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 1-6 of 6 results.

A220393 A modified Engel expansion of Pi.

Original entry on oeis.org

1, 1, 1, 8, 14, 2, 2, 3, 4, 5, 96, 115, 8, 2, 2, 2, 81, 160, 2, 6, 355, 140, 4, 12, 6, 2, 2, 3, 4, 3, 46, 66, 4, 2, 9, 16, 3, 4, 3, 4, 2, 2, 4, 9, 4, 2, 4, 33, 20, 2, 3, 4, 2, 2
Offset: 1

Views

Author

Peter Bala, Dec 13 2012

Keywords

Comments

The Engel expansion of a positive real number x is the unique nondecreasing sequence {e(1), e(2), e(3), ...} of positive integers such that x = 1/e(1) + 1/(e(1)*e(2)) + 1/(e(1)*e(2)*e(3)) + .... The terms in the Engel expansion of x are obtained from the iterates of the mapping g(x) := x*(1 + floor(1/x)) - 1 by means of the formula e(n) = 1 + floor(1/g^(n)(x)).
Let h(x) = floor(1/x)*g(x). This is called the harmonic sawtooth map by Crowley. Then the modified Engel expansion of a real number 0 < x <= 1 is a sequence {a(1), a(2), a(3), ...} of positive integers such that x = 1/a(1) + 1/(a(1)*a(2)) + 1/(a(1)*a(2)*a(3)) + ... whose terms are obtained from the iterates of the harmonic sawtooth map h(x) by the formulas a(1) = 1 + floor(1/x) and, for n >= 2, a(n) = floor(1/h^(n-2)(x))*{1 + floor(1/h^(n-1)(x))}. Here h^(n)(x) = h(h^(n-1)(x)) denotes the n-th iterate of the map h(x), with the convention that h^(0)(x) = x. For further details see the Bala link.
When the real number x > 1 with, say, floor(x) = m, the modified Engel expansion of x is found by first calculating the modified Engel expansion of x - m and then prepending a sequence of m 1's to this.

Crossrefs

Formula

Let x = Pi - 3. Then a(1) = a(2) = a(3) = 1, a(4) = ceiling(1/x) and, for n >= 1, a(n+4) = floor(1/h^(n-1)(x))*ceiling(1/h^(n)(x)).
Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion Pi = Sum_{n>=1} 1/P(n) = 1/1 + 1/1 + 1/1 + 1/8 + 1/(8*14) + 1/(8*14*2) + 1/(8*14*2*2) + .... For n >= 4, the error made in truncating this series to n terms is less than the n-th term.

A028259 Engel expansion of the golden ratio, (1 + sqrt(5))/2 = 1.61803... .

Original entry on oeis.org

1, 2, 5, 6, 13, 16, 16, 38, 48, 58, 104, 177, 263, 332, 389, 4102, 4575, 5081, 9962, 18316, 86613, 233239, 342534, 964372, 1452850, 7037119, 7339713, 8270361, 12855437, 15900982, 19211148, 1365302354, 1565752087, 1731612283
Offset: 1

Views

Author

Naoki Sato (naoki(AT)math.toronto.edu), Dec 11 1999

Keywords

Comments

Cf. A006784 for definition of Engel expansion.
(sqrt(5) - 1)/2 = 1/(golden ratio) has the same Engel expansion starting with a(2). - G. C. Greubel, Oct 16 2016
Each Engel expansion for a constant x (x > 0) starts with floor(x) 1's. After that, the mean growth of the terms is by a factor e for most constants x, i.e., the order of magnitude of the n-th entry is exp(n-floor(c)), for n large enough. This comment is similar to the fact that for the continued fraction terms of most constants, the geometric mean of those terms equals the Khintchine constant for n large enough. Moreover, note that for the golden section, all continued fraction terms are 1 and thus do not obey the Gauss-Kuzmin distribution which leads to the Khintchine constant (i.e., the Khintchine measure for the golden ratio is 1), but the Engel expansion does obey the statistic behavior for most constants. - A.H.M. Smeets, Aug 24 2018

References

  • F. Engel, Entwicklung der Zahlen nach Stammbruechen, Verhandlungen der 52. Versammlung deutscher Philologen und Schulmaenner in Marburg, 1913, pp. 190-191.

Crossrefs

Programs

  • Mathematica
    EngelExp[ A_, n_ ] := Join[ Array[ 1&, Floor[ A ] ], First@Transpose@NestList[ {Ceiling[ 1/Expand[ #[ [ 1 ] ]#[ [ 2 ] ]-1 ] ], Expand[ #[ [ 1 ] ]#[ [ 2 ] ]-1 ]}&, {Ceiling[ 1/(A-Floor[ A ]) ], A-Floor[ A ]}, n-1 ] ]
  • Python
    j = 0
    while j<3100000:
    # to obtain n correct Engel expansion terms about n^2/2 continued fraction steps are needed; 3100000 is a safe bound
        if j == 0:
            p0, q0 = 1, 1
        elif j == 1:
            p1, q1 = p0+1, 1
        else:
            p0, p1 = p1, p1+p0
            q0, q1 = q1, q1+q0
        j = j+1
    i = 0
    while i < 2410:
        i = i+1
        a = q0//p0+1
        print(i, a)
        p0 = a*p0-q0
    # A.H.M. Smeets, Aug 24 2018

Formula

Limit_{n -> oo} log(a(n + floor(golden ratio)))/n = 1.

