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 14 results. Next

A144807 Decimal expansion of the constant 3.0843510490691899023356932... arising in A144783.

Original entry on oeis.org

3, 0, 8, 4, 3, 5, 1, 0, 4, 9, 0, 6, 9, 1, 8, 9, 9, 0, 2, 3, 3, 5, 6, 9, 3, 2, 0, 0, 2, 0, 2, 7, 2, 1, 4, 8, 8, 7, 5, 0, 1, 1, 0, 8, 9, 8, 3, 7, 3, 9, 8, 8, 4, 8, 4, 7, 6, 4, 4, 2, 2, 3, 7, 0, 9, 6, 5, 6, 9, 1, 8, 8, 1, 9, 5, 7, 3, 4, 7, 8, 3, 1, 3, 9, 3, 3, 7, 4, 9, 2, 9, 4, 2, 2, 7, 8, 5, 4, 9, 5, 1, 8, 5, 0, 7
Offset: 1

Views

Author

Artur Jasinski, Sep 21 2008

Keywords

Crossrefs

Cf. A076393, A144783, A144802, A144803, A144804, A144805, A144806, A144808, A144809, A144810, A144811, A144812.

A082732 a(1) = 1, a(2) = 3, a(n) = LCM of all the previous terms + 1.

Original entry on oeis.org

1, 3, 4, 13, 157, 24493, 599882557, 359859081592975693, 129498558604939936868397356895854557, 16769876680757063368089314196389622249367851612542961252860614401811693
Offset: 1

Views

Author

Amarnath Murthy, Apr 14 2003

Keywords

Comments

The LCM is in fact the product of all previous terms. From a(5) onwards the terms alternately end in 57 and 93.

Crossrefs

Cf. A000058, A004168, A144743, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788.

Programs

  • Mathematica
    a[1] = 1; a[2] = 3; a[n_] := Apply[LCM, Table[a[i], {i, 1, n - 1}]] + 1; Table[ a[n], {n, 1, 10}]
c=1.8806785436830780944921917650127503562630617563236301969047995953391479871\
7695395204087358090874194124503892563356447954254847544689332763; Table[c^(2^n),{n,1,6}] or a = {}; k = 4; Do[AppendTo[a, k]; k = k^2 - k + 1, {n, 1, 10}]; a (* Artur Jasinski, Sep 22 2008 *)

Formula

For n>=3, a(n+1) = a(n)^2 - a(n) + 1.
For n>=3, a(n) = A004168(n-3) + 1. - Max Alekseyev, Aug 09 2019
1/3 = Sum_{n=3..oo} 1/a(n) = 1/4 + 1/13 + 1/157 + 1/24493 + ... or 1 = Sum_{n=3..oo} 3/a(n) = 3/4 + 3/13 + 3/157 + 3/24493 + .... If we take segment of length 1 and cut off in each step fragment of maximal length such that numerator of fraction is 3, denominators of such fractions will be successive numbers of this sequence. - Artur Jasinski, Sep 22 2008
a(n+2)=1.8806785436830780944921917650127503562630617563236301969047995953391\
4798717695395204087358090874194124503892563356447954254847544689332763...^(2^n). - Artur Jasinski, Sep 22 2008

Extensions

More terms from Robert G. Wilson v, Apr 15 2003

A144780 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 6.

Original entry on oeis.org

6, 31, 931, 865831, 749662454731, 561993796032558961827631, 315837026779085485103718410756049100028793244531
Offset: 1

Views

Author

Artur Jasinski, Sep 21 2008

Keywords

Links

Crossrefs

Cf. A000058, A082732, A144779, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788, A144804.

Programs

  • Mathematica
    a = {}; k = 6; Do[AppendTo[a, k]; k = k^2 - k + 1, {n, 1, 10}]; a
NestList[#^2-#+1&,6,10] (* Harvey P. Dale, Dec 19 2024 *)

Formula

a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 6.
a(n) ~ c^(2^n) where is c is 2.350117384... (A144804).

Extensions

a(8) moved to b-file by Hugo Pfoertner, Aug 30 2020

A144784 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 11.

Original entry on oeis.org

11, 111, 12211, 149096311, 22229709804712411, 494159998001727075769152612720511, 244194103625066907517263589918036880566782292998362610615987380611
Offset: 1

Views

Author

Artur Jasinski, Sep 21 2008

Keywords

Comments

For the "exact" formula, compare the Aho-Sloane reference in A000058. - N. J. A. Sloane, Apr 07 2014

Links

Crossrefs

Cf. A000058, A082732, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788, A144808.
See A239900 for another version.

Programs

  • Mathematica
    a = {}; r = 11; Do[AppendTo[a, r]; r = r^2 - r + 1, {n, 1, 10}]; a

Formula

a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 11.
a(n) ~ c^(2^n) where c = 3.242214... (see A144808).

A144779 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 5.

Original entry on oeis.org

5, 21, 421, 176821, 31265489221, 977530816197201697621, 955566496615167328821993756200407115362021, 913107329453384594090655605142589591944556891901674138343716072975722193082773842421
Offset: 1

Views

Author

Artur Jasinski, Sep 21 2008

Keywords

Examples

			a(0) = 4, a(1) = 4+1 = 5, a(2) = 4*5+1 = 21, a(3) = 4*5*21+1 = 421, a(4) = 4*5*21*421+1 = 176821, ... - _Philippe Deléham_, Apr 19 2013

Links

Crossrefs

Cf. A000058, A082732, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788.

Programs

  • Mathematica
    a = {}; k = 5; Do[AppendTo[a, k]; k = k^2 - k + 1, {n,1,10}]; a (* Artur Jasinski, Sep 21 2008 *)
NestList[#^2-#+1&,5,8] (* Harvey P. Dale, Jan 17 2012 *)

Formula

a(n) = round(2.127995907464107054577351...)^(2^n) = round(A144803^(2^n)). [corrected by Joerg Arndt, Jan 15 2021]
a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 5.

A144782 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 9.

Original entry on oeis.org

9, 73, 5257, 27630793, 763460694178057, 582872231554839914154126117193, 339740038317718918529575265905277902175236102890836244082057
Offset: 1

Views

Author

Artur Jasinski, Sep 21 2008

Keywords

Links

Crossrefs

Cf. A000058, A082732, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788.

Programs

  • Mathematica
    a = {}; r = 9; Do[AppendTo[a, r]; r = r^2 - r + 1, {n, 1, 10}]; a
NestList[#^2-#+1&,9,10] (* Harvey P. Dale, Aug 31 2014 *)

Formula

a(n) ~ c^(2^n) with c = 2.918012...
a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 9.

A144785 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 12.

Original entry on oeis.org

12, 133, 17557, 308230693, 95006159799029557, 9026170399758739819525199160586693, 81471752085480849000657595909467634426991447160798281416700808089557
Offset: 1

Views

Author

Artur Jasinski, Sep 21 2008

Keywords

References

  • Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330. Solution published in Vol. 43, No. 4, September 2012, pp. 340-342

Crossrefs

A000058, A082732, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788

Programs

  • Mathematica
    a = {}; r = 12; Do[AppendTo[a, r]; r = r^2 - r + 1, {n, 1, 10}]; a or Table[Round[3.39277252592669675143137065018187376847206615308598784654603692312172475924599026837940758013759324881455503678006543568111163817496672898^(2^n)], {n, 1, 8}] (*Artur Jasinski*)
NestList[#^2-#+1&,12,6] (* Harvey P. Dale, Jan 01 2016 *)

Formula

a(n) =3.39277252592669675143137065018187376847206615308598784654603692312172475924599026837940758013759324881455503678006543568111163817496672898^(2^n) a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 11

A144786 If n is an oblong number A002378, then a(n)=a(j) where j is the number of oblong numbers in (0,n], otherwise a(n)=n.

Original entry on oeis.org

1, 1, 3, 4, 5, 1, 7, 8, 9, 10, 11, 3, 13, 14, 15, 16, 17, 18, 19, 4, 21, 22, 23, 24, 25, 26, 27, 28, 29, 5, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 1, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 7, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 8, 73, 74, 75, 76
Offset: 1

Views

Author

Artur Jasinski, Sep 22 2008, Sep 26 2008

Keywords

Comments

As a motivation, consider the greedy decomposition of fractions 1/n into Egyptian fractions,
n=1: 2,3,7,43,1807,3263443,.. A000058
n=2: 3,7,43,1807,3263443,10650056950807,.. A000058
n=3: 4,13,157,24493,599882557,359859081592975693,.. A082732
n=4: 5,21,421,176821,31265489221,977530816197201697621,.. A144779
n=5: 6,31,931,865831,749662454731,561993796032558961827631,.. A144780
n=6: 7,43,1807,3263443,10650056950807,.. A000058
n=7: 8,57,3193,10192057,103878015699193,.. A144781
n=8: 9,73,5257,27630793,763460694178057,.. A144782
n=9: 10,91,8191,67084291,4500302031888391,.. A144783
n=10: 11,111,12211,149096311,22229709804712411,.. A144784
n=11: 12,133,17557,308230693,95006159799029557,.. A144785
n=12: 13,157,24493,599882557,.. A082732
k=13: 14,183,33307,1109322943,..
where the first few denominators of 1/n = 1/b(1)+1/b(2)+... have been tabulated.
For some sets of n, the list b(i) of denominators is essentially the same: consider for example A000058, which represents primarily n=1, then in truncated form also n=2, and then n=6, n=42 etc. Or consider A082732 which represents n=3, then in truncated form n=12, n=156 etc.
The OEIS sequence assigns the primary n to a(n). The interpretation of a(n) with ascending n is: n=1 is primary, a(1)=1.
Decomposition of n=2 is equivalent to n=1, a(2)=1. Cases n=3 to 5 are primary ("original", "new"), and a(n)=n in these cases. n=6 is not new but essentially the same Egyptian series as seen for n=1, so a(6)=1. Cases n=7 to n=11 are "new" sequences, again a(n)=n in these cases, but then n=12 is represented by A082732 as already seen for n=3, so a(12)=3.
Because the first denominator for the decomposition of 1/n is 1/(n+1), n+1 belongs to the sequence of denominators of the expansion of 1/a(n).
The sequences b(.) have recurrences which are essentially 1+b(n-1)*(b(n-1)-1), looking up the oblong number at the position of the previous b(.). This is the reason why reverse look-up of the n via A000194 (number of oblong numbers up to n) as used in the definition is equivalent to the assignment described above.

Examples

			n=1 is not in A002378, so a(1)=1.
n=2 = A000058(2), so a(2)=1 because there is 1 oblong number <=2 and >0.
n=3 is not in A002378, so a(3)=3.
n=6 = A000058(3), so a(6)=a(2) because there are 2 oblong numbers <=6 and >0.

Crossrefs

Cf. A000058, A002378, A078358, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786.

Formula

a(n) = a(A000194(n+1)) if n in A002378. a(n) = n if n in A078358.

Extensions

a(57)=57 inserted, a(61)=61 corrected and better definition provided by Omar E. Pol, Dec 29 2008
I did some further editing of this entry, but many of the lines are still obscure. - N. J. A. Sloane, Dec 29 2008
Comments that connect to Egyptian fractions rephrased by R. J. Mathar, Oct 01 2009

A144787 Recurrence sequence a(n+1)=a(n)^3-a(n)+1 and a(1)=2.

Original entry on oeis.org

2, 7, 337, 38272417, 56060590716839257663297, 176186654453940966415101758343368831005891099500239113100063334235777
Offset: 1

Views

Author

Artur Jasinski, Sep 22 2008

Keywords

Comments

For constant c=1.240554576397679299452... see A144810.

Crossrefs

Cf. A000058, A082732, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788.

Programs

  • Mathematica
    a = {}; k = 2; Do[AppendTo[a, k]; k = k^3 - k + 1, {n, 1, 8}]; a (*Artur Jasinski*)
NestList[#^3-#+1&,2,5] (* Harvey P. Dale, Jun 24 2013 *)

A179090 Variant of Sylvester's sequence: a(n + 1) = a(n)^2 - a(n) + 1, with a(1) = 14.

Original entry on oeis.org

14, 183, 33307, 1109322943, 1230597390756858307, 1514369938137587813730274566118047943, 2293316309534841541915473293317407057146218304394352680966696226728483307
Offset: 1

Views

Author

Ivan Panchenko, Jun 29 2010

Keywords

Comments

The next term (a(8)) has 145 digits. - Harvey P. Dale, Jul 14 2021

Links

Crossrefs

Cf. A144779, A144780, A144781, A144782, A144783, A144784, A144785

Programs

  • Mathematica
    NestList[#^2-#+1&,14,8] (* Harvey P. Dale, Jul 14 2021 *)
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Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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