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A062092
a(n) = 2*a(n-1) - (-1)^n for n > 0, a(0)=2.
14
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
0,1
COMMENTS
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
REFERENCES
T. Koshy, Fibonacci and Lucas numbers with applications, Wiley, 2001, p. 98.
LINKS
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.
FORMULA
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)
MATHEMATICA
LinearRecurrence[{1, 2}, {2, 5}, 40] (* Jean-François Alcover, Aug 02 2021 *)
PROG
(PARI) a(n) = (7*2^n - (-1)^n)/3; \\ Harry J. Smith, Aug 01 2009
(Magma) [(7*2^n-(-1)^n)/3: n in [0..40]]; // G. C. Greubel, Apr 04 2023
(SageMath) [(7*2^n-(-1)^n)/3 for n in range(41)] # G. C. Greubel, Apr 04 2023
CROSSREFS
Cf. A171160 (first differences).
Sequence in context: A122893 A178841 A214319 * A320172 A350243 A079117
KEYWORD
nonn,easy
AUTHOR
Amarnath Murthy, Jun 16 2001
EXTENSIONS
More terms from Jason Earls, Jun 18 2001
Additional comments from Michael Somos, Jun 24 2002
STATUS
approved