A105812 -id:A105812 - cp's OEIS frontend

A066983 a(n+2) = a(n+1) + a(n) + (-1)^n, with a(1) = a(2) = 1.

1, 1, 1, 3, 3, 7, 9, 17, 25, 43, 67, 111, 177, 289, 465, 755, 1219, 1975, 3193, 5169, 8361, 13531, 21891, 35423, 57313, 92737, 150049, 242787, 392835, 635623, 1028457, 1664081, 2692537, 4356619, 7049155, 11405775, 18454929, 29860705, 48315633, 78176339

Offset: 1

Benoit Cloitre, Jan 27 2002

Length of strings given by a successive substitution of a "modified" Kolakoski-(3, 1) sequence. Starting with 1, using the rule "string begins with 1 if previous string ends with 3, string begins with 3 if previous string ends with 1" then applying the classical Kolakoski-(3,1) rule. This gives: 1 -> 3 -> 111 -> 313 -> 1113111 -> 313111313 -> 11131113131113111 and the length of string are 1, 1, 3, 3, 7, 9, 17, ... At step n, length = a(n+1). This substitution leads to two sequences: 1, 1, 1, 3, 1, 1, 1, 3, 1, 3, 1, 1, 1, ... and 3, 1, 3, 1, 1, 1, 3, 1, 3, 1, 1, 1, 3, ... - Benoit Cloitre, Jun 01 2004
Lengths of comparators in subsequent layers of correction network F_n. - Grzegorz Stachowiak (gst(AT)ii.uni.wroc.pl), Nov 28 2004
Convolution of F(n+1) and A105812(n). Action of inverse of sequence array for F(n-1)*(-1)^n on F(n+1). - Paul Barry, Oct 29 2006

Omur Deveci, The Pell-Padovan sequences and the Jacobsthal-Padovan sequences in finite groups, Utilitas Mathematica, 98 (2015), 257-270.

Harry J. Smith, Table of n, a(n) for n = 1..250
K. Atanassov, D. Dimitrov and A. G. Shannon, A remark on psi-function and Pell-Padovan's sequence, Notes Number Theory Discrete Math., 15 (2009), no. 2, 1-44.
Michael Baake and Bernd Sing, Kolakoski-(3,1) is a (deformed) model set, arXiv:math/0206098 [math.MG], 2002-2003.
Taras Goy, S. V. Sharyn, A note on Pell-Padovan numbers and their connection with Fibonacci numbers, Carpathian Math. Publ. (2020) Vol. 12, No. 2, 280-288.
Zehra İşbilir and Nurten Gürses, Pell-Padovan generalized quaternions, Notes on Num. Theory and Disc. Math. (2021) Vol. 27, No. 1, 171—187.
G. Stachowiak, Fibonacci Correction Networks, SWAT 2000, LNCS 1851, 535-548.
G. Stachowiak, Lower Bounds on Correction Networks, ISAAC 2003, LNCS 2906, 221-229.
Dursun Tasci, Gaussian Padovan and Gaussian Pell-Padovan sequences, Commun. Fac. Sci. Univ. Ank. Ser. A1 Math. Stat., 67 (2018), no. 2, 82-88. Sequence R_n.
Index entries for linear recurrences with constant coefficients, signature (0,2,1).

Cf. A000045, A001083, A042942, A064353, A066629.

a:=[1,1];; for n in [3..40] do a[n]:=a[n-1]+a[n-2]+(-1)^n; od; a; # Muniru A Asiru, Aug 09 2018

[n le 2 select 1 else Self(n-1)+Self(n-2)+(-1)^n: n in [1..50]]; // Vincenzo Librandi, Aug 13 2018

seq(coeff(series(x*(1+x-x^2)/((1+x)*(1-x-x^2)), x,n+1),x,n),n=1..40); # Muniru A Asiru, Aug 09 2018

Table[ Floor[ GoldenRatio^(k-1) ] - Floor[ GoldenRatio^(k-1) / Sqrt[5] ], {k, 1, 100} ]  (* Federico Provvedi, Mar 26 2013 *)
LinearRecurrence[{0, 2, 1}, {1, 1, 1}, 40] (* Vincenzo Librandi, Aug 13 2018 *)

{ for (n=1, 250, if (n>2, a=a1 + a2 + (-1)^n; a2=a1; a1=a, a=a1=1; a=a2=1); write("b066983.txt", n, " ", a) ) } \\ Harry J. Smith, Apr 15 2010

vector(40, n, 2*fibonacci(n-2) + (-1)^n) \\ G. C. Greubel, Dec 26 2019

from sympy import fibonacci
def A066983(n): return (fibonacci(n-2)<<1)+(-1 if n&1 else 1) # Chai Wah Wu, May 05 2025

