A105812 -id:A105812

Search: a105812 -id:a105812

Displaying 1-2 of 2 results found.

page 1

A066983

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

+10
17

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,4

COMMENTS

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

REFERENCES

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

LINKS

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).

FORMULA

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

MAPLE

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

MATHEMATICA

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 *)

PROG

(PARI) { 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

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

(GAP) 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

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

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

CROSSREFS

Cf. A000045, A001083, A042942, A064353.

KEYWORD

nonn,easy

AUTHOR

Benoit Cloitre, Jan 27 2002

EXTENSIONS

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

STATUS

approved

A105810

Inverse of a Fibonacci-Pascal matrix A105809.

+10
4

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

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

0,5

COMMENTS

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

LINKS

Table of n, a(n) for n=0..77.

FORMULA

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

EXAMPLE

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.

MAPLE

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

CROSSREFS

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

KEYWORD

easy,sign,tabl

AUTHOR

Paul Barry, May 04 2005

STATUS

approved

page 1

Search completed in 0.005 seconds