A022026 -id:A022026 - cp's OEIS frontend

A022025 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(6,102).

6, 102, 1735, 29513, 502028, 8539699, 145263729, 2470994700, 42032617843, 714991805825, 12162299391068, 206885624804179, 3519208035780561, 59863150041598764, 1018296359995701043, 17321632357467588641, 294647962336362325244, 5012080843035687303187

Offset: 0

R. K. Guy

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..811
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.
Index entries for Pisot sequences

Cf. A022018 - A022024, A022026 - A022032.

a:= proc(n) option remember;
      `if`(n<2, [6, 102][n+1], floor(a(n-1)^2/a(n-2))+1)
    end:
seq(a(n), n=0..20);  # Alois P. Heinz, Sep 18 2015

a[n_] := a[n] = Switch[n, 0, 6, 1, 102, _, 1 + Floor[a[n-1]^2/a[n-2]]];
a /@ Range[0, 20] (* Jean-François Alcover, Nov 16 2020, after Alois P. Heinz *)

a=[6,102];for(n=2,30,a=concat(a,a[n]^2\a[n-1]+1));a \\ M. F. Hasler, Feb 10 2016

The conjectured g.f. (6-5*x^2)/(1-17*x-x^2+14*x^3) yields the same initial terms a(0..271) but from a(272) on a different sequence. - Bruno Berselli and M. F. Hasler, Feb 11 2016

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

An incorrect program was removed by Alois P. Heinz, Apr 27 2019

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

A360194 Array read by antidiagonals: T(m,n) is the number of acyclic spanning subgraphs in the grid graph P_m X P_n.

Original entry on oeis.org

1, 2, 2, 4, 15, 4, 8, 112, 112, 8, 16, 836, 3102, 836, 16, 32, 6240, 85818, 85818, 6240, 32, 64, 46576, 2373870, 8790016, 2373870, 46576, 64, 128, 347648, 65664106, 900013270, 900013270, 65664106, 347648, 128, 256, 2594880, 1816344222, 92146956300, 341008617408, 92146956300, 1816344222, 2594880, 256

Offset: 1

Andrew Howroyd, Jan 29 2023

Acyclic spanning subgraphs are also called spanning forests.

			Table starts:
========================================================
m\n|  1     2        3           4               5
---+----------------------------------------------------
1  |  1     2        4           8              16 ...
2  |  2    15      112         836            6240 ...
3  |  4   112     3102       85818         2373870 ...
4  |  8   836    85818     8790016       900013270 ...
5  | 16  6240  2373870   900013270    341008617408 ...
6  | 32 46576 65664106 92146956300 129187804977182 ...
   ...

Andrew Howroyd, Table of n, a(n) for n = 1..435
Eric Weisstein's World of Mathematics, Acyclic Graph
Eric Weisstein's World of Mathematics, Grid Graph

Rows 1..4 are A000079(n-1), A022026(n-1), A158450, A360195.
Main diagonal is A080691.
Cf. A116469 (spanning trees), A359993 (connected spanning subgraphs), A360202.

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

cp's OEIS Frontend

A022025 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(6,102).

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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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A360194 Array read by antidiagonals: T(m,n) is the number of acyclic spanning subgraphs in the grid graph P_m X P_n.

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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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Maple

Mathematica

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Extensions

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

Mathematica

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Extensions

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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Maple