A022025 -id:A022025 - cp's OEIS frontend

A022026 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(2,15).

2, 15, 112, 836, 6240, 46576, 347648, 2594880, 19368448, 144568064, 1079070720, 8054293504, 60118065152, 448727347200, 3349346516992, 24999862747136, 186601515909120, 1392812676284416, 10396095346638848, 77597512067973120, 579195715157229568

Offset: 0

R. K. Guy

From Alois P. Heinz, Mar 18 2009: (Start)
a(n) is also the number of forests in the 2 X (n+1) grid.
a(0)=2, because there are 2 forests in the 2 X 1 grid: 1.2 and 1-2.
a(1)=15, because there are 15 forests in the 2 X 2 grid:
  1.2 1-2 1.2 1.2 1.2 1-2 1-2 1-2 1.2 1.2 1.2 1.2 1-2 1-2 1-2
  . . . . . | . . | . . | . . | . . | | | | . | | | . | | . |
  4.3 4.3 4.3 4-3 4.3 4.3 4-3 4.3 4-3 4.3 4-3 4-3 4-3 4.3 4-3
a(n) = 8a(n-1) - 4a(n-2) for n>=2, because each of the a(n-1) forests can be extended by 8 patterns:
  .o -o .o -o .o -o .o -o
  .. .. .. .. .| .| .| .|
  .o .o -o -o .o .o -o -o
where 4a(n-2) of these are not forests, namely the extensions of a(n-2) forests by 4 patterns:
  .o-o -o-o .o-o -o-o
  .| | .| | .| | .| |
  .o-o .o-o -o-o -o-o (End)

Alois P. Heinz, Table of n, a(n) for n = 0..1000
M. Desjarlais and R. Molina, Counting Spanning Trees in Grid Graphs
Tomislav Doslic, Planar polycyclic graphs and their Tutte polynomials, Journal of Mathematical Chemistry, Volume 51, Issue 6, 2013, pp. 1599-1607.
Index entries for linear recurrences with constant coefficients, signature (8,-4).

Equals A028859(2n+2)/4.

Cf. A022018 - A022025, A022027 - A022032.

a:= n-> (Matrix([[15,2]]). Matrix([[8, 1], [-4, 0]])^n)[1, 2]:
seq(a(n), n=0..35);  # Alois P. Heinz, Mar 18 2009

CoefficientList[Series[(2-x)/(1-8*x+4*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, May 03 2014 *)

a(n)=([15,2]*[8,1;-4,0]^n)[2] \\ M. F. Hasler, Feb 10 2016

G.f.: (2-x)/(1-8x+4x^2). - David Boyd and Ralf Stephan, Apr 15 2004
From Peter Bala, May 03 2014: (Start)
a(n) = sum of the entries in the 2 X 2 matrix A^n where A is the 2 X 2 matrix [4, 4; 3, 4].
a(n) = (1 + 7*sqrt(3)/12)*(4 + 2*sqrt(3))^n + (1 - 7*sqrt(3)/12)*(4 - 2*sqrt(3))^n. See Desjarlais and Molina. (End)
a(n+1) = ceiling(a(n)^2/a(n-1))-1 for all n > 0. - M. F. Hasler, Feb 10 2016

A022032 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(5,26).

Original entry on oeis.org

5, 26, 135, 700, 3629, 18813, 97527, 505582, 2620947, 13587040, 70435478, 365138879, 1892887004, 9812762803, 50869551972, 263708740319, 1367071205166, 7086923541985, 36738748574433, 190454382472052, 987319198674433, 5118281802804775, 26533271760636405, 137548993480193164

Offset: 0

R. K. Guy

nonn

The empirical g.f. / recurrence agrees with the original definition for at least 2000 terms (and a(2000) ~ 10^1430). - M. F. Hasler, Feb 11 2016

Colin Barker, Table of n, a(n) for n = 0..1000

Cf. A022018 - A022025, A022026 - A022031.

(* This empirical recurrence should not be used to extend the data. *) LinearRecurrence[{5, 1, 0, -1, -1, -1, -1}, {5, 26, 135, 700, 3629, 18813, 97527}, 24] (* Jean-François Alcover, Dec 12 2016 *)

a=[5,26];for(n=2,2000, a=concat(a, ceil(a[n]^2/a[n-1])-1));A022032(n)=a[n+1] \\ M. F. Hasler, Feb 11 2016

Empirical g.f.: -(x^6+x^5+x^4+x^3-x-5) / (x^7+x^6+x^5+x^4-x^2-5*x+1). - Colin Barker, Sep 18 2015

a(n+1) = ceiling(a(n)^2/a(n-1))-1 for all n > 0. - M. F. Hasler, Feb 11 2016

Edited by M. F. Hasler, Feb 11 2016

A022031 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(4,17).

