A329263 Irregular triangle read by rows in which row n is the result of iterating the operation f(n) = n/8 if n == 0 (mod 8), otherwise f(n) = 8*(floor(n/8) + n + 1), terminating at the first occurrence of 1.

1, 2, 24, 3, 32, 4, 40, 5, 48, 6, 56, 7, 64, 8, 1, 3, 32, 4, 40, 5, 48, 6, 56, 7, 64, 8, 1, 4, 40, 5, 48, 6, 56, 7, 64, 8, 1, 5, 48, 6, 56, 7, 64, 8, 1, 6, 56, 7, 64, 8, 1, 7, 64, 8, 1, 8, 1, 9, 88, 11, 104, 13, 120, 15, 136, 17, 160, 20, 184, 23, 208

Offset

1,2

Comments

The operation f(n) can be generalized to C(n,m) = n/m if n == 0 (mod m), m*(floor(n/m) + n + 1) otherwise. The operation for the 3x+1 (Collatz) problem is equivalent to C(n,2) and f(n) = C(n,8).

Conjecture: For any initial value of n >= 1 there is a number k such that f^{k}(n) = 1; in other words, every row of the triangle is finite.

Links

Table of n, a(n) for n=1..71.

Formula

a(n,0) = n, a(n,k + 1) = a(n,k)/8 if a(n,k) == 0 (mod 8), 8*(floor(a(n,k)/8) + a(n,k) + 1) otherwise, for n >= 1.

Example

The irregular array a(n,k) starts:
n\k   0   1   2   3   4    5   6    7   8    9  10  11  12  13 ...
1:    1
2:    2  24   3  32   4   40   5   48   6   56   7  64   8   1
3:    3  32   4  40   5   48   6   56   7   64   8   1
4:    4  40   5  48   6   56   7   64   8    1
5:    5  48   6  56   7   64   8    1
6:    6  56   7  64   8    1
7:    7  64   8   1
8:    8   1
9:    9  88  11 104  13  120  15  136  17  160  20 184  23 208 ...
10:  10  96  12 112  14  128  16    2  24    3  32   4  40   5 ...
a(9,100) = 1 and a(10,20) = 1.

Prog

(PARI) collatz8(n)=N=[n]; while(n>1, N=concat(N, n=if(n%8, 8*(floor(n/8)+n+1), n/8))); N

Crossrefs

Cf. A070165.

Sequence in context: A270562 A100816 A079612 * A227477 A351850 A066585

Adjacent sequences: A329260 A329261 A329262 * A329264 A329265 A329266

Keyword

nonn,easy,tabf

Author

Davis Smith, Nov 09 2019

Status

approved