A027656 Expansion of 1/(1-x^2)^2 (included only for completeness - the policy is always to omit the zeros from such sequences).
1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, 7, 0, 8, 0, 9, 0, 10, 0, 11, 0, 12, 0, 13, 0, 14, 0, 15, 0, 16, 0, 17, 0, 18, 0, 19, 0, 20, 0, 21, 0, 22, 0, 23, 0, 24, 0, 25, 0, 26, 0, 27, 0, 28, 0, 29, 0, 30, 0, 31, 0, 32, 0, 33, 0, 34, 0, 35, 0, 36, 0, 37, 0, 38, 0, 39, 0, 40, 0, 41, 0, 42, 0, 43, 0
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..5000
- Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1)
Programs
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Magma
[(n+2)*(1+(-1)^n)/4: n in [0..75]]; // Vincenzo Librandi, Apr 02 2011
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Mathematica
CoefficientList[Series[1/(1-x^2)^2,{x,0,100}],x] (* Geoffrey Critzer, Jul 12 2013 *)
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PARI
a(n)=if(n%2,0,n/2+1) \\ Charles R Greathouse IV, Jan 18 2012
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SageMath
[(n+2)*((n+1)%2)/2 for n in (0..80)] # G. C. Greubel, Aug 01 2022
Formula
From Paul Barry, May 27 2003: (Start)
Binomial transform is A045891. Partial sums are A008805. The sequence 0, 1, 0, 2, ... has a(n)=floor((n+2)/2)(1-(-1)^n)/2.
a(n) = floor((n+3)/2) * (1+(-1)^n)/2. (End)
a(n) = (n+2)(n+3)/2 mod n+2. - Amarnath Murthy, Jun 17 2004
a(n) = (n+2)*(1 + (-1)^n)/4. - Bruno Berselli, Apr 01 2011
E.g.f.: cosh(x) + x*sinh(x)/2. - Stefano Spezia, Mar 26 2022
Comments