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A085490 Number of pairs with two different elements which can be obtained by selecting unique elements from two sets with n+1 and n^2 elements respectively and n common elements. 5

%I #40 Sep 08 2022 08:45:11

%S 0,1,10,33,76,145,246,385,568,801,1090,1441,1860,2353,2926,3585,4336,

%T 5185,6138,7201,8380,9681,11110,12673,14376,16225,18226,20385,22708,

%U 25201,27870,30721,33760,36993,40426,44065,47916,51985,56278,60801,65560,70561,75810,81313

%N Number of pairs with two different elements which can be obtained by selecting unique elements from two sets with n+1 and n^2 elements respectively and n common elements.

%H Vincenzo Librandi, <a href="/A085490/b085490.txt">Table of n, a(n) for n = 0..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) = n^3 + n^2 - n = n*A028387(n-1).

%F a(n) = A081437(n-1), n>0. - _R. J. Mathar_, Sep 12 2008

%F G.f.: x*(1+6*x-x^2)/(1-x)^4. - _Robert Israel_, Dec 05 2014

%F E.g.f.: x*(1+4*x+x^2)*exp(x). - _Robert Israel_, Dec 05 2014

%F For q a prime power, a(q) is the number of pairs of commuting nilpotent 2*2 matrices with coefficients in GL(q). (Proof: the zero matrix commutes with all q^2 nilpotent matrices, each of the remaining q^2-1 nilpotent matrices commutes with exactly q nilpotent matrices.) - _Mark Wildon_, Jun 18 2017

%e a(2) = 10 because we can write a(2) = 2^3 + 2^2 - 2 = 10.

%p a:=n->sum(n*k, k=0..n):seq(a(n)+sum(n*k, k=2..n), n=0..30); # _Zerinvary Lajos_, Jun 10 2008

%p a:=n->sum(-2+sum(2+sum(2, j=1..n),j=1..n),j=1..n):seq(a(n)/2,n=0..40);# _Zerinvary Lajos_, Dec 06 2008

%p seq(n^3+n^2-n, n=0..100); # _Robert Israel_, Dec 05 2014

%t LinearRecurrence[{4, -6, 4, -1}, {0, 1, 10, 33}, 60] (* _Vincenzo Librandi_, Jun 22 2017 *)

%o (Magma) [n^3+n^2-n: n in [0..50]]; // _Vincenzo Librandi_, Jun 22 2017

%Y Cf. A270109.

%K nonn,easy

%O 0,3

%A Polina S. Dolmatova (polinasport(AT)mail.ru), Aug 15 2003

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