On Mon, May 13, 2019 at 12:04 PM Bill Gosper <billgosper@gmail.com> wrote:
"Hyperfactorial" (= 1^1 2^2 ... n^n) has a particularly simple "Stirling's formula": In[54]:= Hyperfactorial@z | E^(1/12 - z^2/4 - Zeta'[-1]) z^(1/12 + z/2 + z^2/2)
(Zeta'[-1] and Zeta'[2] are connected
In[78]:= Zeta'[-1] // deGlaisher // Simplify // Distribute
Out[78]= 1/12 (1 - EulerGamma - Log[2 π)) + Zeta'[2]/(2 π^2)
via the zeta reflection formula.)
In[55]:= Table[% // Activate, {z, .0, 9}]
Out[55]= {1. | 0., 1. | 0.998755252375579, 4. | 3.99865591643225, 108. | 107.983585892636, 27647.9999999998 | 27645.6209013424, 8.63999999999998*10^7 | 8.63952270335811*10^7, 4.03107840000001*10^12 | 4.03092349178858*10^12, 3.31976639877119*10^18 | 3.31967257400642*10^18, 5.56964379417277*10^25 | 5.56952319419551*10^25, 2.1577941222941*10^34 | 2.15775718825365*10^34}
Unfortunately, an exact expression of the "figure of merit" infinite product of relative errors (~ 1.0021286) requires the "next" higher factorial.
(Mnemonic: Boil Boil, toil and trouble ... ) No, there is a rearrangement trick enabling Product[(E^(k^2/4)*k^(-(1/12) - k/2 - k^2/2)*Hyperfactorial[k])/Glaisher, {k, ∞}] == Sqrt[Glaisher]/(E^((3*Zeta[3])/(8*Pi^2))*(2*Pi)^(1/24)) ~ 1.002128604096867 . after
"Hyperfactorial", 1^1^2 2^2^2 3^3^2 ... n^n^2, for which we still lack a consensual name.
Extremely wrong! As Mathematica has known for years, In[153]:= Product[n^n^s, {n, a - 1}] Out[153]= E^(-Derivative[1][Zeta][-s] + Derivative[1, 0][HurwitzZeta][-s, a]) before which I knew and forgot: d𝜁(s,a)/ds <http://gosper.org/filds.pdf> This ought to get us the general formulas. —rwg I've called it "Second Factorial", with Hyperfactorial being "First
Factorial" and Factorial being "Zeroth Factorial", based on the dubious contention that the dot under the *' *in "*!*" is actually a tiny zero. Replacing that zero with 1, 2, ..., is typographically unattractive, but perhaps encircling those little numerals would help. —rwg
On Sun, May 12, 2019 at 10:10 AM Bill Gosper <billgosper@gmail.com> wrote:
n!/√(2πn)e^n/n^n ~ 1 + O(1/n), whose infinite product over n blows up, giving an infinite figure of demerit. For convergence, we can improve Stirling's approximation: n! ~ √(2π(n+1/6)) n^n/e^n ~ n! (1+1/(144 n^2)+. . .). Then the infinite product of all the relative errors is In[86]:= Product[n!/√(2 π (n + 1/6)) n^n/E^n), {n, ∞}]
Out[87]= E^( 2 Zeta'[-1] - 1/12) √(⅙! √(2π)))
Out[88]= 1.00781097654253
(BtW, Stirling's contribution was "only" the √(2π); de Moivre had already found the rest. This must have been nearly as stunning as Euler's 𝜁(2) = π²/6.)
Stirling's approximation can be improved no end. Perhaps the next better is
E^-z √(2 π z) (1/(12 z) + z)^z ~ z! (1-1/(1440 z^3)+O[1/z]^4) whose "figure of merit" is Product[z!/(E^-z √(2 π z) (1/(12 z) + z)^z),{z,∞}] == E^(2 Zeta'[-1]) √√(2π)/(BarnesG[1 - I/(2√3)] BarnesG[1 + I/(2√3)]) ~ 1.001178221812, which seems to cry out for the BarnesG reflection formula. Which, unfortunately, only applies to G(z)/G(-z), not G(z)G(-z): E.g. BarnesG[1 - I/(2 √3)]/BarnesG[1 + I/(2 √3)] == I^(1/12) E^(-((I PolyLog[2, E^(-(π/√3))])/(2 π))) (π Csch[π/(2 √3)])^(-(I/(2 √3)))==
E^(I(π/24 - Log[π Csch[π/(2√3)]]/(2√3) - PolyLog[2, E^(-π/√3)]/(2π))) —rwg