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 11 results. Next

A270752 (r,1)-greedy sequence, where r(k) = 1/(k*e).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 262, 167395, 42355398928, 2986137074379747535250, 16334453331070842795541380956715272941358931, 334377619479874433401339085661668551899899040409749812309411639875183486098285324762070
Offset: 1

Views

Author

Clark Kimberling, Apr 09 2016

Keywords

Comments

Let x > 0, and let r = (r(k)) be a sequence of positive irrational numbers. Let a(1) be the least positive integer m such that r(1)/m < x, and inductively let a(n) be the least positive integer m such that r(1)/a(1) + ... + r(n-1)/a(n-1) + r(n)/m < x. The sequence (a(n)) is the (r,x)-greedy sequence. We are interested in choices of r and x for which the series r(1)/a(1) + ... + r(n)/a(n) + ... converges to x. See A270744 for a guide to related sequences.

Examples

			a(1) = ceiling(r(1)) = ceiling(1/e) = ceiling(0.367...) = 1;
a(2) = ceiling(r(2)/(1 - r(1)/1)) = 1;
a(3) = ceiling(r(3)/(1 - r(1)/1 - r(2)/1)) = 1.
The first 6 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are 0.367..., 0.551..., 0.674..., 0.766..., 0.839..., 0.901...
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = Infinity; z = 16;
    r[k_] := N[1/(k*E), 1000]; f[x_, 0] = x;
    n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
    f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
    x = 1; Table[n[x, k], {k, 1, z}]
    N[Sum[r[k]/n[x, k], {k, 1, 18}], 200]

Formula

a(n) = ceiling(r(n)/s(n)), where s(n) = 1 - r(1)/a(1) - r(2)/a(2) - ... - r(n-1)/a(n-1).
r(1)/a(1) + ... + r(n)/a(n) + ... = 1.

A376934 Lexicographically earliest sequence of positive integers a(1), a(2), a(3), ... such that for any n > 0, Sum_{k = 1..n} 1/(k*a(k)) < Pi.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 6, 80, 21896, 604163887, 677040508659246685, 447405708743254015046365510044832005, 309471557529368331206803181535934923519436869019793750609292014082198479
Offset: 1

Views

Author

Scott R. Shannon, Oct 11 2024

Keywords

Examples

			a(17) = 604163887 as Sum_{k = 1..17} 1/(k*a(k)) = 1/(1*1) + 1/(2*1) + ... + 1/(16*21896) + 1/(17*604163887) = 3056398076673607759/972881723918332800, which is ~8.2*10^-20 less than Pi.
		

Crossrefs

A270745 (r,1)-greedy sequence, where r(k) = k/tau^k and tau = golden ratio.

Original entry on oeis.org

1, 3, 6, 63, 18292, 822680526, 810795400558456607, 3059608561526409532580548135921262544, 41330361304323764432512049819825720283867354602583921458622085452146872094
Offset: 1

Views

Author

Clark Kimberling, Apr 09 2016

Keywords

Comments

Let x > 0, and let r = (r(k)) be a sequence of positive irrational numbers. Let a(1) be the least positive integer m such that r(1)/m < x, and inductively let a(n) be the least positive integer m such that r(1)/a(1) + ... + r(n-1)/a(n-1) + r(n)/m < x. The sequence (a(n)) is the (r,x)-greedy sequence. We are interested in choices of r and x for which the series r(1)/a(1) + ... + r(n)/a(n) + ... converges to x. See A270744 for a guide to related sequences.

Examples

			a(1) = ceiling(r(1)) = ceiling(1/tau) = ceiling(0.618...) = 1;
a(2) = ceiling(r(2)/(1 - r(1)/1)) = 3;
a(3) = ceiling(r(3)/(1 - r(1)/1 - r(2)/3)) = 6.
The first 6 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are 0.618..., 0.872..., 0.990..., 0.975..., 0.999..., 0.99999999959...
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = Infinity; z = 13;
    r[k_] := N[k/GoldenRatio^k, 1000]; f[x_, 0] = x;
    n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
    f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
    x = 1; Table[n[x, k], {k, 1, z}]
    N[Sum[r[k]/n[x, k], {k, 1, 13}], 200]

Formula

a(n) = ceiling(r(n)/s(n)), where s(n) = 1 - r(1)/a(1) - r(2)/a(2) - ... - r(n-1)/a(n-1).
r(1)/a(1) + ... + r(n)/a(n) + ... = 1.

