Perft(14) estimate:
-------------------

{Perft(0) = 1}

 Perft(1)   20
 Perft(2)   400
 Perft(3)   8,902
 Perft(4)   197,281
 Perft(5)   4,865,609
 Perft(6)   119,060,324
 Perft(7)   3,195,901,860
 Perft(8)   84,998,978,956
 Perft(9)   2,439,530,234,167
Perft(10)   69,352,859,712,417
Perft(11)   2,097,651,003,696,806
Perft(12)   62,854,969,236,701,747
Perft(13)   1,981,066,775,000,396,239

=========================================


   Estimate with Branching Factors:


I only use Branching Factors of even plies for avoiding the 'odd-even effect'.

BF(n) = [Perft(n)]/[Perft(n-1)]

Using Lagrange polinomials for interpolation:

[BF(14)]* = -BF(2) + 6·[BF(4) + BF(12)] - 15·[BF(6) + BF(10)] + 20·BF(8)

[BF(14)]* ~ 31.20194592760110701903111328617

[Perft(14)]* = [Perft(13)]·[BF(14)]* ~ 61813138392529471996.474237957537


         [Perft(14)]* ~ 61,813,138,392,529,471,996


=========================================


   Own estimation:


I define two provisional bounds [Perft'(n)] y [Perft"(n)]:


(n-2)·log[Perft'(n)]   (n-3)·log[Perft(n-1)]
____________________ = _____________________
 n·log[Perft(n-2)]     (n-1)·log[Perft(n-3)]


(n-3)·log[Perft"(n)]   (n-4)·log[Perft(n-1)]
____________________ = _____________________
 n·log[Perft(n-3)]     (n-1)·log[Perft(n-4)]


n = 6, 8, 10, 12 and 14.

I introduce a couple of parameters ß y ß' to be defined as follows:

        [True Perft(n)] - (lower bound)
ß(n) = _________________________________
       (Geometric mean) - (lower bound)

         [True Perft(n)] - (lower bound)
ß'(n) = _________________________________
        (Arithmetic mean) - (lower bound)


Logically |ß(n)| > |ß'(n)|.

----------------------------------------

n = 6:

Perft'(6) ~ 117195628.69126660435006809820962
Perft"(6) ~ 131812282.31006907917506129585269

(Geometric mean) ~ 124289272.64474301871252199740585
(Arithmetic mean) ~ 124503955.5006678417625646970311

ß(6) ~ 0.26286846661081092703699139473544
ß'(6) ~ 0.25514667821568969151958679754055

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n = 8:

Perft'(8) ~ 85448071072.696527647366788601596
Perft"(8) ~ 94286554952.913420579106782235894

(Geometric mean) ~ 89758588718.942393849420014645874
(Arithmetic mean) ~ 89867313012.80497411323678541874

ß(8) ~ -0.10418519387982387798752495688939
ß'(8) ~ -0.10162198014565074994859453052581

----------------------------------------

n = 10:

Perft'(10) ~ 70566039639018.724439241219749495
Perft"(10) ~ 76687869715314.067199418757683525

(Geometric mean) ~ 73563301000993.39648097445973928
(Arithmetic mean) ~ 73626954677166.395819329988716505

ß(10) ~ -0.40476280847341609135325416685804
ß'(10) ~ -0.39634550828169558329350700279514

----------------------------------------

n = 12:

Perft'(12) ~ 64357853234363528.789463136158867
Perft"(12) ~ 69071112659627391.736068618132372

(Geometric mean) ~ 66672847031475177.679584224329504
(Arithmetic mean) ~ 66714482946995460.262765877145615

ß(12) ~ -0.64919569095039774359150530679473
ß'(12) ~ -0.63772598198438680726618342308215

----------------------------------------

n = 14:

Perft'(14) ~ 63540359095620017443.137430353155
Perft"(14) ~ 67503078579525222377.754522031655

(Geometric mean) ~ 65491754084028687211.646630105519
(Arithmetic mean) ~ 65521718837572619910.4459761924

----------------------------------------

[ß(14)]* and [ß'(14)]* are estimated with two Lagrange polinomials:

[ß(14)]* = -ß(6) + 4·[ß(8) + ß(12)] - 6·ß(10)
[ß'(14)]* = -ß'(6) + 4·[ß'(8) + ß'(12)] - 6·ß'(10)

[ß(14)]* ~ -0.847815155091200865233587448319
[ß'(14)]* ~ -0.834465477045666420617656595198

----------------------------------------

I reach the following:

{Perft(14), [ß(14)]*}* = (lower bound) + {[ß(14)]*}·[(geometric mean) - (lower bound)]
{Perft(14), [ß'(14)]*}* = (lower bound) + {[ß'(14)]*}·[(arithmetic mean) - (lower bound)]

{Perft(14), [ß(14)]*}* ~ 61885936850878128968.649632013785
{Perft(14), [ß'(14)]*}* ~ 61886982793352460506.492528666782


      {Perft(14), [ß(14)]*}* ~ 61,885,936,850,878,128,969
      {Perft(14), [ß'(14)]*}* ~ 61,886,982,793,352,460,506


(Arithmetic mean of the last two estimates) ~ 61886459822115294737.571080340281
(Arithmetic mean of the last two estimates) = <P_14> ~ 61,886,459,822,115,294,738

(Half-amplitude of the interval) ~ 522971237165768.92144832650095139
(Half-amplitude of the interval) = |a| ~ 522,971,237,165,769

|a|/<P_14> ~ 8.450495288775327999269466739614e-6 ~ 0.0008450495288775327999269466739614%