cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 13 results. Next

A006274 First differences of A006268.

Original entry on oeis.org

33, 46728, 102266868085272, 1069559300034650646049671038948382825526728
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A006268, A006273, A098316.

Extensions

Definition corrected by N. J. A. Sloane, May 23 2023 using a formula suggested by R. J. Mathar, Apr 26 2007.

A145180 Continued cotangent recurrence a(n+1) = a(n)^3 + 3*a(n) and a(1) = 6.

Original entry on oeis.org

6, 234, 12813606, 2103846732371087589834, 9311985549495522884757461748592522243432897275494229148348315206
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1) = a(n)3 + 3*a(n) and a(1)=k is following:
a(n) = Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))].
The next term (a(6)) has 192 digits. - Harvey P. Dale, Mar 09 2013

Links

Crossrefs

Cf. A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189 (k = 1 to 15 with k=4 being A006267(n+1)).
Cf. A002813.

Programs

  • Mathematica
    a = {}; k = 6; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((6 + Sqrt[40])/2)^(3^(n - 1))], {n, 1, 5}] (* Artur Jasinski *)
NestList[#^3+3#&,6,5] (* Harvey P. Dale, Mar 09 2013 *)

Formula

a(n+1)=a(n)^3 + 3*a(n) and a(1)=6
a(n)=Floor[((6+Sqrt[6^2+4])/2)^(3^(n-1))]
a(n) divides a(n+1) and b(n) = a(n+1)/a(n) satisfies the recurrence b(n+1) = b(n)^3 - 3*b(n-1)^2 + 3. See A002813. - Peter Bala, Nov 23 2012

A145189 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=15.

Original entry on oeis.org

15, 3420, 40001698260, 64008151994095341241755497070780, 262244184463346778261182615794616508638576477409715732397097802610370956164308073990185129764340
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189

Crossrefs

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

Programs

  • Mathematica
    a = {}; k = 15; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((15 + Sqrt[229])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
NestList[#^3+3#&,15,5] (* Harvey P. Dale, Aug 20 2017 *)

Formula

a(n+1)=a(n)3+3*a(n) and a(1)=14
a(n)=Floor[((14+Sqrt[14^2+4])/2)^(3^(n-1))]

A145181 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=7.

Original entry on oeis.org

7, 364, 48229636, 112186849649044142700364, 1411971263214164889494039458947084336929208169473485667118006013929636
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189

Crossrefs

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

Programs

  • Mathematica
    a = {}; k = 7; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((7 + Sqrt[53])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)

Formula

a(n+1)=a(n)^3 + 3*a(n) and a(1)=7
a(n)=Floor[((7+Sqrt[7^2+4])/2)^(3^(n-1))]

A145182 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=8.

Original entry on oeis.org

8, 536, 153992264, 3651713626720249047672536, 48695646535829720063008633136610768101443687873746944465180200686293744264
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189
The next term has 222 digits. - Harvey P. Dale, Mar 02 2018

Crossrefs

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

Programs

  • Mathematica
    a = {}; k = 7; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((8 + Sqrt[68])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
RecurrenceTable[{a[1]==8,a[n]==a[n-1]^3+3a[n-1]},a,{n,5}] (* Harvey P. Dale, Mar 02 2018 *)

Formula

a(n+1)=a(n)3+3*a(n) and a(1)=8
a(n)=Floor[((8+Sqrt[8^2+4])/2)^(3^(n-1))]

A145183 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=9.

Original entry on oeis.org

9, 756, 432083484, 80668317387203269343374356, 524939187888223206865848253384923777974930416670334526494423558665677185033084
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
For k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189

Crossrefs

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

Programs

  • Mathematica
    a = {}; k = 9; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((9 + Sqrt[85])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
NestList[#^3+3#&,9,5] (* Harvey P. Dale, Nov 02 2011 *)

Formula

a(n+1)=a(n)3+3*a(n) and a(1)=9
a(n)=Floor[((9+Sqrt[9^2+4])/2)^(3^(n-1))]

A145184 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=10.

Original entry on oeis.org

10, 1030, 1092730090, 1304784252725333839617919270, 2221345538213703371536935622204403026741331806706388823688859272519059871168740810
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189

Crossrefs

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

Programs

  • Mathematica
    a = {}; k = 10; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((10 + Sqrt[104])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)

Formula

a(n+1)=a(n)3+3*a(n) and a(1)=10
a(n)=Floor[((10+Sqrt[10^2+4])/2)^(3^(n-1))]

A145185 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=11.

Original entry on oeis.org

11, 1364, 2537720636, 16342986943522226847837781364, 4365101043708483494615466932242949707161871659736799144058331102381689400753867700636
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189

Crossrefs

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

Programs

  • Mathematica
    a = {}; k = 11; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((11 + Sqrt[125])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
RecurrenceTable[{a[1]==11,a[n]==a[n-1]^3+3*a[n-1]},a,{n,5}] (* Harvey P. Dale, Jul 23 2012 *)

Formula

a(n+1)=a(n)^3+3*a(n) and a(1)=11
a(n)=Floor[((11+Sqrt[11^2+4])/2)^(3^(n-1))]

A145186 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=12.

Original entry on oeis.org

12, 1764, 5489037036, 165382092777963331246695013764, 4523404750894779548516344022127873154658656755028228436816797201835023951822441803129036
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1)=a(n)^3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189
The next term (a(6)) has 263 digits. - Harvey P. Dale, May 21 2018

Crossrefs

Cf. A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189.

Programs

  • Mathematica
    a = {}; k = 12; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((12 + Sqrt[148])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
NestList[#^3+3#&,12,5] (* Harvey P. Dale, May 21 2018 *)

Formula

a(n+1)=a(n)3+3*a(n) and a(1)=12.
a(n)=Floor[((12+Sqrt[12^2+4])/2)^(3^(n-1))].

A145187 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) with a(1)=13.

Original entry on oeis.org

13, 2236, 11179326964, 1397162674037779847605429310236, 2727350312258670490076364505418491429134385511825631286349491134548728023756939667650354964
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences of the form a(n+1)=a(n)^3+3*a(n) with a(1)=k is a(n)=floor(((k+sqrt(k^2+4))/2)^(3^(n-1))).
For k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189

Crossrefs

Cf. A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

Programs

  • Mathematica
    a = {}; k = 13; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((13 + Sqrt[173])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)

Formula

a(n+1)=a(n)^3+3*a(n) and a(1)=13
a(n)=Floor[((13+Sqrt[13^2+4])/2)^(3^(n-1))]
Showing 1-10 of 13 results. Next

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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