OFFSET
1,4
COMMENTS
b(i,n,k) = (b(i-1,n,k) + b(i-2,n,k)/p^valuation(b(i-1,n,k) + b(i-2,n,k), p), i.e., b(i,n,k) is b(i-1,n,k) + b(i-2,n,k) with all factors of p removed, where p = 3 in this sequence. Therefore, b(i,n,k) is not divisible by 3 for i >= 3.
At least one of b(i,n,k) < b(i-1,n,k) + b(i-2,n,k) and b(i+1,n,k) < b(i-1,n,k) + b(i,n,k) is true for i >= 5.
It appears that all repetends have the form of (1, 1, 2) (the position of 2 possibly changed), multiplied by G = A038502(gcd(n,k)).
Conjecture: T(n,k) >= 0.
This conjecture can be supported by a heuristic argument: Using dynamic programming, we can compute that for p's in A000057, the probability that b(i-1,n,k) + b(i-2,n,k) is not divisible by p is (p-2)/p, and the probability that valuation(b(i-1,n,k) + b(i-2,n,k), p) = x (x >= 1) is 2*(p-1)/p^(x+1). Therefore, the mathematical expectation of a(i) is (a(i-1,n,k) + a(i-2,n,k))*(p-1)/(p+1), which is exactly the average of the earlier two terms when p = 3, and larger when p >= 5.
LINKS
Yifan Xie, Rows n = 1..100 of triangle, flattened
FORMULA
T(n,k) = 0 iff n = k, n = 2k or k = 2n and gcd(x,y) is not divisible by 3.
EXAMPLE
Array begins:
n\k| 1 2 3 4 5 6 7
---+--------------------------------
1 | 0, 0, 12, 3, 6, 11, 17, ...
2 | 0, 0, 24, 0, 12, 12, 4, ...
3 | 4, 13, 14, 11, 23, 19, 4, ...
4 | 7, 0, 5, 0, 2, 24, 10, ...
5 | 1, 5, 15, 8, 0, 27, 11, ...
6 | 18, 4, 5, 13, 23, 14, 10, ...
7 | 3, 1, 12, 10, 9, 7, 29, ...
...
PROG
(PARI) T(n, k)={my(i=-1, z=0); while((z != 2*n || z != 2*k) && (n != 2*z || n != 2*k) && (k != 2*n || k != 2*z), z=n; n=k; k=(z+n)/3^(valuation(z+n, 3)); i++); i; };
KEYWORD
nonn,tabl
AUTHOR
Yifan Xie, Feb 21 2024
STATUS
approved