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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A135004 Decimal expansion of 4/e.

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Nov 15 2007

Keywords

Comments

Leading coefficient of energy difference between the two lowest states of the hydrogen molecular ion (see Holstein-Herring Method link). - Andrea Pinos, Nov 22 2023

Examples

			1.471517764685769...
		

Crossrefs

Cf. A019741 (e/4), A068985 (1/e), A135002 (2/e).

Programs

Formula

Equals lim_{n->oo} (((2*n)!/n!)^(1/n))/n (Godin, 2012). - Amiram Eldar, Apr 02 2022

Extensions

More terms from Harvey P. Dale, Mar 06 2012

A135005 Decimal expansion of 5/e.

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Nov 15 2007

Keywords

Comments

The fraction of substituents which become isolated in a simple polymer model is 1/10 this number, see Flory 1939 (and 1936). - Charles R Greathouse IV, Nov 30 2012

Examples

			5/e = 1.839397205857211607977618850807304337229...
1/(2*e) = 0.1839397205857211607977618850807304337229... (see Greathouse's comment).
		

Crossrefs

Programs

Formula

Equals 5/A001113 and 5*A068985. - Michel Marcus, Sep 17 2016
1/(2*e) = Integral_{x=1..oo} e^(-x^2) * x dx. - Amiram Eldar, Aug 03 2020
1/(2*e) = lim_{n->oo} n * ((n+1)!^(1/(n+1)) - n!^(1/n) - 1/e) (Furdui, 2008). - Amiram Eldar, Apr 20 2023

A181590 Least value of n such that |P(n) - 1/e| < 10^(-i), i=1,2,3... . P(n)=floor(n!/e + 1/2)/n! is the probability of a random permutation on n objects be a derangement.

Original entry on oeis.org

3, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 14, 15, 16, 17, 18, 18, 19, 20, 21, 21, 22, 23, 24, 24, 25, 26, 26, 27, 28, 29, 29, 30, 31, 31, 32, 33, 33, 34, 34, 35, 36, 36, 37, 38, 38, 39, 40, 40, 41, 41, 42, 43, 43, 44, 44, 45, 46, 46, 47, 47, 48, 49, 49, 50, 50, 51, 52, 52, 53, 53
Offset: 1

Views

Author

Washington Bomfim, Oct 31 2010

Keywords

Comments

Both P(n) and the probability that a rooted forest on [n] be a tree tend to 1/e when n rises to infinity. So the events random forest be a tree and random permutation be a derangement become equiprobable as n tends to infinity.
The probability P(n) approaches 1/e quite quickly as this sequence shows. See image clicking the first link.

Examples

			a(2) = 4, a(3) = 6, so for n in the interval 4...5, if we use 1/e as the probability P, we make an error less than 10^(-1).
In general if n is in the interval a(i), ... , a(i+k)-1, k the least positive integer such that a(i+k) > a(i), this error is less than 10(-i-k+1).
For example, a(11) = a(12) = 14, k = 2 and if n is in the interval 14...14, if we use 1/e as the probability P, we make an error less than 10^(-12).
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = 100; f[n_] := Block[{k = 1}, While[ Abs[ Floor[(k!/E + 1/2)]/k! - 1/E] > 1/10^n, k++ ]; k]; Array[f, 71] (* Robert G. Wilson v, Nov 05 2010 *)

Extensions

More terms from Robert G. Wilson v, Nov 05 2010

A182767 Beatty sequence for 1+e^2.

Original entry on oeis.org

8, 16, 25, 33, 41, 50, 58, 67, 75, 83, 92, 100, 109, 117, 125, 134, 142, 151, 159, 167, 176, 184, 192, 201, 209, 218, 226, 234, 243, 251, 260, 268, 276, 285, 293, 302, 310, 318, 327, 335, 343, 352, 360, 369, 377, 385, 394, 402, 411, 419, 427
Offset: 1

Views

Author

Clark Kimberling, Nov 29 2010

Keywords

Comments

Let u=e=A001113 and v=1/e=A068985. Jointly rank {j*u} and {k*v} as in the first comment at A182760; a(n) is the position of n*u.

