, 441, 4096, 1849, 16384, 7225, 65536, 29241, 262144, 116281, 1048576, 466489, 4194304, 1863225, 16777216, 7458361, 67108864, 29822521, 268435456, 119311929, 1073741824, 477204025, 4294967296, 1908903481, 17179869184

Offset: 0

Paul Curtz, Jun 28 2024

Index entries for linear recurrences with constant coefficients, signature (0,3,0,6,0,-8).

Cf. A000302, A003683, A084175, A112387, A139818.

LinearRecurrence[{0, 3, 0, 6, 0, -8}, {1, 1, 4, 1, 16, 9}, 35] (* Amiram Eldar, Jul 01 2024 *)

a(2*n) = A000302(n); a(2*n+1) = A139818(n+1).
(a(2*n) + a(2*n-1))^2 = A084175(n+1)^2 + 16*A003683(n)^2, for n >= 1. - Thomas Scheuerle, Jun 28 2024
G.f. ( 1+x+x^2-2*x^3-2*x^4 ) / ( (x-1)*(2*x+1)*(2*x-1)*(1+x)*(2*x^2+1) ). - R. J. Mathar, Aug 02 2024

A048573 a(n) = a(n-1) + 2*a(n-2), a(0)=2, a(1)=3.

Original entry on oeis.org

2, 3, 7, 13, 27, 53, 107, 213, 427, 853, 1707, 3413, 6827, 13653, 27307, 54613, 109227, 218453, 436907, 873813, 1747627, 3495253, 6990507, 13981013, 27962027, 55924053, 111848107, 223696213, 447392427, 894784853, 1789569707, 3579139413, 7158278827, 14316557653

Offset: 0

Michael Somos, Jun 17 1999

Number of positive integers requiring exactly n signed bits in the modified non-adjacent form representation. - Ralf Stephan, Aug 02 2003
The n-th entry (n>1) of the sequence is equal to the 1,1-entry of the n-th power of the unnormalized 4 X 4 Haar matrix: [1 1 1 0 / 1 1 -1 0 / 1 1 0 1 / 1 1 0 -1]. - Simone Severini, Oct 27 2004
Pisano period lengths:  1, 1, 6, 2, 2, 6, 6, 2, 18, 2, 10, 6, 12, 6, 6, 2, 8, 18, 18, 2, ... - R. J. Mathar, Aug 10 2012
For n >= 1, a(n) is the number of ways to tile a strip of length n+2 with blue squares and blue and red dominos, with the restriction that the first two tiles must be the same color. - Guanji Chen and Greg Dresden, Jul 15 2024

			G.f. = 2 + 3*x + 7*x^2 + 13*x^3 + 27*x^4 + 53*x^5 + 107*x^6 + 213*x^7 + 427*x^8 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wieb Bosma, Signed bits and fast exponentiation, Journal de théorie des nombres de Bordeaux, Vol. 13, No. 1 (2001), pp. 27-41.
Karl Dilcher and Larry Ericksen, Continued fractions and Stern polynomials, Ramanujan Journal 45.3 (2018): 659-681. See Table 2.
Karl Dilcher and Hayley Tomkins, Square classes and divisibility properties of Stern polynomials, Integers, Vol. 18 (2018), Article #A29.
Petro Kosobutskyy, The Collatz problem as a reverse n->0 problem on a graph tree formed from theta*2^n Jacobsthal-type numbers, arXiv:2306.14635 [math.GM], 2023.
Petro Kosobutskyy and Dariia Rebot, Collatz conjecture 3n+/-1 as a Newton binomial problem, Comp. Des. Sys. Theor. Prac., Lviv Nat'l Polytech. Univ. (Ukraine 2023) Vol. 5, No. 1, 137-145. See p. 140.
Saad Mneimneh, Simple Variations on the Tower of Hanoi to Guide the Study of Recurrences and Proofs by Induction, Department of Computer Science, Hunter College, CUNY, 2019.
Sam Northshield, Stern's diatomic sequence 0, 1, 1, 2, 1, 3, 2, 3, 1, 4, ..., Amer. Math. Monthly, Vol. 117, No. 7 (2010), pp. 581-598.
Index entries for linear recurrences with constant coefficients, signature (1,2).

Cf. A084214 (first differences).
Cf. A001045, A014551, A024493, A062510.
Cf. A007283, A007679, A078008, A081254.
Cf. A083322, A083944, A083581, A084170.
Cf. A102713, A127978, A130755, A139763.
Cf. A140360, A140966, A166249, A168642.
Cf. A280560, A290604, A340627.
Cf. A112387, A135318.

