A331272 Irregular triangle in which row n lists numbers m such that A330073(m,n) = 1.

1, 5, 25, 4, 125, 20, 625, 3, 100, 104, 3125, 15, 16, 500, 520, 15625, 2, 75, 80, 83, 86, 2500, 2600, 2604, 78125, 10, 12, 13, 375, 400, 415, 416, 430, 433, 12500, 13000, 13020, 390625, 50, 60, 62, 65, 66, 69, 71, 1875, 2000, 2075, 2080, 2083, 2150, 2165, 2166, 62500, 65000, 65100, 65104, 1953125

Offset

0,2

Comments

The last number in row n is 5^n (A000351(n)).

For numbers m and n such that m is in row n, 5*m is in row n + 1 and if m > 10 and m == 10, 15, 20, or 25 (mod 30), then floor(m/6) is in row n + 1.

The conjecture in A330073 claims that every positive integer appears in this triangle.

Links

Table of n, a(n) for n=0..57.

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.

Formula

If N is the list of numbers in row n, then the list of numbers in row n + 1 is the union of each number in N multiplied by 5 and numbers floor(x/6) where x is in N, congruent to 0 (mod 5), not congruent to 0 or 5 (mod 30), and floor(x/6) > 1.

Example

The irregular triangle starts:
0:   1
1:   5
2:  25
3:   4  125
4:  20  625
5:   3  100  104 3125
6:  15   16  500  520 15625
7:   2   75   80   83    86  2500  2600  2604 78125
8:  10   12   13  375   400   415   416   430   433 12500 13000 13020 390625

Prog

(PARI) A331272(lim)=my(N=[1], b=-1, RC=5*[2..5]); while(b<lim, b++; print(N); N=vecsort(matconcat(apply(X->if(setsearch(RC, X%30)&&(X>RC[1]), [floor(X/6), 5*X], X*5), N))[1, ]))

Crossrefs

Cf. A000351 (5^n), A127824, A330073.

Sequence in context: A175555 A346994 A070387 * A123748 A050108 A070386

Adjacent sequences: A331269 A331270 A331271 * A331273 A331274 A331275

Keyword

nonn,easy,tabf,look

Author

Davis Smith, Jan 13 2020

Status

approved