

A232326


Pierce expansion of 1 to the base Pi.


1



3, 69, 310, 1017, 36745, 214369, 966652, 11159821, 74039764, 550021544, 4481549430, 16543857917, 87205978613, 476981856953, 30989048525367, 203786458494160, 711639924282497, 3174772986229899, 29814569078896025, 100158574806804154
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OFFSET

0,1


COMMENTS

Let r and b be positive real numbers. We define a Pierce expansion of r to the base b to be a (possibly infinite) increasing sequence of positive integers [a(0), a(1), a(2), ...] such that we have the alternating series representation r = b/a(0)  b^2/(a(0)*a(1)) + b^3/(a(0)*a(1)*a(2))  .... Depending on the values of r and b such an expansion may not exist, and if it does exist it may not be unique. When b = 1 and 0 < r < 1 we recover the ordinary Pierce expansion of r.
In the particular case that the base b >= 1 and 0 < r < b then we can find a Pierce expansion of r to the base b as follows:
Define the map f(x) (which depends on the base b) by f(x) = x/b*ceiling(b/x)  1 and let f^(n)(x) denote the nth iterate of the map f(x), with the convention that f^(0)(x) = x.
For n = 0,1,2,... define a(n) = ceiling(b/f^(n)(r)) until f^n(r) = 0.
Then it can be shown that the sequence of positive integers a(n) is a Pierce expansion of r to the base b.
For the present sequence we apply this algorithm with r := 1 and with base b := Pi. See A232325 for an Engel expansion of 1 to the base Pi.


LINKS



FORMULA

a(n) = ceiling(Pi/f^(n)(1)), where f^(n)(x) denotes the nth iterate of the map f(x) = x/Pi*ceiling(Pi/x)  1, with the convention that f^(0)(x) = x.
Pierce series expansion of 1 to the base Pi:
1 = Pi/3  Pi^2/(3*69) + Pi^3/(3*69*310)  Pi^4/(3*69*310*1017) + ....
The associated power series F(z) := 1  ( z/3  z^2/(3*69) + z^3/(3*69*310)  z^4/(3*69*310*1017) + ...) has a zero at z = Pi. Truncating the series F(z) to n terms produces a polynomial F_n(z) with rational coefficients which has a real zero close to Pi.


MAPLE

# Define the nth iterate of the map f(x) = x/b*ceiling(b/x)  1
map_iterate := proc(n, b, x) option remember;
if n = 0 then
x
else
1 + 1/b*thisproc(n1, b, x)*ceil(b/thisproc(n1, b, x))
end if
end proc:
# Define the (signed) terms of the expansion of x to the base b
a := n > ceil(evalf(b/map_iterate(n, b, x))):
Digits:= 500:
# Choose values for x and b
x := 1: b:= Pi:
seq(abs(a(n)), n = 0..19);


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



