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A084214
Inverse binomial transform of a math magic problem.
17
1, 1, 4, 6, 14, 26, 54, 106, 214, 426, 854, 1706, 3414, 6826, 13654, 27306, 54614, 109226, 218454, 436906, 873814, 1747626, 3495254, 6990506, 13981014, 27962026, 55924054, 111848106, 223696214, 447392426, 894784854, 1789569706, 3579139414
OFFSET
0,3
COMMENTS
Inverse binomial transform of A060816.
FORMULA
a(n) = (5*2^n - 3*0^n + 4*(-1)^n)/6.
G.f.: (1+x^2)/((1+x)*(1-2*x)).
E.g.f.: (5*exp(2*x) - 3*exp(0) + 4*exp(-x))/6.
From Paul Barry, May 04 2004: (Start)
The binomial transform of a(n+1) is A020989(n).
a(n) = A001045(n-1) + A001045(n+1) - 0^n/2. (End)
a(n) = Sum_{k=0..n} A001045(n+1)*C(1, k/2)*(1+(-1)^k)/2. - Paul Barry, Oct 15 2004
a(n) = a(n-1) + 2*a(n-2) for n > 2. - Klaus Brockhaus, Dec 01 2009
From Yuchun Ji, Mar 18 2019: (Start)
a(n+1) = Sum_{i=0..n} a(i) + 1 - (-1)^n, a(0)=1.
a(n) = A000975(n-3)*10 + 5 + (-1)^(n-3), a(0)=1, a(1)=1, a(2)=4. (End)
a(n) = A081254(n) + (n-1 mod 2). - Kevin Ryde, Dec 20 2023
MAPLE
A084214 := proc(n)
(5*2^n - 3*0^n + 4*(-1)^n)/6 ;
end proc:
seq(A084214(n), n=0..60) ; # R. J. Mathar, Aug 18 2024
MATHEMATICA
f[n_]:=2/(n+1); x=3; Table[x=f[x]; Numerator[x], {n, 0, 5!}] (* Vladimir Joseph Stephan Orlovsky, Mar 12 2010 *)
LinearRecurrence[{1, 2}, {1, 1, 4}, 50] (* Harvey P. Dale, Mar 05 2021 *)
PROG
(Magma) [(5*2^n-3*0^n+4*(-1)^n)/6: n in [0..35]]; // Vincenzo Librandi, Jun 15 2011
(Haskell)
a084214 n = a084214_list !! n
a084214_list = 1 : xs where
xs = 1 : 4 : zipWith (+) (map (* 2) xs) (tail xs)
-- Reinhard Zumkeller, Aug 01 2011
(PARI) a(n) = 5<<(n-1)\3 + bitnegimply(1, n); \\ Kevin Ryde, Dec 20 2023
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, May 19 2003
STATUS
approved