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Square array of denominators of t(n,k) = (1+1/(k*n))^n, read by descending antidiagonals.
1

%I #10 Sep 20 2013 04:30:51

%S 1,2,4,3,16,27,4,36,216,256,5,64,729,4096,3125,6,100,1728,20736,

%T 100000,46656,7,144,3375,65536,759375,2985984,823543,8,196,5832,

%U 160000,3200000,34012224,105413504,16777216,9,256

%N Square array of denominators of t(n,k) = (1+1/(k*n))^n, read by descending antidiagonals.

%C Limit(t(n,k), n -> infinity) = exp(1/k).

%C 1st row = A000027

%C 2nd row = A016742

%C 3rd row = A016767

%C 4th row = A016804

%C 5th row = A016853

%C 1st column = A000312

%C 2nd column = A062971

%C 3rd column = A091482

%C 4th column = A091483

%e Table of fractions begins:

%e 2, 3/2, 4/3, 5/4, ...

%e 9/4, 25/16, 49/36, 81/64, ...

%e 64/27, 343/216, 1000/729, 2197/1728, ...

%e 625/256, 6561/4096, 28561/20736, 83521/65536, ...

%e ...

%e Table of denominators begins:

%e 1, 2, 3, 4, ...

%e 4, 16, 36, 64, ...

%e 27, 216, 729, 1728, ...

%e 256, 4096, 20736, 65536, ...

%e ...

%e Triangle of antidiagonals begins:

%e 1;

%e 2, 4;

%e 3, 16, 27;

%e 4, 36, 216, 256;

%e ...

%t t[n_, k_] := (1+1/(k*n))^n; Table[t[n-k+1, k], {n, 1, 9}, {k, n, 1, -1}] // Flatten // Denominator

%Y Cf. A229212(numerators), A016742, A016767, A016804, A016853, A000312, A062971, A091482, A091483.

%K frac,tabl,nonn

%O 1,2

%A _Jean-François Alcover_, Sep 16 2013