A270746 (r,1)-greedy sequence, where r(k) = 2/e^k.

Original entry on oeis.org

1, 2, 1, 2, 2, 2, 2, 1, 2, 1, 15, 107, 4536, 9767531, 119608113171152, 27706455600364135685442345886, 683882728856994887705617234665700899371621018916716222985
Offset: 1

Views

Author

Clark Kimberling, Apr 09 2016

Keywords

Comments

Let x > 0, and let r = (r(k)) be a sequence of positive irrational numbers. Let a(1) be the least positive integer m such that r(1)/m < x, and inductively let a(n) be the least positive integer m such that r(1)/a(1) + ... + r(n-1)/a(n-1) + r(n)/m < x. The sequence (a(n)) is the (r,x)-greedy sequence. We are interested in choices of r and x for which the series r(1)/a(1) + ... + r(n)/a(n) + ... converges to x. See A270744 for a guide to related sequences.

Examples

			a(1) = ceiling(r(1)) = ceiling(2/e) = ceiling(0.735...) = 1;
a(2) = ceiling(r(2)/(1 - r(1)/1)) = 2;
a(3) = ceiling(r(3)/(1 - r(1)/1 - r(2)/2)) = 2.
The first 6 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are 0.735..., 0.871..., 0.970..., 0.975..., 0.988..., 0.995...
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = Infinity; z = 19;
    r[k_] := N[2/E^k, 1000]; f[x_, 0] = x;
    n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
    f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
    x = 1; Table[n[x, k], {k, 1, z}]
    N[Sum[r[k]/n[x, k], {k, 1, 19}], 200]

Formula

a(n) = ceiling(r(n)/s(n)), where s(n) = 1 - r(1)/a(1) - r(2)/a(2) - ... - r(n-1)/a(n-1).
r(1)/a(1) + ... + r(n)/a(n) + ... = 1.

A270747 (r,1)-greedy sequence, where r(k) = 4/Pi^k.

Original entry on oeis.org

2, 2, 1, 2, 2, 1, 3, 7, 71, 3131, 5821925, 14364035515947, 451397201144015321568515204, 88020328073777548345010277436911545872870466008026310, 4344173888544359227731947461270153179826227998155726069662805370800638822815760136590246135744249701337368
Offset: 1

Views

Author

Clark Kimberling, Apr 09 2016

Keywords

Comments

Let x > 0, and let r = (r(k)) be a sequence of positive irrational numbers. Let a(1) be the least positive integer m such that r(1)/m < x, and inductively let a(n) be the least positive integer m such that r(1)/a(1) + ... + r(n-1)/a(n-1) + r(n)/m < x. The sequence (a(n)) is the (r,x)-greedy sequence. We are interested in choices of r and x for which the series r(1)/a(1) + ... + r(n)/a(n) + ... converges to x. See A270744 for a guide to related sequences.

Examples

			a(1) = ceiling(r(1)) = ceiling(4/Pi) = ceiling(1.273...) = 2;
a(2) = ceiling(r(2)/(1 - r(1)/2)) = 2;
a(3) = ceiling(r(3)/(1 - r(1)/2 - r(2)/2)) = 1.
The first 6 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are 0.636..., 0.839..., 0.968..., 0.988..., 0.995..., 0.9994...
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = Infinity; z = 16;
    r[k_] := N[4/Pi^k, 1000]; f[x_, 0] = x;
    n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
    f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
    x = 1; Table[n[x, k], {k, 1, z}]
    N[Sum[r[k]/n[x, k], {k, 1, 18}], 200]

Formula

a(n) = ceiling(r(n)/s(n)), where s(n) = 1 - r(1)/a(1) - r(2)/a(2) - ... - r(n-1)/a(n-1).
r(1)/a(1) + ... + r(n)/a(n) + ... = 1.

A270748 (r,1)-greedy sequence, where r(k) = 2/log(k+1).

Original entry on oeis.org

3, 48, 5215, 43930979, 8221176288381971, 237472642129791861355082716048930, 59916111345562665920456160598356741759066440491193682529746704653
Offset: 1

Views

Author

Clark Kimberling, Apr 09 2016

Keywords

Comments

Let x > 0, and let r = (r(k)) be a sequence of positive irrational numbers. Let a(1) be the least positive integer m such that r(1)/m < x, and inductively let a(n) be the least positive integer m such that r(1)/a(1) + ... + r(n-1)/a(n-1) + r(n)/m < x. The sequence (a(n)) is the (r,x)-greedy sequence. We are interested in choices of r and x for which the series r(1)/a(1) + ... + r(n)/a(n) + ... converges to x. See A270744 for a guide to related sequences.

