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A343237
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Triangle T obtained from the array A(n, k) = (k+1)^(n+1) - k^(n+1), n, k >= 0, by reading antidiagonals upwards.
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3
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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, 1, 127, 665, 781, 369, 91, 13, 1, 1, 255, 2059, 3367, 2101, 671, 127, 15, 1, 1, 511, 6305, 14197, 11529, 4651, 1105, 169, 17, 1
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OFFSET
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0,5
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COMMENTS
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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).
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.
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LINKS
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FORMULA
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Array A(n, k) = (k+1)^(n+1) - k^(n+1), n, k >= 0.
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).
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.)
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).
Triangle T(n, m) = A(n-m, m) = (m+1)^(n-m+1) - m^(n-m+1), n >= 0, m = 0, 1, ..., n.
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EXAMPLE
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The array A begins:
n\k 0 1 2 3 4 5 6 7 8 9 ...
-------------------------------------------------------------
0: 1 1 1 1 1 1 1 1 1 1 ...
1: 1 3 5 7 9 11 13 15 17 19 ...
2: 1 7 19 37 61 91 127 169 217 271 ...
3: 1 15 65 175 369 671 1105 1695 2465 3439 ...
4: 1 31 211 781 2101 4651 9031 15961 26281 40951 ...
5: 1 63 665 3367 11529 31031 70993 144495 269297 468559 ...
...
The triangle T begins:
n\m 0 1 2 3 4 5 6 7 8 9 10 ...
-------------------------------------------------------------
0: 1
1: 1 1
2: 1 3 1
3: 1 7 5 1
4: 1 15 19 7 1
5: 1 31 65 37 9 1
6: 1 63 211 175 61 11 1
7: 1 127 665 781 369 91 13 1
8: 1 255 2059 3367 2101 671 127 15 1
9: 1 511 6305 14197 11529 4651 1105 169 17 1
10: 1 1023 19171 58975 61741 31031 9031 1695 217 19 1
...
The three digits numbers with digits from K ={1, 2, 3, 4} having at least one 4 are:
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;
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;
j=3 (thrice 4) 444; 1 = binomial(3, 3)*(4-1)^(3-3).
Together: 27 + 9 + 1 = 37 = A(2, 3) = T(5, 3).
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MAPLE
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egf := exp(exp(x)*y + x)*(exp(x)*y - y + 1): ser := series(egf, x, 12):
cx := n -> series(n!*coeff(ser, x, n), y, 12):
Arow := n -> seq(k!*coeff(cx(n), y, k), k=0..9):
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MATHEMATICA
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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 *)
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CROSSREFS
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Row sequences of array A (nexus numbers): A000012, A005408, A003215, A005917(k+1), A022521, A022522, A022523, A022524, A022525, A022526, A022527, A022528.
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).
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KEYWORD
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AUTHOR
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STATUS
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approved
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