A062275 Array A(n, k) = n^k * k^n, n, k >= 0, read by antidiagonals.
1, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 3, 16, 3, 0, 0, 4, 72, 72, 4, 0, 0, 5, 256, 729, 256, 5, 0, 0, 6, 800, 5184, 5184, 800, 6, 0, 0, 7, 2304, 30375, 65536, 30375, 2304, 7, 0, 0, 8, 6272, 157464, 640000, 640000, 157464, 6272, 8, 0, 0, 9, 16384, 750141, 5308416, 9765625
Offset: 0
Examples
A(3, 2) = 3^2 * 2^3 = 9*8 = 72. The array A(n, k) begins: n\k 0 1 2 3 4 5 6 7 8 9 10 ... 0: 1 0 0 0 0 0 0 0 0 0 0 ... 1: 0 1 2 3 4 5 6 7 8 9 10 ... 2: 0 2 16 72 256 800 2304 6272 16384 41472 102400 ... 3: 0 3 72 729 5184 30375 157464 750141 3359232 14348907 59049000 ... ... The triangle T(n, k) begins: n\k 0 1 2 3 4 5 6 7 8 9 ... 0: 1 1: 0 0 2: 0 1 0 3: 0 2 2 0 4: 0 3 16 3 0 5: 0 4 72 72 4 0 6: 0 5 256 729 256 5 0 7: 0 6 800 5184 5184 800 6 0 8: 0 7 2304 30375 65536 30375 2304 7 0 9: 0 8 6272 157464 640000 640000 157464 6272 8 0 ... - _Wolfdieter Lang_, May 22 2018
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Crossrefs
Programs
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Mathematica
{{1}}~Join~Table[(#^k k^#) &[n - k], {n, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, May 24 2018 *)
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PARI
t1(n)=n-binomial(round(sqrt(2+2*n)), 2) t2(n)=binomial(floor(3/2+sqrt(2+2*n)), 2)-(n+1) a(n)=t1(n)^t2(n)*t2(n)^t1(n) \\ Eric Chen, Jun 09 2018
Formula
From Wolfdieter Lang, May 22 2018: (Start)
As a triangle: T(n, k) = (n-k)^k * k^(n-k), for n >= 1 and k = 1..n. (End)
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