%I #20 Feb 04 2023 09:53:16
%S 1,1,6,1,10,27,1,14,69,108,1,18,123,404,405,1,22,195,892,2155,1458,1,
%T 26,273,1716,5845,10830,5103,1,30,375,2732,13525,36042,52241,17496,1,
%U 34,477,4324,24575,99774,213647,244648,59049,1,38,603,6060,44545,208146,705215,1232504,1120599,196830
%N Table read by antidiagonals: T(n,k) is the number of idempotent n X n 0..k matrices of rank 1.
%C Table starts
%C 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
%C 6, 10, 14, 18, 22, 26, 30, 34, 38, ...
%C 27, 69, 123, 195, 273, 375, 477, 603, ...
%C 108, 404, 892, 1716, 2732, 4324, 6060, ...
%C 405, 2155, 5845, 13525, 24575, 44545, ...
%C 1458, 10830, 36042, 99774, 208146, ...
%C 5103, 52241, 213647, 705215, ...
%C 17496, 244648, ...
%C 59049, ...
%C ...
%H Robert Israel, <a href="/A224524/b224524.txt">Table of n, a(n) for n = 1..10011</a>
%e Some solutions for n=3, k=4:
%e 1 0 0 0 4 4 0 0 0 0 4 2 1 2 1 0 0 0 0 1 0
%e 0 0 0 0 1 1 3 1 0 0 0 0 0 0 0 0 0 0 0 1 0
%e 1 0 0 0 0 0 0 0 0 0 2 1 0 0 0 1 4 1 0 0 0
%p f:= proc(n,k)
%p local tot, a1, a0, a2, m,u;
%p tot:= 0;
%p for a1 from 1 to n do
%p for a0 from 0 to n-a1 do
%p a2:= n-a1-a0;
%p if a0 = 0 then tot:= tot + n!/(a1!*a2!)*a1*(k-1)^a2
%p elif a2 = 0 then tot:= tot + n!/(a0!*a1!)*a1*(k+1)^a0
%p else
%p u:= n!/(a0!*a1!*a2!)*a1;
%p for m from 2 to k do
%p tot:= tot + u*((m-1)^a2 - (m-2)^a2)*(floor(k/m)+1)^a0
%p od
%p fi
%p od od;
%p tot
%p end proc:
%p seq(seq(f(i,j-i),i=1..j-1),j=2..20); # _Robert Israel_, Dec 15 2019
%t Unprotect[Power]; 0^0 = 1; Protect[Power];
%t f[n_, k_] := Module[{tot, a1, a0, a2, m, u}, tot = 0; For[a1 = 1, a1 <= n, a1++, For[a0 = 0, a0 <= n - a1, a0++, a2 = n - a1 - a0; Which[a0 == 0, tot = tot + n!/(a1!*a2!)*a1*(k - 1)^a2, a2 == 0, tot = tot + n!/(a0!*a1!)*a1*(k + 1)^a0, True, u = n!/(a0!*a1!*a2!)*a1; For[m = 2, m <= k, m++, tot = tot + u*((m - 1)^a2 - (m - 2)^a2)*(Floor[k/m] + 1)^a0]]]]; tot];
%t Table[Table[f[i, j - i], {i, 1, j - 1}], {j, 2, 20}] // Flatten (* _Jean-François Alcover_, Feb 04 2023, after _Robert Israel_ *)
%Y Column 1 is A027471(n+1).
%K nonn,tabl
%O 1,3
%A _R. H. Hardin_, Apr 09 2013