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A174390
a(2n) = -n, a(2n+1) = 2*n+4.
1
0, 4, -1, 6, -2, 8, -3, 10, -4, 12, -5, 14, -6, 16, -7, 18, -8, 20, -9, 22, -10, 24, -11, 26, -12, 28, -13, 30, -14, 32, -15, 34, -16, 36, -17, 38, -18, 40, -19, 42, -20, 44, -21, 46, -22, 48, -23, 50, -24, 52, -25
OFFSET
0,2
COMMENTS
Given a first row 1/A026741(k+1), k >= 0, of an array, namely
1/1, 1/1, 1/3, 1/2, 1/5, 1/3, 1/7, ...,
the next row generated by the Akiyama-Tanigawa transform is
0, 4/3, -1/2, 6/5, -2/3, 8/7, -3/4, 10/9, -4/5, 12/11, -5/6, 14/13, ...
The current sequence contains the numerators of these fractions; the denominators are A026741(n+2).
LINKS
D. Merlini, R. Sprugnoli, M. C. Verri, The Akiyama-Tanigawa Transformation, Integers, 5 (1) (2005) #A05.
FORMULA
a(n) = +2*a(n-2) -a(n-4).
a(n) = (1/4)*(n + 6 - 3*(-1)^n*(n + 2)).
G.f.: x*(4 - x - 2*x^2) / ( (1-x)^2*(1+x)^2 ).
E.g.f.: (1/4)*((6+x)*exp(x) - 3*(2-x)*exp(-x)). - G. C. Greubel, Dec 04 2022
MATHEMATICA
LinearRecurrence[{0, 2, 0, -1}, {0, 4, -1, 6}, 60] (* G. C. Greubel, Dec 04 2022 *)
PROG
(Magma) [(1/4)*(n+6 -3*(-1)^n*(n+2)): n in [0..50]]; // G. C. Greubel, Dec 04 2022
(SageMath) [(1/4)*(n + 6 - 3*(-1)^n*(n + 2)) for n in range(51)] # G. C. Greubel, Dec 04 2022
CROSSREFS
Sequence in context: A202521 A247362 A098987 * A153017 A038457 A141649
KEYWORD
easy,sign
AUTHOR
Paul Curtz, Mar 18 2010
STATUS
approved