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A018917 Define the generalized Pisot 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). This is T(3,5). 2
3, 5, 8, 12, 17, 24, 33, 45, 61, 82, 110, 147, 196, 261, 347, 461, 612, 812, 1077, 1428, 1893, 2509, 3325, 4406, 5838, 7735, 10248, 13577, 17987, 23829, 31568, 41820, 55401, 73392, 97225, 128797, 170621, 226026, 299422, 396651, 525452, 696077, 922107 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
COMMENTS
Not to be confused with the Pisot T(3,5) sequence, which is A020745. - R. J. Mathar, Feb 13 2016
Is 1 followed by this sequence equal to A167385? - Bruno Berselli, Feb 17 2016
LINKS
D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
FORMULA
Conjecture: a(n)=a(n-1)+a(n-2)-a(n-4). G.f.: (3+2*x-x^3)/(1-x)/(1-x^2-x^3). [Colin Barker, Feb 16 2012]
Conjecture: a(n) = a(n-1) + A000931(n+8). - Reinhard Zumkeller, Dec 30 2012
MATHEMATICA
RecurrenceTable[{a[1] == 3, a[2] == 5, a[n] == Ceiling[a[n-1]^2/a[n-2]] - 1}, a, {n, 50}] (* Bruno Berselli, Feb 17 2016 *)
PROG
(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
T(3, 5, 60) \\ Colin Barker, Feb 14 2016
(Magma) Tiv:=[3, 5]; [n le 2 select Tiv[n] else Ceiling(Self(n-1)^2/Self(n-2))-1: n in [1..50]]; // Bruno Berselli, Feb 17 2016
CROSSREFS
Sequence in context: A133263 A238531 A038088 * A167385 A265061 A265062
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified August 30 07:09 EDT 2024. Contains 375532 sequences. (Running on oeis4.)