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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A210116 Floor of the expected value of number of trials until exactly five cells are empty in a random distribution of n balls in n cells.

Original entry on oeis.org

7776, 311, 51, 16, 7, 4, 3, 3, 2, 3, 3, 4, 5, 8, 11, 16, 25, 40, 66, 110, 187, 325, 574, 1032, 1885, 3492, 6557, 12467, 23988, 46667, 91731, 182078, 364734, 736972, 1501318, 3082136, 6374007, 13273719, 27825438, 58697777, 124566798
Offset: 6

Views

Author

Washington Bomfim, Mar 18 2012

Keywords

Comments

Also floor of the expected value of number of trials until we have n-5 distinct symbols in a random sequence on n symbols of length n. A055775 corresponds to zero cells empty.

Examples

			For n=6, there are 6^6 = 46656 sequences on 6 symbols of length 6. Only 6 sequences has a unique symbol, so a(6) = floor(46656/6) = 7776.
		

References

  • W. Feller, An Introduction to Probability Theory and its Applications, 2nd ed, Wiley, New York, 1965, (2.4) p. 92. (Occupancy problems)

Crossrefs

Formula

With m = 5, a(n) = floor(n^n/(binomial(n,m)*_Sum{v=0..n-m-1}((-1)^v*binomial(n-m,v)*(n-m-v)^n)))

A121563 Numerator of Sum_{i=1..n} i!/(i^i).

Original entry on oeis.org

0, 1, 3, 31, 523, 333787, 3029083, 357534985867, 1466339436008107, 86629054728081110243, 54153614078975016066611, 1404709830784309760237846406097211, 1404749977001540310126014452972211
Offset: 0

Views

Author

Jonathan Vos Post, Aug 07 2006

Keywords

Examples

			A121563/A121564 = 0, 1, 3/2, 31/18, 523/288, 333787/180000, ... .
		

Crossrefs

Denominators are in A121564.

Programs

  • Mathematica
    Numerator[Table[Sum[i!/(i^i),{i,n}],{n,12}]] (* James C. McMahon, Oct 19 2024 *)

Formula

Numerator of Sum_{i=1..n} A000142(i)/A000312(i).
Limit_{n -> oo} A121563(n)/A121564(n) = A094082.

A121564 Denominator of Sum_{i=1..n} i!/(i^i).

Original entry on oeis.org

1, 1, 2, 18, 288, 180000, 1620000, 190591380000, 780662292480000, 46097327708651520000, 28810829817907200000000, 747278726094210559615027200000000, 747278726094210559615027200000000, 17410163370762081276553474535120146483200000000
Offset: 0

Views

Author

Jonathan Vos Post, Aug 07 2006

Keywords

Examples

			A121563/A121564 = 0, 1, 3/2, 31/18, 523/288, 333787/180000, ... .
		

Crossrefs

Numerators are in A121563.

Programs

  • Mathematica
    Denominator[Table[Sum[i!/(i^i),{i,n}],{n,12}]] (* James C. McMahon, Oct 19 2024 *)

Formula

Denominator of Sum_{i=1..n} A000142(i)/A000312(i).
Limit_{n -> oo} A121563(n)/A121564(n) = A094082.

A208847 A056915(n) mod 5228905 mod 17.

Original entry on oeis.org

3, 4, 13, 15, 8, 14, 9, 5, 0, 11, 16, 10, 2, 12, 7, 1, 6, 16, 3, 10, 5, 8, 7, 16, 6, 11, 13, 6, 10, 6, 11, 16, 9, 1, 1, 15, 5, 1, 14, 7, 15, 2, 14, 9, 2, 6, 14, 3, 3, 14, 12, 6, 2, 4, 10, 16, 6, 10, 9, 3, 3, 1, 7, 9, 11, 5
Offset: 1

Views

Author

Washington Bomfim, Mar 02 2012

Keywords

Comments

A056915(n) mod 5228905 mod 17 is a bijection from the set of the first 17 terms of A056915 to {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}.
From an algorithm based on strong pseudoprimes to bases 2,3 and 5, and a table T with the first 17 terms of A056915, we can test if n is prime, odd n, 1 < n < 42550716781. When n is a prime, we check if n belongs to T. A fast way to do that is to compute i = n mod 5228905 mod 17 and compare n with T[i]. If n is not equal to T[i], n is prime.
Terms computed using table by Charles R Greathouse IV. See A056915.

