A200441 Expansion of g.f. 1/(1 - 33*x + x^2).

1, 33, 1088, 35871, 1182655, 38991744, 1285544897, 42383989857, 1397386120384, 46071357982815, 1518957427312511, 50079523743330048, 1651105326102579073, 54436396237641779361, 1794749970516076139840, 59172312630792870835359, 1950891566845648661427007

Offset: 0

Author

Bruno Berselli, Nov 18 2011

Keywords

• nonn
• easy

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).

Crossrefs

Cf. A004190, A056594, A077421; A029547, A029548; A200442.

Cf. A101950.

Programs

Magma

Z:=PolynomialRing(Integers()); N:=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]];

Mathematica

LinearRecurrence[{33, -1}, {1, 33}, 17]

Maxima

makelist(sum((-1)^k*binomial(n-k,k)*33^(n-2*k),k,0,floor(n/2)),n,0,16);

PARI

Vec(1/(1-33*x+x^2)+O(x^17))

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