OFFSET
0,2
COMMENTS
A Diophantine property of these numbers: (a(n+1)-a(n-1))^2 - 1085*a(n)^2 = 4.
More generally, for t(m)=(m+sqrt(m^2-4))/2 and u(n)=(t(m)^(n+1)-1/t(m)^(n+1))/(t(m)-1/t(m)), we can verify that (u(n+1)-u(n-1))^2-(m^2-4)*u(n)^2=4.
a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,32}. - 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 (33,-1).
FORMULA
a(n) = 33*a(n-1)-a(n-2) with a(0)=1, a(1)=33.
a(n) = -a(-n-2) = (t^(n+1)-1/t^(n+1))/(t-1/t), where t=(33+sqrt(1085))/2.
a(n) = sum((-1)^k*binomial(n-k, k)*33^(n-2k), k=0..floor(n/2)).
a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*32^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 0} (1 + 1/a(n)) = 1/31*(31 + sqrt(1085)).
Product {n >= 1} (1 - 1/a(n)) = 1/66*(31 + sqrt(1085)). (End)
E.g.f.: exp(33*x/2)*cosh(sqrt(1085)*x/2) + 33*exp(33*x/2)*sinh(sqrt(1085)*x/2)/sqrt(1085). - Stefano Spezia, Apr 16 2023
MATHEMATICA
LinearRecurrence[{33, -1}, {1, 33}, 17]
PROG
(PARI) Vec(1/(1-33*x+x^2)+O(x^17))
(Magma) Z<x>:=PolynomialRing(Integers()); N<r>:=NumberField(x^2-1085); S:=[(((33+r)/2)^n-1/((33+r)/2)^n)/r: n in [1..17]]; [Integers()!S[j]: j in [1..#S]];
(Maxima) makelist(sum((-1)^k*binomial(n-k, k)*33^(n-2*k), k, 0, floor(n/2)), n, 0, 16);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bruno Berselli, Nov 18 2011
STATUS
approved