A085540 -id:A085540 - cp's OEIS frontend

A085537 a(n) = n^4 - n^3.

0, 0, 8, 54, 192, 500, 1080, 2058, 3584, 5832, 9000, 13310, 19008, 26364, 35672, 47250, 61440, 78608, 99144, 123462, 152000, 185220, 223608, 267674, 317952, 375000, 439400, 511758, 592704, 682892, 783000, 893730, 1015808, 1149984, 1297032, 1457750, 1632960

Offset: 0

N. J. A. Sloane, Jul 05 2003

For n>=1, a(n) is equal to the number of functions f:{1,2,3,4}->{1,2,...,n} such that for a fixed x in {1,2,3,4} and a fixed y in {1,2,...,n} we have f(x)<>y. - Aleksandar M. Janjic and Milan Janjic, Mar 13 2007
Let K_n denote the complete graph on n (n>1) vertices. The sequence corresponds to the Wiener index of K_n X K_n (Cartesian product of K_n with itself). - K.V.Iyer, Mar 12 2009
Lewis proved that the order of a solvable nonabelian finite group |G| is less than or equal to e^4 - e^3, where when d is an irreducible character degree of G, then there is a positive integer e such that |G| = d(d+e). - Jonathan Vos Post, Jun 21 2012

Michael B. Porter, Table of n, a(n) for n = 0..100000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
Mark L. Lewis, Bounding group orders by large character degrees: A question of Snyder, arXiv:1206.4334 [math.GR], Jun 19 2012.
Eric Weisstein's World of Mathematics, Rook Graph.
Eric Weisstein's World of Mathematics, Wiener Index.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

A diagonal of A228273.
Cf. A000578, A000583, A092364, A146489.
Cf. A085540 (same sequence with initial 0 dropped).

Table[(n - 1) n^3, {n, 0, 20}] (* Eric W. Weisstein, Sep 08 2017 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 8, 54, 192, 500}, {0, 20}] (* Eric W. Weisstein, Sep 08 2017 *)
CoefficientList[Series[2 x^2 (4 + 7 x + x^2)/(1 - x)^5, {x, 0, 20}], x] (* Eric W. Weisstein, Sep 08 2017 *)

PARI
```
A085537(n) = n^4-n^3
```

From R. J. Mathar, Sep 12 2008: (Start)
a(n) = A085540(n-1).
G.f.: 2*x^2*(4 + 7*x + x^2)/(1-x)^5. (End)
a(n) = A000583(n) - A000578(n). - Omar E. Pol, Jun 23 2012
Sum_{n>=2} 1/a(n) = 3 - zeta(2) - zeta(3) = A152419. - Daniel Suteu, Feb 06 2017
a(n) = 2*A092364(n+1). - Bruno Berselli, Sep 08 2017
Sum_{n>=2} (-1)^n/a(n) = Pi^2/12 + 2*log(2) + 3*zeta(3)/4 - 3. - Amiram Eldar, Jul 05 2020
E.g.f.: exp(x)*x^2*(4 + 5*x + x^2). - Stefano Spezia, Jul 06 2021
Product_{n>=2} (1 - 1/a(n)) = A146489. - Amiram Eldar, Nov 22 2022

A092364 a(n) = n^2*binomial(n,2).

Original entry on oeis.org

0, 4, 27, 96, 250, 540, 1029, 1792, 2916, 4500, 6655, 9504, 13182, 17836, 23625, 30720, 39304, 49572, 61731, 76000, 92610, 111804, 133837, 158976, 187500, 219700, 255879, 296352, 341446, 391500, 446865, 507904, 574992, 648516, 728875, 816480

Offset: 1

Jon Perry, Mar 19 2004

Coefficient of x^2 in expansion of (1+n*x)^n.
For n>3, a(n) is twice the area of a triangle with vertices at points (C(n-1,3),C(n,3)), (C(n,3),C(n+1,3)), and (C(n+1,3),C(n+2,3)). - J. M. Bergot, Jun 05 2014
Also the Harary index of the n X n rook complement graph for n != 2. - Eric W. Weisstein, Sep 14 2017

Eric Weisstein's World of Mathematics, Harary Index
Eric Weisstein's World of Mathematics, Rook Complement Graph
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A085540.

[n^3*(n-1)/2: n in [1..50]]; // Wesley Ivan Hurt, Jun 04 2014

A092364 := proc(n) n^3*(n-1)/2 ; end proc: # R. J. Mathar, Mar 10 2011

f[n_]:=(n^4-n^3)/2; lst={};Do[AppendTo[lst,f[n]],{n,5!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 04 2009 *)
Table[n^2 Binomial[n, 2], {n, 20}] (* Eric W. Weisstein, Sep 14 2017 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 4, 27, 96, 250}, 20] (* Eric W. Weisstein, Sep 14 2017 *)
CoefficientList[Series[-((x (4 + 7 x + x^2))/(-1 + x)^5), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 14 2017 *)

z(n)=n^2*binomial(n,2); for(i=1,40,print1(","z(i)))

a(n) = n^3*(n-1)/2. Equals A085540(n-1)/2. - Zerinvary Lajos, May 09 2007, corrected Mar 10 2011
G.f.: -x^2*(4+7*x+x^2) / (x-1)^5. - R. J. Mathar, Mar 10 2011
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Eric W. Weisstein, Sep 14 2017
From Amiram Eldar, May 14 2022: (Start)
Sum_{n>=2} 1/a(n) = 6 - Pi^2/3 - 2*zeta(3).
Sum_{n>=2} (-1)^n/a(n) = Pi^2/6 + 4*log(2) + 3*zeta(3)/2 - 6. (End)
E.g.f.: exp(x)*x^2*(4 + 5*x + x^2)/2. - Stefano Spezia, Jun 10 2023

cp's OEIS Frontend

A085537 a(n) = n^4 - n^3.

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