OFFSET
0,1
COMMENTS
Let x(n) = (1/2)*(-(2*n + 1) + sqrt((2*n + 1)^2 + 4)) and f(n,k) = (-1)*Sum_{i=1..k} Sum_{j=1..i} (-1)^floor(j*x(n)). Then a(n) = k is the least integer k > 0 such that f(n, k) = 0. In other words, f(n, a(n)) = 0, and if f(n,k) = 0, then a(n) <= k. [Edited by Petros Hadjicostas, Jul 12 2020]
LINKS
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n >= 5. - Harvey P. Dale, Jul 23 2013
MATHEMATICA
Table[16n^4+32n^3+36n^2+20n+3, {n, 0, 30}] (* or *) LinearRecurrence[ {5, -10, 10, -5, 1}, {3, 107, 699, 2547, 6803}, 30] (* Harvey P. Dale, Jul 23 2013 *)
PROG
(PARI) a(n) = 16*n^4+32*n^3+36*n^2+20*n+3
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Benoit Cloitre, May 25 2004
STATUS
approved