Extensions

Corrected and extended by Vladeta Jovovic, Aug 16 2001

A220394 A modified Engel expansion of exp(1).

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 8, 2, 10, 99, 20, 2, 2, 2, 2, 2, 2, 3, 6, 4, 8, 14, 2, 2, 4, 6, 10, 252, 81, 30, 28, 31, 60, 4, 6, 3, 4, 2, 2, 2, 2, 19, 54, 8, 6, 22, 63, 4, 2, 4, 6, 2, 2, 5, 12, 4, 2, 2, 2, 2, 6, 15, 10, 348, 172, 2, 2, 4, 6, 4, 30, 207, 220
Offset: 1

Views

Author

Peter Bala, Dec 13 2012

Keywords

Comments

See A220393 for a description of the modified Engel expansion of a positive real number. For further details see the Bala link.
The Engel expansion for exp(1) is the sequence of positive integers A000027.

Crossrefs

Formula

Let h(x) = x*(floor(1/x) + (floor(1/x))^2) - floor(1/x) where x = exp(1) - 2, then a(1) = a(2) = 1, a(3) = ceiling(1/x) and, for n >= 1, a(n+3) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).
Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion exp(1) = Sum_{n>=1} 1/P(n) = 1/1 + 1/1 + 1/2 + 1/(2*3) + 1/(2*3*4) + 1/(2*3*4*5) + 1/(2*3*4*5*8) + .... For n >= 3, the error made in truncating this series to n terms is less than the n-th term.

A220395 A modified Engel expansion of log(2).

Original entry on oeis.org

2, 3, 8, 6, 2, 4, 93, 60, 2, 2, 2, 2, 3, 12, 10, 2, 2, 14, 52, 6, 5, 8, 2, 2, 5, 8, 2, 2, 3, 4, 14, 273, 40, 2, 3, 4, 4, 12, 27, 16, 14, 26, 4, 6, 4, 6, 2, 3, 12, 10, 4, 6, 14, 65, 12, 8, 6, 2, 7, 90, 294, 40, 2, 2, 32, 155, 8, 7, 12, 2, 2, 2, 2, 4, 6, 3, 10
Offset: 1

Views

Author

Peter Bala, Dec 13 2012

Keywords

Comments

See A220393 for the definition of the modified Engel expansion of a positive real number. For further details see the Bala link.

Crossrefs

Formula

Let h(x) = x*(floor(1/x) + (floor(1/x))^2) - floor(1/x). Let x = log(2). Then a(1) = 1 + floor(1/x) and, for n >= 1, a(n+1) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).
Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion log(2) = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*3) + 1/(2*3*8) + 1/(2*3*8*6) + 1/(2*3*8*6*2) + .... The error made in truncating this series to n terms is less than the n-th term.

A220396 A modified Engel expansion of the Euler-Mascheroni constant gamma.

Original entry on oeis.org

2, 7, 18, 4, 2, 2, 3, 1466, 1464, 9, 24, 4, 2, 9, 104, 60, 8, 2, 3, 6, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 32, 30, 2, 13, 36, 6, 4, 3, 6, 6, 4, 4, 6, 2, 4, 6, 2, 4, 6, 9, 24, 4, 5, 8, 2, 2, 2, 2, 2, 3, 20
Offset: 1

Views

Author

Peter Bala, Dec 13 2012

Keywords

Comments

See A220393 for the definition of the modified Engel expansion of a positive real number. For further details see the Bala link.

Crossrefs

Formula

Let h(x) = x*(floor(1/x) + (floor(1/x))^2) - floor(1/x). Let x = gamma (see A001620). Then a(1) = 1 + floor(1/x) and, for n >= 1, a(n+1) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).
Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion sqrt(2) = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*7) + 1/(2*7*18) + 1/(2*7*18*4) + 1/(2*7*18*4*2) + .... The error made in truncating this series to n terms is less than the n-th term.

A220397 A modified Engel expansion of sqrt(2).

Original entry on oeis.org

1, 3, 6, 4, 2, 2, 4, 6, 23, 66, 108, 7738, 290, 9, 24, 32, 30, 4, 6, 3, 6, 24, 22, 2, 6, 20, 6, 9, 16, 5, 12, 4, 12, 22, 5, 8, 3, 6, 4, 2, 2, 4, 6, 2, 2, 2, 2, 13, 24, 2, 3, 4, 2, 2, 2, 2, 23, 44, 21, 40, 8, 14, 3, 6, 12, 10, 11, 30, 4, 4, 9, 4, 3, 4, 2, 16, 45, 46, 528
Offset: 1

Views

Author

Peter Bala, Dec 13 2012

Keywords

Comments

See A220393 for a definition of the modified Engel expansion of a positive real number. For further details see the Bala link.

Crossrefs

Formula

Let h(x) = x*(floor(1/x) + (floor(1/x))^2) - floor(1/x). Let x = sqrt(2) - 1. Then a(1) = 1, a(2) = ceiling(1/x) and, for n >= 1, a(n+2) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).
Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion sqrt(2) = Sum_{n>=1} 1/P(n) = 1 + 1/3 + 1/(3*6) + 1/(3*6*4) + 1/(3*6*4*2) + 1/(3*6*4*2*2) + .... For n >= 2, the error made in truncating this series to n terms is less than the n-th term.
Showing 1-6 of 6 results.