OFFSET
1,2
COMMENTS
In general, for sequences that multiply the first k natural numbers, and then add the product of the next k natural numbers (preserving the order of operations up to n), we have a(n) = Sum_{i=1..floor(n/k)} (k*i)!/(k*i-k)! + Sum_{j=1..k-1} (1-sign((n-j) mod k)) * (Product_{i=1..j} n-i+1). Here, k=2. - Wesley Ivan Hurt, Sep 10 2018
a(2n) is the total area of the family of n rectangles, where the k-th rectangle has dimensions (2k) X (2k-1). - Wesley Ivan Hurt, Oct 01 2018
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).
FORMULA
a(n) = (1/12)*(2*n^3+4*n+3/2+(3*n^2-6*n-3/2)*(-1)^n). [based on Alcover program]
G.f.: x*(x^5 - x^4 + 6*x^3 + x + 1)/((x-1)^4*(x+1)^3). [Joerg Arndt, Sep 13 2013]
E.g.f.: (x*(9 + 9*x + 2*x^2)*cosh(x) + (3 + 3*x + 3*x^2 + 2*x^3)*sinh(x))/12. - Stefano Spezia, Apr 18 2023
EXAMPLE
1 = 1
1*2 = 2
1*2 + 3 = 5
1*2 + 3*4 = 14
1*2 + 3*4 + 5 = 19
1*2 + 3*4 + 5*6 = 44
1*2 + 3*4 + 5*6 + 7 = 51
1*2 + 3*4 + 5*6 + 7*8 = 100
1*2 + 3*4 + 5*6 + 7*8 + 9 = 109
1*2 + 3*4 + 5*6 + 7*8 + 9*10 = 190
...
MATHEMATICA
a[n_?OddQ] := (2*n^3-3*n^2+10*n+3)/12; a[n_?EvenQ] := n*(n+2)*(2*n-1)/12; Table[a[n], {n, 1, 40}] (* Jean-François Alcover, Sep 10 2013 *)
CoefficientList[Series[x(x^5 - x^4 + 6*x^3 + x + 1)/((x-1)^4*(x+1)^3), {x, 0, 40}], x] (* Stefano Spezia, Sep 23 2018 *)
LinearRecurrence[{1, 3, -3, -3, 3, 1, -1}, {1, 2, 5, 14, 19, 44, 51}, 50] (* Harvey P. Dale, Mar 11 2023 *)
PROG
(PARI) Vec( x*(x^5 - x^4 + 6*x^3 + x + 1)/((x-1)^4*(x+1)^3) + O(x^66) ) \\ Joerg Arndt, Sep 17 2013
(Magma) [(1/12)*(2*n^3+4*n+3/2+(3*n^2-6*n-3/2)*(-1)^n): n in [1..50]]; // Vincenzo Librandi, Sep 11 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Robert Pfister, Sep 09 2013
EXTENSIONS
Definition corrected by Ivan Panchenko, Dec 02 2013
STATUS
approved