Crossrefs

Programs

  • Maple
    A182767 := proc(n) floor(n*(1+exp(2))) ; end proc:

Formula

a(n)=floor(n*(1+e^2)) = floor(n+n*A072334).

A278327 Decimal expansion of 1/e - 1/e^2.

Original entry on oeis.org

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

Views

Author

Alois P. Heinz, Nov 18 2016

Keywords

Comments

The probability, as n = 2^k increases without bound, that a randomized skip list with n elements and p = 1/2 has exactly k levels.

Examples

			0.232544157934829629701524275188976464...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[1/E - 1/E^2, 10, 120][[1]] (* Amiram Eldar, May 27 2023 *)

Formula

Equals 1/A001113 - 1/A072334 = A068985 - A092553.

A282822 a(n) = (n - 4)*n! for n>=0.

Original entry on oeis.org

-4, -3, -4, -6, 0, 120, 1440, 15120, 161280, 1814400, 21772800, 279417600, 3832012800, 56043187200, 871782912000, 14384418048000, 251073478656000, 4623936565248000, 89633231880192000, 1824676506132480000, 38926432130826240000, 868546016919060480000
Offset: 0

Views

Author

Bruno Berselli, Feb 22 2017

Keywords

Crossrefs

Cf. A034865.
Cf. sequences with formula (n + k)*n! listed in A282466.

Programs

  • Mathematica
    Table[(n - 4) n!, {n, 0, 30}] (* or *)
    RecurrenceTable[{a[0] == -4, a[n] == n a[n - 1] + n!}, a, {n, 0, 30}]

Formula

E.g.f.: -(4 - 5*x)/(1 - x)^2.
a(n) = n*a(n-1) + n!, with n>0, a(0)=-4.
a(n) = 2*A034865(n) for n>3.
From Amiram Eldar, Jan 14 2021: (Start)
Sum_{n>=5} 1/a(n) = 313/288 - 5*e/12 - gamma/24 + Ei(1)/24 = 313/288 - (5/12)*A001113 - (1/24)*A001620 + A091725/24.
Sum_{n>=5} (-1)^(n+1)/a(n) = -25/288 + 1/(6*e) + gamma/24 - Ei(-1)/24 = -25/288 - (1/6)*A068985 + (1/24)*A001620 + (1/24)*A099285. (End)

A325748 First term of n-th difference sequence of (floor(k/e)), k >= 0.

Original entry on oeis.org

0, 0, 1, -3, 6, -9, 9, 0, -27, 81, -161, 231, -165, -363, 2079, -6435, 16071, -35343, 70686, -130339, 222717, -352715, 515944, -692967, 849299, -950455, 996435, -1042415, 1042415, 1, -6926071, 36516481, -137549665, 436362817, -1233496132, 3197432557
Offset: 1

Views

Author

Clark Kimberling, Jun 07 2019

Keywords

Crossrefs

Cf. A068985.

Programs

  • Mathematica
    Table[First[Differences[Table[Floor[n/E], {n, 0, 50}], n]], {n, 1, 50}]

A337986 Prime numbers p such that v_p(A000166(k)) = v_p(k-1) for all k > 1, where v_p(k) is the p-adic valuation of k.

Original entry on oeis.org

2, 5, 7, 17, 19, 23, 29, 43, 59, 61, 71, 73, 107, 113, 131, 137, 149, 157, 173, 181, 191, 197, 199, 229, 233, 239, 241, 251, 257, 311, 317, 331, 349, 383, 401, 409, 421, 461, 491, 499, 541, 547, 557, 599, 601, 613, 619, 641, 653, 673, 719, 751, 761, 787, 797, 809
Offset: 1

Views

Author

Amiram Eldar, Jan 29 2021

Keywords

Comments

Miska (2016) proved that the complement of this sequence within the primes is infinite, and conjectured that this sequence is also infinite, and that its asymptotic density within the primes is 1/e (A068985). Numerically, he found that there are 28990 terms below 10^6, which are about 37% of all the primes less than 10^6.

