OFFSET
0,2
COMMENTS
For A049453(n+1), the corresponding formula would be a(n)=(n+1)*(6*n+7) and its partial sums would be given by a(n)=(n+1)*(n+2)*(4*n+7)/2.
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
a(n) = n*(n+1)*(4*n+3)/2.
G.f.: x*(7+5*x)/(1-x)^4. - Colin Barker, Jan 12 2012
a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4). - Vincenzo Librandi, Apr 27 2012
MATHEMATICA
Table[(n(4n-1)(n-1))/2, {n, 40}] (* Harvey P. Dale, Mar 11 2011 *)
CoefficientList[Series[x*(7+5*x)/(1-x)^4, {x, 0, 50}], x] (* Vincenzo Librandi, Apr 27 2012 *)
PROG
(Magma) I:=[0, 7, 33, 90]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Apr 27 2012
(PARI) a(n) = n*(n+1)*(4*n+3)/2; \\ Altug Alkan, Apr 20 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Barry E. Williams, Dec 17 1999
STATUS
approved