%I #49 Jun 12 2021 06:36:04
%S 1,6,21,52,105,186,301,456,657,910,1221,1596,2041,2562,3165,3856,4641,
%T 5526,6517,7620,8841,10186,11661,13272,15025,16926,18981,21196,23577,
%U 26130,28861,31776,34881,38182,41685,45396,49321,53466,57837,62440,67281,72366
%N q-factorial numbers 3!_q.
%C Number of proper n-colorings of the 4-cycle with one vertex color fixed (offset 2). - _Michael Somos_, Jul 19 2002
%C n such that x^3 + x^2 + x + n factors over the integers. - _James R. Buddenhagen_, Apr 19 2005
%C If Y is a 4-subset of an n-set X then, for n>=5, a(n-5) is the number of 5-subsets of X having at least two elements in common with Y. - _Milan Janjic_, Dec 08 2007
%C Equals row sums of the Connell (A001614) sequence read as a triangle. - _Gary W. Adamson_, Sep 01 2008
%C Binomial transform of 1, 5, 10, 6, 0, 0, 0 (0 continued). - _Philippe Deléham_, Mar 17 2014
%C Digital root is A251780. - _Peter M. Chema_, Jul 11 2016
%D T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.
%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 + 1)*(n^2 + n + 1).
%F a(n) = (n+1)^3-2*T(n) where T(n) =n*(n+1)/2= A000217(n) is the n-th triangular number. - Herman Jamke (hermanjamke(AT)fastmail.fm), Sep 14 2006
%F a(n) = n^8 mod (n^3+n), with offset 1..a(1)=1. - _Gary Detlefs_, May 02 2010
%F a(n) = 4*a(n-1)-6*a(n-2)+ 4*a(n-3)- a(n-4), n>3. - _Harvey P. Dale_, Jul 11 2011
%F G.f.: (1+2*x+3*x^2)/(1-x)^4. - _Harvey P. Dale_, Jul 11 2011
%F For n>0 a(n) = Sum_{k=A000217(n-1)...A000217(n+1)} k. - _J. M. Bergot_, Feb 11 2015
%F E.g.f.: (1 + 5*x + 5*x^2 + x^3)*exp(x). - _Ilya Gutkovskiy_, Jul 11 2016
%e For 2-colorings only 1212 is proper so a(2-2)=1. The proper 3-colorings are: 1212,1313,1213,1312,1232,1323 so a(3-2)=6.
%e a(0) = 1*1 = 1;
%e a(1) = 1*1 + 5*1 = 6;
%e a(2) = 1*1 + 5*2 + 10*1 = 21;
%e a(3) = 1*1 + 5*3 + 10*3 + 6*1 = 52;
%e a(4) = 1*1 + 5*4 + 10*6 + 6*4 = 105; etc. - _Philippe Deléham_, Mar 17 2014
%p A069778 := proc(n)
%p (n+1)*(n^2+n+1) ;
%p end proc: # _R. J. Mathar_, Aug 24 2013
%t LinearRecurrence[{4, -6, 4, -1}, {1, 6, 21, 52}, 41] (* or *) Table[(n + 1) (n^2 + n + 1), {n, 0, 41}] (* _Harvey P. Dale_, Jul 11 2011 *)
%t Table[QFactorial[3, n], {n, 0, 41}] (* _Arkadiusz Wesolowski_, Oct 31 2012 *)
%o (PARI) a(n)=(n+1)*(n^2+n+1)
%Y Cf. A069777, A069779, A218503, A056108 (first differences).
%Y Cf. A001614. - _Gary W. Adamson_, Sep 01 2008
%Y Cf. A226449. - _Bruno Berselli_, Jun 09 2013
%K nonn,easy
%O 0,2
%A _Franklin T. Adams-Watters_, Apr 07 2002