[seqfan] Re: two dumb questions

Jamie Morken jmorken at shaw.ca
Mon Jul 24 21:32:43 CEST 2017


Hi,The triangle you found, identified as A054521 by Brad Klee seems pretty interesting I think.It can be constructed with a simple rule:Put all zero's on the diagonal, and all 1's on the first column:1                                    
1    0                                
1        0                            
1            0                        
1                0                    
1                    0                
1                        0            
1                            0        
1                                0    
1                                    0Then to fill out the triangle, column n is just reversed and appendedrow n, ie for the second column, copy in the second row reversed 01,01,01multiple times.1                                    
1    0                                
1    1    0                            
1    0        0                        
1    1            0                    
1    0                0                
1    1                    0            
1    0                        0        
1    1                            0    
1    0                                0Can do the same for the third column, whichis simply the reversed third row 011 copiedin multiple times.1                                    
1    0                                
1    1    0                            
1    0    1    0                        
1    1    1        0                    
1    0    0            0                
1    1    1                0            
1    0    1                    0        
1    1    0                        0    
1    0    1                            0And one more subsequent row 4th to 10th fills out the triangle1                                    
1    0                                
1    1    0                            
1    0    1    0                        
1    1    1    1    0                    
1    0    0    0    1    0                
1    1    1    1    1    1    0            
1    0    1    0    1    0    1    0        
1    1    0    1    1    0    1    1    0    
1    0    1    0    0    0    1    0    1    0Also there is a very similar triangle I found by checking the triangle "repeating pattern in GCD(x,n)" which is already on OEIS A050873.A050873 can be constructed with the same technique as A054521 above except starting from this form:1                                    
1    2                                
1        3                            
1            4                        
1                5                    
1                    6                
1                        7            
1                            8        
1                                9    
1                                    10
For the triangle up to row 28 and column 14:1    1                                                    
2    1    2                                                
3    1    1    3                                            
4    1    2    1    4                                        
5    1    1    1    1    5                                    
6    1    2    3    2    1    6                                
7    1    1    1    1    1    1    7                            
8    1    2    1    4    1    2    1    8                        
9    1    1    3    1    1    3    1    1    9                    
10    1    2    1    2    5    2    1    2    1    10                
11    1    1    1    1    1    1    1    1    1    1    11            
12    1    2    3    4    1    6    1    4    3    2    1    12        
13    1    1    1    1    1    1    1    1    1    1    1    1    13    
14    1    2    1    2    1    2    7    2    1    2    1    2    1    14
15    1    1    3    1    5    3    1    1    3    5    1    3    1    1
16    1    2    1    4    1    2    1    8    1    2    1    4    1    2
17    1    1    1    1    1    1    1    1    1    1    1    1    1    1
18    1    2    3    2    1    6    1    2    9    2    1    6    1    2
19    1    1    1    1    1    1    1    1    1    1    1    1    1    1
20    1    2    1    4    5    2    1    4    1    10    1    4    1    2
21    1    1    3    1    1    3    7    1    3    1    1    3    1    7
22    1    2    1    2    1    2    1    2    1    2    11    2    1    2
23    1    1    1    1    1    1    1    1    1    1    1    1    1    1
24    1    2    3    4    1    6    1    8    3    2    1    12    1    2
25    1    1    1    1    5    1    1    1    1    5    1    1    1    1
26    1    2    1    2    1    2    1    2    1    2    1    2    13    2
27    1    1    3    1    1    3    1    1    9    1    1    3    1    1
28    1    2    1    4    1    2    7    4    1    2    1    4    1    14The triangle has a copy of itself embedded, ie for the rightmost column above:
14
1
2
1
2
1
2
7
2
1
2
1
2
1
14
That is also the 14th row:
14    1    2    1    2    1    2    7    2    1    2    1    2    1    14Which is also a palindrome.A couple more properties I noticed:Rows n, have n prime if the row is all 1's up to the last column in the row.Rows n, have n a square of a prime if the row has a record count less thanthe prime rows n.
cheers,Jamiein reply to:Peter Lawrence 
    peterl95124 at sbcglobal.net
       

    Sat Jul 22 03:43:47 CEST 2017
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A051731: characteristic triangle of divisors of n

A000005: = number of divisors of n   =   row sums of A051731

1.  How is A051731 different from Redheffer matrix A143104, 
which seems to have same FORMULA definition ?
a(i,j) = 1 if j=1 or j|i; 0 otherwise.





Axxxxxx:  characteristic triangle of numbers with gcd( I, j ) = 1

A000010 = number of relative-primes of n   =   row sums of Axxxxxx


a(i,j) = 1 if gcd(j,i) = 1; 0 otherwise.

 1:   1,                                               1
 2:   1, 0,                                            1
 3:   1, 1, 0,                                         2
 4:   1, 0, 1, 0,                                      2
 5:   1, 1, 1, 1, 0,                                   4
 6:   1, 0, 0, 1, 0, 0,                                2
 7:   1, 1, 1, 1, 1, 1, 0,                             6
 8:   1, 0, 1, 0, 1, 0, 1, 0,                          4
 9:   1, 1, 0, 1, 1, 0, 1, 1, 0,                       6
10:   1, 0, 1, 0, 0, 0, 1, 0, 1, 0,                    4
11:   1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0,                 10
12:   1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0,              4
13:   1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0,           12
14:   1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0,        6
15:   1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,     8
16:   1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0,  8


2.    I suspect I computed this triangle incorrectly, because I can’t find it in the OEIS  _?_



Peter Lawrence.



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