A331460 Irregular triangle read by rows in which row n is the result of iterating the operation f(n) = n/7 if n == 0 (mod 7), otherwise f(n) = 7*(n + ceiling(n/7)), terminating at the first occurrence of 1.

1, 2, 21, 3, 28, 4, 35, 5, 42, 6, 49, 7, 1, 3, 28, 4, 35, 5, 42, 6, 49, 7, 1, 4, 35, 5, 42, 6, 49, 7, 1, 5, 42, 6, 49, 7, 1, 6, 49, 7, 1, 7, 1, 8, 70, 10, 84, 12, 98, 14, 2, 21, 3, 28, 4, 35, 5, 42, 6, 49, 7, 1, 9, 77, 11, 91, 13, 105, 15, 126, 18

Offset: 1

Author

Davis Smith, Jan 23 2020

Keywords

• nonn
• easy
• tabf

Comments

f(n) is the operation C(n,m) = n/m if n == 0 (mod m) and m*(n + ceiling(n/m)) otherwise, where m = 7. The operations in the Collatz (3x + 1) problem (A070165), A330073, and A329263 are C(n,2), C(n,5), and C(n,8) respectively.

Conjecture: For any number n >= 1, there exists a k such that f^{k}(n) = 1, where f^{0}(n) = n and f^{k + 1}(n) = f(f^{k}(n)).

For any numbers n and k such that f^{k}(n) = 1, f^{k + 1}(7*n) = 1 and if n == 0 (mod 7) and n !== 0 or 7 (mod 56), then f^{k + 1}(floor(n/8)) = 1.

Examples

			The irregular array T(n,k) starts:
n\k   0   1   2   3   4    5   6    7   8    9  10  11  12  13  14  15  16 ...
1:    1
2:    2  21   3  28   4   35   5   42   6   49   7   1
3:    3  28   4  35   5   42   6   49   7    1
4:    4  35   5  42   6   49   7    1
5:    5  42   6  49   7    1
6:    6  49   7   1
7:    7   1
8:    8  70  10  84  12   98  14    2  21    3  28   4  35   5  42   6  49 ...
9:    9  77  11  91  13  105  15  126  18  147  21   3  28   4  35   5  42 ...
10:  10  84  12  98  14    2  21    3  28    4  35   5  42   6  49   7   1
...
T(8,18) = 1 and T(9,20) = 1.
		

Links

• T. Ahmed and H. Snevily, Are There an Infinite Number of Collatz Integers?, 2013.
• M. Bruschi, A generalization of the Collatz problem and conjecture, arXiv:0810.5169 [math.NT], 2008.
• W. Carnielli, Some Natural Generalizations Of The Collatz Problem, Applied Mathematics E-Notes, 15 (2015), 197-206.

Crossrefs

Cf. A070165, A329263, A330073.

Programs

Mathematica

f[n_] := NestWhileList[If[Mod[#, 7] == 0, #/7, 7 (Floor[#/7] + # + 1)] &, n, # > 1 &]; Flatten[Table[f[n], {n, 10}]]

PARI

row(n)=my(N=List([n])); while(n>1, listput(N, n=if(n%7, 7*(n+ceil(n/7)), n/7))); Vec(N)

Formula

T(n,0) = n and T(n,k + 1) = T(n,k)/7 if T(n,k) == 0 (mod 7), 7*(T(n,k) + ceiling(T(n,k)/7)) otherwise, for n >= 1.