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 1-5 of 5 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

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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.

A220398 A modified Engel expansion of the golden ratio (1/2)*(1 + sqrt(5)) (A001622).

Original entry on oeis.org

1, 2, 5, 8, 3, 4, 4, 6, 2, 162, 322, 2, 51842, 103682, 2, 5374978562, 10749957122, 2, 57780789062419261442, 115561578124838522882, 2, 6677239169351578707225356193679818792962, 13354478338703157414450712387359637585922, 2
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 = 1/2*(1 + sqrt(5)) - 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))).
Recurrence equations: For n >= 3, a(3*n) = 2. For n >= 4 we have a(3*n+2) = 2*a(3*n+1) - 2 and a(3*n+1) = 2*(a(3*n-2) - 1)^2.
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 + 1/2 + 1/(2*5) + 1/(2*5*8) + 1/(2*5*8*3) + 1/(2*5*8*3*4) + .... For n >= 2, the error made in truncating this series to n terms is less than the n-th term.

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.

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 1-5 of 5 results.