A188160 For an unordered partition of n with k parts, remove 1 from each part and append the number k to get a new partition until a partition is repeated. a(n) gives the maximum steps to reach a period considering all unordered partitions of n.

0, 1, 2, 4, 5, 6, 7, 8, 10, 12, 12, 12, 13, 18, 20, 20, 17, 18, 21, 28, 30, 30, 24, 24, 25, 32, 40, 42, 42, 35, 31, 32, 36, 45, 54, 56, 56, 48, 40, 40, 41, 50, 60, 70, 72, 72, 63, 54, 49, 50, 55, 66, 77, 88, 90, 90, 80, 70, 60, 60, 61

Offset

1,3

Comments

Alternatively, if one iteratively removes the largest part z(1) and adds 1 to the next z(1) parts to get a new partition until a partition recurs, one gets the same maximum number of steps to reach a period.

The two shuffling operations are isomorphic for unordered partitions.

The two operations have the same length and number of periods for ordered and unordered partitions.

The steps count the operations including any pre-periodic part up to the end of first period, that is, the number of distinct partitions without including the first return.

References

R. Baumann LOG IN, 4 (1987)

Halder, Heise Einführung in Kombinatorik, Hanser Verlag (1976) 75 ff.

Links

Table of n, a(n) for n=1..61.

Ethan Akin, Morton Davis, Bulgarian solitaire, Am. Math. Monthly 92 (4) (1985) 237-250

J. Brandt, Cycles of partitions, Proc. Am. Math. Soc. 85 (3) (1982) 483-486

Formula

a((k^2+k-2)/2-j) = k^2-k-2-(k+1)*j with 0<=j<=(k-4)/2 and 4<=k.

a((k^2+k+2)/2+j) = k^2-k-k*j with 0<=j<=(k-4)/2 and 4<=k,

a((k^2+2*k-(k mod 2))/2+j) = (k^2+2*k-(k mod 2))/2+j with 0 <= j <= 1 and 2 <= k.

a(T(k)) = 2*T(k-1) = k^2-k with 1 <= k for the triangular numbers T(k)=A000217(k).

Example

For k=6 and 0 <= j <= 1:
a(19)=21;  a(20)=28;  a(21)=30;  a(22)=30;  a(23)=24;  a(24)=24;  a(25)=25.
For n=4: (1+1+1+1)->(4)->(3+1)->(2+2)->(2+1+1)--> a(4)=4.
For n=5: (1+1+1+1+1)->(5)->(4+1)->(3+2)->(2+2+1)->(3+1+1)-->a(5)=5.

Maple

A188160 := proc(n)
        local k, j, T ;
        if n <= 2 then
                return n-1 ;
        end if;
        for k from 0 do
                T := k*(k+1) /2 ;
                if n = T and k >= 1 then
                        return k*(k-1) ;
                end if;
                if k>=4 then
                        j := T-1-n ;
                        if j>= 0 and j <= (k-4)/2 then
                                return k^2-k-2-(k+1)*j ;
                        end if;
                        j := n-T-1 ;
                        if j>= 0 and j <= (k-4)/2 then
                                return k^2-k-k*j ;
                        end if;
                end if;
                if k >= 2 then
                        j := n-(k^2+2*k-(k mod 2))/2 ;
                        if j>=0 and j <= 1 then
                                return (k^2+2*k-(k mod 2))/2+j
                        end if;
                end if;
        end do:
        return -1 ;
end proc: # R. J. Mathar, Apr 22 2011

Crossrefs

Cf. A185700, A177922, A184996, A037306.

Sequence in context: A039079 A260375 A361144 * A047571 A190237 A190225

Adjacent sequences: A188157 A188158 A188159 * A188161 A188162 A188163

Keyword

nonn

Author

Paul Weisenhorn, Mar 28 2011

Status

approved