OFFSET
0,2
COMMENTS
A Diophantine property of these numbers: (a(n+1)-a(n-1))^2 - 1221*a(n)^2 = 4. (See also comment in A200441.)
a(n) equals the number of 01-avoiding words of length n on alphabet {0,1,...,34}. - Milan Janjic, Jan 26 2015
LINKS
Bruno Berselli, Table of n, a(n) for n = 0..500
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (35,-1).
FORMULA
G.f.: 1/(1-35*x+x^2).
a(n) = 35*a(n-1)-a(n-2) with a(0)=1, a(1)=35.
a(n) = -a(-n-2) = (t^(n+1)-1/t^(n+1))/(t-1/t) where t=(35+sqrt(1221))/2.
a(n) = sum((-1)^k*binomial(n-k, k)*35^(n-2k), k=0..floor(n/2)).
a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*34^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = 1/33*(33 + sqrt(1221)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 1/70*(33 + sqrt(1221)). - Peter Bala, Dec 23 2012
MATHEMATICA
LinearRecurrence[{35, -1}, {1, 35}, 17]
PROG
(PARI) Vec(1/(1-35*x+x^2)+O(x^17))
(Magma) Z<x>:=PolynomialRing(Integers()); N<r>:=NumberField(x^2-1221); S:=[(((35+r)/2)^n-1/((35+r)/2)^n)/r: n in [1..17]]; [Integers()!S[j]: j in [1..#S]];
(Maxima) makelist(sum((-1)^k*binomial(n-k, k)*35^(n-2*k), k, 0, floor(n/2)), n, 0, 16);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bruno Berselli, Nov 21 2011
STATUS
approved