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a(n) = A000129(n)*A000129(n+1)/2.
16

%I #56 Sep 28 2024 21:16:26

%S 0,1,5,30,174,1015,5915,34476,200940,1171165,6826049,39785130,

%T 231884730,1351523251,7877254775,45912005400,267594777624,

%U 1559656660345,9090345184445,52982414446326,308804141493510,1799842434514735,10490250465594899,61141660359054660,356359711688733060

%N a(n) = A000129(n)*A000129(n+1)/2.

%C May be called Pell triangles.

%H Vincenzo Librandi, <a href="/A084158/b084158.txt">Table of n, a(n) for n = 0..500</a>

%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Barry4/barry64.html">Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays</a>, JIS 12 (2009) 09.8.6.

%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 Ronald Orozco López, <a href="https://arxiv.org/abs/2408.08943">Generating Functions of Generalized Simplicial Polytopic Numbers and (s,t)-Derivatives of Partial Theta Function</a>, arXiv:2408.08943 [math.CO], 2024. See p. 11.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,5,-1).

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

%F a(n) = 5*a(n-1) + 5*a(n-2) - a(n-3). - _Mohamed Bouhamida_, Sep 02 2006; corrected by _Antonio Alberto Olivares_, Mar 29 2008

%F a(n) = (-1/8)*(-1)^n + (( sqrt(2)+1)/16)*(3+2*sqrt(2))^n + ((-sqrt(2)+1)/16)*(3-2*sqrt(2))^n. - _Antonio Alberto Olivares_, Mar 30 2008

%F sqrt(a(n) - a(n-1)) = A000129(n). - _Antonio Alberto Olivares_, Mar 30 2008

%F O.g.f.: x/((1+x)(1-6*x+x^2)). - _R. J. Mathar_, May 18 2008

%F a(n) = A041011(n)*A041011(n+1). - _R. K. Guy_, May 18 2008

%F From _Mohamed Bouhamida_, Aug 30 2008: (Start)

%F a(n) = 6*a(n-1) - a(n-2) - (-1)^n.

%F a(n) = 7*(a(n-1) - a(n-2)) + a(n-3) - 2*(-1)^n. (End)

%F In general, for n>k+1, a(n+k) = A003499(k+1)*a(n-1) - a(n-k-2) - (-1)^n A000129(k+1)^2. - _Charlie Marion_, Jan 04 2012

%F For n>0, a(2n-1)*a(2n+1) = oblong(a(2n)); a(2n)*a(2n+2) = oblong(a(2n+1)-1). - _Charlie Marion_, Jan 09 2012

%F a(n) = A046729(n)/4. - _Wolfdieter Lang_, Mar 07 2012

%F a(n) = sum of squares of first n Pell numbers A000129 (A079291). - _N. J. A. Sloane_, Jun 18 2012

%F a(n) = (A002315(n) - (-1)^n)/8. - _Adam Mohamed_, Sep 05 2024

%p with(combinat): a:=n->fibonacci(n,2)*fibonacci(n-1,2)/2: seq(a(n), n=1..22); # _Zerinvary Lajos_, Apr 04 2008

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

%o (Magma) [Floor(((Sqrt(2)+1)^(2*n+1)-(Sqrt(2)-1)^(2*n+1)-2*(-1)^n)/16): n in [0..35]]; // _Vincenzo Librandi_, Jul 05 2011

%o (PARI) Pell(n)=([2, 1; 1, 0]^n)[2, 1];

%o a(n)=Pell(n)*Pell(n+1)/2 \\ _Charles R Greathouse IV_, Mar 21 2016

%o (PARI) a(n)=([0,1,0; 0,0,1; -1,5,5]^n*[0;1;5])[1,1] \\ _Charles R Greathouse IV_, Mar 21 2016

%o (SageMath) [(lucas_number2(2*n+1,2,-1) -2*(-1)^n)/16 for n in (0..30)] # _G. C. Greubel_, Aug 18 2022

%Y Cf. A000129, A001652, A001654, A002203, A002315, A003499.

%Y Cf. A041011, A046729, A079291, A084159, A084175.

%K easy,nonn

%O 0,3

%A _Paul Barry_, May 18 2003