A000689 -id:A000689 - cp's OEIS frontend

A354782 Second digit from left in decimal expansion of 2^n (n >= 4).

6, 2, 4, 2, 5, 1, 0, 0, 0, 1, 6, 2, 5, 3, 6, 2, 0, 0, 1, 3, 6, 3, 7, 3, 6, 3, 0, 1, 2, 5, 7, 4, 8, 3, 7, 4, 0, 1, 3, 7, 7, 5, 0, 4, 8, 6, 1, 2, 5, 0, 8, 6, 2, 4, 8, 7, 1, 3, 6, 2, 8, 6, 3, 4, 9, 9, 1, 3, 7, 4, 8, 7, 5, 5, 0, 0, 2, 4, 8, 6, 9, 8, 7, 5, 0, 1, 2, 4, 9, 9, 9, 9, 9, 5, 1, 3, 2, 5, 0, 0, 0, 0, 1, 6, 2, 4, 2, 5, 1

Offset: 4

N. J. A. Sloane, Jul 07 2022, following a suggestion from Alexander Wajnberg

			2^4 = 16, so a(4) = 6. 2^5 = 32, so a(5) = 2.

Alois P. Heinz, Table of n, a(n) for n = 4..10000

Cf. A000079, A000689, A008952, A034887, A160590.

a:= n-> parse(""||(2^n)[2]):
seq(a(n), n=4..112);  # Alois P. Heinz, Jul 07 2022

A354782[n_]:=IntegerDigits[2^n][[2]];Array[A354782,100,4] (* Paolo Xausa, Oct 22 2023 *)

def A354782(n): return int(str(1<Chai Wah Wu, Jul 07 2022

A361390 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) is carryless n^k base 10.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 4, 3, 1, 0, 1, 8, 9, 4, 1, 0, 1, 6, 7, 6, 5, 1, 0, 1, 2, 1, 4, 5, 6, 1, 0, 1, 4, 3, 6, 5, 6, 7, 1, 0, 1, 8, 9, 4, 5, 6, 9, 8, 1, 0, 1, 6, 7, 6, 5, 6, 3, 4, 9, 1, 0, 1, 2, 1, 4, 5, 6, 1, 2, 1, 10, 1, 0, 1, 4, 3, 6, 5, 6, 7, 6, 9, 100, 11, 1, 0, 1, 8, 9, 4, 5, 6, 9, 8, 1, 1000, 121, 12, 1

Offset: 0

Seiichi Manyama, Mar 10 2023

			4 * 4 = 16, so T(4,2) = 6. 6 * 4 = 24, so T(4,3) = 4.
Square array begins:
  1, 0, 0, 0, 0, 0, 0, 0, ...
  1, 1, 1, 1, 1, 1, 1, 1, ...
  1, 2, 4, 8, 6, 2, 4, 8, ...
  1, 3, 9, 7, 1, 3, 9, 7, ...
  1, 4, 6, 4, 6, 4, 6, 4, ...
  1, 5, 5, 5, 5, 5, 5, 5, ...
  1, 6, 6, 6, 6, 6, 6, 6, ...
  1, 7, 9, 3, 1, 7, 9, 3, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
David Applegate, Marc LeBrun and N. J. A. Sloane, Carryless Arithmetic (I): The Mod 10 Version.
Index entries for sequences related to carryless arithmetic

Columns k=0..4 give A000012, A001477, A059729, A169885, A169886.
Rows n=0..4 give A000007, A000012, A000689, A001148, A168428.
T(11,k) gives A059734.
Main diagonal gives A361351.
Cf. A004248, A359697.

T(n, k) = fromdigits(Vec(Pol(digits(n))^k)%10);

A167280 Period length 12: 0,0,1,2,4,7,4,8,7,4,8,5 (and repeat).

Original entry on oeis.org

0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7

Offset: 0

Paul Curtz, Nov 01 2009

The sum of the terms in the period is 50, so the partial sums of the sequence are also 12-periodic if reduced modulo 50 or modulo 10.
The weighted partial sums b(n) = sum_{i=0..n} a(i)*2^i obey b(n) = b(n+12) (mod 10).
Third column is A000689. (Which table or array is this referring to? R. J. Mathar, Nov 01 2009)
The set of digits in the period is the same as in A141425.
A derived sequence with terms a(n)+a(n+6) has period length 6: 4, 8, 8, 6, 12, 12 (repeat).

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,1).

a(n) = A113405(n+1) mod 10.

G.f.: x^2*(1+2*x+4*x^2+7*x^3+4*x^4+8*x^5+7*x^6+4*x^7+8*x^8+5*x^9)/( (1-x)*(1+x+x^2)*(1+x)*(1-x+x^2)*(1+x^2)*(x^4-x^2+1)) [R. J. Mathar, Nov 03 2009]

Edited by R. J. Mathar, Nov 05 2009

A352378 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.

Original entry on oeis.org

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

Offset: 1

Davis Smith, Mar 14 2022

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.

			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
...

Cf. A005054, A123384, A216099.

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

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))))

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

cp's OEIS Frontend

A354782 Second digit from left in decimal expansion of 2^n (n >= 4).

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A361390 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) is carryless n^k base 10.

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A167280 Period length 12: 0,0,1,2,4,7,4,8,7,4,8,5 (and repeat).

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A352378 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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