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A184996
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.
1
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, 38, 46, 48, 48, 41, 38, 39, 43, 52, 61, 63, 63, 55, 47, 48, 49, 58, 68, 78, 80, 80, 71, 62, 58, 59, 64, 75, 86, 97, 99, 99, 89, 79, 69, 70, 71, 82, 94, 106, 118, 120, 120, 109, 98, 87
OFFSET
1,3
COMMENTS
If one plays with p(n,n) unordered partitions, one gets the same number and length of periods.
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)
REFERENCES
R. Baumann, Computer-Knobelei, LOGIN, 4 (1987), pages ?.
H. R. Halder and W. Heise, Einführung in Kombinatorik, Hanser Verlag, Munich, 1976, pp. 75ff.
FORMULA
a((k^2+k-2)/2-j)=k^2-3-(k+1)*j with 0<=j<=(k-4) div 2 and 4<=k.
a((k^2+k+2)/2+j)=k^2-1-k*j with 0<=j<=(k-5) div 2 and 5<=k.
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.
a(T(k))=k^2-1 with 1<= k for all triangular numbers T(k).
EXAMPLE
For k=6: a(19)=26; a(20)=3; a(21)=35; a(22)=35; a(23)=29; a(24)=30; a(25)=31.
For n=4: (1+1+1+1)->(4)->(3+1)->(2+2)->(1+1+2)->(1+3)--> a(4)=5 steps.
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.
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul Weisenhorn, Mar 28 2011
EXTENSIONS
Partially edited by N. J. A. Sloane, Apr 08 2011
STATUS
approved