OFFSET
0,2
COMMENTS
A Diophantine property of these numbers: (a(n+1)-a(n-1))^2 - 957*a(n)^2 = 4. (See also comment in A200441.)
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,30}. - 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 (31,-1).
FORMULA
G.f.: 1/(1-31*x+x^2).
a(n) = 31*a(n-1)-a(n-2) with a(0)=1, a(1)=31.
a(n) = -a(-n-2) = (t^(n+1)-1/t^(n+1))/(t-1/t) where t=(31+sqrt(957))/2.
a(n) = sum((-1)^k*binomial(n-k, k)*31^(n-2k), k=0..floor(n/2)).
a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*30^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = 1/29*(29 + sqrt(957)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 1/62*(29 + sqrt(957)). - Peter Bala, Dec 23 2012
MATHEMATICA
LinearRecurrence[{31, -1}, {1, 31}, 17]
PROG
(PARI) Vec(1/(1-31*x+x^2)+O(x^17))
(Magma) /* By the closed form: */ Z<x>:=PolynomialRing(Integers()); N<r>:=NumberField(x^2-957); S:=[(((31+r)/2)^n-1/((31+r)/2)^n)/r: n in [1..17]]; [Integers()!S[j]: j in [1..#S]];
(Maxima) makelist(sum((-1)^k*binomial(n-k, k)*31^(n-2*k), k, 0, floor(n/2)), n, 0, 16);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bruno Berselli, Nov 18 2011
STATUS
approved