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A145888 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} in which k is the largest entry in the cycle containing 1 (1 <= k <= n). 1
1, 1, 1, 2, 1, 3, 6, 2, 4, 12, 24, 6, 10, 20, 60, 120, 24, 36, 60, 120, 360, 720, 120, 168, 252, 420, 840, 2520, 5040, 720, 960, 1344, 2016, 3360, 6720, 20160, 40320, 5040, 6480, 8640, 12096, 18144, 30240, 60480, 181440, 362880, 40320, 50400, 64800 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

1,4

COMMENTS

Row sums are the factorials (A000142).

T(n,n) = n!/2 = A001710(n) (n>=2).

Sum_{k=1..n} k*T(n,k) = A121586(n).

REFERENCES

Solution to Problem 1831 by J. W. Grossman. Mathematics Magazine, 83, No. 5, 2010, pp. 392-393.

LINKS

Table of n, a(n) for n=1..49.

J. W. Grossman, Solution to Problem 1831 proposed by Emeric Deutsch, Mathematics Magazine, 83, No. 5, 2010, pp. 392-393.

FORMULA

T(n,1)=(n-1)!; T(n,k)=n!/((n-k+1)(n-k+2)) for 2 <= k <= n.

EXAMPLE

T(4,3)=4 because we have (132)(4), (13)(24), (123)(4), (13)(2)(4).

Triangle starts:

    1;

    1,   1;

    2,   1,   3;

    6,   2,   4,  12;

   24,   6,  10,  20,  60;

  120,  24,  36,  60, 120, 360;

MAPLE

T:=proc(n, k) if k=1 then factorial(n-1) elif k <= n then factorial(n)/((n-k+1)*(n-k+2)) else 0 end if end proc: for n to 10 do seq(T(n, k), k=1..n) end do; # yields sequence in triangular form

CROSSREFS

Cf. A000142, A001710, A121586.

Sequence in context: A182928 A141476 A212360 * A213935 A106578 A238960

Adjacent sequences:  A145885 A145886 A145887 * A145889 A145890 A145891

KEYWORD

nonn,tabl

AUTHOR

Emeric Deutsch, Nov 10 2008

STATUS

approved

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Last modified March 22 01:06 EDT 2019. Contains 321406 sequences. (Running on oeis4.)