Examples

			a(1) = ceiling(r(1)) = ceiling(2/log(2)) = ceiling(2.885...) = 3;
a(2) = ceiling(r(2)/(1 - r(1)/3)) = 48;
a(3) = ceiling(r(3)/(1 - r(1)/3 - r(2)/48)) = 5215.
The first 3 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are 0.961..., 0.997..., 0.99999997...
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = Infinity; z = 16;
    r[k_] := N[2/Log[k + 1], 1000]; f[x_, 0] = x;
    n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
    f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
    x = 1; Table[n[x, k], {k, 1, z}]
    N[Sum[r[k]/n[x, k], {k, 1, 18}], 200]

Formula

a(n) = ceiling(r(n)/s(n)), where s(n) = 1 - r(1)/a(1) - r(2)/a(2) - ... - r(n-1)/a(n-1).
r(1)/a(1) + ... + r(n)/a(n) + ... = 1.

A270749 (r,1)-greedy sequence, where r(k) = k/log(k+1).

Original entry on oeis.org

2, 7, 117, 28231, 934841727, 1391154929853413822, 3358221400639080017571595039208647108, 84149630763494298099512446622134485046922136023978562834130778814722933257
Offset: 1

Views

Author

Clark Kimberling, Apr 09 2016

Keywords

Comments

Let x > 0, and let r = (r(k)) be a sequence of positive irrational numbers. Let a(1) be the least positive integer m such that r(1)/m < x, and inductively let a(n) be the least positive integer m such that r(1)/a(1) + ... + r(n-1)/a(n-1) + r(n)/m < x. The sequence (a(n)) is the (r,x)-greedy sequence. We are interested in choices of r and x for which the series r(1)/a(1) + ... + r(n)/a(n) + ... converges to x. See A270744 for a guide to related sequences.

Examples

			a(1) = ceiling(r(1)) = ceiling(1/log(2)) = ceiling(1.442...) = 2;
a(2) = ceiling(r(2)/(1 - r(1)/2)) = 7;
a(3) = ceiling(r(3)/(1 - r(1)/2 - r(2)/7)) = 117.
The first 3 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are 0.721..., 0.981..., 0.99991...
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = Infinity; z = 16;
    r[k_] := N[k/Log[k + 1], 1000]; f[x_, 0] = x;
    n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
    f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
    x = 1; Table[n[x, k], {k, 1, z}]
    N[Sum[r[k]/n[x, k], {k, 1, 18}], 200]

Formula

a(n) = ceiling(r(n)/s(n)), where s(n) = 1 - r(1)/a(1) - r(2)/a(2) - ... - r(n-1)/a(n-1).
r(1)/a(1) + ... + r(n)/a(n) + ... = 1.

A270750 (r,1)-greedy sequence, where r(k) = 1/(k*log(k+1)).

Original entry on oeis.org

2, 2, 5, 52, 7132, 657650603, 642344866115572775, 833790618410287382945149122154404558, 1229679779588111283437146138551802288646488858072438842199407751052675116
Offset: 1

Views

Author

Clark Kimberling, Apr 09 2016

Keywords

Comments

Let x > 0, and let r = (r(k)) be a sequence of positive irrational numbers. Let a(1) be the least positive integer m such that r(1)/m < x, and inductively let a(n) be the least positive integer m such that r(1)/a(1) + ... + r(n-1)/a(n-1) + r(n)/m < x. The sequence (a(n)) is the (r,x)-greedy sequence. We are interested in choices of r and x for which the series r(1)/a(1) + ... + r(n)/a(n) + ... converges to x. See A270744 for a guide to related sequences.

