OFFSET
0,3
LINKS
Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (11,-33,33,-11,1).
FORMULA
G.f.: (4 / (1 - x) + (1 - 6*x) / (1 - 7*x + x^2) + (5 - 10*x) / (1 - 3*x + x^2)) / 10.
a(0)=1, a(1)=1, a(2)=2, a(3)=7, a(4)=35, a(n)=11*a(n-1)-33*a(n-2)+ 33*a(n-3)- 11*a(n-4)+a(n-5). - Harvey P. Dale, Nov 18 2013
a(n) = a(1-n) for all n in Z. - Michael Somos, Sep 22 2014
0 = a(n)*(+a(n+1) + 4*a(n+2)) + a(n+1)*(-11*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 22 2014
a(n) = b(n+1) * b(n) * b(n-1) * b(n-2) / 6 for all n in Z where b = A005247. - Michael Somos, Sep 22 2014
EXAMPLE
G.f. = 1 + x + 2*x^2 + 7*x^3 + 35*x^4 + 210*x^5 + 1365*x^6 + 9165*x^7 + ...
MATHEMATICA
RecurrenceTable[{a[0]==a[1]==1, a[n]==a[n-1] (11a[n-1]-a[n-2])/(a[n-1]+ 4a[n-2])}, a, {n, 30}] (* or *) LinearRecurrence[{11, -33, 33, -11, 1}, {1, 1, 2, 7, 35}, 30] (* Harvey P. Dale, Nov 18 2013 *)
PROG
(PARI) {a(n) = (4 + fibonacci(4*n - 1)/3 + fibonacci(4*n - 3)/3 + 5 * fibonacci(2*n - 1)) / 10};
(PARI) {a(n) = my(A); if( n<1, n = 1-n); if( n<3, n, A = vector(n, k, k); for(k=3, n, A[k] = A[k-1] * (11*A[k-1] - A[k-2]) / (A[k-1] + 4*A[k-2])); A[n])}; /* Michael Somos, Sep 22 2014 */
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Michael Somos, Feb 05 2010
STATUS
approved