No, convergence is quite rapid (note that Warren has defined h(n) to be (-1)^(what we would call h(n))). And the 1/(x+1) constant is: 0.39876108810841881240743054440027306033 whereas 3^(1/6)/Zeta[3]^(5/237)/3 is: 0.39876108811024652417160627287477934253 so unfortunately that's not an identity. (If it were an identity, that would be *really* strange and exciting.) Sincerely, Adam P. Goucher

Sent: Monday, July 07, 2014 at 8:26 AM From: "Bill Gosper" <billgosper@gmail.com> To: math-fun@mailman.xmission.com Subject: [math-fun] (no subject)

Whoa, how are you getting these? Convergence is absent or useless. WDS>

h(n) = (-1)^(parity of bit-sum of binary representation of n)

f(x) S=SUM[ f(n) * h(n), for n=0..2^k-1 with k large] ln(x+1) S=-log(2)/2 ln(x+2) S = -0.1379330125 ln(x+3) -0.07070756527 x^j 0 for any fixed integer j>=0 1/(x+1) 0.398761088108

[http://isc.carma.newcastle.edu.au/advancedCalc identifies this as 3^(1/6)/Zeta[3]^(5/237)/3 . I'm surprised you didn't notice.]

1/(x+2) 0.1049709156499 sqrt(x) -0.63407426 1/sqrt(x+1) 0.1983140804979 -- Warren D. Smith

With f(x):= 2^(x+1), the quantities (1+S)/2 and (1-S)/2

are the only solutions t to Peano[t] = {t,1}. I.e., in the square-fill, they map straight to the top edge.

--rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun