Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #25 May 22 2024 02:13:43
%S 0,1,7,33,139,565,2271,9097,36403,145629,582535,2330161,9320667,
%T 37282693,149130799,596523225,2386092931,9544371757,38177487063,
%U 152709948289,610839793195,2443359172821,9773436691327,39093746765353,156374987061459,625499948245885,2501999792983591
%N a(n) = 4*a(n-1) + 2*n - 1.
%H Stefano Spezia, <a href="/A141291/b141291.txt">Table of n, a(n) for n = 0..1500</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-9,4).
%F a(n) = 4*a(n-1) + 2*n-1, given a(0) = 0, a(1) = 1.
%F Row sums of triangle A141290 starting with offset 1.
%F From _R. J. Mathar_, Feb 02 2010: (Start)
%F a(n) = 6*a(n-1) -9*a(n-2) +4*a(n-3).
%F G.f.: x*(1+x)/((1-4*x)*(x-1)^2). (End)
%F E.g.f.: exp(x)*(5*(exp(3*x) - 1) - 6*x) /9. - _Stefano Spezia_, May 21 2024
%e a(4) = 139 = 4*a(3) + 7 = 4*33 + 7.
%e a(4) = 139 = sum of row 4 terms of triangle A141290 = (64, + 48 + 20 + 7).
%t LinearRecurrence[{6,-9,4},{0,1,7},27] (* _Stefano Spezia_, May 21 2024 *)
%Y Cf. A141290.
%K nonn,easy
%O 0,3
%A _Gary W. Adamson_, Jun 22 2008
%E Definition and formula corrected by _Paolo P. Lava_, Oct 07 2008
%E More terms from _R. J. Mathar_, Feb 02 2010