A256095 Triangle of greatest common divisors of two triangular numbers (A000217).
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
Offset: 0
Examples
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 ...
Links
- Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)
Programs
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Maple
T:= (i,j) -> igcd(i*(i+1)/2,j*(j+1)/2): seq(seq(T(i,j),j=0..i),i=0..20); # Robert Israel, Jan 20 2020
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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
Formula
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)