A333378 a(n) = F(n) * (-1)^(n*(n-1)/2) where F(n) = A000045(n) Fibonacci numbers.

0, 1, -1, -2, 3, 5, -8, -13, 21, 34, -55, -89, 144, 233, -377, -610, 987, 1597, -2584, -4181, 6765, 10946, -17711, -28657, 46368, 75025, -121393, -196418, 317811, 514229, -832040, -1346269, 2178309, 3524578, -5702887, -9227465, 14930352, 24157817, -39088169

Offset: 0

Author

Michael Somos, Mar 17 2020

Keywords

• sign
• easy

Comments

This is a strong elliptic divisibility sequence t_n as given in [Kimberling, p. 16] where x = -1, y = -2, z = -3.

The Hankel transform of -A026726(n+k) is a(2*n+2+k) for k = 0, 1.

Let a(n) := F(n) * (-1)^binomial(n, 2). Then a(m - n) * a(m + n) = a(m + 1) * a(m - 1) * a(n)^2 - a(n + 1) * a(n - 1) * a(m)^2. This plus gcd(f[n], f[m]) = |f[gcd(n, m)]| makes a[] a strong elliptic divisibility sequence. Likewise F(n) * (-1)^binomial(n - 1, 2), but no other asSIGNation (mod scaling). - Bill Gosper, May 28 2008

Examples

			G.f. = x - x^2 - 2*x^3 + 3*x^4 + 5*x^5 - 8*x^6 - 13*x^7 + 21*x^8 + ...
		

Links

• Clark Kimberling, Strong divisibility sequences and some conjectures, Fib. Quart., 17 (1979), 13-17.
• Index entries for linear recurrences with constant coefficients, signature (0,-3,0,-1).

Crossrefs

Cf. A000045, A026726.

Programs

Mathematica

a[ n_] := Fibonacci[n] (-1)^(n (n - 1) / 2);
a[ n_] := With[{m=n-1}, I^m^2 ChebyshevU[m, I/2]]; (* Michael Somos, Mar 19 2020 *)

PARI

{a(n) = fibonacci(n) * (-1)^(n*(n-1)/2)};

Sage

def A333378():
    a, b, c, d = False, True, True, False
    x, y = 0, 1
    while True:
        yield x if a else -x
        x, y = y, x - y
        a, b, c, d = b, c, d, a
a = A333378()
print([next(a) for  in range(39)]) # _Peter Luschny, Mar 19 2020

Formula

G.f.: (x^3 - x^2 + x)/(x^4 + 3*x^2 + 1).

a(n) = -a(-n) = -3*a(n+2) -a(n+4) for all n in Z.

0 = a(n)^2 -a(n+1)^2 +a(n+2)^2 +2*a(n)*a(n+2) for all n in Z.

0 = a(n)*(+a(n+2)) +a(n+1)*(+a(n+1) +a(n+3)) +a(n+2)*(+a(n+2)) for all n in Z.

0 = a(n)*a(n+4) - a(n+1)*a(n+3) - 2*a(n+2)^2 for all n in Z.

0 = a(n)*a(n+5) + 2*a(n+1)*a(n+4) - 3*a(n+2)*a(n+3) for all n in Z.

a(n+1) = i^(n^2) * U(n, i/2) for all n in Z [From Gosper, Mar 19 2020]. - Michael Somos, Mar 19 2020