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a(n) = (n - 1)*(4*n^2 - 8*n + 5).
5

%I #70 Sep 03 2024 01:19:50

%S 0,5,34,111,260,505,870,1379,2056,2925,4010,5335,6924,8801,10990,

%T 13515,16400,19669,23346,27455,32020,37065,42614,48691,55320,62525,

%U 70330,78759,87836,97585,108030,119195,131104,143781,157250,171535,186660,202649,219526,237315,256040,275725,296394,318071

%N a(n) = (n - 1)*(4*n^2 - 8*n + 5).

%C Conjecture: For n > 1, a(n) is the maximum eigenvalue of a 2*(n-1) X 2*(n-1) square matrix M defined as M[i,j,n] = j + n*(i-1) if i is odd and M[i,j,n] = n*i - j + 1 if i is even (see A317614). - _Stefano Spezia_, Dec 27 2018

%C Connections can be made to A022144 and A010014. Namely, a formula for A022144 is (2*n+1)^2 - (2*n-1)^2. A formula for A010014 is (2*n+1)^3 - (2*n-1)^3. The general form can be represented by (2*n+1)^d - (2*n-1)^d, where d designates the number of dimensions. When d is 4, a(n) = ((2*(n-1)+1)^4 - (2*(n-1)-1)^4)/16, namely the general form shifted by 1 and divided by 16 is a(n). - _Yigit Oktar_, Aug 16 2024

%H Colin Barker, <a href="/A317297/b317297.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F a(n) = 4*n^3 - 12*n^2 + 13*n - 5 = A033430(n) - A135453(n) + A008595(n) - 5.

%F G.f.: x^2*(5 + 14*x + 5*x^2)/(1 - x)^4. - _Colin Barker_, Sep 01 2018

%F a(n) = 4*a(n - 1) - 6*a(n - 2) + 4*a(n - 3) - a(n - 4) for n > 4. - _Stefano Spezia_, Sep 01 2018

%F E.g.f.: exp(x)*(5*x + 12*x^2 + 4*x^3). - _Stefano Spezia_, Jan 15 2019

%F a(n) = ((2*(n-1)+1)^4 - (2*(n-1)-1)^4)/16. - _Yigit Oktar_, Aug 16 2024

%t Table[(n - 1) (4 n^2 - 8 n + 5), {n, 1, 50}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {0, 5, 34, 111}, 50] (* or *) CoefficientList[Series[x (5 + 14 x + 5 x^2)/(1 - x)^4, {x, 0, 50}], x] (* _Stefano Spezia_, Sep 01 2018 *)

%o (PARI) a(n) = (n - 1)*(4*n^2 - 8*n + 5)

%o (PARI) concat(0, Vec(x^2*(5 + 14*x + 5*x^2)/(1 - x)^4 + O(x^50))) \\ _Colin Barker_, Sep 01 2018

%Y First bisection of A006003.

%Y Nonzero terms give the row sums of A007607.

%Y Conjecture: 0 together with a bisection of A246697.

%Y Cf. A219086 (partial sums).

%Y Cf. A008595, A033430, A135453, A317614.

%Y Cf. A010014, A022144 (see comments)

%K nonn,easy

%O 1,2

%A _Omar E. Pol_, Sep 01 2018