Original entry on oeis.org

4, 17, 72, 304, 1283, 5414, 22845, 96397, 406757, 1716352, 7242319, 30559689, 128949662, 544115986, 2295951781, 9687997993, 40879475731, 172495033261, 727860031657, 3071278144467, 12959565068034, 54684179957837, 230745362360740, 973653116715681, 4108426630946045

Offset: 0

R. K. Guy

nonn

The empirical g.f. / recurrence agrees with the original definition for at least 2000 terms (and a(2000) ~ 10^1250). - M. F. Hasler, Feb 11 2016

Colin Barker, Table of n, a(n) for n = 0..1000

Cf. A022018 - A022025, A022026 - A022032.

a=[4,17];for(n=2,2000,a=concat(a,ceil(a[n]^2/a[n-1])-1));A022031(n)=a[n+1] \\ M. F. Hasler, Feb 11 2016

Empirical g.f.: -(x^6+x^5+x^4+x^3-x-4) / ((x-1)*(x^6+2*x^5+3*x^4+4*x^3+4*x^2+3*x-1)). - Colin Barker, Sep 18 2015

a(n+1) = ceiling(a(n)^2/a(n-1))-1 for all n > 0. a(n+1)/a(n) ~ 4.219599938... as n -> oo. - M. F. Hasler, Feb 11 2016

Edited by M. F. Hasler, Feb 11 2016

A022024 Define the sequence S(a(0),a(1)) by a(n+2) is the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n) for n >= 0. This is S(6,66).

Original entry on oeis.org

6, 66, 727, 8009, 88232, 972018, 10708349, 117969769, 1299627646, 14317498734, 157730385799, 1737655093709, 19143078927992, 210891949829430, 2323315631208341, 25595076182769253, 281971126093205254, 3106367622527151978, 34221659288246953735, 377006879658404795777

Offset: 0

R. K. Guy

nonn

This coincides with the linearly recurrent sequence defined by the expansion of (6 - 5*x^2)/(1 - 11*x - x^2 + 9*x^3) only up to n <= 169. - Bruno Berselli, Feb 11 2016

Colin Barker, Table of n, a(n) for n = 0..950

Cf. A022018 - A022025, A022026 - A022032.

A022024 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[6,66]) ;
    else
        a := procname(n-1)^2/procname(n-2) ;
        if type(a,'integer') then
            a+1 ;
        else
            ceil(a) ;
        fi;
    end if;
end proc: # R. J. Mathar, Feb 10 2016

a[n_] := a[n] = Switch[n, 0, 6, 1, 66, _, Floor[a[n-1]^2/a[n-2]]+1];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Feb 08 2024 *)

a=[6,66];for(n=2,30,a=concat(a,a[n]^2\a[n-1]+1));a \\ M. F. Hasler, Feb 10 2016

def a(n):
    if n == 0: return 6
    prev_1, prev_2 = 66, 6
    for i in range(2, n + 1):
        prev_2, prev_1 = prev_1, (prev_1 ** 2) // prev_2 + 1
    return prev_1 # Paul Muljadi, Feb 12 2024

a(n+1) = floor(a(n)^2/a(n-1))+1 for all n > 0. - M. F. Hasler, Feb 10 2016

Double-checked and extended to 3 lines of data by M. F. Hasler, Feb 10 2016

A022027 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(2,16).

Original entry on oeis.org

2, 16, 127, 1008, 8000, 63492, 503904, 3999232, 31739888, 251903488, 1999230976, 15866888256, 125927492096, 999423012864, 7931916549888, 62951622430720, 499615287394304, 3965194632954880, 31469750573916160, 249759543441752064, 1982215569002196992, 15731845549721911296

Offset: 0

R. K. Guy

nonn

Not to be confused with the Pisot T(2,16) sequence, which is A013730. - R. J. Mathar, Feb 13 2016

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for Pisot sequences

Cf. A022018 - A022025, A022026 - A022032.

