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A220417 Table T(n,k) = k^n - n^k, n, k > 0, read by descending antidiagonals.

%I A220417 #48 Oct 04 2019 13:13:44
%S A220417 0,1,-1,2,0,-2,3,1,-1,-3,4,0,0,0,-4,5,-7,-17,17,7,-5,6,-28,-118,0,118,
%T A220417 28,-6,7,-79,-513,-399,399,513,79,-7,8,-192,-1844,-2800,0,2800,1844,
%U A220417 192,-8,9,-431,-6049,-13983,-7849,7849,13983,6049,431,-9,10,-924,-18954,-61440,-61318,0,61318,61440,18954,924,-10
%N A220417 Table T(n,k) = k^n - n^k, n, k > 0, read by descending antidiagonals.
%H A220417 Boris Putievskiy, <a href="/A220417/b220417.txt">Rows n = 1..76 of triangle, flattened</a>
%F A220417 As a linear array, the sequence is a(n) = A004736(n)^A002260(n) - A002260(n)^A004736(n) or
%F A220417 a(n) = ((t*t + 3*t + 4)/2 - n)^(n - t*(t + 1)/2) - (n - t*(t + 1)/2)^((t*t + 3*t + 4)/2 - n) where t = floor((-1 + sqrt(8*n - 7))/2).
%e A220417 The table T(n,k) (with rows n >= 1 and columns k >= 1) begins as follows:
%e A220417    0   1     2      3       4       5 ...
%e A220417   -1   0     1      0      -7     -28 ...
%e A220417   -2  -1     0    -17    -118    -513 ...
%e A220417   -3   0    17      0    -399   -2800 ...
%e A220417   -4   7   118    399       0   -7849 ...
%e A220417   -5  28   513   2800    7849       0 ...
%e A220417   ...
%e A220417 The start of the sequence as a triangular array, read by rows (i.e., descending antidiagonals of T(n,k)), is as follows:
%e A220417   0;
%e A220417   1,  -1;
%e A220417   2,   0,   -2;
%e A220417   3,   1,   -1, -3;
%e A220417   4,   0,    0,  0,  -4;
%e A220417   5,  -7,  -17, 17,   7, -5;
%e A220417   6, -28, -118,  0, 118, 28, -6;
%e A220417   ...
%e A220417 In the above triangle, row number m contains m numbers: m^1 - 1^m, (m-1)^2 - 2^(m-1), ..., 1^m - m^1.
%o A220417 (Python)
%o A220417 t=int((math.sqrt(8*n-7) - 1)/ 2)
%o A220417 m=((t*t+3*t+4)/2-n)**(n-t*(t+1)/2)-(n-t*(t+1)/2)**((t*t+3*t+4)/2-n)
%o A220417 (PARI) matrix(9, 9, n, k, k^n - n^k) \\ _Michel Marcus_, Oct 04 2019
%Y A220417 Cf. A002260, A004736, A051128, A051129, A055651, A156353.
%K A220417 sign,tabl
%O A220417 1,4
%A A220417 _Boris Putievskiy_, Dec 14 2012