OFFSET
1,7
COMMENTS
Let b(n) = 2^a(n+1). Then b(1)=b(2)=b(3)=b(4)=1 and b(n)*b(n-4) = b(n-1)*b(n-3) + c(n)*b(n-2)^2, c(3*n)=2, c(3*n+1)=c(3*n+2)=1 for all n in Z. - Michael Somos, Oct 18 2018
LINKS
Michael Somos, Somos Polynomials
Index entries for linear recurrences with constant coefficients, signature (2, -1, 1, -2, 1).
FORMULA
a(n) = 1 + a(n-1) + a(n-3) - a(n-4) for all n in Z.
G.f.: x*(1-2*x+x^2-x^3+2*x^4)/((1+x+x^2)* (1-x)^3). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
a(n) = a(7-n) for all n in Z. - Michael Somos, Oct 18 2018
MATHEMATICA
e[1] = 1; e[2] = e[3] = e[4] = e[5] = 0; e[n_] := e[n] = 1 + e[n - 1] + e[n - 3] - e[n - 4]; Table[e[n], {n, 1, 70}]
a[ n_] := Quotient[ Binomial[n - 3, 2], 3]; (* Michael Somos, Oct 18 2018 *)
PROG
(Sage) [floor(binomial(n, 2)/3) for n in range(-2, 59)] # Zerinvary Lajos, Dec 01 2009
(PARI) {a(n) = binomial(n-3, 2)\3}; /* Michael Somos, Oct 18 2018 */
CROSSREFS
KEYWORD
nonn
AUTHOR
Robert G. Wilson v, Jan 11 2001
EXTENSIONS
G.f. proposed by Maksym Voznyy checked and corrected by R. J. Mathar, Sep 16 2009
STATUS
approved