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%I #17 Feb 05 2022 09:15:01
%S 0,2,5,12,35,122,465,1832,7295,29142,116525,466052,1864155,7456562,
%T 29826185,119304672,477218615,1908874382,7635497445,30541989692,
%U 122167958675,488671834602,1954687338305,7818749353112,31274997412335,125099989649222,500399958596765
%N a(n) = (4^n + 15*n - 1)/9.
%C Conjecture: satisfies a linear recurrence having signature (6, -9, 4). (This is correct, see Formula section.)
%D Roman Andronov, How can I prove that 4^n+15n-1 is divisible by 9?, Quora Digest (Nov. 17, 2018).
%H Colin Barker, <a href="/A323397/b323397.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-9,4).
%F a(n+1) = 4*a(n) - 5*n + 2, with a(0)=0, a(1)=2. This implies a(n+2) = 5*a(n+1) - 4*a(n) - 5, and also that a(n+3) = 6*a(n+2) - 9*a(n+1) + 4*a(n). - _N. J. A. Sloane_, Jan 13 2019
%F G.f.: x*(2 - 7*x) / ((1 - x)^2*(1 - 4*x)). - _Colin Barker_, Jan 19 2019
%t Table[(4^n+15n-1)/9,{n,0,40}]
%o (PARI) concat(0, Vec(x*(2 - 7*x) / ((1 - x)^2*(1 - 4*x)) + O(x^30))) \\ _Colin Barker_, Jan 19 2019
%K nonn,easy
%O 0,2
%A _Harvey P. Dale_, Jan 13 2019
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