Examples

			a(1) = ceiling(r(1)) = ceiling(1/log(2)) = ceiling(1.442...) = 2;
a(2) = ceiling(r(2)/(1 - r(1)/2)) = 2;
a(3) = ceiling(r(3)/(1 - r(1)/2 - r(2)/2)) = 5.
The first 3 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are 0.721..., 0.948..., 0.996...
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = Infinity; z = 16;
    r[k_] := N[1/(k*Log[k + 1]), 1000]; f[x_, 0] = x;
    n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
    f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
    x = 1; Table[n[x, k], {k, 1, z}]
    N[Sum[r[k]/n[x, k], {k, 1, 18}], 200]

Formula

a(n) = ceiling(r(n)/s(n)), where s(n) = 1 - r(1)/a(1) - r(2)/a(2) - ... - r(n-1)/a(n-1).
r(1)/a(1) + ... + r(n)/a(n) + ... = 1.

A270751 (r,1)-greedy sequence, where r(k) = 1/(k*tau) and tau = golden ratio.

Original entry on oeis.org

1, 1, 3, 37, 1204, 21029921, 425355555167420, 439183524292095499600664584581, 240317442633783387248198509182959563857071128274317237128901
Offset: 1

Views

Author

Clark Kimberling, Apr 09 2016

Keywords

Comments

Let x > 0, and let r = (r(k)) be a sequence of positive irrational numbers. Let a(1) be the least positive integer m such that r(1)/m < x, and inductively let a(n) be the least positive integer m such that r(1)/a(1) + ... + r(n-1)/a(n-1) + r(n)/m < x. The sequence (a(n)) is the (r,x)-greedy sequence. We are interested in choices of r and x for which the series r(1)/a(1) + ... + r(n)/a(n) + ... converges to x. See A270744 for a guide to related sequences.

Examples

			a(1) = ceiling(r(1)) = ceiling(1/tau) = ceiling(0.618...) = 1;
a(2) = ceiling(r(2)/(1 - r(1)/1)) = 1;
a(3) = ceiling(r(3)/(1 - r(1)/1 - r(2)/1)) = 3.
The first 3 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are 0.618..., 0.927..., 0.995...
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = Infinity; z = 16;
    r[k_] := N[1/(k*GoldenRatio), 1000]; f[x_, 0] = x;
    n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
    f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
    x = 1; Table[n[x, k], {k, 1, z}]
    N[Sum[r[k]/n[x, k], {k, 1, 18}], 200]

Formula

a(n) = ceiling(r(n)/s(n)), where s(n) = 1 - r(1)/a(1) - r(2)/a(2) - ... - r(n-1)/a(n-1).
r(1)/a(1) + ... + r(n)/a(n) + ... = 1.
Conjecture: a(n) = A270584(n-1) for n>1. - R. J. Mathar, Jun 21 2025

A270916 (r,1)-greedy sequence, where r(k) = 1/(k*sqrt(2)).

Original entry on oeis.org

1, 2, 3, 5, 65, 6529, 136091233, 41625259047416909, 2189507051227161558033650829868135, 75931290362065676573711484986356332365619562746656079489987281066955
Offset: 1

Views

Author

Clark Kimberling, Apr 09 2016

Keywords

Comments

Let x > 0, and let r = (r(k)) be a sequence of positive irrational numbers. Let a(1) be the least positive integer m such that r(1)/m < x, and inductively let a(n) be the least positive integer m such that r(1)/a(1) + ... + r(n-1)/a(n-1) + r(n)/m < x. The sequence (a(n)) is the (r,x)-greedy sequence. We are interested in choices of r and x for which the series r(1)/a(1) + ... + r(n)/a(n) + ... converges to x. See A270744 for a guide to related sequences.

Examples

			a(1) = ceiling(r(1)) = ceiling(1/sqrt(2)) = ceiling(0.707...) = 1;
a(2) = ceiling(r(2)/(1 - r(1)/1)) = 2;
a(3) = ceiling(r(3)/(1 - r(1)/1 - r(2)/2)) = 3.
The first 6 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are 0.707..., 0.883..., 0.962..., 0.997..., 0.99998..., 0.9999999992...
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = Infinity; z = 16;
    r[k_] := N[1/(k*Sqrt[2]), 1000]; f[x_, 0] = x;
    n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
    f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
    x = 1; Table[n[x, k], {k, 1, z}]
    N[Sum[r[k]/n[x, k], {k, 1, 18}], 200]

Formula

a(n) = ceiling(r(n)/s(n)), where s(n) = 1 - r(1)/a(1) - r(2)/a(2) - ... - r(n-1)/a(n-1).
r(1)/a(1) + ... + r(n)/a(n) + ... = 1.
Conjecture: a(n) = A270582(n-1). - R. J. Mathar, Jun 02 2016
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