%I #17 Jan 21 2020 09:13:38

%S 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,

%T 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,

%U 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

%N Triangle of greatest common divisors of two triangular numbers (A000217).

%H Robert Israel, <a href="/A256095/b256095.txt">Table of n, a(n) for n = 0..10010</a> (rows 0 to 140, flattened)

%F T(n, m) = gcd(Tri(n), Tri(m)), 0 <= m <= n, with the triangular numbers Tri = A000217.

%F T(n, 0) = Tri(n) = T(n, n). T(n, 1) = 1, n >= 0.

%F 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), ...

%F 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)

%e The triangle T(n, m) begins:

%e n\m 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

%e 0: 0

%e 1: 1 1

%e 2: 3 1 3

%e 3: 6 1 3 6

%e 4: 10 1 1 2 10

%e 5: 15 1 3 3 5 15

%e 6: 21 1 3 3 1 3 21

%e 7: 28 1 1 2 2 1 7 28

%e 8: 36 1 3 6 2 3 3 4 36

%e 9: 45 1 3 3 5 15 3 1 9 45

%e 10: 55 1 1 1 5 5 1 1 1 5 55

%e 11: 66 1 3 6 2 3 3 2 6 3 11 66

%e 12: 78 1 3 6 2 3 3 2 6 3 1 6 78

%e 13: 91 1 1 1 1 1 7 7 1 1 1 1 13 91

%e 14: 105 1 3 3 5 15 21 7 3 15 5 3 3 7 105

%e ...

%p T:= (i,j) -> igcd(i*(i+1)/2,j*(j+1)/2):

%p seq(seq(T(i,j),j=0..i),i=0..20); # _Robert Israel_, Jan 20 2020

%o (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

%Y Cf. A000217, A144437.

%Y T(2n,n) gives A026741.

%K nonn,easy,tabl

%O 0,4

%A _Wolfdieter Lang_, Mar 17 2015