login

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”).

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