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 A022034 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(6,31). 1
 6, 31, 160, 825, 4253, 21924, 113017, 582596, 3003248, 15481566, 79806558, 411398091, 2120732851, 10932252540, 56355115894, 290507292601, 1497547928268, 7719771085196, 39794963809107, 205140687086569, 1057488120864155, 5451288779669969, 28101071561042234 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS This coincides with the Pisot T(6,31) sequence as defined in A008776 at least up to n <=18000. - R. J. Mathar, Feb 13 2016 The recurrence 5*a(n-1)+a(n-2)-a(n-3)-a(n-5) starts to fail at n=22. - R. J. Mathar, Feb 13 2016 LINKS Alois P. Heinz, Table of n, a(n) for n = 0..1402 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. MAPLE a:= proc(n) option remember;       `if`(n<2, [6, 31][n+1], ceil(a(n-1)^2/a(n-2))-1)     end: seq(a(n), n=0..30);  # Alois P. Heinz, Sep 18 2015 MATHEMATICA RecurrenceTable[{a[1] == 6, a[2] == 31, a[n] == Ceiling[a[n-1]^2/a[n-2]] - 1}, a, {n, 30}] (* 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(6, 31, 40) \\ Colin Barker, Feb 14 2016 (MAGMA) Tiv:=[6, 31]; [n le 2 select Tiv[n] else Ceiling(Self(n-1)^2/Self(n-2))-1: n in [1..30]]; // Bruno Berselli, Feb 17 2016 CROSSREFS Sequence in context: A065096 A077352 A038223 * A277669 A047665 A003128 Adjacent sequences:  A022031 A022032 A022033 * A022035 A022036 A022037 KEYWORD nonn AUTHOR EXTENSIONS Incorrect conjectures deleted by Alois P. Heinz, Sep 18 2015 STATUS approved

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