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A242849
Triangle read by rows: T(n,k) = A060828(n)/(A060828(k) * A060828(n-k)).
4
1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 3, 3, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 9, 9, 3, 9, 9, 3, 9, 9, 1, 1, 1, 9, 3, 3, 9, 3, 3, 9, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 3, 3, 1, 9, 9, 3, 9, 9, 1
OFFSET
0,8
COMMENTS
This triangle can be obtained by replacing each entry of Pascal's Triangle by the largest power of 3 dividing that entry.
The exponent of T(n,k) is the number of 'carries' that occur when adding k and n-k in base 3 using the traditional addition algorithm.
If T(n,k) != 0 mod 3, then n dominates k in base 3.
LINKS
Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.
Tyler Ball and Daniel Juda, Dominance over N, Rose-Hulman Undergraduate Mathematics Journal, Vol. 13, No. 2, Fall 2013.
E. Burlachenko, Fractal generalized Pascal matrices, arXiv:1612.00970 [math.NT], 2016. See p. 7.
Tom Edgar and Michael Z. Spivey, Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.
FORMULA
T(n,k) = A060828(n)/(A060828(k) * A060828(n-k)).
T(n,k) = Product_{i=1..n} A038500(i)/(Product_{i=1..k} A038500(i)*Product_{i=1..n-k} A038500(i)).
T(n,k) = A038500(n)/n*(k/A038500(k)*T(n-1,k-1)+(n-k)/A038500(n-k)*T(n-1,k)).
EXAMPLE
The triangle begins
1
1 1
1 1 1
1 3 3 1
1 1 3 1 1
1 1 1 1 1 1
1 3 3 1 3 3 1.
MATHEMATICA
s3[n_] := 3^IntegerExponent[n!, 3];
T[n_, k_] := s3[n]/(s3[k] s3[n-k]);
Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 06 2018 *)
PROG
(Sage)
m=50
T=[0]+[3^valuation(i, 3) for i in [1..m]]
Table=[[prod(T[1:i+1])/(prod(T[1:j+1])*prod(T[1:i-j+1])) for j in [0..i]] for i in [0..m-1]]
[x for sublist in Table for x in sublist]
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Tom Edgar, May 23 2014
STATUS
approved