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 1-3 of 3 results.

A073229 Decimal expansion of e^(1/e).

Original entry on oeis.org

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

Views

Author

Rick L. Shepherd, Jul 22 2002

Keywords

Comments

e^(1/e) = 1/((1/e)^(1/e)) (reciprocal of A072364).
Let w(n+1)=A^w(n); then w(n) converges if and only if (1/e)^e <= A <= e^(1/e) (see the comments in A073230) for initial value w(1)=A. If A=e^(1/e) then lim_{n->infinity} w(n) = e. - Benoit Cloitre, Aug 06 2002; corrected by Robert FERREOL, Jun 12 2015
x^(1/x) is maximum for x = e and the maximum value is e^(1/e). This gives an interesting and direct proof that 2 < e < 4 as 2^(1/2) < e^(1/e) > 4^(1/4) while 2^(1/2) = 4^(1/4). - Amarnath Murthy, Nov 26 2002
For large n, A234604(n)/A234604(n-1) converges to e^(1/e). - Richard R. Forberg, Dec 28 2013
Value of the unique base b > 0 for which the exponential curve y=b^x and its inverse y=log_b(x) kiss each other; the kissing point is (e,e). - Stanislav Sykora, May 25 2015
Actually, there is another base with such property, b=(1/e)^e with kiss point (1/e,1/e). - Yuval Paz, Dec 29 2018
The problem of finding the maximum of f(x) = x^(1/x) was posed and solved by the Swiss mathematician Jakob Steiner (1796-1863) in 1850. - Amiram Eldar, Jun 17 2021

Examples

			1.44466786100976613365833910859...
		

References

  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 35.

Crossrefs

Cf. A001113 (e), A068985 (1/e), A073230 ((1/e)^e), A072364 ((1/e)^(1/e)), A073226 (e^e).

Programs

  • Maple
    evalf[110](exp(exp(-1))); # Muniru A Asiru, Dec 29 2018
  • Mathematica
    RealDigits[ E^(1/E), 10, 110] [[1]]
  • PARI
    exp(1)^exp(-1)

Formula

Equals 1 + Integral_{x = 1/e..1} (1 + log(x))/x^x dx = 1 - Integral_{x = 0..1/e} (1 + log(x))/x^x dx. - Peter Bala, Oct 30 2019
Equals Sum_{k>=0} exp(-k)/k!. - Amiram Eldar, Aug 13 2020
Equals lim_{x->oo} (Sum_{n>=1} (x/n)^n)^(1/x) (Furdui, 2017). - Amiram Eldar, Mar 26 2022

A277473 E.g.f.: -exp(x)*LambertW(-x).

Original entry on oeis.org

0, 1, 4, 18, 116, 1060, 12702, 187810, 3296120, 66897288, 1540762010, 39693752494, 1130866726596, 35300006582620, 1198036854980630, 43921652697277170, 1729775120233353968, 72831210167041246480, 3264674481128340280242, 155220967397580333229270
Offset: 0

Views

Author

Vaclav Kotesovec, Oct 17 2016

Keywords

Crossrefs

Partial sums of A038051.

Programs

  • Mathematica
    CoefficientList[Series[-Exp[x]*LambertW[-x], {x, 0, 20}], x] * Range[0, 20]!
    Table[Sum[Binomial[n, k]*k^(k-1), {k, 1, n}], {n, 0, 20}]
  • PARI
    x='x+O('x^50); concat([0], Vec(serlaplace(-exp(x)*lambertw(-x)))) \\ G. C. Greubel, Jun 11 2017

Formula

a(n) = Sum_{k=1..n} binomial(n,k) * k^(k-1).
a(n) ~ exp(exp(-1)) * n^(n-1).

A377535 First term of n-th differences of the sequence x^(x-1) for x >= 1.

Original entry on oeis.org

1, 1, 6, 42, 416, 5210, 79212, 1417094, 29168624, 679100562, 17645739500, 506235093782, 15893604725352, 542039221415354, 19954673671286564, 788708093950072830, 33312472504166976992, 1497371019734704549538, 71368260385615670087388, 3595248209512068272420582, 190872048208819769608101080
Offset: 0

Views

Author

Harri Aaltonen, Oct 31 2024

Keywords

Comments

Inverse binomial transform of A000169.
It appears that a(n) is the number of partial functions f on [n] such that every point in [n] is either in the domain of f or in the image of f. Cf. A377763. - Geoffrey Critzer, Nov 06 2024

Crossrefs

Programs

  • Maple
    a:= n-> add((j+1)^j*(-1)^(n-j)*binomial(n,j), j=0..n):
    seq(a(n), n=0..20);  # Alois P. Heinz, Oct 31 2024
  • Mathematica
    With[{t = Table[n^(n - 1), {n, 1, 21}]}, Table[Sum[(-1)^(i - j) * Binomial[i, j] * t[[j + 1]], {j, 0, i}], {i, 0, Length[t] - 1}]] (* Amiram Eldar, Oct 31 2024 *)
  • PARI
    lista(nn) = my(v = vector(nn+1, n, n^(n-1)), vv=vector(nn+1)); vv[1] = v[1]; for (n=1, nn, my(w = vector(#v-1, k, v[k+1] - v[k])); vv[n+1] = w[1]; v = w;); vv; \\ Michel Marcus, Oct 31 2024

Formula

G.f.: Sum_{j>=1} A000169(j)*x^(j-1)/(1+x)^j. - Alois P. Heinz, Oct 31 2024
Showing 1-3 of 3 results.