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 24 results. Next

A115240 a(n) = A001672(n) - A115239(n).

Original entry on oeis.org

0, 0, 3, 10, 33, 104, 328, 1031, 3241, 10183, 31991, 100503, 315740, 991928, 3116234, 9789941, 30756009, 96622854, 303549648, 953629346, 2995914948, 9411944394, 29568495367, 92892167824, 291829352014, 916808948392, 2880240257014
Offset: 1

Views

Author

Hieronymus Fischer, Jan 17 2006

Keywords

Examples

			a(5) = A001672(5) - A115239(5) = 306 - 273 = 33.
		

Crossrefs

Programs

  • Mathematica
    a[1] = Floor[Pi]; a[n_] := a[n] = Floor[a[n - 1]*Pi]; Array[ Floor[Pi^# ] - a[ # ] &, 27] (* Robert G. Wilson v *)

Extensions

More terms from Robert G. Wilson v, Jan 18 2006

A119604 Merged values of A014217 = {floor(((1+sqrt(5))/2)^n)}, A000149 = {floor(e^n)}, and A001672 = {floor(Pi^n)}, with multiplicity.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 3, 4, 6, 7, 9, 11, 17, 20, 29, 31, 46, 54, 76, 97, 122, 148, 199, 306, 321, 403, 521, 842, 961, 1096, 1364, 2206, 2980, 3020, 3571, 5777, 8103, 9349, 9488, 15126, 22026, 24476
Offset: 0

Views

Author

Bryan David Levenson, Jun 03 2006

Keywords

Examples

			a(8)=6 because floor(((1+sqrt(5))/2)^4)=6, a(9)=7 because floor(e^2)=7 and a(10)=9 because floor(Pi^2)=9.
		

Crossrefs

Extensions

Edited by M. F. Hasler, May 29 2018

A137994 a(n) is the smallest integer > a(n-1) such that {Pi^a(n)} < {Pi^a(n-1)}, where {x} = x - floor(x), a(1)=1.

Original entry on oeis.org

1, 3, 81, 264, 281, 472, 1147, 2081, 3207, 3592, 10479, 12128, 65875, 114791, 118885
Offset: 1

Views

Author

Leroy Quet and M. F. Hasler, Mar 14 2008

Keywords

Comments

The sequence was suggested by Leroy Quet on Pi day 2008, cf. A138324.
The next such number must be greater than 100000. - Hieronymus Fischer, Jan 06 2009
a(16) > 300,000. - Robert Price, Mar 25 2019

Examples

			a(3)=81, since {Pi^81}=0.0037011283.., but {Pi^k}>=0.0062766802... for 1<=k<=80; thus {Pi^81}<{Pi^k} for 1<=k<81. - _Hieronymus Fischer_, Jan 06 2009
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = 10000;
    p = .999;
    Select[Range[1, 5000],
    If[FractionalPart[Pi^#] < p, p = FractionalPart[Pi^#]; True] &] (* Robert Price, Mar 12 2019 *)
  • PARI
    default(realprecision,10^4); print1(a=1); for(i=1,100, f=frac(Pi^a); until( frac(Pi^a++)
    				

Extensions

a(11)-a(13) from Hieronymus Fischer, Jan 06 2009
Edited by R. J. Mathar, May 21 2010
a(14)-a(15) from Robert Price, Mar 12 2019

A002160 Nearest integer to Pi^n.

Original entry on oeis.org

1, 3, 10, 31, 97, 306, 961, 3020, 9489, 29809, 93648, 294204, 924269, 2903677, 9122171, 28658146, 90032221, 282844564, 888582403, 2791563950, 8769956796, 27551631843, 86556004192, 271923706894, 854273519914, 2683779414318, 8431341691876, 26487841119104, 83214007069230
Offset: 0

Views

Author

Keywords

Examples

			a(0) = 1 because Pi^0 = 1;
a(2) = 10 because Pi^2 = 9.8696...;
a(10) = 93648 because Pi^10 = 93648.047476...
		

References

  • A. Fletcher, J. C. P. Miller, L. Rosenhead, and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 122.
  • J. T. Peters, Ten-Place Logarithm Table. Vols. 1 and 2, rev. ed. Ungar, NY, 1957, Vol. 1 (Appendix), p. 1.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000227 (e^n), A001672 (floor(Pi^n)), A001673 (ceiling(Pi^n)).

Programs

  • Maple
    a := []: Digits := 1000: for n from 0 to 50 do: a := [op(a),round(Pi^n)]: od: seq(a[i+1],i=0..50);
  • Mathematica
    Round[Pi^Range[0,40]] (* Harvey P. Dale, Jun 10 2024 *)
  • PARI
    apply( A002160(n)=Pi^n\/1, [0..50]) \\ An error message will say so if default(realprecision) must be increased. - M. F. Hasler, May 27 2018
    
  • Sage
    [round(pi^n) for n in range(0,29)] # Stefano Spezia, Jan 15 2025

Extensions

More terms from Mark Hudson (mrmarkhudson(AT)hotmail.com), Jan 29 2003
Edited by M. F. Hasler, May 27 2018

A153710 Numbers k such that the fractional part of Pi^k is less than 1/k.