Crossrefs

A215460 a(n) = floor(n!^2 / n^n).

Original entry on oeis.org

1, 1, 1, 2, 4, 11, 30, 96, 339, 1316, 5584, 25733, 128025, 683949, 3905083, 23731187, 152934464, 1041782238, 7479469995, 56448098958, 446768591341, 3700276748921, 32007269639380, 288630046441757, 2708888570942365, 26419890078249485, 267389254029561667, 2804508541393392135, 30446382653772707171, 341729529206733994569
Offset: 1

Views

Author

Alex Ratushnyak, Aug 11 2012

Keywords

Examples

			a(4) = floor((4!)^2 / (4^4)) = floor(24^2 / 256) = floor(2.25) = 2.
		

Crossrefs

Programs

  • Mathematica
    A215460[n_] := Quotient[n!^2, n^n];
    Array[A215460, 30] (* Paolo Xausa, Aug 05 2025 *)
  • Python
    f = 1
    for n in range(1,33):
        print(f*f // n**n)
        f *= n+1

Extensions

More terms from Paolo Xausa, Aug 05 2025

A235496 a(n) = round(n^n/n!) where round(1/2)=1.

Original entry on oeis.org

1, 1, 2, 5, 11, 26, 65, 163, 416, 1068, 2756, 7148, 18614, 48639, 127463, 334865, 881658, 2325751, 6145597, 16263866, 43099804, 114356611, 303761260, 807692035, 2149632061, 5726042115, 15264691107, 40722913454, 108713644516, 290404350963, 776207020880
Offset: 0

Views

Author

Vincenzo Librandi, Jan 15 2014

Keywords

Comments

We have two versions of this sequence, this and A240571, because there is no universal agreement on how to round a number like 9/2. - N. J. A. Sloane, Dec 13 2015

Crossrefs

See A240571 for another version.

Programs

  • Magma
    [Round(n^n/Factorial(n)): n in [1..40]];
    
  • Maple
    a:= n-> round(n^n/n!):
    seq(a(n), n=0..32);  # Alois P. Heinz, Dec 13 2015
  • Mathematica
    Table[Floor[n^n/n! + 1/2], {n, 40}]
  • PARI
    s=[]; for(n=1, 30, s=concat(s, round(n^n/n!))); s \\ Colin Barker, Jan 19 2014

Formula

a(n) = Round(A000312(n)/A000142(n)).

Extensions

Name clarified by Sean A. Irvine, Jan 12 2025

A114853 a(n) = floor(n^n/n!!).

Original entry on oeis.org

1, 2, 9, 32, 208, 972, 7843, 43690, 409968, 2604166, 27447010, 193491763, 2241278030, 17224712961, 216027868615, 1787142709274, 24006211998207, 211773735868781, 3021737893128258, 28218694885361552, 424936725846414486
Offset: 1

Views

Author

Jonathan Vos Post, Feb 20 2006

Keywords

Comments

This is to double factorial A006882 as A055775 "Floor(n^n/n!)" is to factorial. This sequence is a weak first approximation of a double factorial analog to Stirling's approximation to factorial. Note that a(n) is exact for n = 1, 2, 3, 4, 6.

Examples

			a(10) = floor((10^10)/3840) = floor(2604166.67) = 2604166.
		

Crossrefs

Programs

  • Maple
    A114853 := proc(n)
        n^n/doublefactorial(n) ;
        floor(%) ;
    end proc:
    seq(A114853(n),n=1..25) ; # R. J. Mathar, Jun 23 2014
  • Mathematica
    Table[Floor[n^n/n!!],{n,30}] (* Harvey P. Dale, Jul 29 2023 *)

Formula

a(n) = floor(n^n/n!!). a(n) = floor(A000312(n)/A006882(n)).

