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%I #36 Sep 08 2022 08:44:50
%S 1,0,7,30,77,156,275,442,665,952,1311,1750,2277,2900,3627,4466,5425,
%T 6512,7735,9102,10621,12300,14147,16170,18377,20776,23375,26182,29205,
%U 32452,35931,39650,43617,47840,52327,57086,62125,67452,73075,79002,85241,91800,98687,105910
%N Values of Newton-Gregory forward interpolating polynomial (1/3)*(n-1)*(2*n+3)*(2*n-1).
%C Starting at a(2)=7, partial sums of A073577. - _J. M. Bergot_, Apr 20 2016
%D 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
%H Vincenzo Librandi, <a href="/A030440/b030440.txt">Table of n, a(n) for n = 0..2000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F G.f.: (1+13*x^2-2*x^3-4*x)/(1-x)^4. - _R. J. Mathar_, May 18 2014
%F a(n) = (1/6) * (A106734(2n) - 1), n > 0. - _Mathew Englander_, Jun 06 2014
%F E.g.f.: (3 - 3*x + 12*x^2 + 4*x^3)*exp(x)/3. - _Ilya Gutkovskiy_, Apr 20 2016
%F 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
%t LinearRecurrence[{4, -6, 4, -1}, {1, 0, 7, 30}, 40] (* _Vincenzo Librandi_, Apr 20 2018 *)
%o (PARI) a(n) = (n-1)*(2*n-1)*(2*n+3)/3; \\ _Altug Alkan_, Apr 19 2018
%o (Magma) [(1/3)*(n-1)*(2*n-1)*(2*n+3):n in [0..50]]; // _Vincenzo Librandi_, Apr 20 2018
%Y Cf. A073577, A000384 (trinomial k=2 column), A106734, A127672, A027907.
%K nonn,easy
%O 0,3
%A Ilias.Kotsireas(AT)lip6.fr (Ilias Kotsireas)
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