A336246 Array read by upwards antidiagonals: T(n,k) is the number of ways to place n persons on different seats such that each person number p, 1 <= p <= n, differs from the seat number s(p), 1 <= s(p) <= n+k, k >= 0.
0, 1, 1, 2, 3, 2, 9, 11, 7, 3, 44, 53, 32, 13, 4, 265, 309, 181, 71, 21, 5, 1854, 2119, 1214, 465, 134, 31, 6, 14833, 16687, 9403, 3539, 1001, 227, 43, 7, 133496, 148329, 82508, 30637, 8544, 1909, 356, 57, 8, 1334961, 1468457, 808393, 296967, 81901, 18089, 3333, 527, 73, 9
Offset: 1
Examples
For k=1, the n-tuples of seat numbers are: - for n=1: 2 => T(1,1) = 1. - for n=2: 21, 23, 31 => T(2,1) = 3, 21: person 1 sits on seat 2 and vice versa. A counterexample is 13 because person 1 would sit on seat 1. - for n=3: 214,231,234,241,312,314,341,342,412,431,432 => T(3,1) = 11. Array begins: 0 1 2 3 4 ... 1 3 7 13 21 ... 2 11 32 71 134 ... 9 53 181 465 1001 ... 44 309 1214 3539 8544 ... .. ... .... .... ....
Links
- Gerhard Kirchner, Fixed-point-free variations
Crossrefs
Programs
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Maxima
block(nr: 0, k: -1, mmax: 55, /*First mmax terms are returned, recurrence used*/ a: makelist(0, n, 1, mmax), while nr
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Maxima
block(n: 1, k: 0, mmax: 55, /*First mmax terms are returned, explicit formula used*/ a: makelist(0, n, 1, mmax), for m from 1 thru mmax do (su: 0, for r from 0 thru n do su: su+(-1)^r*binomial(n,r)*(n+k-r)!/k!, a[m]: su, if n=1 then (n: k+2, k: 0) else (n: n-1, k: k+1)), return(a));
Formula
T(n,k) = (n+k-1)*T(n-1,k) + (n-1)*T(n-2,k) for n >= 2, k >= 0 with T(0,k)=1 and T(1,k)=k.
For n=0, there is one empty variation. T(0,k) is used for the recurrence only, not in the table. For n=1, the person can be placed on seat number 2..k+1 (if k > 0).
You also find the recurrence in the formula section of A000166 (k=0) and in the name section of the other sequences listed above (1 <= k <= 10). Some sequences have a different offset.
T(n,k) = Sum_{r=0..n} (-1)^r*binomial(n,r)*(n+k-r)!/k!.
Proofs see link.
Comments