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

Author

Paul Weisenhorn, Mar 28 2011

Keywords

• nonn

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)

Examples

			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.
		

References

• R. Baumann, Computer-Knobelei, LOGIN, 4 (1987), pages ?.
• H. R. Halder and W. Heise, Einführung in Kombinatorik, Hanser Verlag, Munich, 1976, pp. 75ff.

Crossrefs

Cf. A185700, A092964, A037306.

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).

Extensions

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