A329263 -id:A329263 - cp's OEIS frontend

A330073 Irregular triangle read as rows in which row n is the result of iterating the operation f(n) = n/5 if n == 0 (mod 5), otherwise f(n) = 5*(floor(n/5) + n + 1), terminating at the first occurrence of 1.

1, 2, 15, 3, 20, 4, 25, 5, 1, 3, 20, 4, 25, 5, 1, 4, 25, 5, 1, 5, 1, 6, 40, 8, 50, 10, 2, 15, 3, 20, 4, 25, 5, 1, 7, 45, 9, 55, 11, 70, 14, 85, 17, 105, 21, 130, 26, 160, 32, 195, 39, 235, 47, 285, 57, 345, 69, 415, 83, 500, 100, 20, 4, 25, 5, 1, 8, 50

Offset: 1

Davis Smith, Nov 30 2019

f(n) is the operation C(n,m) = n/m if n == 0 (mod m), m*(floor(n/m) + n + 1) otherwise where m = 5. C(n,2) is the operation in the Collatz problem (A070165) and C(n,8) is the operation in A329263.
Conjecture: For any initial value n >= 1, there is a number k such that f^{k}(n) = 1, where f^{0}(n) = n and f^{k}(n) = f(f^{k - 1}(n)).
For any number n, if n is a power of 5 multiplied by 1, 2, 3, or 4, then there is a number k such that f^{k}(n) = 1. If n is congruent to 10, 15, 20, or 25 (mod 30) and there is a k such that f^{k}(n) = 1, then f^{k + 1}(floor(n/6)) = 1.

			The irregular array T(n,k) starts:
n\k   0   1   2   3   4   5   6   7   8   9   10   11   12   13 ...
1:    1
2:    2  15   3  20   4  25   5   1
3:    3  20   4  25   5   1
4:    4  25   5   1
5:    5   1
6:    6  40   8  50  10   2  15   3  20   4   25    5    1
7:    7  45   9  55  11  70  14  85  17 105   21  130   26  160 ...
8:    8  50  10   2  15   3  20   4  25   5    1
9:    9  55  11  70  14  85  17 105  21 130   26  160   32  195 ...
10:  10   2  15   3  20   4  25   5   1
T(7,31) = 1 and T(9,29) = 1.

Cf. A070165, A329263.

Array[NestWhileList[If[Mod[#, 5] == 0, #/5, 5 (Floor[#/5] + # + 1)] &, #, # > 1 &] &, 8] // Flatten (* Michael De Vlieger, Dec 01 2019 *)

row(n)=my(N=[n],m=5);while(n>1,N=concat(N,n=if(n%m,m*(n+floor(n/m)+1),n/m)));N

T(n,0) = n, T(n,k + 1) = T(n,k)/5 if T(n,k) == 0 (mod 5), 5*(T(n,k) + floor(T(n,k)/5) + 1) otherwise, for n >= 1.

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.

Original entry on oeis.org

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

Davis Smith, Jan 23 2020

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.

			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.

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.

Cf. A070165, A329263, A330073.

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

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

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.

cp's OEIS Frontend

A330073 Irregular triangle read as rows in which row n is the result of iterating the operation f(n) = n/5 if n == 0 (mod 5), otherwise f(n) = 5*(floor(n/5) + n + 1), terminating at the first occurrence of 1.

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