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A129455
An analog of Pascal's triangle based on A129454. T(n, k) = A129454(n+1)/(A129454(n-k+1)*A129454(k+1)).
4
1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 256, 384, 256, 1, 1, 5, 640, 640, 5, 1, 1, 1146617856, 2866544640, 244611809280, 2866544640, 1146617856, 1, 1, 7, 4013162496, 6688604160, 6688604160, 4013162496, 7, 1, 1, 35184372088832, 123145302310912, 47066867504069920948224, 919274755938865643520, 47066867504069920948224, 123145302310912, 35184372088832, 1
OFFSET
0,5
COMMENTS
It appears that the T(n,k) are always integers. This would follow from the conjectured prime factorization given in A129454. Calculation suggests that the binomial coefficients C(n,k) divide T(n,k) and that T(n,k)/C(n,k) are perfect sixth powers.
FORMULA
T(n, k) = Product_{h=1..n} Product_{i=1..n} Product_{j=1..n} gcd(h,i,j)/( (Product_{h=1..n-k} Product_{i=1..n-k} Product_{j=1..n-k} gcd(h,i,j))*(Product_{h=1..k} Product_{i=1..k} Product_{j=1..k} gcd(h,i,j)) ).
T(n, n-k) = T(n, k). - G. C. Greubel, Feb 07 2024
EXAMPLE
Triangle starts:
1;
1, 1;
1, 2, 1;
1, 3, 3, 1;
1, 256, 384, 256, 1;
1, 5, 640, 640, 5, 1;
MATHEMATICA
A129454[n_]:= Product[GCD[j, k, m], {j, n-1}, {k, n-1}, {m, n-1}];
A129455[n_, k_]:= A129454[n+1]/(A129454[k+1]*A129454[n-k+1]);
Table[A129455[n, k], {n, 0, 9}, {k, 0, n}]//Flatten (* G. C. Greubel, Feb 07 2024 *)
PROG
(Magma)
A129454:= func< n | n le 1 select 1 else (&*[(&*[(&*[GCD(GCD(j, k), m): k in [1..n-1]]): j in [1..n-1]]): m in [1..n-1]]) >;
A129455:= func< n, k | A129454(n+1)/(A129454(n-k+1)*A129454(k+1)) >;
[A129455(n, k): k in [0..n], n in [0..9]]; // G. C. Greubel, Feb 07 2024
(SageMath)
def A129454(n): return product(product(product(gcd(gcd(j, k), m) for k in range(1, n)) for j in range(1, n)) for m in range(1, n))
def A129455(n, k): return A129454(n+1)/(A129454(n-k+1)*A129454(k+1))
flatten([[A129455(n, k) for k in range(n+1)] for n in range(10)]) # G. C. Greubel, Feb 07 2024
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Peter Bala, Apr 16 2007
STATUS
approved