cp's OEIS Frontend

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.

%I A177922 #21 Jan 19 2023 09:35:24
%S A177922 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,
%T A177922 53,61,63,63,56,49,55,64,73,82,84,84,76,68,68,76,86,96,106,108,99,90,
%U A177922 81,89,100,111,122,133,135,135,125,115,105,105,115,127,139
%N 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.
%C A177922 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).
%H A177922 R. Baumann, <a href="https://www.yumpu.com/de/document/read/1860388/nr-163-164">Computer-Knobelei, </a>, LOGIN, 163/164 (2010), 141-142 (in German).
%F A177922 a((k^2+k-2)/2-j) = (3k^2-3k-4)/2-(k+1)*j with 0<=j<=(k-2) div 2, for k>1.
%F A177922 a((k^2+k)/2) = (3k^2-3k)/2, for k>1.
%F A177922 a((k^2+k+2)/2) = (3k^2-3k)/2-k*j with 0<=j<=(k-3) div 2, for k>1.
%F A177922 a(2u^2+2u) = 4u^2+u with 1<=u and k=2u.
%e A177922 For k=10 and j=2 the formula gives; a(52)=111; a(55)=135; a(58)=115; a(60)=105;
%e A177922 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.
%K A177922 nonn
%O A177922 1,2
%A A177922 _Paul Weisenhorn_, Dec 16 2010