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A256095
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Triangle of greatest common divisors of two triangular numbers (A000217).
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3
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0, 1, 1, 3, 1, 3, 6, 1, 3, 6, 10, 1, 1, 2, 10, 15, 1, 3, 3, 5, 15, 21, 1, 3, 3, 1, 3, 21, 28, 1, 1, 2, 2, 1, 7, 28, 36, 1, 3, 6, 2, 3, 3, 4, 36, 45, 1, 3, 3, 5, 15, 3, 1, 9, 45, 55, 1, 1, 1, 5, 5, 1, 1, 1, 5, 55, 66, 1, 3, 6, 2, 3, 3, 2, 6, 3, 11, 66, 78, 1, 3, 6, 2, 3, 3, 2, 6, 3, 1, 6, 78, 91, 1, 1, 1, 1, 1, 7, 7, 1, 1, 1, 1, 13, 91, 105, 1, 3, 3, 5, 15, 21, 7, 3, 15, 5, 3, 3, 7, 105
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OFFSET
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0,4
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LINKS
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FORMULA
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T(n, m) = gcd(Tri(n), Tri(m)), 0 <= m <= n, with the triangular numbers Tri = A000217.
T(n, 0) = Tri(n) = T(n, n). T(n, 1) = 1, n >= 0.
Columns m=2: A144437(n-1), m=3: repeat(6, 2, 3, 3, 2, 6, 3, 1, 6, 6, 1, 3) (guess), m=4: repeat(10, 5, 1, 2, 2, 5, 5, 2, 2, 1, 5, 10, 2, 1, 1, 10, 10, 1, 1, 2) (guess), m=5 repeat(15, 3, 1, 3, 15, 5, 3, 3, 1, 15, 15, 1, 3, 3, 5) (guess), ...
From Robert Israel, Jan 21 2020: (Start) The guesses are correct. More generally, for each k>=1, T(n,k) is periodic in n with period 2*A000217(k) if k == 0 or 3 (mod 4), A000217(k) if k == 1 or 2 (mod 4). (End)
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EXAMPLE
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The triangle T(n, m) begins:
n\m 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
0: 0
1: 1 1
2: 3 1 3
3: 6 1 3 6
4: 10 1 1 2 10
5: 15 1 3 3 5 15
6: 21 1 3 3 1 3 21
7: 28 1 1 2 2 1 7 28
8: 36 1 3 6 2 3 3 4 36
9: 45 1 3 3 5 15 3 1 9 45
10: 55 1 1 1 5 5 1 1 1 5 55
11: 66 1 3 6 2 3 3 2 6 3 11 66
12: 78 1 3 6 2 3 3 2 6 3 1 6 78
13: 91 1 1 1 1 1 7 7 1 1 1 1 13 91
14: 105 1 3 3 5 15 21 7 3 15 5 3 3 7 105
...
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MAPLE
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T:= (i, j) -> igcd(i*(i+1)/2, j*(j+1)/2):
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PROG
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(PARI) tabl(nn) = {for (n=0, nn, trn = n*(n+1)/2; for (k=0, n, print1(gcd(trn, k*(k+1)/2), ", "); ); print(); ); } \\ Michel Marcus, Mar 17 2015
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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