[(5*2^n+(-1)^n)/3: n in [0..35]]; // Vincenzo Librandi, Jul 05 2011

LinearRecurrence[{1,2},{2,3},40] (* Harvey P. Dale, Dec 11 2017 *)

{a(n) = if( n<0, 0, (5*2^n + (-1)^n) / 3)};

{a(n) = if (n<0 ,0, if( n<2, n+2, a(n-1) + 2*a(n-2)))};

[(5*2^n+(-1)^n)/3 for n in range(35)] # G. C. Greubel, Apr 10 2019

G.f.: (2 + x) / (1 - x - 2*x^2).
a(n) = (5*2^n + (-1)^n) / 3.
a(n) = 2^(n+1) - A001045(n).
a(n) = A084170(n)+1 = abs(A083581(n)-3) = A081254(n+1) - A081254(n) = A084214(n+2)/2.
a(n) = 2*A001045(n+1) + A001045(n) (note that 2 is the limit of A001045(n+1)/A001045(n)). - Paul Barry, Sep 14 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-3, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=-charpoly(A,-1). - Milan Janjic, Jan 27 2010
Equivalently, with different offset, a(n) = b(n+1) with b(0)=1 and b(n) = Sum_{i=0..n-1} (-1)^i (1 + (-1)^i b(i)). - Olivier Gérard, Jul 30 2012
a(n) = A000975(n-2)*10 + 5 + 2*(-1)^(n-2), a(0)=2, a(1)=3. - Yuchun Ji, Mar 18 2019
a(n+1) = Sum_{i=0..n} a(i) + 1 + (1-(-1)^n)/2, a(0)=2. - Yuchun Ji, Apr 10 2019
a(n) = 2^n + J(n+1) = J(n+2) + J(n+1) - J(n), where J is A001045. - Yuchun Ji, Apr 10 2019
a(n) = A001045(n+2) + A078008(n) = A062510(n+1) - A078008(n+1) = (A001045(n+2) + A062510(n+1))/2 = A014551(n) + 2*A001045(n). - Paul Curtz, Jul 14 2021
From Thomas Scheuerle, Jul 14 2021: (Start)
a(n) = A083322(n) + A024493(n).
a(n) = A127978(n) - A102713(n).
a(n) = A130755(n) - A166249(n).
a(n) = A007679(n) + A139763(n).
a(n) = A168642(n) XOR A007283(n).
a(n) = A290604(n) + A083944(n). (End)
From Paul Curtz, Jul 21 2021: (Start)
a(n) = 5*A001045(n) - A280560(n+1) = abs(A140360(n+1)) - A280560(n+1).
a(n) = 2^n + A001045(n+1) = A001045(n+3) - A000079(n).
a(n) = A001045(n+4) - A340627(n). (End)
a(n) = A001045(n+5) - A005010(n).
a(n+1) + a(n) = a(n+2) - a(n) = 5*2^n. - Michael Somos, Feb 22 2023
a(n) = A135318(2*n) + A135318(2*n+1) = A112387(2*n) + A112387(2*n+1). - Paul Curtz, Jun 26 2024
E.g.f.: (cosh(x) + 5*cosh(2*x) - sinh(x) + 5*sinh(2*x))/3. - Stefano Spezia, May 18 2025

Formula of Milan Janjic moved here from wrong sequence by Paul D. Hanna, May 29 2010

A135318 The Kentucky-2 sequence: a(n) = a(n-2) + 2*a(n-4), with a[0..3] = [1, 1, 1, 2].

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 5, 8, 11, 16, 21, 32, 43, 64, 85, 128, 171, 256, 341, 512, 683, 1024, 1365, 2048, 2731, 4096, 5461, 8192, 10923, 16384, 21845, 32768, 43691, 65536, 87381, 131072, 174763, 262144, 349525, 524288, 699051, 1048576, 1398101, 2097152, 2796203

Offset: 0

Paul Curtz, Feb 16 2008

Shifted Jacobsthal recurrence.
From L. Edson Jeffery, Apr 21 2011: (Start)
Let U be the unit-primitive matrix (see [Jeffery])
U=U_(6,2)=
(0 0 1)
(0 2 0)
(2 0 1),
let i in {0,1}, m>=0 an integer and n=2*m+i. Then a(n)=a(2*m+i)=Sum_{j=0..2} (U^m)_(i,j). (End)
a(n) is also the pebbling number of the cycle graph C_{n+1} for n > 1. - Eric W. Weisstein, Jan 07 2021
From Greg Dresden and Ziyi Xie, Aug 25 2023: (Start)
a(n) is the number of ways to tile a zig-zag strip of n cells using squares (of 1 cell) and triangles (of 3 cells). Here is the zig-zag strip corresponding to n=11, with 11 cells:
       _     _
   _|   |_|   |_
  |   |_|   |_|   |_
  |_|   |_|   |_|   |
  |   |_|   |_|   |_|
  |_|   |_|   |_|,
and here are the two types of triangles (where one is just a reflection of the other):
   _               _
  |   |_       _|   |
  |       |     |       |
  |    _| and |_    |
  |_|             |_|.
As an example, here is one of the a(11) = 32 ways to tile the zig-zag strip of 11 cells:
       _     _
   _|   |_|   |_
  |   |_|       |   |_
  |       |_    |       |
  |    _|   |_|    _|
  |_|   |_|   |_|.   (End)

			Let i=0 and m=3. Then U^3 = (2,0,3;0,8,0;6,0,5), and the first-row sum (corresponding to i=0) is 2 + 0 + 3 = 5. Hence a(n) = a(2*m+i) = a(2*3+0) = a(6) = 2 + 3 = 5.

