A232326 - OEIS

%I #11 Nov 27 2013 13:59:57

%S 3,69,310,1017,36745,214369,966652,11159821,74039764,550021544,

%T 4481549430,16543857917,87205978613,476981856953,30989048525367,

%U 203786458494160,711639924282497,3174772986229899,29814569078896025,100158574806804154

%N Pierce expansion of 1 to the base Pi.

%C 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.

%C See A058635, A192223 and A230600 for some predictable Pierce expansions to a base b other than 1.

%C 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:

%C 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 n-th iterate of the map f(x), with the convention that f^(0)(x) = x.

%C For n = 0,1,2,... define a(n) = ceiling(b/f^(n)(-r)) until f^n(-r) = 0.

%C Then it can be shown that the sequence of positive integers |a(n)| is a Pierce expansion of r to the base b.

%C 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.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PierceExpansion.html">Pierce Expansion</a>

%F a(n) = ceiling(Pi/f^(n)(-1)), where f^(n)(x) denotes the n-th iterate of the map f(x) = x/Pi*ceiling(Pi/x) - 1, with the convention that f^(0)(x) = x.

%F Pierce series expansion of 1 to the base Pi:

%F 1 = Pi/3 - Pi^2/(3*69) + Pi^3/(3*69*310) - Pi^4/(3*69*310*1017) + ....

%F 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.

%p # Define the n-th iterate of the map f(x) = x/b*ceiling(b/x) - 1

%p map_iterate := proc(n,b,x) option remember;

%p if n = 0 then

%p    x

%p else

%p    -1 + 1/b*thisproc(n-1,b,x)*ceil(b/thisproc(n-1,b,x))

%p end if

%p end proc:

%p # Define the (signed) terms of the expansion of x to the base b

%p a := n -> ceil(evalf(b/map_iterate(n,b,x))):

%p Digits:= 500:

%p # Choose values for x and b

%p x := -1: b:= Pi:

%p seq(abs(a(n)), n = 0..19);

%Y Cf. A014014, A006784, A058635, A061233, A192223, A230600, A232325, A232327, A232328.

%K nonn,easy

%O 0,1

%A _Peter Bala_, Nov 26 2013