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 1-4 of 4 results.

A381456 Decimal expansion of Product_{p prime} p^(1/(p^2-1)).

Original entry on oeis.org

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

Views

Author

Jwalin Bhatt, Feb 24 2025

Keywords

Comments

The geometric mean of the zeta distribution with parameter value 2 (A381522) approaches this constant.
In general, for parameter value `s` it approaches e^(-zeta'(s)/zeta(s)). - Jwalin Bhatt, Feb 26 2025

Examples

			1.768198078153244984130853077...

Crossrefs

Cf. A000040, A001620, A074962, A013661, A085548, A306016, A381522, A381898.

Programs

  • Mathematica
    N[Exp[-Zeta'[2]/Zeta[2]], 120]
  • PARI
    exp(-zeta'(2)/zeta(2)) \\ Amiram Eldar, Feb 24 2025
  • Python
    from mpmath import zeta, diff, exp, mp
mp.dps = 120
const = exp(-diff(zeta, 2)/zeta(2))
A381456 = [int(d) for d in mp.nstr(const, n=mp.dps)[:-1] if d != '.']  # Jwalin Bhatt, Apr 08 2025
  • Sage
    N(exp(-diff(zeta(s:=var('s')), s).subs(s==2) / zeta(2)), 120)

Formula

Equals Product_{p>=2} p^(1/(p^2-1)) where p is prime.
Equals (A^12)/(2*Pi*(e^gamma)) where A = A074962 is the Glaisher-Kinkelin constant and gamma = A001620 is the Euler-Mascheroni constant.
Equals e^(-zeta'(2)/zeta(2)).
Equals exp((Sum_{k>=2} log(k)/(k^2))*(6/(Pi^2))).
Equals (Product_{k>=2} k^(1/(k^2)))^(6/(Pi^2)).
Equals exp(A306016). - Hugo Pfoertner, Feb 24 2025

A381900 Sequence where k is appended after every (2^(k-1))*k occurrences of 1, with multiple values following a 1 listed in order.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 4, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 2, 4, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 5
Offset: 1

Views

Author

Jwalin Bhatt, Mar 09 2025

Keywords

Comments

The frequencies of the terms follow the logarithmic distribution with parameter value 1/2.
The geometric mean approaches A381898 in the limit.

Examples

			After every (2*2=4) ones we see a 2,
after every (4*3=12) ones we see a 3,
after every (8*4=32) ones we see a 4 and so on.

Links

Crossrefs

Cf. A381522, A381898.

Programs

  • Python
    from itertools import islice
def logarithmic_distribution_generator():
    num_ones, num_reached = 0, 1
    while num_ones := num_ones+1:
        yield 1
        for num in range(2, num_reached+2):
            if num_ones % ((2**(num-1))*(num)) == 0:
                yield num
                num_reached += num == num_reached+1
A381900 = list(islice(logarithmic_distribution_generator(), 120))

A382095 Decimal expansion of exp((Sum_{k>=2} log(k)/(k-1)!)/e).

Original entry on oeis.org

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

Views

Author

Jwalin Bhatt, Mar 25 2025

Keywords

Comments

The geometric mean of the Poisson distribution with parameter value 1 (A382093) approaches this constant.

Examples

			1.774294375788813063406286573197109...

Crossrefs

Cf. A193424, A381456, A381898, A382093.

Programs

  • Mathematica
    N[Exp [Sum[Log[i]/Factorial[i-1], {i, 2, Infinity}] / E ], 120]

Formula

Equals exp((Sum_{k>=2} log(k)/(k-1)!)/e) = exp(A193424/e).
Equals (Product_{k>=2} k^(1/(k-1)!)) ^ (1/e).

A382961 A sequence constructed by greedily sampling the logarithmic distribution for parameter value 1/2 so as to minimize discrepancy.

Original entry on oeis.org

1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 4, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 5, 1, 1, 2, 1, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 2, 1, 1, 4, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 1, 1, 4, 1, 1, 2, 1, 1, 1, 2
Offset: 1

Views

Author

Jwalin Bhatt, Apr 10 2025

Keywords

Comments

The geometric mean approaches A381898 = exp(-PolyLog'(1,1/2)/log(2)) in the limit.
The logarithmic distribution PDF is p(i) = 1/(log(2)*(2^i)*i).

Examples

			Let p(k) denote the probability of k and c(k) denote the number of occurrences of k among the first n-1 terms; then the expected number of occurrences of k among n random terms is given by n*p(k).
We subtract the actual occurrences c(k) from the expected occurrences and pick the one with the highest value.
| n | n*p(1) - c(1) | n*p(2) - c(2) | n*p(3) - c(3) | choice |
|---|---------------|---------------|---------------|--------|
| 1 |     0.721     |       -       |       -       |   1    |
| 2 |     0.442     |     0.360     |       -       |   1    |
| 3 |     0.164     |     0.541     |       -       |   2    |
| 4 |     0.885     |    -0.278     |     0.240     |   1    |
| 5 |     0.606     |    -0.098     |     0.300     |   1    |
| 6 |     0.328     |     0.082     |     0.360     |   3    |

Links

Crossrefs

Cf. A381617, A381898, A381900, A383238.

Programs

  • Mathematica
    probCountDiff[j_, k_, count_] := k/(Log[2]*(2^j)*j) - Lookup[count, j, 0]
samplePDF[n_] := Module[{coeffs, unreachedVal, counts, k, probCountDiffs, mostProbable},
  coeffs = ConstantArray[0, n]; unreachedVal = 1; counts = <||>;
  Do[probCountDiffs = Table[probCountDiff[i, k, counts], {i, 1, unreachedVal}];
    mostProbable = First@FirstPosition[probCountDiffs, Max[probCountDiffs]];
    If[mostProbable == unreachedVal, unreachedVal++]; coeffs[[k]] = mostProbable;
    counts[mostProbable] = Lookup[counts, mostProbable, 0] + 1; , {k, 1, n}]; coeffs]
A382961 = samplePDF[120]
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.