[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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