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Minerva Catral et al., Generalizing Zeckendorf's Theorem: The Kentucky Sequence, arXiv:1409.0488 [math.NT], 2014. See 1.3 p. 2, same sequence without the first 2 terms.
Shaoshi Chen, Hanqian Fang, Sergey Kitaev, and Candice X.T. Zhang, Patterns in Multi-dimensional Permutations, arXiv:2411.02897 [math.CO], 2024. See pp. 2, 17.
L. E. Jeffery, Unit-primitive matrices.
Eric Weisstein's World of Mathematics, Cycle Graph
Eric Weisstein's World of Mathematics, Pebbling Number
Index entries for linear recurrences with constant coefficients, signature (0,1,0,2).

Cf. A000079, A001045, A112387.

Cf. A048573.

[(2^Floor(n/2)*(5-(-1)^n)+(-1)^Floor(n/2)*(1+(-1)^n))/6: n in [0..50]]; // Vincenzo Librandi, Aug 10 2011

a:= n-> (<<0|1>, <2|1>>^(iquo(n, 2, 'm')). <<1, 1+m>>)[1,1]:
seq(a(n), n=0..50);  # Alois P. Heinz, May 30 2022

LinearRecurrence[{0,1,0,2},{1,1,1,2},40] (* Harvey P. Dale, Oct 14 2015 *)

From R. J. Mathar, Feb 19 2008: (Start)
O.g.f.: (1/(1+x^2)+(-2-3*x)/(2*x^2-1))/3.
a(2n) = A001045(n+1).
a(2n+1) = A000079(n). (End)
From L. Edson Jeffery, Apr 21 2011: (Start)
G.f.: (1+x+x^3)/((1+x^2)*(1-2*x^2)).
a(n) = (((-i)^(n+1)-i^(n+1))*2*i*sqrt(2)+3*(1+(-1)^(n+1))*2^((n+2)/2)+(1-(-1)^(n+1))*2^((n+5)/2))/(12*sqrt(2)), where i=sqrt(-1). (End)
a(n) = (2^floor(n/2)*(5-(-1)^n)+(-1)^floor(n/2)*(1+(-1)^n))/6 = (A016116(n)*A010711(n)+2*A056594(n))/6. - Bruno Berselli, Apr 21 2011
a(2n) = 2*a(2n-1) - a(2n-2); a(2n+1) = a(2n) + a(2n-2). - Richard R. Forberg, Aug 19 2013
a(n) = A112387(n + (-1)^n). - Alois P. Heinz, Sep 28 2023
E.g.f.: (2*cos(x) + 4*cosh(sqrt(2)*x) + 3*sqrt(2)*sinh(sqrt(2)*x))/6. - Stefano Spezia, Nov 09 2024
a(2*n) + a(2*n+1) = A048573(n) for n >= 0. - Paul Curtz, May 18 2025

More terms from R. J. Mathar, Feb 19 2008

A340627 a(n) = (112^n - 2(-1)^n)/3 for n >= 0.

Original entry on oeis.org

3, 8, 14, 30, 58, 118, 234, 470, 938, 1878, 3754, 7510, 15018, 30038, 60074, 120150, 240298, 480598, 961194, 1922390, 3844778, 7689558, 15379114, 30758230, 61516458, 123032918, 246065834, 492131670, 984263338, 1968526678, 3937053354, 7874106710, 15748213418, 31496426838

Offset: 0

Paul Curtz, Apr 25 2021

Based on A112387.
Prepended with 0, 1, its difference table is
0,  1,  1,  2,  1,   4,  3,   8, ... = mix A001045(n), 2^n.
1,  0,  1, -1,  3,  -1,  5,  -3, ... = mix A001045(n+1), -A001045(n).
-1, 1, -2,  4, -4,   6, -8,  14, ... = mix -2^n, A084214(n+1).
2, -3,  6, -8, 10, -14, 22, -30, ... = mix 2*A001045(n+2), -a(n).

Index entries for linear recurrences with constant coefficients, signature (1,2).

Cf. A000079, A001045, A062510, A078008, A112387.

