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A177922 For each composition (ordered partition) of n, remove the first part z(1) and add 1 to the next z(1) parts to get a new composition until a partition is repeated. Among all compositions of n, a(n) gives the maximum of steps needed to reach a period. 1
0, 2, 3, 5, 7, 9, 9, 11, 16, 18, 18, 18, 22, 28, 30, 30, 25, 29, 36, 43, 45, 45, 39, 39, 45, 53, 61, 63, 63, 56, 49, 55, 64, 73, 82, 84, 84, 76, 68, 68, 76, 86, 96, 106, 108, 99, 90, 81, 89, 100, 111, 122, 133, 135, 135, 125, 115, 105, 105, 115, 127, 139 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

n=4 has 2^3=8 compositions: 4; 3+1; 1+3; 1+1+2; 1+2+1: 2+1+1; 2+2; 1+1+1+1; the period is [(2+1+1),(2+2),(3+1)]; to reach the repetition from each composition one needs at most 5 steps; (1+3)->(4)->(1+1+1+1)->(2+1+1)->(2+2)->(3+1)->(2+1+1).

REFERENCES

R. Baumann, Computer-Knobelei, LOG IN, 163/164 (2010), 141-142.

LINKS

Table of n, a(n) for n=1..62.

FORMULA

for k>1: a[(k^2+k-2)/2-j]=(3k^2-3k-4)/2-(k+1)j with 0<=j<=(k-2) div 2; a[(k^2+k)/2]=(3k^2-3k)/2;

a[(k^2+k+2)/2]=(3k^2-3k)/2-kj with 0<=j<=(k-3) div 2;

a[2u^2+2u]=4u^2+u with 1<=u and k=2u

EXAMPLE

For k=10 and j=2 the formula gives; a[52]=111; a[55]=135; a[58]=115; a[60]=105;

For n=4: (4)->(1+1+1+1)->(2+1+1)->(2+2)->(3+1) [4 steps]; (3+1)->(2+1+1)->(2+2) [2 steps]; (1+3)->(4)->(1+1+1+1)->(2+1+1)-(2+2)->(3+1) [5 steps]; (1+1+2)->(2+2)->(3+1)->(2+1+1) [3 steps]; (1+2+1)->(3+1)->(2+1+1)->(2+2) [3 steps]; (2+1+1)->(2+2)->(3+1) [2 steps]; (2+2)->(3+1)->(2+1+1) [2 steps]; (1+1+1+1)->(2+1+1)->(2+2)->(3+1) [3 steps]; so at most 5 steps are needed, a(4)=5.

CROSSREFS

Sequence in context: A139790 A118784 A193889 * A102246 A089743 A080587

Adjacent sequences:  A177919 A177920 A177921 * A177923 A177924 A177925

KEYWORD

nonn

AUTHOR

Paul Weisenhorn, Dec 16 2010

STATUS

approved

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Last modified August 11 18:37 EDT 2020. Contains 336428 sequences. (Running on oeis4.)