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Define the sequence T(a(0),a(1)) by a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n) for n >= 0. This is T(3,7).
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%I #28 Feb 16 2016 06:08:35

%S 3,7,16,36,80,177,391,863,1904,4200,9264,20433,45067,99399,219232,

%T 483532,1066464,2352161,5187855,11442175,25236512,55660880,122763936,

%U 270764385,597189651,1317143239,2905050864,6407291380,14131726000,31168502865,68744297111

%N Define the sequence T(a(0),a(1)) by a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n) for n >= 0. This is T(3,7).

%C a(n) = A077852(n+1) (Barker's recurrence) is correct at least up to n=32000. - _R. J. Mathar_, Feb 11 2016

%C Not to be confused with the Pisot T(3,7) sequence, which is A020746. - _R. J. Mathar_, Feb 13 2016

%H Colin Barker, <a href="/A019489/b019489.txt">Table of n, a(n) for n = 0..1000</a>

%H D. W. Boyd, <a href="http://www.researchgate.net/publication/258834801">Linear recurrence relations for some generalized Pisot sequences</a>, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.

%H <a href="/index/Ph#Pisot">Index entries for Pisot sequences</a>

%F Empirical G.f.: -(x^3-x^2+2*x-3) / ((x-1)*(x^3+2*x-1)). [_Colin Barker_, Dec 21 2012]

%F a(n+1) = ceiling(a(n)^2/a(n-1))-1 for n>0. - _Bruno Berselli_, Feb 15 2016

%p A019489 := proc(n)

%p option remember;

%p if n <= 1 then

%p op(n+1,[3,7]) ;

%p else

%p a := procname(n-1)^2/procname(n-2) ;

%p if type(a,'integer') then

%p a-1 ;

%p else

%p floor(a) ;

%p fi;

%p end if;

%p end proc: # _R. J. Mathar_, Feb 11 2016

%o (PARI) T(a0, a1, maxn) = a=vector(maxn); a[1]=a0; a[2]=a1; for(n=3, maxn, a[n]=ceil(a[n-1]^2/a[n-2])-1); a

%o T(3, 7, 30) \\ _Colin Barker_, Feb 16 2016

%K nonn

%O 0,1

%A _R. K. Guy_