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A212853
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Number of n X 6 arrays with rows being permutations of 0..5 and no column j greater than column j-1 in all rows.
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13
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1, 90921, 179781181, 191740223841, 164481310134301, 128645361626874561, 96426023622482278621, 70816637331790329140481, 51492108377805402906874141, 37256471170472317193421713601, 26890352949868734582700237312861
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OFFSET
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1,2
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COMMENTS
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Let P_6 be the set of all lists b = (b_1, b_2, b_3, b_4, b_5, b_6) of integers b_i >= 0, i = 1, ..., 6, such that 1*b_1 + 2*b_2 + 3*b_3 + 4*b_4 + 5*b_5 + 6*b_6 = 6; i.e., P_6 is the set all integer partitions of 6. Then |P_6| = A000041(6) = 11.
From Eq. (6), p. 248, in Abramson and Promislow (1978), with t=0, we get a(n) = A212855(n,6) = Sum_{b in P_6} (-1)^(6-Sum_{j=1..6} b_j) * (b_1 + b_2 + b_3 + b_4 + b_5 + b_6)!/(b_1! * b_2! * b_3! * b_4! * b_5! * b_6!) * (6! / ((1!)^b_1 * (2!)^b_2 * (3!)^b_3 * (4!)^b_4 * (5!)^b_5 * (6!)^b_6))^n.
The integer partitions of 6 are listed on p. 831 of Abramowitz and Stegun (1964). We see that the corresponding multinomial coefficients 6! / ((1!)^b_1 * (2!)^b_2 * (3!)^b_3 * (4!)^b_4 * (5!)^b_5 * (6!)^b_6) are all distinct; that is, A070289(6) = A000041(6) = 11 and A309951(6,s) = A325305(6,s) for s = 0..11. (Compare with the comments for A212854.)
Using the information about partitions of 6 in Eq. (6) (with t=0), p. 248, of Abramson and Promislow (1978), we may derive the explicit equation for a(n) shown below.
Using standard results from the theory of difference equations (since the solution is known explicitly), we may derive R. H. Hardin's empirical recurrence. The recurrence is equivalent to Sum_{s = 0..11} (-1)^s * A325305(6,s) * a(n-s) = 0 for n >= 12.
(End)
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LINKS
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Morton Abramson and David Promislow, Enumeration of arrays by column rises, J. Combinatorial Theory Ser. A 24(2) (1978), 247-250; see Eq. (6) (with t=0), p. 248, and the comments above.
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FORMULA
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Empirical: a(n) = 1602*a(n-1) - 929171*a(n-2) + 260888070*a(n-3) - 39883405500*a(n-4) + 3492052425000*a(n-5) - 177328940580000*a(n-6) + 5153150631600000*a(n-7) - 82577533320000000*a(n-8) + 669410956800000000*a(n-9) - 2224399449600000000*a(n-10) + 1632586752000000000*a(n-11) for n >= 12. [It is correct; see the comments above.]
a(n) = -1 + 2*6^n + 2*15^n + 20^n - 3*30^n - 6*60^n - 90^n + 4*120^n + 6*180^n - 5*360^n + 720^n for n >= 1. - Petros Hadjicostas, Sep 08 2019
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EXAMPLE
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Some solutions for n=3:
0 3 1 4 2 5 0 3 1 4 2 5 0 3 1 4 2 5 0 3 1 4 2 5
3 0 2 4 5 1 1 3 0 4 5 2 4 0 3 1 2 5 0 1 5 2 3 4
1 2 4 0 3 5 5 0 4 2 3 1 2 1 5 4 3 0 3 1 5 0 4 2
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MATHEMATICA
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T[n_, k_] := T[n, k] = If[k == 0, 1, -Sum[Binomial[k, j]^n*(-1)^j*T[n, k - j], {j, 1, k}]];
a[n_] := T[n, 6];
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CROSSREFS
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Cf. A000041, A070289, A212850, A212851, A212852, A212854, A212855, A212856, A212857, A309951, A325305.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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