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.

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A114869 s(n) = floor(n^(n/5)/n!!!!!).

Original entry on oeis.org

1, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 4, 6, 5, 5, 7, 10, 16, 12, 14, 18, 26, 39, 31, 35, 45, 64, 98, 79, 88, 114, 163, 249, 200, 223, 291, 416, 636, 511, 572, 745, 1067, 1634, 1316, 1474, 1922, 2755, 4222, 3405, 3817, 4982, 7147, 10961, 8848, 9925, 12966
Offset: 1

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Author

Jonathan Vos Post, Feb 20 2006

Keywords

Comments

This sequence is an approximation of a quintuple factorial analog to Stirling's approximation to factorial. Note that a(n) is exact for n = 1, 5, 10.

Examples

			a(10) = floor(10^2/10!!!!!) = floor(10^2/50) = floor(2) = 2.
a(15) = floor(15^3/15!!!!!) = floor((15^3)/750) = floor(4.5) = 4.
a(20) = floor(20^4/20!!!!!) = floor((20^4)/15000) = floor(10.6666667) = 10.
a(25) = floor(25^5/25!!!!!) = floor((25^5)/375000) = floor(26.0416667) = 26.
a(30) = floor(30^6/30!!!!!) = floor((30^6)/11250000) = floor(64.8) = 64.
a(35) = floor(35^7/35!!!!!) = floor((35^7)/393750000) = floor(163.401389) = 163.
		

Crossrefs

Programs

  • Mathematica
    fac[n_Integer, m_Integer] := Block[{t = n, f = Max[1, n]}, While[t > m, t -= m; f *= t]; f]; a[n_] := Floor[n^(n/5)/fac[n, 5]]; Array[a, 65] (* Giovanni Resta, Jun 15 2016 *)

Formula

a(n) = floor(n^(n/5)/n!!!). a(n) = floor((A000312(n)^(1/5))/A085157(n)).

Extensions

Corrected and extended by Giovanni Resta, Jun 15 2016

A114968 a(n) = floor(n^(n/6)/n!!!!!!).

Original entry on oeis.org

1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 4, 6, 4, 4, 5, 7, 10, 15, 11, 12, 14, 18, 26, 37, 28, 29, 35, 46, 64, 94, 71, 74, 89, 117, 163, 238, 180, 188, 226, 297, 416, 608, 461, 481, 580, 763, 1067, 1563, 1187, 1240, 1496, 1969, 2755, 4038, 3070
Offset: 1

Views

Author

Jonathan Vos Post, Feb 22 2006

Keywords

Comments

This sequence is an approximation of a sextuple factorial analog to Stirling's approximation to factorial. Note that a(n) is exact for n = 1, 6, 8, 9, 12.

Examples

			a(12) = floor(12^2/12!!!!!) = floor(12^2/72) = floor(2) = 2.
a(18) = floor(18^3/18!!!!!) = floor((18^3)/1296) = floor(4.5) = 4.
a(24) = floor(24^4/20!!!!!) = floor((24^4)/31104) = floor(10.6666667) = 10.
a(30) = floor(30^5/25!!!!!) = floor((30^5)/933120) = floor(26.0416667) = 26.
a(36) = floor(36^6/30!!!!!) = floor((36^6)/33592320) = floor(64.8) = 64.
		

Crossrefs

Programs

  • Mathematica
    fac[n_, m_] := Block[{t = n, f = Max[1, n]}, While[t > m, t -= m; f *= t]; f]; a[n_] := Floor[n^(n/6)/fac[n, 6]]; Array[a, 65] (* Giovanni Resta, Jun 15 2016 *)

Formula

a(n) = floor(n^(n/6)/n!!!!!!). a(n) = floor(n^(n/6)/n!6). a(n) = floor((A000312(n)^(1/6))/A085158(n)).

Extensions

Corrected and extended by Giovanni Resta, Jun 15 2016

A121050 Partial sums of floor(n!/e^n).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 5, 18, 62, 226, 892, 3835, 17910, 90401, 490421, 2844970, 17570207, 115078228, 796630963, 5811215750, 44551231383, 358087648032, 3010990488143, 26433752431402, 241852566790848, 2302304545944415, 22768248457537798, 233580248716343376
Offset: 1

Views

Author

Jonathan Vos Post, Aug 08 2006

Keywords

Crossrefs

Programs

  • Mathematica
    Accumulate[Table[Floor[n!/Exp[n]],{n,30}]] (* Harvey P. Dale, Aug 25 2016 *)

Formula

a(n) = Sum_{i=1..n} floor(i!/exp(i)).

Extensions

Corrected by and incorrect examples removed by Harvey P. Dale, Aug 25 2016

A127585 Exponential error term from Stirling's Approximation.

Original entry on oeis.org

1, 1, 18, 345, 10243, 437769, 25260317, 1873346813, 172254143084, 19114537903943, 2506628271002200, 382005168783773474, 66734799966312471195, 13212509243902296154744, 2936153006332857671962341, 726345521215072990990045577, 198595552305314906351047196508
Offset: 0

Views

Author

Jonathan Vos Post, Apr 02 2007

Keywords

Examples

			a(1) = Floor[(sqrt(2*pi) * (1^1) * (1^(1/2))) - 1! ] = Floor(1.50662827) = 1.
a(2) = Floor[(sqrt(2*pi) * (2^2) * (2^(2/2))) - 2! ] = Floor(18.0530262) = 18.
		

Crossrefs

Formula

a(n) = floor(sqrt(2*Pi)*(n^n)*(n^(n/2))) - n!.

Extensions

More terms from Alois P. Heinz, Jan 24 2024
Previous Showing 21-24 of 24 results.