%I #22 Mar 21 2023 07:20:39
%S 1,1,2,4,8,16,32,65,136,293,642,1410,3072,6606,14004,29295,60592,
%T 124187,252742,511672,1031912,2075452,4166408,8353165,16732664,
%U 33498977,67040458,134134046,268333872,536748474,1073595228,2147309211,4294760928,8589691767
%N Binomial transform of (1,0,1,0,1,0,1,1,1,1,1,...).
%C The sequence starting 1,2,4,... is the binomial transform of (1, 1, 1, 1, 1, 1, 2, 2, 2, ...) with A035038(n) = Sum_{k=0..5} C(n,k) + 2*Sum_{k=6..n} C(n,k) = 2^n - (n^5 - 5*n^4 + 25*n^3 + 5*n^2 + 94*n + 120)/120. This gives the partial sums of A084636.
%H Colin Barker, <a href="/A084637/b084637.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (8,-27,50,-55,36,-13,2).
%F a(n) = Sum_{k=0..2} C(n, 2*k) + Sum_{k=6..n} C(n, k).
%F a(n) = 2^n - n*(n^4 - 10*n^3 + 55*n^2 - 110*n + 184)/120.
%F G.f.: (1-7*x+21*x^2-35*x^3+35*x^4-21*x^5+7*x^6) / ((1-x)^6*(1-2*x)). - _Colin Barker_, Mar 17 2016
%t Table[2^n -n*(n^4-10*n^3+55*n^2-110*n+184)/120, {n,0,50}] (* _G. C. Greubel_, Mar 19 2023 *)
%o (PARI) Vec((1-7*x+21*x^2-35*x^3+35*x^4-21*x^5+7*x^6)/((1-x)^6*(1-2*x)) + O(x^50)) \\ _Colin Barker_, Mar 17 2016
%o (Magma) [2^n -n*(n^4-10*n^3+55*n^2-110*n+184)/120: n in [0..50]]; // _G. C. Greubel_, Mar 19 2023
%o (SageMath) [2^n -n*(n^4-10*n^3+55*n^2-110*n+184)/120 for n in range(51)] # _G. C. Greubel_, Mar 19 2023
%Y Cf. A000225, A000325, A035038, A084634, A084635, A084636.
%K easy,nonn
%O 0,3
%A _Paul Barry_, Jun 06 2003