Examples

			2 is a term since A007814(A000166(k)) = A007814(k-1) for all k > 1.
		

Crossrefs

Programs

  • Mathematica
    e[n_] := e[n] = Subfactorial[n]/(n - 1); q[p_] := PrimeQ[p] && AllTrue[Table[e[n], {n, 2, p + 1}], ! Divisible[#, p] &]; Select[Range[1000], q]

Formula

A prime p is a term if and only if p does not divide any of the numbers A000255(k), k in {2, ..., p+1}.

A365594 The denominators of a series that converges to 1/e obtained using Whittaker's Root Series Formula.

Original entry on oeis.org

3, 42, 154, 3817, 1141283, 119706444, 1396550916, 20958700652, 2359646218028, 324742403298918, 107268957934572210, 41877140987048387615, 19073758392921536694655, 10024177256513161424322680, 376301673554116445531842536, 10673126660749797308728534491
Offset: 1

Views

Author

Raul Prisacariu, Sep 10 2023

Keywords

Comments

The Whittaker's Root Series Formula is applied to 1+x-x^2/2+x^3/3-x^4/4+x^5/5-x^6/6 +..., which is 1 + the Taylor expansion of log(1+x). The series obtained after applying Whittaker's Root Series Formula: 1/e-1=(-1)/1+(1/2)/(1*3/2)+(1/12)/((3/2)*(7/3))+(1/18)/((7/3)*(11/3))+(1/20)/((11/3)*(347/60))+(563/10800)/((347/60)*(3289/360))+ ... . The series can be simplified to: 1/e=1/3+1/42+1/154+9/3817+1126/1141283+ ... . The sequence is formed by the denominators of the simplified series.
The fractions in the denominators of the non-simplified series seem to be equal to terms from A323339 divided by the corresponding terms from A323340. Thus, the Whittaker's Root Series for 1 + the Taylor expansion of log(1+x) offers an alternative method for obtaining the terms of A323339 and A323340 using the determinants of Toeplitz matrices (formed using the coefficients of 1 + the Taylor expansion of log(1+x)).

Examples

			Whittaker's Root Series Formula is applied to 1 + the Taylor expansion of log(1+x) and the terms are simplified. The sequence is formed by the denominators of the simplified terms, starting with the second term in the Whittaker's Root Series.
a(1) is the denominator of -(-1/2)/(1*det((1,-1/2),(1,1))) = (1/2)/(3/2) = 1/3.
a(2) is the denominator of -det((-1/2,1/3),(1,-1/2))/(det((1,-1/2),(1,1))*det((1,-1/2,1/3),(1,1,-1/2),(0,1,1))) = (1/12)/((3/2)*(7/3)) = 1/42.
		

Crossrefs

Cf. A068985 (1/e), A365595 (numerators), A323339, A323340.

Programs

  • Mathematica
    c[k_] := If[k < 0, 0, SeriesCoefficient[1 + Log[1 + x], {x, 0, k}]]; Table[-Det[ToeplitzMatrix[Table[c[3 - j], {j, 1, n}], Table[c[j + 1], {j, 1, n}]]] / (Det[ToeplitzMatrix[Table[c[2 - j], {j, 1, n}], Table[c[j], {j, 1, n}]]] * Det[ToeplitzMatrix[Table[c[2 - j], {j, 1, n + 1}], Table[c[j], {j, 1, n + 1}]]]), {n, 1, 20}] // Denominator (* Vaclav Kotesovec, Oct 09 2023 *)

Formula

a(n) is the denominator of the simplified fraction -det ToeplitzMatrix((c(2),c(1),c(0),0,0,...,0),(c(2),c(3),c(4),...,c(n+1)))/(det ToeplitzMatrix((c(1),c(0),0,...,0),(c(1),c(2),c(3),...,c(n)))*det ToeplitzMatrix((c(1),c(0),0,...,0),(c(1),c(2),c(3),...,c(n+1)))), where c(0)=1, c(1)=1, c(2)=-1/2, c(3)=1/3, c(4)=-1/4, c(n)=(1/n)*(-1)^(n+1).c(n) is simply the coefficient of x^n in the series formed by 1+ the Taylor expansion of log(1+x).

