OFFSET
0,2
COMMENTS
a(n) equals n!^3 times the determinant of the n X n matrix whose (i,j)-entry is KroneckerDelta[i, j] (((i^3 + 1)/(i^3)) - 1) + 1. - John M. Campbell, May 20 2011
Let b(0)=b(1)=1; b(n)=max(b(n-1)+(n-1)^3, b(n-2)+(n-2)^3); then a(n)=b(n+1). - Yalcin Aktar, Jul 28 2011
a(n-1) is the number of sets of n words of length n over binary alphabet where the first letter occurs n times. a(2) = 10: {aab,abb,bbb}, {aab,bab,bbb}, {aab,bba,bbb}, {aba,abb,bbb}, {aba,bab,bbb}, {aba,bba,bbb}, {abb,baa,bbb}, {abb,bab,bba}, {baa,bab,bbb}, {baa,bba,bbb}. - Alois P. Heinz, Feb 16 2023
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
a(n) = (n^4 + 2*n^3 + n^2 + 4)/4.
G.f.: (1 - 3*x + 10*x^2 - 3*x^3 + x^4)/(1-x)^5.
a(n) = 1 + C(n+1,2)^2 = 1 + A000537(n).
From G. C. Greubel, Sep 11 2016: (Start)
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
E.g.f.: (1/4)*(4 + 4*x + 14*x^2 + 8*x^3 + x^4)*exp(x). (End)
a(n) = a(n-1)+n^3. - Charles U. Lonappan, Jun 09 2021
MATHEMATICA
s = 1; lst = {s}; Do[s += n^3; AppendTo[lst, s], {n, 1, 42, 1}]; lst (* Zerinvary Lajos, Jul 12 2009 *)
Table[n!^3*Det[Array[KroneckerDelta[#1, #2](((#1^3+1)/(#1^3))-1)+1&, {n, n}]], {n, 1, 30}] (* John M. Campbell, May 20 2011 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {1, 2, 10, 37, 101}, 25] (* or *) Table[(n^4 + 2*n^3 + n^2 + 4)/4, {n, 0, 25}] (* G. C. Greubel, Sep 11 2016 *)
PROG
(Magma) [(n^4 + 2*n^3 + n^2 + 4)/4: n in [0..40]]; // Vincenzo Librandi, Feb 13 2015
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Jan 07 2009
STATUS
approved