|
%I #38 May 14 2023 02:10:59
%S 1,5,16,44,113,281,688,1672,4049,9789,23648,57108,137889,332913,
%T 803744,1940432,4684641,11309749,27304176,65918140,159140497,
%U 384199177,927538896,2239277016,5406092977,13051463021,31509019072,76069501220,183648021569,443365544417
%N First partial sums of A005409; second partial sums of A001333.
%C Form an array having the first column all 1's and the first row the squares 1, 4, 9, ..., so m(n,1) = 1 and m(1,n) = n^2 for n = 1, 2, 3, ..., and let the interior terms be m(i,j) = m(i,j-1) + m(i-1,j-1) + m(i-1,j). Then the sums of the terms in the antidiagonals are the terms of this sequence. - _J. M. Bergot_, Nov 16 2012
%C Define a triangle with T(n,n)=n+1 and T(n,0)=n*(n+1)+1 for n >= 0. Define the interior terms via T(r,c) = T(r-2,c-1) + T(r-1,c-1) + T(r-1,c). Then the row sums are a(n) = Sum_{k=0..n} T(n,k). - _J. M. Bergot_, Feb 27 2013
%H G. C. Greubel, <a href="/A048777/b048777.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-4,0,1).
%F a(n) = 2*a(n-1) + a(n-2) + 2*n+1 with a(0)=1, a(1)=5.
%F a(n) = ( {(5+(7/2)*sqrt(2))*(1+sqrt(2))^n - (5-(7/2)*sqrt(2))*(1-sqrt(2))^n}/2*sqrt(2) ) - (2*n+5)/2.
%F a(n) = (1/2)*( Pell(n+3) + Pell(n+2) -2*n -5 ), with Pell(n) = A000129(n). - _Ralf Stephan_, May 15 2007
%F From _Colin Barker_, Sep 20 2012: (Start)
%F a(n) = 4*a(n-1) - 4*a(n-2) + a(n-4).
%F G.f.: (1+x)/((1-x)^2*(1-2*x-x^2)). (End)
%F a(n) = A048776(n-1) + A048776(n). - _R. J. Mathar_, Feb 28 2013
%F a(n) = (A002203(n+3) - 2*(2*n+5))/4. - _G. C. Greubel_, Apr 23 2021
%F E.g.f.: exp(x)*(7*cosh(sqrt(2)*x) + 5*sqrt(2)*sinh(sqrt(2)*x) - 2*x - 5)/2. - _Stefano Spezia_, May 13 2023
%t LinearRecurrence[{4,-4,0,1},{1,5,16,44},40] (* _Harvey P. Dale_, Nov 12 2017 *)
%t Table[(LucasL[n+3, 2] -2*(2n+5))/4, {n,0,35}] (* _G. C. Greubel_, Apr 23 2021 *)
%o (Magma) I:=[1,5,16,44]; [n le 4 select I[n] else 4*Self(n-1) -4*Self(n-2) +Self(n-4): n in [1..36]]; // _G. C. Greubel_, Apr 23 2021
%o (Sage) [(lucas_number2(n+3,2,-1) -2*(2*n+5))/4 for n in (0..35)] # _G. C. Greubel_, Apr 23 2021
%Y Cf. A000129, A001333, A002203, A005409, A048776.
%K easy,nonn
%O 0,2
%A _Barry E. Williams_
%E More terms from _Harvey P. Dale_, Nov 12 2017
|
