A030440 Values of Newton-Gregory forward interpolating polynomial (1/3)*(n-1)*(2*n+3)*(2*n-1).
1, 0, 7, 30, 77, 156, 275, 442, 665, 952, 1311, 1750, 2277, 2900, 3627, 4466, 5425, 6512, 7735, 9102, 10621, 12300, 14147, 16170, 18377, 20776, 23375, 26182, 29205, 32452, 35931, 39650, 43617, 47840, 52327, 57086, 62125, 67452, 73075, 79002, 85241, 91800, 98687, 105910
Offset: 0
References
- S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234; http://www.scirp.org/journal/am; http://dx.doi.org/10.4236/am.2014.515216
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..2000
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Programs
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Magma
[(1/3)*(n-1)*(2*n-1)*(2*n+3):n in [0..50]]; // Vincenzo Librandi, Apr 20 2018
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Mathematica
LinearRecurrence[{4, -6, 4, -1}, {1, 0, 7, 30}, 40] (* Vincenzo Librandi, Apr 20 2018 *)
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PARI
a(n) = (n-1)*(2*n-1)*(2*n+3)/3; \\ Altug Alkan, Apr 19 2018
Formula
G.f.: (1+13*x^2-2*x^3-4*x)/(1-x)^4. - R. J. Mathar, May 18 2014
a(n) = (1/6) * (A106734(2n) - 1), n > 0. - Mathew Englander, Jun 06 2014
E.g.f.: (3 - 3*x + 12*x^2 + 4*x^3)*exp(x)/3. - Ilya Gutkovskiy, Apr 20 2016
a(n+1) = trinomial(2*n+1, 3) = binomial(2*n+1, 3) + (2*n+1)*(2*n) = n*(2*n+1)*(2*n+5)/3, for n >= 0, with the trinomial irregular triangle A027907. a(n+1) = trinomial(2*n+1, 4*n-1), for n >= 1 (symmetry). a(n+1) = Integral_{x=0..2} (1/sqrt(4 - x^2))*(x^2 - 1)^(2*n+1)*R(4*(n-1), x)/Pi with the R polynomial coefficients given in A127672. [Comtet, p. 77, the integral formula for q=3, n -> 2*n+1, k = 3, rewritten with x = 2*cos(phi)]. The g.f. of {a(n+1)}{n>= 0} is x*(7 + 2*x - x^2)/(1 - x)^4. - _Wolfdieter Lang, Apr 19 2018
Comments