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Sum of the areas of the first n+1 Pell triangles.
4

%I #26 Sep 25 2017 11:03:21

%S 0,1,6,36,210,1225,7140,41616,242556,1413721,8239770,48024900,

%T 279909630,1631432881,9508687656,55420693056,323015470680,

%U 1882672131025,10973017315470,63955431761796,372759573255306,2172602007770041

%N Sum of the areas of the first n+1 Pell triangles.

%C Convolution of A059841(n) and A001109(n+1).

%C Partial sums of A084158.

%H S. Falcon, <a href="https://www.researchgate.net/publication/298789400_On_the_Sequences_of_Products_of_Two_k-Fibonacci_Numbers">On the Sequences of Products of Two k-Fibonacci Numbers</a>, American Review of Mathematics and Statistics, March 2014, Vol. 2, No. 1, pp. 111-120.

%H Roger B. Nelson, <a href="http://www.jstor.org/stable/10.4169/math.mag.89.3.159">Multi-Polygonal Numbers</a>, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (6,0,-6,1).

%F G.f.: x/((1-x)*(1+x)*(1-6*x+x^2)).

%F a(n) = 6*a(n-1)-6*a(n-3)+a(n-4).

%F a(n) = (3-2*sqrt(2))^n*(3/32-sqrt(2)/16)+(3+2*sqrt(2))^n*(sqrt(2)/16+3/32)-(-1)^n/16-1/8.

%F a(n) = sum{k=0..n, (sqrt(2)*(sqrt(2)+1)^(2*k)/8-sqrt(2)*(sqrt(2)-1)^(2*k)/8)*((1+(-1)^(n-k))/2.

%F a(n) = sum{k=0..n, A001029(k)*A001029(k+1)/2}.

%F a(n) = (A001333(n+1)^2 - 1)/8 = ((A000129(n) + A000129(n+1))^2 - 1)/8. - _Richard R. Forberg_, Aug 25 2013

%F a(n) = A002620(A000129(n+1)) = A000217(A048739(n-1)), n > 0. - _Ivan N. Ianakiev_, Jun 21 2014

%t Accumulate[LinearRecurrence[{5,5,-1},{0,1,5},30]] (* _Harvey P. Dale_, Sep 07 2011 *)

%t LinearRecurrence[{6, 0, -6, 1},{0, 1, 6, 36},22] (* _Ray Chandler_, Aug 03 2015 *)

%Y Cf. A096977, A064831, A096978.

%K easy,nonn

%O 0,3

%A _Paul Barry_, Jul 17 2004