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A modified Pascal's triangle, read by rows, and modified as follows: binomial(n,j) is replaced by gcd(2^n, binomial(n,j)), i.e., the largest power of 2 dividing binomial(n,j).
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%I #26 Dec 25 2016 02:13:51

%S 1,1,1,1,2,1,1,1,1,1,1,4,2,4,1,1,1,2,2,1,1,1,2,1,4,1,2,1,1,1,1,1,1,1,

%T 1,1,1,8,4,8,2,8,4,8,1,1,1,4,4,2,2,4,4,1,1,1,2,1,8,2,4,2,8,1,2,1,1,1,

%U 1,1,2,2,2,2,1,1,1,1,1,4,2,4,1,8,4,8,1,4,2,4,1,1,1,2,2,1,1,4,4,1,1,2,2,1,1

%N A modified Pascal's triangle, read by rows, and modified as follows: binomial(n,j) is replaced by gcd(2^n, binomial(n,j)), i.e., the largest power of 2 dividing binomial(n,j).

%C If N is a power of 2, then the first N rows are invariant under all 6 symmetries of an equilateral triangle. - _Paul Boddington_, Dec 17 2003

%H G. C. Greubel, <a href="/A082907/b082907.txt">Table of n, a(n) for the first 50 rows, flattened</a>

%H Tyler Ball, Tom Edgar, and Daniel Juda, <a href="http://dx.doi.org/10.4169/math.mag.87.2.135">Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem</a>, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.

%H E. Burlachenko, <a href="https://arxiv.org/abs/1612.00970">Fractal generalized Pascal matrices</a>, arXiv:1612.00970 [math.NT], 2016. See p. 5.

%H Tom Edgar and Michael Z. Spivey, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Edgar/edgar3.html">Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers</a>, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.

%F From _Paul Boddington_, Dec 17 2003: (Start)

%F T(n, j) = c(n)/(c(j)*c(n-j)) where c(n)=A060818(n).

%F T(n, j) = (b(j)*b(n-j))/b(n) where b(n)=A001316(n) (Gould's sequence). (End)

%e Triangle read by rows:

%e 1,

%e 1,1,

%e 1,2,1,

%e 1,1,1,1,

%e 1,4,2,4,1,

%e 1,1,2,2,1,1,

%e 1,2,1,4,1,2,1,

%e 1,1,1,1,1,1,1,1,

%e 1,8,4,8,2,8,4,8,1,

%e 1,1,4,4,2,2,4,4,1,1,

%e ...

%e For n = -1 + 2^k, such rows consist of all 1's since all binomial coefficients C(n,j) are odd.

%t Flatten[Table[Table[GCD[2^n, Binomial[n, j]], {j, 0, n}], {n, 0, 25}], 1]

%t f[n_] := Denominator[CatalanNumber[n - 1]/2^(n - 1)]; T[n_, k_] := f[n]/(f[k]*f[n - k]); Table[T[n, k], {n, 0, 7}, {k, 0, n}]//Flatten (* _G. C. Greubel_, Dec 24 2016 *)

%Y Cf. A000005, A000079, A001316, A007318, A060818.

%K nonn,tabl

%O 0,5

%A _Labos Elemer_, Apr 23 2003

%E Edited by _Jon E. Schoenfield_, Dec 24 2016