I:=[2,16]; [n le 2 select I[n] else Ceiling(Self(n-1)^2/Self(n-2))-1: n in [1..30]]; // Vincenzo Librandi, Feb 12 2016

RecurrenceTable[{a[1] == 2, a[2] == 16, a[n] == Ceiling[a[n-1]^2 / a[n-2] - 1]}, a, {n, 30}] (* Vincenzo Librandi, Feb 12 2016 *)

a=[2, 16]; for(n=2, 1000, a=concat(a, ceil(a[n]^2/a[n-1])-1)); A022027(n)=a[n+1] \\ M. F. Hasler, Feb 11 2016

Conjectures: a(n) = 8*a(n-1)-4*a(n-3). G.f.: -(x^2-2) / (4*x^3-8*x+1). - Colin Barker, Sep 18 2015

a(n+1) = ceiling(a(n)^2/a(n-1))-1 for n>0. - M. F. Hasler, Feb 11 2016

Double-checked (original definition agrees with g.f. / recurrence for n=0..1000), extended and edited by M. F. Hasler, Feb 11 2016

A022021 Define the sequence S(a(0),a(1)) by a(n+2) is the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n) for n >= 0. This is S(5,20).

Original entry on oeis.org

5, 20, 81, 329, 1337, 5434, 22086, 89767, 364852, 1482917, 6027219, 24497237, 99567416, 404685244, 1644816681, 6685249720, 27171759829, 110437838993, 448867366641, 1824392026070, 7415121953942, 30138277741915, 122495056843392, 497873139253657, 2023572780632275

Offset: 0

R. K. Guy

nonn

This coincides with the linearly recurrent sequence defined by the expansion of (5 - 4*x^2)/(1 - 4*x - x^2 + 3*x^3) only up to n <= 39. - Bruno Berselli, Feb 11 2016

Colin Barker, Table of n, a(n) for n = 0..1000

Cf. A022018 - A022025, A022026 - A022032.

A022021 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[5,20]) ;
    else
        a := procname(n-1)^2/procname(n-2) ;
        if type(a,'integer') then
            a+1 ;
        else
            ceil(a) ;
        fi;
    end if;
end proc: # R. J. Mathar, Feb 10 2016

a=[5,20];for(n=2,30,a=concat(a,a[n]^2\a[n-1]+1));a \\ M. F. Hasler, Feb 10 2016

a(n+1) = floor(a(n)^2/a(n-1))+1 for all n > 0. - M. F. Hasler, Feb 10 2016

Double-checked and extended to 3 lines of data by M. F. Hasler, Feb 10 2016

A022022 Define the sequence S(a(0),a(1)) by a(n+2) is the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n) for n >= 0. This is S(5,45).

Original entry on oeis.org

5, 45, 406, 3664, 33067, 298425, 2693244, 24306152, 219359637, 1979690177, 17866428166, 161242026212, 1455186832835, 13132858524565, 118522219370436, 1069646525028644, 9653410934956277, 87120689404042085, 786252089896134534, 7095815621924558952, 64038747861388870507

Offset: 0

R. K. Guy

nonn

This coincides with the linearly recurrent sequence defined by the expansion of (5 - 4*x^2)/(1 - 9*x - x^2 + 7*x^3) only up to n <= 103. - Bruno Berselli, Feb 11 2016

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A022018 - A022025, A022026 - A022032.

a:= proc(n) option remember;
      `if`(n<2, [5, 45][n+1], floor(a(n-1)^2/a(n-2))+1)
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 18 2015

nxt[{a_,b_}]:=Module[{c=Ceiling[b^2/a]},c=If[c<=b^2/a,c+1,c];{b,c}]; Transpose[NestList[nxt,{5,45},20]][[1]] (* Harvey P. Dale, Feb 11 2014 *)

a=[5,45];for(n=2,30,a=concat(a,a[n]^2\a[n-1]+1));a \\ M. F. Hasler, Feb 10 2016

a(n+1) = floor(a(n)^2/a(n-1))+1 for all n > 0. - M. F. Hasler, Feb 10 2016

Double-checked and edited by M. F. Hasler, Feb 10 2016

A022023 Define the sequence S(a(0),a(1)) by a(n+2) is the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n) for n >= 0. This is S(6,30).

Original entry on oeis.org

6, 30, 151, 761, 3836, 19337, 97477, 491378, 2477019, 12486565, 62944332, 317300149, 1599498817, 8063016906, 40645382751, 204891935393, 1032852992092, 5206575364849, 26246162074765, 132305973770306, 666949729466899, 3362069972805741, 16948075698414380

Offset: 0

R. K. Guy

nonn

This coincides with the linearly recurrent sequence defined by the expansion of (6 - 5*x^2)/(1 - 5*x - x^2 + 4*x^3) only up to n <= 69. - Bruno Berselli, Feb 11 2016

Colin Barker, Table of n, a(n) for n = 0..1000

Cf. A022018 - A022025, A022026 - A022032.

