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a(n) = n^4 - n^3.
11

%I #71 Nov 22 2022 02:35:08

%S 0,0,8,54,192,500,1080,2058,3584,5832,9000,13310,19008,26364,35672,

%T 47250,61440,78608,99144,123462,152000,185220,223608,267674,317952,

%U 375000,439400,511758,592704,682892,783000,893730,1015808,1149984,1297032,1457750,1632960

%N a(n) = n^4 - n^3.

%C 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

%C 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

%C 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

%H Michael B. Porter, <a href="/A085537/b085537.txt">Table of n, a(n) for n = 0..100000</a>

%H Milan Janjic, <a href="https://pmf.unibl.org/wp-content/uploads/2017/10/enumfun.pdf">Enumerative Formulas for Some Functions on Finite Sets</a>.

%H Mark L. Lewis, <a href="http://arxiv.org/abs/1206.4334">Bounding group orders by large character degrees: A question of Snyder</a>, arXiv:1206.4334 [math.GR], Jun 19 2012.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/RookGraph.html">Rook Graph</a>.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/WienerIndex.html">Wiener Index</a>.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F From _R. J. Mathar_, Sep 12 2008: (Start)

%F a(n) = A085540(n-1).

%F G.f.: 2*x^2*(4 + 7*x + x^2)/(1-x)^5. (End)

%F a(n) = A000583(n) - A000578(n). - _Omar E. Pol_, Jun 23 2012

%F Sum_{n>=2} 1/a(n) = 3 - zeta(2) - zeta(3) = A152419. - _Daniel Suteu_, Feb 06 2017

%F a(n) = 2*A092364(n+1). - _Bruno Berselli_, Sep 08 2017

%F Sum_{n>=2} (-1)^n/a(n) = Pi^2/12 + 2*log(2) + 3*zeta(3)/4 - 3. - _Amiram Eldar_, Jul 05 2020

%F E.g.f.: exp(x)*x^2*(4 + 5*x + x^2). - _Stefano Spezia_, Jul 06 2021

%F Product_{n>=2} (1 - 1/a(n)) = A146489. - _Amiram Eldar_, Nov 22 2022

%t Table[(n - 1) n^3, {n, 0, 20}] (* _Eric W. Weisstein_, Sep 08 2017 *)

%t LinearRecurrence[{5, -10, 10, -5, 1}, {0, 8, 54, 192, 500}, {0, 20}] (* _Eric W. Weisstein_, Sep 08 2017 *)

%t CoefficientList[Series[2 x^2 (4 + 7 x + x^2)/(1 - x)^5, {x, 0, 20}], x] (* _Eric W. Weisstein_, Sep 08 2017 *)

%o (PARI) A085537(n) = n^4-n^3

%Y A diagonal of A228273.

%Y Cf. A000578, A000583, A092364, A146489.

%Y Cf. A085540 (same sequence with initial 0 dropped).

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_, Jul 05 2003