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%I #38 Nov 13 2023 05:19:50
%S 3,10,38,150,598,2390,9558,38230,152918,611670,2446678,9786710,
%T 39146838,156587350,626349398,2505397590,10021590358,40086361430,
%U 160345445718,641381782870,2565527131478,10262108525910,41048434103638,164193736414550,656774945658198,2627099782632790
%N a(n) = (2 + 7*4^n)/3.
%C Consider A092808 and its differences:
%C 1, 0, 3, 1, 11, 5, 43, 21, 171, ...
%C -1, 3, -2, 10, -6, 38, -22, 150, ... = b(n).
%C a(n) is the second bisection of b(n). The first is A047849.
%C a(n) mod 9 is the period 9 sequence: repeat {3, 1, 2, 6, 4, 5, 0, 7, 8].
%C b(n) + b(n+1) = A135520{n).
%H Colin Barker, <a href="/A308124/b308124.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5,-4).
%F a(n) = 4*a(n-1) - 2 for n=1,2,... , a(0) = 3.
%F a(n+1) = a(n) + A002042(n).
%F Binomial transform of A141495(n+1) = 3, 7, 21, ....
%F From _Colin Barker_, Jul 23 2019: (Start)
%F G.f.: (3 - 5*x) / ((1 - x)*(1 - 4*x)).
%F a(n) = 5*a(n-1) - 4*a(n-2) for n>1.
%F (End)
%F a(n+2) = a(n) + 35*A000302(n) for n=0,1,2, ... .
%t LinearRecurrence[{5,-4},{3,10},30] (* _Paolo Xausa_, Nov 13 2023 *)
%o (PARI) a(n) = (2 + 7*4^n)/3; \\ _Stefano Spezia_, Jul 23 2019
%o (PARI) Vec((3 - 5*x) / ((1 - x)*(1 - 4*x)) + O(x^40)) \\ _Colin Barker_, Jul 23 2019
%Y Cf. A000302, A001045, A002042, A092808, A047849, A135520, A141495.
%K nonn,easy
%O 0,1
%A _Paul Curtz_, Jul 23 2019
%E a(14)-a(25) from _Stefano Spezia_, Jul 23 2019
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