A317297 a(n) = (n - 1)*(4*n^2 - 8*n + 5).

0, 5, 34, 111, 260, 505, 870, 1379, 2056, 2925, 4010, 5335, 6924, 8801, 10990, 13515, 16400, 19669, 23346, 27455, 32020, 37065, 42614, 48691, 55320, 62525, 70330, 78759, 87836, 97585, 108030, 119195, 131104, 143781, 157250, 171535, 186660, 202649, 219526, 237315, 256040, 275725, 296394, 318071

Offset

1,2

Comments

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

Links

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Formula

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

G.f.: x^2*(5 + 14*x + 5*x^2)/(1 - x)^4. - Colin Barker, Sep 01 2018

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

E.g.f.: exp(x)*(5*x + 12*x^2 + 4*x^3). - Stefano Spezia, Jan 15 2019

Mathematica

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 *)

Prog

(PARI) a(n) = (n - 1)*(4*n^2 - 8*n + 5)
(PARI) concat(0, Vec(x^2*(5 + 14*x + 5*x^2)/(1 - x)^4 + O(x^50))) \\ Colin Barker, Sep 01 2018

Crossrefs

First bisection of A006003.

Nonzero terms give the row sums of A007607.

Conjecture: 0 together with a bisection of A246697.

Cf. A219086 (partial sums).

Cf. A008595, A033430, A135453, A317614.

Sequence in context: A039773 A289947 A272955 * A135973 A186636 A034224

Adjacent sequences: A317294 A317295 A317296 * A317298 A317299 A317300

Keyword

nonn,easy

Author

Omar E. Pol, Sep 01 2018

Status

approved