cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

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

Original entry on oeis.org

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

Views

Author

Davis Smith, Jan 13 2020

Keywords

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.

Examples

			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
		

Crossrefs

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

Programs

  • PARI
    A331272(lim)=my(N=[1], b=-1, RC=5*[2..5]); while(bif(setsearch(RC,X%30)&&(X>RC[1]),[floor(X/6),5*X],X*5),N))[1,]))

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.

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

Views

Author

Davis Smith, Jan 23 2020

Keywords

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.
		

Crossrefs

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.
Showing 1-2 of 2 results.