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%I #10 Feb 27 2021 21:36:07
%S 0,3,8,7,7,6,5,5,4,3,3,2,1,1,0,8,8,8,8,8,8,8,8,8,8,8,8,8,8,7,7,7,7,7,
%T 7,7,7,7,7,7,7,7,7,7,6,6,6,6,6,6,6,6,6,6,6,6,6,6,5,5,5,5,5,5,5,5,5,5,
%U 5,5,5,5,5,4,4,4,4,4,4,4,4,4,4,4,4,4,4
%N First digit other than 9 in the fractional part of the decimal expansion of (1/1000^n)^(1/1000^n).
%C From _Jon E. Schoenfield_, Feb 21 2021: (Start)
%C Even if each term of this sequence were incremented by 1 (to give them a minimum of 1 and a maximum of 9), their distribution would not follow Benford's law (nor does the related sequence whose n-th term is the first nonzero digit in the fractional part of the decimal expansion of 1 - (1/1000^n)^(1/1000^n)).
%C For n > 1, a(n) - a(n-1) = 0 or -1 except when a(n) = 8 and a(n-1) != 8; this situation occurs at n = 2, 15, 145, 1448, ..., and from each such value of n until the next, all non-9 digits occur with very nearly equal frequency. E.g., the digits 0..8 occur with frequencies
%C [ 1, 0, 0, 1, 0, 0, 0, 0, 0] in a(0)..a(1),
%C [ 1, 2, 1, 2, 1, 2, 1, 2, 1] in a(2)..a(14),
%C [ 14, 15, 14, 15, 14, 15, 14, 15, 14] in a(15)..a(144),
%C [145,144,145,145,145,144,145,145,145] in a(145)..a(1447).
%C (End)
%H <a href="/index/Be#Benford">Index entries for sequences related to Benford's law</a>
%e From _Jon E. Schoenfield_, Feb 21 2021: (Start)
%e In each of the decimal expansions in the table below, each 9 has been replaced by an underscore character (to make it easy to see at a glance the first non-9 digit in the fractional part):
%e .
%e decimal expansion of (1/1000^n)^(1/1000^n)
%e n with each 9 replaced by "_" a(n)
%e -- ------------------------------------------------- ----
%e 0 1.00000000000000000000000000000000000000000000... 0
%e 1 0.__3116048420_337715764260768851547466351_162... 3
%e 2 0.____8618458487576222544_06332_28167145404344... 8
%e 3 0._______7_276734377780460834_3251023824_06354... 7
%e 4 0.__________72368_78884453188455735031275_4061... 7
%e 5 0._____________6546122360508__11203267556__264... 6
%e 6 0.________________5855346832610717854658364771... 5
%e 7 0.___________________516457130471250406367_124... 5
%e 8 0.______________________44737_57768142_0358356... 4
%e 9 0._________________________3783020248_16076653... 3
%e 10 0.____________________________30_2244721017862... 3
%e 11 0._______________________________240146_1_311_... 2
%e 12 0.__________________________________17106_3665... 1
%e 13 0._____________________________________101__18... 1
%e 14 0.________________________________________032_... 0
%e 15 0.__________________________________________8_... 8
%e (End)
%o (PARI) zerotozero(n) = { local(x,y,z,v,j); for(x=0,n, y=1000^x; v=(1./y)^(1/y); z=Vec(Str(v)); for(j=3,n, if(z[j]<>"9", print1(z[j]",");break) ) ) }
%K base,nonn
%O 0,2
%A _Cino Hilliard_, May 06 2005
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