A114854 a(n) = floor(n^(n/2)/n!!).

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 8, 10, 20, 26, 51, 64, 128, 163, 326, 416, 834, 1067, 2148, 2755, 5559, 7147, 14449, 18613, 37696, 48638, 98650, 127463, 258857, 334864, 680822, 881657, 1794294, 2325750, 4737361
Offset: 1

Views

Author

Jonathan Vos Post, Feb 20 2006

Keywords

Comments

This sequence is a second approximation of a double factorial analog to Stirling's approximation to factorial. Note that a(n) is exact for n = 1, 2, 4.

Examples

			a(10) = floor((10^5)/3840) = floor(26.0416667) = 26.
a(11) = floor((11^5.5)/10395) = floor(51.3848715) = 51.
		

Crossrefs

Programs

  • Maple
    A114854 := proc(n)
        n^(n/2)/doublefactorial(n) ;
        floor(%) ;
    end proc:
    seq(A114854(n),n=1..35) ; # R. J. Mathar, Jun 23 2014
  • Mathematica
    Table[Floor[n^(n/2)/n!!],{n,40}] (* Harvey P. Dale, Apr 04 2019 *)

Formula

a(n) = floor(n^(n/2)/n!!). a(n) = floor(sqrt(A000312(n))/A006882(n)).

A114863 a(n) = floor(n^(n/3)/n!!!).

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 3, 3, 4, 7, 7, 10, 18, 18, 26, 45, 44, 64, 113, 112, 163, 287, 285, 416, 733, 731, 1067, 1885, 1885, 2755, 4873, 4881, 7147, 12659, 12697, 18613, 33007, 33143, 48638, 86337, 86777, 127463, 226454, 227795, 334864, 595382, 599342, 881657
Offset: 1

Views

Author

Jonathan Vos Post, Feb 20 2006

Keywords

Comments

This sequence is an approximation of a triple factorial analog to Stirling's approximation to factorial. Note that a(n) is exact for n = 1, 3, 6.

Examples

			a(9) = floor((9^3)/162) = floor(4.5) = 4.
a(10) = floor((10^3.33333)/280) = floor(7.69381906) = 7.
a(27) = floor((27^9)/7142567040) = floor(1067.62701) = 1067.
		

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/3)/fac[n, 3]]; Array[a, 40] (* Giovanni Resta, Jun 15 2016 *)
    Table[Floor[n^(n/3)/Times@@Range[n,1,-3]],{n,50}] (* Harvey P. Dale, Aug 06 2021 *)

Formula

a(n) = floor(n^(n/3)/n!!!). a(n) = floor((A000312(n)^(1/3))/A007661(n)).

Extensions

More terms from Giovanni Resta, Jun 15 2016

A114868 a(n) = floor(n^(n/4)/n!!!!).

Original entry on oeis.org

1, 0, 0, 1, 1, 1, 1, 2, 3, 2, 3, 4, 7, 6, 7, 10, 17, 14, 18, 26, 41, 36, 44, 64, 104, 91, 112, 163
Offset: 1

Views

Author

Jonathan Vos Post, Feb 20 2006

Keywords

Comments

This sequence is an approximation to a quadruple factorial analog of Stirling's approximation to the factorial function. Note that a(n) is exact for n = 1, 4, 8.

Examples

			a(8) = floor((8^2)/8!!!!) = floor((8^2)/32) = floor(2) = 2.
a(9) = floor((9^2.25)/9!!!!) = floor((9^2.25)/45) = floor(3.11769145) = 3.
a(16) = floor((16^4)/16!!!!) = floor((16^4)/6144) = floor(10.6666667) = 10.
a(20) = floor((20^5)/20!!!!) = floor((20^5)/122880) = floor(26.0416667) = 26.
		

Crossrefs

Formula

a(n) = floor(n^(n/4)/n!!!). a(n) = floor((A000312(n)^(1/4))/A007662(n)).
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