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%I #25 Sep 08 2022 08:46:00
%S 1,35,1224,42805,1496951,52350480,1830769849,64024594235,
%T 2239030028376,78302026398925,2738331893933999,95763314261291040,
%U 3348977667251252401,117118455039532542995,4095796948716387752424,143235774750034038791845,5009156319302474969962151
%N Expansion of 1/(1-35*x+x^2).
%C A Diophantine property of these numbers: (a(n+1)-a(n-1))^2 - 1221*a(n)^2 = 4. (See also comment in A200441.)
%C a(n) equals the number of 01-avoiding words of length n on alphabet {0,1,...,34}. - _Milan Janjic_, Jan 26 2015
%H Bruno Berselli, <a href="/A200724/b200724.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/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (35,-1).
%F G.f.: 1/(1-35*x+x^2).
%F a(n) = 35*a(n-1)-a(n-2) with a(0)=1, a(1)=35.
%F a(n) = -a(-n-2) = (t^(n+1)-1/t^(n+1))/(t-1/t) where t=(35+sqrt(1221))/2.
%F a(n) = sum((-1)^k*binomial(n-k, k)*35^(n-2k), k=0..floor(n/2)).
%F a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*34^k. - _Philippe Deléham_, Feb 10 2012
%F Product {n >= 0} (1 + 1/a(n)) = 1/33*(33 + sqrt(1221)). - _Peter Bala_, Dec 23 2012
%F Product {n >= 1} (1 - 1/a(n)) = 1/70*(33 + sqrt(1221)). - _Peter Bala_, Dec 23 2012
%t LinearRecurrence[{35, -1}, {1, 35}, 17]
%o (PARI) Vec(1/(1-35*x+x^2)+O(x^17))
%o (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]];
%o (Maxima) makelist(sum((-1)^k*binomial(n-k,k)*35^(n-2*k),k,0,floor(n/2)),n,0,16);
%Y Cf. A029547, A144128.
%K nonn,easy
%O 0,2
%A _Bruno Berselli_, Nov 21 2011
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