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%I #51 Dec 11 2022 02:14:48
%S 1,7,23,59,135,291,607,1243,2519,5075,10191,20427,40903,81859,163775,
%T 327611,655287,1310643,2621359,5242795,10485671,20971427,41942943,
%U 83885979,167772055,335544211,671088527,1342177163,2684354439
%N a(n) = 5*2^n - 4*n - 5.
%C Row sums of A125233.
%C A triangle with left and right borders being the odd numbers 1,3,5,7,... will give the same partial sums for the sum of its rows. - _J. M. Bergot_, Sep 29 2012
%C The triangle in the above comment is constructed the same way as Pascal's triangle, i.e., C(n, k) = C(n-1, k) + C(n-1, k-1). - _Michael B. Porter_, Oct 03 2012
%H Vincenzo Librandi, <a href="/A126284/b126284.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-5,2).
%F a(1) = 1; a(2) = 7; a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3), n > 2.
%F The 6th diagonal from the right of A126277.
%F G.f.: x*(1+3*x)/(1-4*x+5*x^2-2*x^3). - _Colin Barker_, Feb 12 2012
%F E.g.f.: 5*exp(2*x) - (5+4*x)*exp(x). - _G. C. Greubel_, Oct 23 2018
%p A126284:=n->5*2^n-4*n-5; seq(A126284(n), n=1..50); # _Wesley Ivan Hurt_, Mar 27 2014
%t CoefficientList[Series[(1 + 3 x)/(1 - 4 x + 5 x^2 - 2 x^3), {x, 0, 50}], x] (* _Vincenzo Librandi_, Mar 28 2014 *)
%o (PARI) a(n)=5<<n-4*n-5 \\ _Charles R Greathouse IV_, Oct 03 2012
%o (Magma) [5*2^n - 4*n - 5: n in [1..30]]; // _G. C. Greubel_, Oct 23 2018
%o (GAP) List([1..30],n->5*2^n-4*n-5); # _Muniru A Asiru_, Oct 24 2018
%Y Cf. A000384, A125233, A126277.
%K nonn,easy
%O 1,2
%A _Gary W. Adamson_, Dec 24 2006
%E More terms from _Vladimir Joseph Stephan Orlovsky_, Oct 18 2008
%E New definition from _R. J. Mathar_, Sep 29 2012
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