login
Expansion of 1/(1 - 37*x + x^2).
1

%I #32 Sep 08 2022 08:46:01

%S 1,37,1368,50579,1870055,69141456,2556363817,94516319773,

%T 3494547467784,129203739988235,4777043832096911,176621418047597472,

%U 6530215423929009553,241441349267325755989,8926799707467123962040

%N Expansion of 1/(1 - 37*x + x^2).

%C Chebyshev polynomials S(n, 37).

%C A Diophantine property of these numbers: (a(n+1)-a(n-1))^2 - 1365*a(n)^2 = 4. - _Bruno Berselli_, Feb 09 2012

%C a(n) equals the number of 01-avoiding words of length n on alphabet {0,1,...,36}. - _Milan Janjic_, Jan 26 2015

%H Bruno Berselli, <a href="/A206565/b206565.txt">Table of n, a(n) for n = 0..500</a>

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>.

%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials</a>.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (37,-1).

%F G.f.: 1/(1 - 37*x + x^2).

%F a(n) = Sum_{k=0..n} A049310(n,k)*37^k.

%F a(n) = 37*a(n-1) - a(n-2), n>=1; a(0)=1, a(1) = 37, a(-1) = 0.

%F a(n) = -a(-n-2) = (t^(n+1)-1/t^(n+1))/(t-1/t) where t=(37+sqrt(1365))/2. - _Bruno Berselli_, Feb 09 2012

%F a(n) = Sum_{k=0..n} A101950(n,k)*36^k. - _Philippe Deléham_, Feb 10 2012

%F Product {n >= 0} (1 + 1/a(n)) = 1/35*(35 + sqrt(1365)). - _Peter Bala_, Dec 23 2012

%F Product {n >= 1} (1 - 1/a(n)) = 1/74*(35 + sqrt(1365)). - _Peter Bala_, Dec 23 2012

%t CoefficientList[Series[1/(1-37x+x^2),{x,0,20}],x] (* or *) LinearRecurrence[ {37,-1},{1,37},20] (* _Harvey P. Dale_, Dec 13 2017 *)

%o (PARI) Vec(1/(1-37*x+x^2)+O(x^15))

%o (Magma) Z<x>:=PolynomialRing(Integers()); N<r>:=NumberField(x^2-1365); S:=[(((37+r)/2)^n-1/((37+r)/2)^n)/r: n in [1..15]]; [Integers()!S[j]: j in [1..#S]];

%o (Maxima) makelist(sum((-1)^k*binomial(n-k, k)*37^(n-2*k), k, 0, floor(n/2)), n, 0, 14);

%Y Cf. A144128, A078987.

%K nonn,easy

%O 0,2

%A _Philippe Deléham_, Feb 09 2012