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A352378
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Irregular triangle read by rows: T(n,k) is the (n-th)-to-last digit of 2^p such that p == k + A123384(n-1) (mod A005054(n)); k >= 0.
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0
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2, 4, 8, 6, 1, 3, 6, 2, 5, 1, 2, 4, 9, 9, 8, 6, 3, 7, 4, 8, 7, 5, 0, 0, 1, 2, 5, 0, 0, 0, 1, 3, 7, 5, 0, 1, 2, 5, 1, 3, 6, 2, 4, 8, 7, 4, 9, 8, 6, 2, 5, 1, 3, 7, 4, 9, 8, 7, 5, 1, 2, 4, 8, 6, 3, 6, 3, 6, 2, 4, 9, 9, 9, 9, 8, 7, 4, 9, 9, 9, 8, 6, 2, 4, 9, 8, 7, 4, 8, 6, 3, 7, 5, 1, 2, 5, 0, 1, 3, 7, 4, 8, 6, 2, 5, 0, 1
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OFFSET
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1,1
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COMMENTS
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The n-th row of this triangle is the cycle of the (n-th)-to-last digit of powers of 2.
The period of the last n digits of powers of 2 where the exponent is greater than or equal to n is A005054(n). As a result, this triangle can be used to get the (n-th)-to-last digit of a large power of 2; if p == k + A123384(n-1) (mod A005054(n)), then the (n-th)-to-last digit (base 10) of 2^p is T(n,k). For example, for n = 1, if p == 1 (mod 4), then 2^p == 2 (mod 10) and if p == 3 (mod 4), then 2^p == 8 (mod 10). For n = 2, if p == 4 (mod 20), then the second-to-last digit of 2^p (base 10) is 1 and if p == 7 (mod 20), then the second-to-last digit of 2^p (base 10) is 2.
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LINKS
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FORMULA
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For n > 1, T(n,0) = 1.
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EXAMPLE
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Irregular triangle begins:
n/k| 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ... | Number of terms:
---+---------------------------------------+-----------------
1 | 2, 4, 8, 6; | 4
2 | 1, 3, 6, 2, 5, 1, 2, 4, 9, 9, 8, ... | 20
3 | 1, 2, 5, 0, 0, 0, 1, 3, 7, 5, 0, ... | 100
4 | 1, 2, 4, 8, 6, 2, 5, 1, 2, 4, 8, ... | 500
5 | 1, 3, 6, 3, 6, 2, 4, 9, 9, 8, 7, ... | 2500
6 | 1, 2, 5, 0, 0, 1, 3, 7, 5, 1, 2, ... | 12500
...
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PROG
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(PARI) A352378_rows(n)=my(N=logint(10^(n-1), 2), k=4*5^(n-1)); vector(k, v, floor(lift(Mod(2, 10^n)^(v+N))/(10^(n-1))))
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CROSSREFS
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The (n-th)-to-last digit of a power of 2: A000689 (n=1), A160590 (n=2).
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KEYWORD
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nonn,tabf,base
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AUTHOR
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STATUS
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approved
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