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A178969
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Last nonzero decimal digit of (10^10^n)!.
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1
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8, 2, 6, 4, 2, 2, 6, 2, 6, 4, 2, 2
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internal format)
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OFFSET
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0,1
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COMMENTS
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It is possible to find a(10) using the program from the second Bomfim link, or a similar GMP program. Reserve 312500001 words instead of 31250001. Use a computer with at least 6 GB of RAM. - Washington Bomfim, Jan 06 2011
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LINKS
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FORMULA
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a(n) = A008904(2^(10^n)), if n > 0.
For n >= 1, with N = 10^n, 2^N represented in base 5 as (a_h, ..., a_0)_5, t = Sum_{i = h, h-1, ..., 0} (a_i even), x = Sum_{i = h, h-1, ..., 1}(Sum_{k = h, h-1, ..., i} (a_i)), z = (x + t/2) mod 4, and y = 2^z, a(n) = 6*(y mod 2) + y*(1 - (y mod 2)).
(End)
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MATHEMATICA
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f[n_] := Mod[6Times @@ (Rest[FoldList[{ 1 + #1[[1]], #2!2^(#1[[1]]#2)} &, {0, 0}, Reverse[IntegerDigits[n, 5]]]]), 10][[2]]; (* Jacob A. Siehler *) Table[ f[10^10^n], {n, 0, 4}]
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PROG
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(PARI)\\ L is the list of the N digits of 2^(10^n) in base 5.
\\ With 2^(10^n) in base 5 as (a_h, ... , a_0)5,
\\ L[1] = a_0, ... , L[N] = a_h.
convert(n)={n=2^(10^n); x=n; N=floor(log(n)/log(5)) + 1;
L = listcreate(N);
while(x, n=floor(n/5); r=x-5*n; listput(L, r); x=n; );
L; N
};
print("0 8"); for(n=1, 5, print1(n, " "); convert(n); q=0; t=0; x=0; forstep(i=N, 2, -1, a_i=L[i]; q+=a_i; x+=q; t+=a_i*(1-a_i%2); ); a_i=L[1]; t+=a_i*(1-a_i%2); z=(x+t/2)%4; y=2^z; an=6*(y%2)+y*(1-(y%2)); print(an)); \\ Washington Bomfim, Jan 06 2011
(Python)
from functools import reduce
from sympy.ntheory.factor_ import digits
def A178969(n): return reduce(lambda x, y:x*y%10, (((6, 2, 4, 8, 6, 2, 4, 8, 2, 4, 8, 6, 6, 2, 4, 8, 4, 8, 6, 2)[(a<<2)|(i*a&3)] if i*a else (1, 1, 2, 6, 4)[a]) for i, a in enumerate(sympydigits(1<<10**n, 5)[-1:0:-1])), 6) if n else 8 # Chai Wah Wu, Dec 07 2023
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CROSSREFS
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KEYWORD
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nonn,base,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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