A100329 -id:A100329 - cp's OEIS frontend

A073937 a(n) = a(n-1) - a(n-2) + a(n-3) + a(n-4), a(0)=4, a(1)=1, a(2)=-1, a(3)=1.

4, 1, -1, 1, 7, 6, -1, 1, 15, 19, 4, 1, 31, 53, 27, 6, 63, 137, 107, 39, 132, 337, 351, 185, 303, 806, 1039, 721, 791, 1915, 2884, 2481, 2303, 4621, 7683, 7846, 7087, 11545, 19987, 23375, 22020, 30177, 51519, 66737, 67415, 82374, 133215, 184993, 201567, 232163, 348804

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 13 2002

From Kai Wang, Nov 03 2020: (Start)
Let f(x) = x^4 - x^3 - x^2 - x - 1 and {x1,x2,x3,x4} be the roots of f(x). Then a(n) = (x1*x2*x3)^n + (x1*x2*x4)^n + (x1*x3*x4)^n + (x2*x3*x4)^n.
Let g(y) = y^4 - y^3 + y^2 - y - 1 and {y1,y2,y3,y4} be the roots of g(y). Then a(n) = y1^n + y2^n + y3^n + y4^n. (End)

Michael De Vlieger, Table of n, a(n) for n = 0..9024
Kai Wang, Identities, generating functions and Binet formula for generalized k-nacci sequences, 2020.
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for linear recurrences with constant coefficients, signature (1,-1,1,1).

Cf. A000078, A001630, A073817, A074058, A100329.

CoefficientList[Series[(4-3*x+2*x^2-x^3)/(1-x+x^2-x^3-x^4), {x, 0, 50}], x]
LinearRecurrence[{1,-1,1,1},{4,1,-1,1},60] (* Harvey P. Dale, Sep 05 2021 *)

polsym(y^4 - y^3 + y^2 - y - 1, 55) \\ Joerg Arndt, Nov 07 2020

G.f.: (4 - 3*x + 2*x^2 - x^3)/(1 - x + x^2 - x^3 - x^4).
From Kai Wang, Nov 03 2020: (Start)
For n >= 1, a(n) = Sum_{j1,j2,j3,j4>=0; j1+2*j2+3*j3+4*j4=n} (-1)^j2*n*(j1+j2+j3+j4-1)!/(j1!*j2!*j3!*j4!).
For n > 1, a(n) = (-1)^n*(4*A100329(n+1) + 3*A100329(n) + 2*A100329(n-1) + A100329(n-2)). (End)
From Peter Bala, Jan 19 2023: (Start)
a(n) = (-1)^n*A074058(n).
a(n) = trace of M^n, where M is the 4 X 4 matrix [[0, 0, 0, -1], [-1, 0, 0, 1], [0, -1, 0, 1], [0, 0, -1, 1]].
The Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^k) for positive integers n and r and all primes p. See Zarelua. (End)

A180046 a(n+1) = a(n-3) + a(n-2) - a(n-1) + a(n) starting with 1, 2, 3, 4.

Original entry on oeis.org

1, 2, 3, 4, 4, 5, 8, 11, 12, 14, 21, 30, 35, 40, 56, 81, 100, 115, 152, 218, 281, 330, 419, 588, 780, 941, 1168, 1595, 2148, 2662, 3277, 4358, 5891, 7472, 9216, 11993, 16140, 20835, 25904, 33202, 44273, 57810, 72643, 92308, 121748, 159893, 203096, 257259

Offset: 1

Ian Stewart, Aug 08 2010

			1 2 3 4 (by definition); 1 + 2 - 3 + 4 = 4, 2 + 3 - 4 + 4 = 5, 3 + 4 - 4 + 5 = 8.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,-1,1,1).

Cf. A100329.

import Data.List (zipWith4)
a180046 n = a180046_list !! (n-1)
a180046_list = [1..4] ++ zipWith4 (((((+) .) . (+)) .) . (-))
                          (drop 3 a180046_list) (drop 2 a180046_list)
                            (tail a180046_list) a180046_list
-- Reinhard Zumkeller, Oct 08 2014

LinearRecurrence[{1,-1,1,1},{1,2,3,4},50] (* Harvey P. Dale, Mar 18 2015 *)

Vec(x*(1+x)*(2*x^2+1)/(1-x+x^2-x^3-x^4)+O(x^99)) \\ Charles R Greathouse IV, Jul 06 2011

G.f.: -x*(1+x)*(2*x^2+1)/(-1+x-x^2+x^3+x^4). - R. J. Mathar, Aug 14 2010

a(n) = (-1)^(n)*(A100329(n-1)-A100329(n)-2*A100329(n-2)+2*A100329(n-3)) with A100329(-1) = A100329(-2) = 0. - Johannes W. Meijer, Jul 06 2011

More terms from R. J. Mathar, Aug 14 2010

Name edited by Michel Marcus, Feb 20 2025

cp's OEIS Frontend

A073937 a(n) = a(n-1) - a(n-2) + a(n-3) + a(n-4), a(0)=4, a(1)=1, a(2)=-1, a(3)=1.

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