

A220393


A modified Engel expansion of Pi.


6



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
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OFFSET

1,4


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^(n2)(x))*{1 + floor(1/h^(n1)(x))}. Here h^(n)(x) = h(h^(n1)(x)) denotes the nth 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.


LINKS



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^(n1)(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 nth term.


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



