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Coefficients in g.f. for certain marked mesh patterns.
4

%I #88 Sep 05 2024 08:07:59

%S 1,4,19,107,702,5274,44712,422568,4407120,50292720,623471040,

%T 8344624320,119938250880,1842662908800,30136443724800,522780938265600,

%U 9587900602828800,185371298306611200,3768248516336640000,80349669847157760000,1793238207723325440000,41806479141525288960000

%N Coefficients in g.f. for certain marked mesh patterns.

%C See Kitaev and Remmel for precise definition.

%C The listed terms a(3)-a(10) of this sequence can be produced by the formula (n-1)!*(H(n-1)-1/2)/2, where H(n) = A001008(n)/A002805(n) is the n-th harmonic number. - _Gary Detlefs_, May 28 2012

%C a(n) is also the number of nonzero elements left in the matrix where all the rows consist of permutations of 11...n after we delete for each element with the value of 'k' k elements of this type, and repeat this operation until no more elements with the value of k can be deleted. The whole operation should be done for all the values of k from 1 to n. - _Anton Zakharov_, Jun 28 2016

%H Joerg Arndt, <a href="/A182541/b182541.txt">Table of n, a(n) for n = 3..101</a>

%H Sergey Kitaev and Jeffrey Remmel, <a href="http://arxiv.org/abs/1201.1323">Simple marked mesh patterns</a>, arXiv preprint arXiv:1201.1323 [math.CO], 2012.

%H Anton Zakharov, <a href="/A182541/a182541_2.pdf">Matrix-related sequences</a>

%F a(n) = A001710(n+1) * (1 + Sum_{k=2..n} 1/(k+1) ). - _Anton Zakharov_, Jun 28 2016

%F a(n) ~ sqrt(Pi/2)*exp(-n)*n^(n-1/2)*log(n). - _Ilya Gutkovskiy_, Jul 12 2016

%F From _Pedro Caceres_, Apr 19 2019: (Start)

%F a(n) = (n-3)! * Sum_{i=1..n-2} (Sum_{j=1..i} (i/j)).

%F a(n) = (1/4) * (n-1)! * (2*harmonic(n-1)-1). (End)

%F a(n) = (-(n-1)! + 2 * |Stirling1(n,2)|)/4. - _Seiichi Manyama_, Sep 05 2024

%e a(1) corresponds to the 1 X 2 matrix 11 -> 1 element is left and there are no more ones to delete => n(1) = 1. a(2) corresponds to the 3 X 3 matrix 112 121 211 -> 102 120 210 -> 102 100 010 only 4 nonzero elements are left and a(2) = 4 = 3 + 3/3. a(3) = 12 + 12/3 + 12/4 = 19 = 19 nonzero elements left in the 4 X 12 matrix after the deletion for each element with the value of 1 one element with the value of 1, for every element with the value of 2 - two elements with the value of 2 and for each element with the value of 3 - three elements with the value of 3). - _Anton Zakharov_, Jun 28 2016

%t Table[Numerator[(n+1)!/2] *(1 + Sum[1/(k+1), {k, 2, n}]),{n, 1, 22}] (* _Indranil Ghosh_, Mar 12 2017 *)

%o (PARI) for(n=1, 22, print1(numerator((n + 1)!/2) * (1 + sum(k=2, n, 1/(k+1))),", ")) \\ _Indranil Ghosh_, Mar 12 2017

%Y Cf. A001710, A307642.

%K nonn

%O 3,2

%A _N. J. A. Sloane_, May 04 2012

%E More terms from _Anton Zakharov_, Jun 28 2016