A022023 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[6,30]) ;
    else
        a := procname(n-1)^2/procname(n-2) ;
        if type(a,'integer') then
            a+1 ;
        else
            ceil(a) ;
        fi;
    end if;
end proc: # R. J. Mathar, Feb 10 2016

a=[6,30];for(n=2,30,a=concat(a,a[n]^2\a[n-1]+1));a \\ M. F. Hasler, Feb 10 2016

a(n+1) = floor(a(n)^2/a(n-1))+1 for all n > 0. - M. F. Hasler, Feb 10 2016

Double-checked and extended to 3 lines of data by M. F. Hasler, Feb 10 2016

A022028 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(2,32).

Original entry on oeis.org

2, 32, 511, 8160, 130304, 2080776, 33227136, 530591744, 8472821696, 135299330048, 2160544546816, 34500930175488, 550932488167424, 8797635454304256, 140486159827464192, 2243371097334087680, 35823556473710968832, 572053014300755787776, 9134901260033419902976

Offset: 0

R. K. Guy

nonn

Not to be confused with the Pisot T(2,32) sequence as defined in A008776, which is A013776. - R. J. Mathar, Feb 13 2016

Colin Barker, Table of n, a(n) for n = 0..830
Index entries for Pisot sequences

Cf. A022018 - A022025, A022026 - A022032.

a=[2,32];for(n=2,2000,a=concat(a,ceil(a[n]^2/a[n-1])-1));A022028(n)=a[n+1] \\ M. F. Hasler, Feb 11 2016

Conjecture: a(n) = 16*a(n-1)-8*a(n-3). G.f.: -(x^2-2) / (8*x^3-16*x+1). - Colin Barker, Sep 18 2015

a(n+1) = ceiling(a(n)^2/a(n-1))-1 for all n > 0. a(n+1)/a(n) ~ 15.968627... as n -> oo. - M. F. Hasler, Feb 11 2016

A022030 For even n, a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n); for odd n, the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n); a(0) = 4, a(1) = 16.

Original entry on oeis.org

4, 16, 63, 249, 984, 3889, 15370, 60745, 240075, 948819, 3749901, 14820274, 58572352, 231488326, 914882931, 3615779646, 14290202610, 56477415835, 223208766625, 882160643536, 3486455360919, 13779090092886, 54457408494633, 215225339261149, 850608722312629, 3361756570848769

Offset: 0

R. K. Guy

nonn

Original definition: a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n).

This original definition would lead to sequence 4, 16, 63, 248, 976, 3841, ... which agrees to over 2000 terms with the conjectured g.f. = (4 - x^2)/(1 - 4*x + x^3). - M. F. Hasler, Feb 11 2016

Index entries for Pisot sequences

Cf. A022026 - A022032, A022018 - A022025.

a=[4,16];for(n=2,2000,a=concat(a,if(bittest(n,0),a[n]^2\a[n-1]+1,ceil(a[n]^2/a[n-1])-1)));A022030(n)=a[n+1] \\ M. F. Hasler, Feb 11 2016

Conjecture: a(n) = 4*a(n-1)-a(n-3)+a(n-4). G.f. = (4-x^2+x^3)/(1-4*x+x^3-x^4). - Colin Barker, Feb 16 2012

a(n) = ceiling(a(n-1)^2/a(n-2))-1 for even n > 0, a(n) = floor(a(n-1)^2/a(n-2))+1 for even n > 0. - M. F. Hasler, Feb 11 2016

Edited (definition changed to fit data, extended to 3 lines) by M. F. Hasler, Feb 11 2016

cp's OEIS Frontend

A022026 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(2,15).

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A022032 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(5,26).

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A022031 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(4,17).

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A022024 Define the sequence S(a(0),a(1)) by a(n+2) is the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n) for n >= 0. This is S(6,66).

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A022027 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(2,16).

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Magma

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A022021 Define the sequence S(a(0),a(1)) by a(n+2) is the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n) for n >= 0. This is S(5,20).

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A022022 Define the sequence S(a(0),a(1)) by a(n+2) is the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n) for n >= 0. This is S(5,45).

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