Extensions

a(6)-a(7) corrected and extended by Vaclav Kotesovec, Oct 09 2023

A365595 The numerators of a series that converges to 1/e obtained using Whittaker's Root Series Formula.

Original entry on oeis.org

1, 1, 1, 9, 1126, 53825, 302989, 2285199, 133296721, 9731109349, 1737376806937, 372236638394027, 94229801087550639, 27818002500902930641, 591930814558449521261, 9591188150350759241842, 2816408483135723327055984, 1394771058490469072473603553, 385768133102988434073147277769
Offset: 1

Views

Author

Raul Prisacariu, Sep 10 2023

Keywords

Comments

The Whittaker's Root Series Formula is applied to 1+x-x^2/2+x^3/3-x^4/4+x^5/5-x^6/6 +..., which is 1 + the Taylor expansion of log(1+x). The series obtained after applying Whittaker's Root Series Formula: 1/e-1=(-1)/1+(1/2)/(1*3/2)+(1/12)/((3/2)*(7/3))+(1/18)/((7/3)*(11/3))+(1/20)/((11/3)*(347/60))+(563/10800)/((347/60)*(3289/360))+ ... . The series can be simplified to: 1/e=1/3+1/42+1/154+9/3817+1126/1141283+ ... . The sequence is formed by the numerators of the simplified series.
The fractions in the denominators of the non-simplified series seem to be equal to terms from A323339 divided by the corresponding terms from A323340. Thus, the Whittaker's Root Series for 1 + the Taylor expansion of log(1+x) offers an alternative method for obtaining the terms of A323339 and A323340 using the determinants of Toeplitz matrices (formed using the coefficients of 1 + the Taylor expansion of log(1+x)).

Examples

			Whittaker's Root Series Formula is applied to 1 + the Taylor expansion of log(1+x) and the terms are simplified. The sequence is formed by the numerators of the simplified terms, starting with the second term in the Whittaker's Root Series.
a(1) is the numerator of -(-1/2)/(1*det((1,-1/2),(1,1))) = (1/2)/(3/2) = 1/3.
a(2) is the numerator of -det((-1/2,1/3),(1,-1/2))/(det((1,-1/2),(1,1))*det((1,-1/2,1/3),(1,1,-1/2),(0,1,1))) = (1/12)/((3/2)*(7/3)) = 1/42.
		

Crossrefs

Cf. A068985 (1/e), A365594 (denominators), A323339, A323340.

Programs

  • Mathematica
    c[k_] := If[k < 0, 0, SeriesCoefficient[1 + Log[1 + x], {x, 0, k}]]; Table[-Det[ToeplitzMatrix[Table[c[3 - j], {j, 1, n}], Table[c[j + 1], {j, 1, n}]]] / (Det[ToeplitzMatrix[Table[c[2 - j], {j, 1, n}], Table[c[j], {j, 1, n}]]] * Det[ToeplitzMatrix[Table[c[2 - j], {j, 1, n + 1}], Table[c[j], {j, 1, n + 1}]]]), {n, 1, 20}] // Numerator (* Vaclav Kotesovec, Oct 09 2023 *)

Formula

a(n) is the numerator of the simplified fraction -det ToeplitzMatrix((c(2),c(1),c(0),0,0,...,0),(c(2),c(3),c(4),...,c(n+1)))/(det ToeplitzMatrix((c(1),c(0),0,...,0),(c(1),c(2),c(3),...,c(n)))*det ToeplitzMatrix((c(1),c(0),0,...,0),(c(1),c(2),c(3),...,c(n+1)))), where c(0)=1, c(1)=1, c(2)=-1/2, c(3)=1/3, c(4)=-1/4, c(n)=(1/n)*(-1)^(n+1).c(n) is simply the coefficient of x^n in the series formed by 1+ the Taylor expansion of log(1+x).

Extensions

More terms from Vaclav Kotesovec, Oct 09 2023
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