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a(n) = A054413(n-1)^2, n >= 1.
7

%I #37 Mar 23 2024 08:06:33

%S 0,1,49,2500,127449,6497401,331240000,16886742601,860892632649,

%T 43888637522500,2237459621014849,114066552034234801,

%U 5815156694124960000,296458924848338725201,15113590010571150025249,770496631614280312562500

%N a(n) = A054413(n-1)^2, n >= 1.

%C See the comment in A099279. This is example a=7.

%H Michael A. Allen and Kenneth Edwards, <a href="https://www.fq.math.ca/Papers1/60-5/allen.pdf">Fence tiling derived identities involving the metallonacci numbers squared or cubed</a>, Fib. Q. 60:5 (2022) 5-17.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (50,50,-1).

%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>

%F a(n) = A054413(n-1)^2, n >= 1. a(0)=0.

%F a(n) = 50*a(n-1) + 50*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=49.

%F a(n) = 51*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.

%F a(n) = 2*(T(n, 51/2) - (-1)^n)/53 with twice the Chebyshev polynomials of the first kind: 2*T(n, 51/2) = A099368(n).

%F G.f.: x*(1-x)/((1-51*x+x^2)*(1+x)) = x*(1-x)/(1-50*x-50*x^2+x^3).

%F a(n+1) = (1 + (-1)^n)/2 + 49*Sum_{k=1..n} k*a(n+1-k). - _Michael A. Allen_, Feb 21 2023

%t LinearRecurrence[{50,50,-1},{0,1,49},20] (* _Harvey P. Dale_, Jul 27 2023 *)

%Y Cf. A054413.

%Y Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, A099365, A099366, this sequence, A099369, A099372, A099374.

%K nonn,easy

%O 0,3

%A _Wolfdieter Lang_, Oct 18 2004