A352378 - OEIS

%I #21 Apr 26 2022 21:13:44

%S 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,

%T 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,

%U 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

%N 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.

%C The n-th row of this triangle is the cycle of the (n-th)-to-last digit of powers of 2.

%C 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.

%F For n > 1, T(n,0) = 1.

%e Irregular triangle begins:

%e n/k| 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ... | Number of terms:

%e ---+---------------------------------------+-----------------

%e 1  | 2, 4, 8, 6;                           |                4

%e 2  | 1, 3, 6, 2, 5, 1, 2, 4, 9, 9,  8, ... |               20

%e 3  | 1, 2, 5, 0, 0, 0, 1, 3, 7, 5,  0, ... |              100

%e 4  | 1, 2, 4, 8, 6, 2, 5, 1, 2, 4,  8, ... |              500

%e 5  | 1, 3, 6, 3, 6, 2, 4, 9, 9, 8,  7, ... |             2500

%e 6  | 1, 2, 5, 0, 0, 1, 3, 7, 5, 1,  2, ... |            12500

%e ...

%o (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))))

%Y Cf. A005054, A123384, A216099.

%Y The (n-th)-to-last digit of a power of 2: A000689 (n=1), A160590 (n=2).

%K nonn,tabf,base

%O 1,1

%A _Davis Smith_, Mar 14 2022