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A343237 Triangle T obtained from the array A(n, k) = (k+1)^(n+1) - k^(n+1), n, k >= 0, by reading antidiagonals upwards.

%I A343237 #27 Nov 04 2022 12:42:01
%S A343237 1,1,1,1,3,1,1,7,5,1,1,15,19,7,1,1,31,65,37,9,1,1,63,211,175,61,11,1,
%T A343237 1,127,665,781,369,91,13,1,1,255,2059,3367,2101,671,127,15,1,1,511,
%U A343237 6305,14197,11529,4651,1105,169,17,1
%N A343237 Triangle T obtained from the array A(n, k) = (k+1)^(n+1) - k^(n+1), n, k >= 0, by reading antidiagonals upwards.
%C A343237 This is the row reversed version of the triangle A047969(n, m). The corresponding array A047969 is a(n, k) = A(k, n) (transposed of array A).
%C A343237 A(n-1, k-1) = k^n - (k-1)^n gives the number of n-digit numbers with digits from K = {1, 2, 3, ..., k} such that any digit from K, say k, appears at least once. Motivated by a comment in A005061 by _Enrique Navarrete_, the instance k=4 for n >= 1 (the column 3 in array A), and the d = 3 (sub)-diagonal sequence of T for m >= 0.
%F A343237 Array A(n, k) = (k+1)^(n+1) - k^(n+1), n, k >= 0.
%F A343237 A(n-1, k-1) = Sum_{j=1} binomial(n, j)*(k-1)^(n-j) = Sum_{j=0} binomial(n, j)*(k-1)^(n-j) - (k-1)^n = (1+(k-1))^n - (k-1)^n = k^n - (k-1)^n (from the combinatorial comment on A(n-1, k-1) above).
%F A343237 O.g.f. row n of array A: RA(n, x) = P(n, x)/(1 - x)^n, with P(n, x) = Sum_{m=0..n} A008292(n+1, m+1)*x^m, (the Eulerian number triangle  A008292 has offset 1) for n >= 0. (See the Oct 26 2008 comment in A047969 by _Peter Bala_). RA(n, x) = polylog(-(n+1), x)*(1-x)/x. (For P(n, x) see the formula by _Vladeta Jovovic_, Sep 02 2002.)
%F A343237 E.g.f. of e.g.f.s of the rows of array A: EE(x, y) = exp(x)*(1 + y*(exp(x) - 1))*exp(y*exp(x)), that is A(n, k) = [y^k/k!][x^n/n!] EE(x, y).
%F A343237 Triangle T(n, m) = A(n-m, m) = (m+1)^(n-m+1) - m^(n-m+1), n >= 0, m = 0, 1, ..., n.
%F A343237 E.g.f.: -(exp(x)-1)/(x*exp(x)*y-x). - _Vladimir Kruchinin_, Nov 02 2022
%e A343237 The array A begins:
%e A343237 n\k  0  1   2    3     4     5     6      7      8      9 ...
%e A343237 -------------------------------------------------------------
%e A343237 0:   1  1   1    1     1     1     1      1      1      1 ...
%e A343237 1:   1  3   5    7     9    11    13     15     17     19 ...
%e A343237 2:   1  7  19   37    61    91   127    169    217    271 ...
%e A343237 3:   1 15  65  175   369   671  1105   1695   2465   3439 ...
%e A343237 4:   1 31 211  781  2101  4651  9031  15961  26281  40951 ...
%e A343237 5:   1 63 665 3367 11529 31031 70993 144495 269297 468559 ...
%e A343237 ...
%e A343237 The triangle T begins:
%e A343237 n\m   0    1     2     3     4     5    6    7   8  9 10 ...
%e A343237 -------------------------------------------------------------
%e A343237 0:    1
%e A343237 1:    1    1
%e A343237 2:    1    3     1
%e A343237 3:    1    7     5     1
%e A343237 4:    1   15    19     7     1
%e A343237 5:    1   31    65    37     9     1
%e A343237 6:    1   63   211   175    61    11    1
%e A343237 7:    1  127   665   781   369    91   13    1
%e A343237 8:    1  255  2059  3367  2101   671  127   15   1
%e A343237 9:    1  511  6305 14197 11529  4651 1105  169  17  1
%e A343237 10:   1 1023 19171 58975 61741 31031 9031 1695 217 19  1
%e A343237 ...
%e A343237 Combinatorial interpretation (cf. A005061 by _Enrique Navarrete_)
%e A343237 The three digits numbers with digits from K ={1, 2, 3, 4} having at least one 4 are:
%e A343237 j=1 (one 4): 114, 141, 411; 224, 242, 422; 334, 343, 433; 124, 214, 142, 241, 412, 421; 134, 314, 143, 341, 413, 431; 234, 243, 423. That is,  3*3 + 3!*3 = 27 = binomial(3, 1)*(4-1)^(3-1) = 3*3^2;
%e A343237 j=2 (twice 4):  144, 414, 441;  244, 424, 442;  344, 434, 443; 3*3 = 9 = binomial(3, 2)*(4-1)^(3-2) = 3*3;
%e A343237 j=3 (thrice 4) 444; 1 = binomial(3, 3)*(4-1)^(3-3).
%e A343237 Together: 27 + 9 + 1 = 37 = A(2, 3) = T(5, 3).
%p A343237 egf := exp(exp(x)*y + x)*(exp(x)*y - y + 1): ser := series(egf, x, 12):
%p A343237 cx := n -> series(n!*coeff(ser, x, n), y, 12):
%p A343237 Arow := n -> seq(k!*coeff(cx(n), y, k), k=0..9):
%p A343237 for n from 0 to 5 do Arow(n) od; # _Peter Luschny_, May 10 2021
%t A343237 A[n_, k_] := (k + 1)^(n + 1) - k^(n + 1); Table[A[n - k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* _Amiram Eldar_, May 10 2021 *)
%Y A343237 Cf. A005061, A008292, A047969 (reversed), A045531 (central diagonal), A047970 (row sums of triangle).
%Y A343237 Row sequences of array A (nexus numbers): A000012, A005408, A003215, A005917(k+1), A022521, A022522, A022523, A022524, A022525, A022526, A022527, A022528.
%Y A343237 Column sequences of array A: A000012, A000225(n+1), A001047(n+1), A005061(n+1), A005060(n+1), A005062(n+1), A016169(n+1), A016177(n+1), A016185(n+1), A016189(n+1), A016195(n+1), A016197(n+1).
%K A343237 nonn,tabl,easy
%O A343237 0,5
%A A343237 _Wolfdieter Lang_, May 10 2021