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.

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.

Links

Table of n, a(n) for n=1..70.

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

Cf. A185700, A092964, A037306.

Sequence in context: A300737 A062958 A295075 * A153309 A047486 A229838

Adjacent sequences: A184993 A184994 A184995 * A184997 A184998 A184999

Keyword

nonn

Author

Paul Weisenhorn, Mar 28 2011

Extensions

Partially edited by N. J. A. Sloane, Apr 08 2011

Status

approved