[2*fibonacci(n-2) + (-1)^n for n in (1..40)] # G. C. Greubel, Dec 26 2019

For n > 4, a(n-2) = floor(2 * phi^n/sqrt(5)) + (1 + (-1)^n)/2.
a(n) = 2 * Fibonacci(n-2) + (-1)^n. - Vladeta Jovovic, Mar 19 2003
G.f.: x*(1+x-x^2)/((1+x)*(1-x-x^2)). - Paul Barry, Oct 29 2006
a(n) = A066629(n-2) - A066629(n-3), n > 2. - R. J. Mathar, Jan 14 2009
a(n) = floor(phi^(n-1)) - floor(phi^(n-1)/sqrt(5)). - Federico Provvedi, Mar 26 2013
a(1) = a(2) = a(3) = 1; for n > 3, a(n) = 2*a(n-2) + a(n-3). - Taras Goy, Aug 03 2018
a(n) = (-1)^n + (-1 - 3/sqrt(5))*((1/2)*(1 - sqrt(5)))^n + (-1 + 3/sqrt(5))*((1/2)*(1 + sqrt(5)))^n. - Stefano Spezia, Jul 22 2019

Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A105810 Inverse of a Fibonacci-Pascal matrix A105809.

Original entry on oeis.org

1, -1, 1, 0, -2, 1, 1, 2, -3, 1, -2, -1, 5, -4, 1, 3, -1, -6, 9, -5, 1, -4, 4, 5, -15, 14, -6, 1, 5, -8, -1, 20, -29, 20, -7, 1, -6, 13, -7, -21, 49, -49, 27, -8, 1, 7, -19, 20, 14, -70, 98, -76, 35, -9, 1, -8, 26, -39, 6, 84, -168, 174, -111, 44, -10, 1, 9, -34, 65, -45, -78, 252, -342, 285, -155, 54, -11, 1

Offset: 0

Paul Barry, May 04 2005

First column is A105811, row sums are A105812, antidiagonal sums are (-1)^n.

			The triangle T(n,m) begins:
n\m   0   1   2    3   4    5    6    7     8    9   10  11  12 13 ...
0:    1
1:   -1   1
2:    0  -2   1
3:    1   2  -3    1
4:   -2  -1   5   -4   1
5:    3  -1  -6    9  -5    1
6:   -4   4   5  -15  14   -6    1
7:    5  -8  -1   20 -29   20   -7    1
8:   -6  13  -7  -21  49  -49   27   -8     1
9:    7 -19  20   14 -70   98  -76   35    -9    1
10:  -8  26 -39    6  84 -168  174 -111    44  -10    1
11:   9 -34  65  -45 -78  252 -342  285  -155   54  -11   1
12: -10  43 -99  110  33 -330  594 -627   440 -209   65 -12   1
13:  11 -53 142 -209  77  363 -924 1221 -1067  649 -274  77 -13  1
... Reformatted and extended - _Wolfdieter Lang_, Oct 04 2014
-----------------------------------------------------------------------
Recurrence for T(n, 0) with row n-1 entries from Z-sequence (see a link given above): 3 = T(5, 0) = -(1*(-2) + 1*(-1) + 1*5 + 2*(-4) + 3*1) = 3.

Cf. A105809, A105811, A105812, A248155 (alternating row sum). - Wolfdieter Lang, Oct 04 2014

C := proc (n, k) if 0 <= k and k <= n then factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if
end proc:
for n from 0 to 10 do
    seq((-1)^(n+k)*(C(n, n-k) - add(C(n-i, n-k-i), i = 2..n)), k = 0..n);
end do; # Peter Bala, Mar 21 2018

Riordan array ((1+x-x^2)/(1+x)^2, x/(1+x)); Number triangle T(n, 0)=A105811(n), T(n, m)=-T(n-1, m-1)+T(n-1, m).
From Wolfdieter Lang, Oct 04 2014: (Start)
O.g.f. for row polynomials R(n,x) = sum(T(n,m)*x^m,m=0..n): (1 + z - z^2)/((1+z)*(1+(1-x)*z)) (Riordan property).
O.g.f. column m: x^m*(1 + x - x^2)/(1 + x)^(m+2), m >= 0.
The A-sequence of this Riordan triangle is [1, -1]. See the above given recurrence for T(n,m) for n>=1. The Z-sequence has o.g.f. -(1 - x^2)/(1 - x - x^2) and is -A132916(n+5) = -[1, 1, 1, 2, 3, 5, 8, 13, 21, 34,...]. See the W. Lang link under A006232 for Riordan A- and Z-sequences. (End)
T(n,k) = (-1)^(n+k)*(C(n, n-k) - Sum_{i = 2..n} C(n-i, n-k-i)), where C(n,k) = n!/(k!*(n-k)!) for 0 <= k <= n, otherwise 0. - Peter Bala, Mar 21 2018

cp's OEIS Frontend

A066983 a(n+2) = a(n+1) + a(n) + (-1)^n, with a(1) = a(2) = 1.

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