

A212854


Number of n X 7 arrays with rows being permutations of 0..6 and no column j greater than column j1 in all rows.


13



1, 3081513, 53090086057, 429966316953825, 2675558106868421881, 14895038886845467640193, 78785944892341703819175577, 406643086764765052892275303425, 2073826171428339544452057104498041
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OFFSET

1,2


COMMENTS

We have a(m) = R(m,n=7,t=0) = A212855(m,7) for m >= 1, where R(m,n,t) = LHS of Eq. (6) of Abramson and Promislow (1978, p. 248).
Let P_7 be the set of all lists b = (b_1, b_2,..., b_7) of integers b_i >= 0, i = 1, ..., 7 such that 1*b_1 + 2*b_2 + ... + 7*b_7 = 7; i.e., P_7 is the set all integer partitions of 7. Then P_7 = A000041(7) = 15.
We have a(m) = A212855(m,7) = Sum_{b in P_7} (1)^(7  Sum_{j=1..7} b_j) * (b_1 + b_2 + ... + b_7)!/(b_1! * b_2! * ... * b_7!) * (7! / ((1!)^b_1 * (2!)^b_2 * ... * (7!)^b_7))^m.
The integer partitions of 7 are listed on p. 831 of Abramowitz and Stegun (1964). We see that, when (b_1, b_2, ..., b_7) = (0, 2, 1, 0, 0, 0, 0) or (3, 0, 0, 1, 0, 0, 0) (i.e., we have the partitions 2+2+3 and 1+1+1+4), the corresponding multinomial coefficients are 210 = 7!/(2!2!3!) = 7!/(1!1!1!4!), so the number of terms in the expression for a(m) is P_7  1 = 15  1 = 14 (see below in the Formula section).
Let M_7 := [1, 7, 21, 35, 42, 105, 140, 210, 420, 630, 840, 1260, 2520, 5040] be the A070289(7) = 15  1 = 14 distinct multinomial coefficients corresponding to the 15 integer partitions of 7 in P_7. The characteristic equation of the recurrence for a(m) is f(x) := Product_{r in M_7} (xr) = Sum_{i = 0..14} (1)^{14i} * c_i * x^i. It turns out that c_{14} = 1, c_{13} = 11271, c_{12} = 46169368, c_{11} = 92088653622, and so on (see R. H. Hardin's recurrence below), and c_0 = 2372695722072874920960000000000 = product of elements in M_7.
It follows that a(m) satisfies the recurrence Sum_{i = 0..14} (1)^{14i} * c_i * a(mi) = 0, which is equivalent to R. H. Hardin's empirical recurrence below.
If we count the multinomial coefficient 210 twice in the characteristic equation (since it corresponds to two different integer partitions of 7) then we get (x210)*f(x) = Sum_{i = 0..15} (1)^{15i} * d_i * x^i, where (d_0, d_1, ..., d_15) is row k = 7 in irregular triangular array A309951. We have d_{15} = 1, d_{14} = 11481, ..., d_0 = 498266101635303733401600000000000 (see Alois P. Heinz's bfile for A309951 with entries 37 to 52). Note that d_0 = 210 * c_0.
We then have Sum_{s = 0..15} (1)^s * A309951(7, s) * a(ms) = 0 for m >= 16. The latter recurrence is of order 15, and it is not minimal (as opposed to the one below by R. H. Hardin, which is of order 14 and minimal).
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LINKS

Morton Abramson and David Promislow, Enumeration of arrays by column rises, J. Combinatorial Theory Ser. A 24(2) (1978), 247250; see Eq. (6) (with t=0), p. 248, and the comments above.


FORMULA

Empirical: a(n) = 11271*a(n1) 46169368*a(n2) +92088653622*a(n3) 100896701243149*a(n4) +64220064122517975*a(n5) 24283767237355832850*a(n6) +5479502670227877007500*a(n7) 734487423806273666445000*a(n8) +57519812656973505919500000*a(n9) 2547756421856270328438000000*a(n10) +60760702040873540340600000000*a(n11) 700874827794270417254400000000*a(n12) +3015300813467611878720000000000*a(n13) 2372695722072874920960000000000*a(n14). [It is correct; see the comments above and one of the formulas below.]
a(n) = 1  2*7^n  2*21^n  2*35^n + 3*42^n + 6*105^n + 3*140^n  210^n  12*420^n  4*630^n + 5*840^n + 10*1260^n  6*2520^n + 5040^n.  Petros Hadjicostas, Aug 25 2019
Sum_{s = 0..14} (1)^s * A325305(7, s) * a(ns) = 0 for n >= 15. (This is the same as R. H. Hardin's recurrence above, and it follows from Eq. (6), p. 248, in Abramson and Promislow (1978) with t=0.)  Petros Hadjicostas, Sep 06 2019


EXAMPLE

Some solutions for n=3
..0..3..4..1..5..2..6....0..3..4..1..5..2..6....0..3..4..1..5..2..6
..1..0..3..5..2..6..4....1..0..3..2..4..5..6....1..0..4..2..5..6..3
..5..2..1..0..6..3..4....4..6..5..1..0..3..2....2..4..0..6..3..5..1


MATHEMATICA

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, 7];


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



