A171160 -id:A171160 - cp's OEIS frontend

A062092 a(n) = 2*a(n-1) - (-1)^n for n > 0, a(0)=2.

2, 5, 9, 19, 37, 75, 149, 299, 597, 1195, 2389, 4779, 9557, 19115, 38229, 76459, 152917, 305835, 611669, 1223339, 2446677, 4893355, 9786709, 19573419, 39146837, 78293675, 156587349, 313174699, 626349397, 1252698795, 2505397589

Offset: 0

Amarnath Murthy, Jun 16 2001

Let A be the Hessenberg matrix of order n, defined by: A[1,j] = A[i,i] = 1, A[i,i-1] = -1, and A[i,j] = 0 otherwise. Then, for n>=1, a(n-1) = charpoly(A,3). - Milan Janjic, Jan 24 2010

T. Koshy, Fibonacci and Lucas numbers with applications, Wiley, 2001, p. 98.

Harry J. Smith, Table of n, a(n) for n = 0..200
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.
Index entries for linear recurrences with constant coefficients, signature (1,2).

Cf. A000032, A001045, A002487, A005009, A078008, A154879, A201630.

Cf. A171160 (first differences).

[(7*2^n-(-1)^n)/3: n in [0..40]]; // G. C. Greubel, Apr 04 2023

LinearRecurrence[{1, 2}, {2, 5}, 40] (* Jean-François Alcover, Aug 02 2021 *)

a(n) = (7*2^n - (-1)^n)/3; \\ Harry J. Smith, Aug 01 2009

[(7*2^n-(-1)^n)/3 for n in range(41)] # G. C. Greubel, Apr 04 2023

a(n) = a(n-1) + 2*a(n-2).
a(n) = (7*2^n - (-1)^n)/3.
a(n) = 2^(n+1) + A001045(n).
A002487(a(n)) = A000032(n+1).
G.f.: (2+3*x)/(1-x-2*x^2).
E.g.f.: (7*exp(2*x) - exp(-x))/3.
a(n) = Sum_{j=0..2} A001045(n-j) (sum of 3 consecutive elements of the Jacobsthal sequence). - Alexander Adamchuk, May 16 2006
From Paul Curtz, Jun 03 2022: (Start)
a(n) = A001045(n+3) - A078008(n).
a(n) = A078008(n+3) - A001045(n).
a(n) = A005009(n-1) - a(n-1) for n >= 1.
a(n) = a(n-2) + A005009(n-2) for n >= 2.
a(n) = A154879(n-2) + 3*A201630(n-2) for n >= 2. (End)

More terms from Jason Earls, Jun 18 2001

Additional comments from Michael Somos, Jun 24 2002

A355668 Array read by upwards antidiagonals T(n,k) = J(k) + n*J(k+1) where J(n) = A001045(n) is the Jacobsthal numbers.

Original entry on oeis.org

0, 1, 1, 2, 2, 1, 3, 3, 4, 3, 4, 4, 7, 8, 5, 5, 5, 10, 13, 16, 11, 6, 6, 13, 18, 27, 32, 21, 7, 7, 16, 23, 38, 53, 64, 43, 8, 8, 19, 28, 49, 74, 107, 128, 85, 9, 9, 22, 33, 60, 95, 150, 213, 256, 171, 10, 10, 25, 38, 71, 116, 193, 298, 427, 512, 341

Offset: 0

Paul Curtz, Jul 13 2022

			Row n=0 is A001045(k), then for further rows we successively add A001045(k+1).
       k=0  k=2  k=3  k=4  k=5  k=6  k=7  k=8  k=9 k=10
  n=0:  0    1    1    3    5   11   21   43   85  171 ... = A001045
  n=1:  1    2    4    8   16   32   64  128  256  512 ... = A000079
  n=2:  2    3    7   13   27   53  107  213  427  853 ... = A048573
  n=3:  3    4   10   18   38   74  150  298  598 1194 ... = A171160
  n=4:  4    5   13   23   49   95  193  383  769 1535 ... = abs(A140683)
  ...

Cf. A001477, A000027, A016777, A016885, A017449, A321373.

Antidiagonal sums give A320933(n+1).

T[n_, k_] := (2^k - (-1)^k + n*(2^(k + 1) + (-1)^k))/3; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Jul 13 2022 *)

T(n, k) = (2^k - (-1)^k + n*(2^(k + 1) + (-1)^k))/3.

G.f.: (x*(y-1) - y)/((x - 1)^2*(y + 1)*(2*y - 1)). - Stefano Spezia, Jul 13 2022

cp's OEIS Frontend

A062092 a(n) = 2*a(n-1) - (-1)^n for n > 0, a(0)=2.

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