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%I #73 Feb 03 2021 23:01:36
%S 1,1,3,1,5,8,1,7,21,19,1,9,40,79,43,1,11,65,205,281,94,1,13,96,421,
%T 991,963,201,1,15,133,751,2569,4612,3217,423,1,17,176,1219,5531,15085,
%U 20905,10547,880,1,19,225,1849,10513,39186,86241,92935,34089,1815
%N Square array read by upward antidiagonals: T(n, k) is the number of n-ary strings of length k containing 00.
%H Robert P. P. McKone, <a href="/A340156/b340156.txt">Antidiagonals n = 2..100, flattened</a>
%F T(n, k) = n^k - A180165(n+1,k-1), where A180165 in the number of strings not containing 00.
%F m(2) = [1 - 1/n, 1/n, 0; 1 - 1/n, 0, 1/n; 0, 0, 1], is the probability/transition matrix for two consecutive "0" -> "containing 00".
%e For n = 3 and k = 4, there are 21 strings: {0000, 0001, 0002, 0010, 0011, 0012, 0020, 0021, 0022, 0100, 0200, 1000, 1001, 1002, 1100, 1200, 2000, 2001, 2002, 2100, 2200}.
%e Square table T(n,k):
%e k=2: k=3: k=4: k=5: k=6: k=7:
%e n=2: 1 3 8 19 43 94
%e n=3: 1 5 21 79 281 963
%e n=4: 1 7 40 205 991 4612
%e n=5: 1 9 65 421 2569 15085
%e n=6: 1 11 96 751 5531 39186
%e n=7: 1 13 133 1219 10513 87199
%e n=8: 1 15 176 1849 18271 173608
%e n=9: 1 17 225 2665 29681 317817
%t m[r_] := Normal[With[{p = 1/n}, SparseArray[{Band[{1, 2}] -> p, {i_, 1} /; i <= r -> 1 - p, {r + 1, r + 1} -> 1}]]];
%t T[n_, k_, r_] := MatrixPower[m[r], k][[1, r + 1]]*n^k;
%t Reverse[Table[T[n, k - n + 2, 2], {k, 2, 11}, {n, 2, k}], 2] // Flatten (* _Robert P. P. McKone_, Jan 26 2021 *)
%Y Cf. A008466 (row 2), A186244 (row 3), A000567 (column 4).
%Y Cf. A180165 (not containing 00), A340242 (containing 000).
%Y Cf. A000045, A028859, A125145, A086347, A180033, A180167, A322054.
%Y Cf. A005563, A033445.
%K nonn,tabl
%O 2,3
%A _Robert P. P. McKone_, Dec 29 2020
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