A377280 Given n cards, each time you reversing the order of the top 1, 2, 3, ..., n-1, n cards, then repeat reversing 1, 2, 3, ... cards. Do reversing at least once. the minimum number of steps required to return all the cards to their original position.

1, 4, 9, 12, 25, 36, 28, 32, 81, 60, 121, 120, 117, 196, 75, 80, 204, 324, 228, 200, 147, 264, 529, 504, 200, 676, 540, 252, 841, 900, 186, 192, 1089, 748, 1225, 324, 740, 1140, 1521, 1080, 1681, 336, 1204, 484, 540, 460, 1692, 1152, 735, 2500, 2601, 624, 2809, 972, 1980, 784, 2508, 696, 1416, 3300

Offset

1,2

Comments

The sequence is only restored after one of the full length n reversal, so a(n) >= n.

The process of reversing blocks from 1 to n corresponds to the order transformation of numbers in sequence A130517.

Links

Table of n, a(n) for n=1..60.

Formula

a(n) = n * A003558(n).

Example

For example, using "abc" to represent three cards, the card positions at the end of each step are: abc, bac, cab, cab, acb, bca, bca, cba, abc. Therefore, it takes 9 steps. If there are 4 cards "abcd", the sequence of changes is: abcd, bacd, cabd, dbac, dbac, bdac, adbc, cbda, cbda, bcda, dcba, abcd, so it takes 12 steps

Prog

(PARI) A377280(n)=my(M=Mod(2, 2*n+1), o=znorder(M)); if(o%2==0&&M^(o/2)==-1, n*o/2, o*n) \\ Kevin Ryde
(Python)
from sympy.ntheory import n_order
def A377280(n):
    modular = 2*n + 1
    order = n_order(2, 2*n+1)
    if order % 2 == 0 and pow(2, order//2, modular) == modular - 1:
        return (order//2) * n
    else:
        return order * n # after Kevin Ryde

Crossrefs

Cf. A003558. A130517.

Sequence in context: A212875 A069082 A344415 * A297414 A076794 A254520

Adjacent sequences: A377277 A377278 A377279 * A377281 A377282 A377283

Keyword

nonn

Author

Youhua Li, Oct 22 2024

Status

approved