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%I #21 Jun 07 2022 11:15:23
%S 0,0,1,1,2,3,3,4,4,6,5,7,6,9,7,10,8,12,9,13,10,15,11,16,12,18,13,19,
%T 14,21,15,22,16,24,17,25,18,27,19,28,20,30,21,31,22,33,23,34,24,36,25,
%U 37,26,39,27,40,28,42,29,43,30,45,31,46,32,48,33,49,34,51
%N Number of strict partitions of n with Cookie Monster number 2.
%C Given a set of integers representing the number of cookies in jars, The Cookie Monster number is the minimum number of moves Cookie Monster must use to empty the jars when in one move he may choose any subset of jars and take the same number of cookies from each of those jars.
%C Partitions have Cookie Monster number 2 if either they have two distinct values, or they have three distinct values, where the largest value is the sum of the other two. These are the partitions of n into distinct numbers with Cookie Monster number 2.
%C Three distinct values are only possible when n is even, in which case the largest value will be n/2. The number of strict partitions of n into two parts is just floor((n-1)/2). - _Andrew Howroyd_, Apr 29 2020
%H Andrew Howroyd, <a href="/A226107/b226107.txt">Table of n, a(n) for n = 1..1000</a>
%H L. M. Braswell and T. Khovanova, <a href="http://arxiv.org/abs/1305.4305">Cookie Monster Devours Naccis</a> arXiv:1305.4305 [math.HO], 2013.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,1,0,1,0,-1).
%F From _Colin Barker_, Apr 10 2020: (Start)
%F G.f.: x^3*(1 + x + x^2 + 2*x^3) / ((1 - x)^2*(1 + x)^2*(1 + x^2)).
%F a(n) = a(n-2) + a(n-4) - a(n-6) for n>6.
%F (End)
%F a(2*n+1) = n; a(2*n) = n - 1 + floor((n-1)/2). - _Andrew Howroyd_, Apr 29 2020
%e If there are 7 cookies, the total number of partitions is 15. 5 partitions of 7 are strict: (7), (1,6), (2,5), (3,4), (1,2,4). One of these partitions, (7), corresponds to Cookie Monster number 1 (it has one value). One of these partitions, (1,2,4), has Cookie Monster number 3 (it has three values and the largest is not the sum of the other two). The remaining 3 partitions, (1,6), (2,5) and (3,4) have Cookie Monster number 2, so a(7)= 3.
%t Table[Length[
%t Select[Union[Map[Union, IntegerPartitions[n]]],
%t Total[#] ==
%t n && (Length[#] ==
%t 2 || (Length[#] == 3 && #[[3]] == #[[1]] + #[[2]])) &]], {n, 50}]
%t LinearRecurrence[{0,1,0,1,0,-1},{0,0,1,1,2,3},70] (* _Harvey P. Dale_, Jun 07 2022 *)
%o (PARI) a(n) = {(n-1)\2 + if(n%2==0, (n/2-1)\2)} \\ _Andrew Howroyd_, Apr 29 2020
%o (PARI) concat([0,0], Vec(x^3*(1 + x + x^2 + 2*x^3) / ((1 - x)^2*(1 + x)^2*(1 + x^2)) + O(x^40))) \\ _Colin Barker_, Apr 29 2020
%Y Cf. A000041, A226084, A000009.
%K nonn,easy
%O 1,5
%A _Leigh Marie Braswell_ and _Tanya Khovanova_, May 26 2013
%E Terms a(51) and beyond from _Andrew Howroyd_, Apr 29 2020
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