Original entry on oeis.org

1, 3, 5, 9, 10, 11, 59, 81, 264, 281, 472, 3592, 10479, 12128, 65875, 118885
Offset: 1

Views

Author

Hieronymus Fischer, Jan 08 2009

Keywords

Comments

Numbers k such that fract(Pi^k) < 1/k, where fract(x) = x-floor(x).
The next such number must be greater than 100000.
a(17) > 300000. - Robert Price, Mar 25 2019

Examples

			a(4) = 9 since fract(Pi^9) = 0.0993... < 1/9, but fract(Pi^k) = 0.3891..., 0.2932..., 0.5310... for 6 <= k <= 8, which all are greater than 1/k.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[1000], N[FractionalPart[Pi^#], 100] < (1/#) &]  (* G. C. Greubel, Aug 25 2016 *)
  • PARI
    isok(k) = frac(Pi^k) < 1/k; \\ Michel Marcus, Feb 11 2014

Extensions

a(16) from Robert Price, Mar 25 2019

A153712 Numbers k such that the fractional part of Pi^k is greater than 1-(1/k).

Original entry on oeis.org

1, 2, 15, 22, 58, 109, 157, 1030, 1071, 1274, 2008, 2322, 5269, 151710
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Comments

Numbers k such that fract(Pi^k) > 1-(1/k), where fract(x) = x-floor(x).
The next such number must be greater than 100000.
a(15) > 300000. - Robert Price, Mar 25 2019

Examples

			a(3) = 15, since fract(Pi^15) = 0.969... > 0.933... = 1 - (1/15), but fract(Pi^k) <= 1 - (1/k) for 3 <= k <= 14.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[1000], N[FractionalPart[Pi^#], 100] > 1 - (1/#) &]  (* G. C. Greubel, Aug 25 2016 *)

Extensions

a(14) from Robert Price, Mar 25 2019

A153711 Minimal exponents m such that the fractional part of Pi^m obtains a maximum (when starting with m=1).

Original entry on oeis.org

1, 2, 15, 22, 58, 157, 1030, 5269, 145048, 151710
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Comments

Recursive definition: a(1)=1, a(n) = least number m>a(n-1) such that the fractional part of Pi^m is greater than the fractional part of Pi^k for all k, 1<=k
The next such number must be greater than 100000.
a(11) > 300000. - Robert Price, Mar 25 2019

Examples

			a(3)=15, since fract(Pi^15)= 0.9693879984..., but fract(Pi^k)<=0.8696... for 1<=k<=14; thus fract(Pi^15)>fract(Pi^k) for 1<=k<15 and 15 is the minimal exponent > 2 with this property.
		

Programs

  • Mathematica
    $MaxExtraPrecision = 100000;
    p = 0; Select[Range[1, 10000],
    If[FractionalPart[Pi^#] > p, p = FractionalPart[Pi^#]; True] &] (* Robert Price, Mar 25 2019 *)

Formula

Recursion: a(1):=1, a(k):=min{ m>1 | fract(Pi^m) > fract(Pi^a(k-1))}, where fract(x) = x-floor(x).

Extensions

a(9)-a(10) from Robert Price, Mar 25 2019

A153715 Greatest number m such that the fractional part of Pi^A153711(m) >= 1-(1/m).

Original entry on oeis.org

1, 7, 32, 53, 189, 2665, 10810, 26577, 128778, 483367
Offset: 1

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(3) = 32, since 1-(1/33) = 0.9696... > fract(Pi^A153711(3)) = fract(Pi^15) = 0.96938... >= 0.96875 = 1-(1/32).
		

Programs

  • Mathematica
    $MaxExtraPrecision = 100000;
    A153711 = {1, 2, 15, 22, 58, 157, 1030, 5269, 145048, 151710};
    Floor[1/(1-FractionalPart[Pi^A153711])] (* Robert Price, Apr 18 2019 *)

Formula

a(n) = floor(1/(1-fract(Pi^A153711(n)))), where fract(x) = x-floor(x).

Extensions

a(9)-a(10) from Robert Price, Apr 18 2019

A153714 Greatest number m such that the fractional part of Pi^A153710(n) <= 1/m.

Original entry on oeis.org

7, 159, 50, 10, 21, 55, 117, 270, 307, 744, 757, 7804, 13876, 62099, 70718, 154755
Offset: 1

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(2)=159 since 1/160<fract(Pi^A153710(2))=fract(Pi^3)=0.0062766...<=1/159.
		

Programs

  • Mathematica
    A153710 = {1, 3, 5, 9, 10, 11, 59, 81, 264, 281, 472, 3592, 10479,
       12128, 65875, 118885};
    Table[fp = FractionalPart[Pi^A153710[[n]]]; m = Floor[1/fp];
    While[fp <= 1/m, m++]; m - 1, {n, 1, Length[A153710]}] (* Robert Price, May 10 2019 *)

Formula

a(n) = floor(1/fract(Pi^A153710(n))), where fract(x) = x-floor(x).

Extensions

a(16) from Robert Price, May 10 2019

A153716 Greatest number m such that the fractional part of Pi^A153712(n) >= 1-(1/m).

Original entry on oeis.org

1, 7, 32, 53, 189, 131, 2665, 10810, 2693, 1976, 3697, 4289, 26577, 483367
Offset: 1

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(3) = 32, since 1-(1/33) = 0.9696... > fract(Pi^A153712(3)) = fract(Pi^15) = 0.96938... >= 0.96875 = 1-(1/32).
		

Programs

  • Mathematica
    A153712 = {1, 2, 15, 22, 58, 109, 157, 1030, 1071, 1274, 2008, 2322,
       5269, 151710};
    Table[Floor[1/(1 - FractionalPart[Pi^A153712[[n]]])], {n, 1,
    Length[A153712]}] (* Robert Price, May 10 2019 *)

Formula

a(n) = floor(1/(1-fract(Pi^A153712(n)))), where fract(x) = x-floor(x).

Extensions

a(14) from Robert Price, May 10 2019
Showing 1-10 of 24 results. Next