Cf. also A052997.

LinearRecurrence[{1, 2}, {3, 8}, 35] (* Amiram Eldar, Apr 25 2021 *)

a(n) = (11*2^n - 2*(-1)^n)/3 \\ Felix Fröhlich, Apr 25 2021

a(n) = 2^(n+2) - A078008(n), n>=0.
a(n) = (A062510(n) = 3*A001045(n)) + A001045(n+3), n>=0.
a(0)=3, a(2*n+1) = 2*a(2*n) + 2, a(2*n+2) = 2*a(2*n+1) - 2, n>=0.
a(n) = 4*A052997(n-1) + 2, n>=2. - Hugo Pfoertner, Apr 25 2021
a(n+1) = 11*2^n - a(n) for n>=0.
a(n+3) = 33*2^n - a(n) for n>=0.
a(n+5) = 121*2^n - a(n) for n>=0.
etc.
a(n+2) = a(n) + 11*2^n for n>=0.
a(n+4) = a(n) + 55*2^n for n>=0.
a(n+6) = a(n) + 231*2^n for n>=0.
etc.
G.f.: (3 + 5*x)/(1 - x - 2*x^2). - Stefano Spezia, Apr 26 2021
E.g.f: (11*exp(2*x) - 2*exp(-x))/3. - Jianing Song, Apr 26 2021

More terms from Michel Marcus, Apr 25 2021

New name from Jianing Song, Apr 25 2021

A140946 Triangle T(n,k) = (-2)^n*(-1)^k if kA001045(n+1).

Original entry on oeis.org

1, -2, -1, 4, -4, 3, -8, 8, -8, -5, 16, -16, 16, -16, 11, -32, 32, -32, 32, -32, -21, 64, -64, 64, -64, 64, -64, 43, -128, 128, -128, 128, -128, 128, -128, -85, 256, -256, 256, -256, 256, -256, 256, -256, 171, -512, 512, -512, 512, -512, 512, -512, 512, -512, -341, 1024, -1024, 1024, -1024, 1024

Offset: 0

Paul Curtz, Jul 24 2008

The sequence appears if the values b(n+1)-2*b(n) are computed from the (flattened) sequence b(.)=A140944.

Reading the triangle by rows, taking absolute values and removing duplicates we obtain A112387.

			1;
-2,-1;
4,-4,3;
-8,8,-8,-5;
16,-16,16,-16,11;
-32,32,-32,32,-32,-21;
64,-64,64,-64,64,-64,43;
-128,128,-128,128,-128,128,-128,-85;

Cf. A140513, A140589.

(* A = A140944 *) A[0, 0] = 0; A[1, 0] = A[0, 1] = 1; A[0, k_] := A[0, k] = A[0, k-1] + 2*A[0, k-2]; A[n_, n_] = 0; A[n_, k_] := A[n, k] = A[n-1, k+1] - A[n-1, k];  T[n_, n_] := T[n, n] = A[n+1, 0] - 2*A[n, n]; T[n_, k_] := T[n, k] = A[n, k+1] - 2*A[n, k]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 17 2014 *)

T(n,k) = A140944(n,k+1)-2*A140944(n,k), k

T(n,n) = A140944(n+1,0) -2*A140944(n,n).

Edited by R. J. Mathar, Jul 06 2011

A374927 a(n) = A135318(n)^2.

Original entry on oeis.org

1, 1, 1, 4, 9, 16, 25, 64, 121, 256, 441, 1024, 1849, 4096, 7225, 16384, 29241, 65536, 116281, 262144, 466489, 1048576, 1863225, 4194304, 7458361, 16777216, 29822521, 67108864, 119311929, 268435456, 477204025, 1073741824, 1908903481, 4294967296, 7635439161

Offset: 0

Paul Curtz, Jul 24 2024

A374098 terms swapped by pairs.

Index entries for linear recurrences with constant coefficients, signature (0,3,0,6,0,-8).

Cf. A000302, A112387, A135318, A139818, A374098.

a(2*n) = A139818(n+1).
a(2*n+1) = 4^n = A000302(n).
a(n) = 3*a(n-2) +6*a(n-4) -8*a(n-6).

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A374098 a(n) = A112387(n)^2.

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A048573 a(n) = a(n-1) + 2*a(n-2), a(0)=2, a(1)=3.

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A135318 The Kentucky-2 sequence: a(n) = a(n-2) + 2*a(n-4), with a[0..3] = [1, 1, 1, 2].

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A340627 a(n) = (11*2^n - 2*(-1)^n)/3 for n >= 0.

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A140946 Triangle T(n,k) = (-2)^n*(-1)^k if kA001045(n+1).

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A374927 a(n) = A135318(n)^2.

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A340627 a(n) = (112^n - 2(-1)^n)/3 for n >= 0.