OFFSET
0,3
COMMENTS
If x^3 = 2 and n >= 0, then there are unique integers a, b, c such that (1 + x)^n = a + b*x + c*x^2. The coefficient b is a(n).
REFERENCES
Maribel Díaz Noguera [Maribel Del Carmen Díaz Noguera], Rigoberto Flores, Jose L. Ramirez, and Martha Romero Rojas, Catalan identities for generalized Fibonacci polynomials, Fib. Q., 62:2 (2024), 100-111. See Table 3.
R. Schoof, Catalan's Conjecture, Springer-Verlag, 2008, pp. 17-18.
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
A. Kumar Gupta and A. Kumar Mittal, Integer Sequences associated with Integer Monic Polynomial, arXiv:math/0001112 [math.GM], Jan 2000.
Index entries for linear recurrences with constant coefficients, signature (3,-3,3).
FORMULA
a(n) = 3*a(n-1) - 3*a(n-2) + 3*a(n-3), n > 2.
a(n) = Sum_{0..floor(n/3)}, 2^k * binomial(n, 3*k+1). - Ralf Stephan, Aug 30 2004
From R. J. Mathar, Apr 01 2008: (Start)
O.g.f.: x*(1 - x)/(1 - 3*x + 3*x^2 - 3*x^3).
a(n+1) - a(n) = A052101(n). (End)
EXAMPLE
G.f.: = x + 2*x^2 + 3*x^3 + 6*x^4 + 15*x^5 + 36*x^6 + 81*x^7 + 180*x^8 + ...
MAPLE
A052102:= n-> add(2^j*binomial(n, 3*j+1), j=0..floor(n/3)); seq(A052102(n), n=0..40); # G. C. Greubel, Apr 15 2021
MATHEMATICA
LinearRecurrence[{3, -3, 3}, {0, 1, 2}, 32] (* Ray Chandler, Sep 23 2015 *)
PROG
(PARI) {a(n) = polcoeff( lift( Mod(1 + x, x^3 - 2)^n ), 1)} /* Michael Somos, Aug 05 2009 */
(PARI) {a(n) = sum(k=0, n\3, 2^k * binomial(n, 3*k + 1))} /* Michael Somos, Aug 05 2009 */
(PARI) {a(n) = if( n<0, 0, polcoeff( (x - x^2) / (1 - 3*x + 3*x^2 - 3*x^3) + x * O(x^n), n))} /* Michael Somos, Aug 05 2009 */
(Magma) [n le 3 select n-1 else 3*(Self(n-1) -Self(n-2) +Self(n-3)): n in [1..40]]; // G. C. Greubel, Apr 15 2021
(Sage) [sum(2^j*binomial(n, 3*j+1) for j in (0..n//3)) for n in (0..40)] # G. C. Greubel, Apr 15 2021
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ashok K. Gupta and Ashok K. Mittal (akgjkiapt(AT)hotmail.com), Jan 20 2000
STATUS
approved