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A048776 First partial sums of A048739; second partial sums of A000129.

%I A048776 #38 May 13 2023 20:37:02
%S A048776 1,4,12,32,81,200,488,1184,2865,6924,16724,40384,97505,235408,568336,
%T A048776 1372096,3312545,7997204,19306972,46611168,112529329,271669848,
%U A048776 655869048,1583407968,3822685009,9228778012,22280241060,53789260160,129858761409,313506783008
%N A048776 First partial sums of A048739; second partial sums of A000129.
%H A048776 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-4,0,1).
%F A048776 a(n) = 2*a(n-1) + a(n-2) + n + 1; a(0)=1, a(1)=4.
%F A048776 a(n) = (((7/2 + (5/2)*sqrt(2))*(1+sqrt(2))^n - (7/2 - (5/2)*sqrt(2))*(1-sqrt(2))^n)/2*sqrt(2)) - (n+3)/2.
%F A048776 a(n) = (A000129(n+3) - (n+3))/2 = Sum_{j} A047662(n-j+1, j+1). - _Henry Bottomley_, Jul 09 2001
%F A048776 From _R. J. Mathar_, Feb 06 2010: (Start)
%F A048776 a(n) = 4*a(n-1) - 4*a(n-2) + a(n-4).
%F A048776 G.f.: -1/((x^2+2*x-1) * (x-1)^2). (End)
%F A048776 Define an array with m(n,1)=1 and m(1,k) = k*(k+1)/2 for n=1,2,3,...  The interior terms are m(n,k) = m(n,k-1) + m(n-1,k-1) + m(n-1,k). The sum of the terms in each antidiagonal=a(n). - _J. M. Bergot_, Dec 01 2012 [This is A154948 without the first column. The diagonal is m(n,n) = A161731(n-1). _R. J. Mathar_, Dec 06 2012]
%F A048776 E.g.f.: exp(x)*(10*cosh(sqrt(2)*x) + 7*sqrt(2)*sinh(sqrt(2)*x) - 2*(3 + x))/4. - _Stefano Spezia_, May 13 2023
%p A048776 with(combinat):seq((fibonacci(n+3, 2)-n-3)/2, n=0..25); # _Zerinvary Lajos_, Jun 02 2008
%t A048776 a=b=0;Table[c=2*b+a+n;a=b;b=c,{n,1,40}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 02 2011*)
%t A048776 LinearRecurrence[{4,-4,0,1},{1,4,12,32},30] (* _Harvey P. Dale_, Aug 27 2014 *)
%Y A048776 Cf. A001333, A000129, A048739.
%K A048776 easy,nonn
%O A048776 0,2
%A A048776 _Barry E. Williams_
%E A048776 More terms from _Harvey P. Dale_, Aug 27 2014