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%I #11 Mar 14 2016 03:32:34
%S 0,1,3,5,7,8,9,11,13,15,15,16,17,22,24,24,22,23,26,33,35,35,29,30,31,
%T 38,46,48,48,41,38,39,43,52,61,63,63,55,47,48,49,58,68,78,80,80,71,62,
%U 58,59,64,75,86,97,99,99,89,79,69,70,71,82,94,106,118,120,120,109,98,87
%N For each ordered partition of n with k numbers, remove 1 from each part and add the number k to get a new partition, until a partition is repeated. Among all ordered partitions of n, a(n) gives the maximum number of steps needed to reach a period.
%C If one plays with p(n,n) unordered partitions, one gets the same number and length of periods.
%C If one removes the first part z(1) of each partition and adds 1 to the next z(1) parts to get a new partition, until a partition is repeated, one gets the same length and number of periods, playing with 2^(n-1) ordered or p(n,n) unordered partitions (A185700, A092964, A037306)
%D R. Baumann, Computer-Knobelei, LOGIN, 4 (1987), pages ?.
%D H. R. Halder and W. Heise, Einführung in Kombinatorik, Hanser Verlag, Munich, 1976, pp. 75ff.
%F a((k^2+k-2)/2-j)=k^2-3-(k+1)*j with 0<=j<=(k-4) div 2 and 4<=k.
%F a((k^2+k+2)/2+j)=k^2-1-k*j with 0<=j<=(k-5) div 2 and 5<=k.
%F a((k^2+2*k-2+k mod 2)/2+j)=(k^2+4*k-2+k mod 2)/2+j with 0<=j<=2-k mod 2 and 4<=k.
%F a(T(k))=k^2-1 with 1<= k for all triangular numbers T(k).
%e For k=6: a(19)=26; a(20)=3; a(21)=35; a(22)=35; a(23)=29; a(24)=30; a(25)=31.
%e For n=4: (1+1+1+1)->(4)->(3+1)->(2+2)->(1+1+2)->(1+3)--> a(4)=5 steps.
%e For n=5: (1+1+1+1+1)->(5)->(4+1)->(3+2)->(2+1+2)->(1+1+3)->(2+3)->(1+2+2)--> a(5)=7 steps.
%Y Cf. A185700, A092964, A037306.
%K nonn
%O 1,3
%A _Paul Weisenhorn_, Mar 28 2011
%E Partially edited by _N. J. A. Sloane_, Apr 08 2011
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