I did wonder about Red (embarassment) or White (fear); though I think we may safely disregard Orange (excess of Ulster protestant zeal? too long spent in tanning parlour?). Could possibly turn out to be Green (envy) or Blue (acute depression). But I have little doubt that much time spent in Brown (study) would result only in Black (rage). Till a consensus emerges, I shall have to hold my trousers up. WFL On 3/30/14, Bill Gosper <billgosper@gmail.com> wrote:

BTW, I'm having difficulty solving (c7) for F , wot does not appear in the equation ... Has my belt fallen down? WFL

Thanks for reminding. I've yet to figure out how to stop html from randomly changing ϕ to f and Φ to F. NOW what color is your belt? --rwg

On 3/30/14, Bill Gosper <billgosper@gmail.com> wrote:

in http://www.tweedledum.com/rwg/idents.htm that Dick Askey once deemed "wild". Unfortunately, using that bleeping useless Glaisher symbol instead of Zeta'[-1], which it bashes to 1/12 - Log[Glaisher]. This is like bashing Zeta[3] to Apéry. Zeta'[-1] is also simply related to Porter's constant, so why doesn't Mma call it (some function of) Porter? Since it comes up in Hyperfactorial[1/2] and BarnesG[1/2], I used to advocate the symbol π_1, where π_0 := π, but, so far at least, Mma doesn't support Hyperfactorial[n,k] := 1^1^k 2^2^k ... n^n^k. In the past, and on that page, I used Zeta'[2], but that is almost always clumsier. Best for that identity is probably Product[n!*(n/E)^-n/Sqrt[2*π*(n + 1/6)], {n, ∞}] == E^(-(1/12)+2 (Zeta'[-1]) (2 π)^(1/4) Sqrt[(1/6)!]. Can someone emit a rule to melt Glaishers? --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun