OFFSET
0,1
COMMENTS
From Creighton Dement, Dec 18 2004: (Start)
Define the following sequences:
b(2n) = c(2n+1), b(2n+1) = c(2n); (c(n)) = (1, -3, -7, 17, 41, -99, -239, 577, 1393, -3363, -8119, 19601, 47321). This is the sequence A001333, apart from signs. Then c(2n) = ((-1)^n)*A002315(n) and c(2n+1) = ((-1)^(n+1))*A001541(n+1).
(d(n)) = (2, 4, -10, -24, 58, 140, -338, -816, 1970, 4756, -11482, -27720). This is A052542, apart from signs. Also, d(2n) = ((-1)^n)*A075870(n), d(2n+1) = ((-1)^n)*A005319(n+1).
(e(n)) = (1, -1, -5, 5, 29, -29, -169, 169, 985, -985, -5741, 5741, 33461, -33461), e(2n) = d(2n)/2, e(2n+1) = - d(2n)/2.
(f(n)) = (2, 2, -12, -12, 70, 70, -408, -408, 2378, 2378, -13860, -13860, ) f(2n) = f(2n+1) = d(2n+1)/2.
(g(n)) = (0, -3, 0, 17, 0, -99, 0, 577, 0, -3363, 0, 19601, 0, -114243, 0, 665857), g(2n) = 0, g(2n+1) = c(2n+1).
Then a(2n) = - c(2n+1), a(2n+1) = d(2n+1) and we have the following conjectures: c(n) + d(n) = e(n) + f(n) = g(n) + a(n); c(n) + d(n) = b(n). In other words, the sequences (c(n) + d(n)) = (e(n) + f(n)) = (g(n) + h(n)) all represent the sequence c with even- and odd-indexed terms reversed. (End)
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,-6,0,-1).
FORMULA
a(n) = -2*a(n-1) - 3*a(n-2) if n is even; a(n) = (4*a(n-1) - a(n-2))/3 if n is odd. - R. J. Mathar, Jun 18 2014
MATHEMATICA
LinearRecurrence[{0, -6, 0, -1}, {3, 4, -17, -24}, 41] (* G. C. Greubel, Apr 09 2023 *)
PROG
(Magma) I:=[3, 4, -17, -24]; [n le 4 select I[n] else -6*Self(n-2)-Self(n-4): n in [1..40]]; // G. C. Greubel, Apr 09 2023
(SageMath)
@CachedFunction
def a(n): # a = A100434
if (n<4): return (3, 4, -17, -24)[n]
else: return -6*a(n-2) - a(n-4)
[a(n) for n in range(41)] # G. C. Greubel, Apr 09 2023
CROSSREFS
KEYWORD
sign,easy
AUTHOR
N. J. A. Sloane, Nov 21 2004, suggested by correspondence from Creighton Dement
STATUS
approved