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A060802 To weigh from 1 to n, make the heaviest weight as small as possible, under the condition of using fewest pieces of different, single weights; a(n) = weight of the heaviest weight.

%I A060802 #38 Mar 13 2015 14:09:01
%S A060802 1,2,2,3,3,3,4,4,4,4,5,5,6,7,8,6,6,6,7,7,8,8,9,9,10,11,12,13,14,15,16,
%T A060802 10,10,10,11,11,12,12,13,13,14,14,15,15,16,16,17,17,18,19,20,21,22,23,
%U A060802 24,25,26,27,28,29,30,31,32,18,18,18,19,19,20,20,21
%N A060802 To weigh from 1 to n, make the heaviest weight as small as possible, under the condition of using fewest pieces of different, single weights; a(n) = weight of the heaviest weight.
%C A060802 Starting at n = 2^x (x > 2) you get: 3 entries of 2^floor(log_2(x)-2)+2, then 2 entries of each subsequent integer until you reach the halfway point between 2^x and 2^(x+1), then 1 entry of each subsequent integer until you reach 2^(x+1)-1. Proved (see link). - _David Consiglio, Jr._, Jan 08 2015
%H A060802 David Consiglio, Jr., <a href="/A060802/b060802.txt">Table of n, a(n) for n = 1..1024</a>
%H A060802 David Consiglio, Jr., <a href="/A060802/a060802_1.py.txt">Python that quickly provides weights given the proof by Schoenfield</a>
%H A060802 David Consiglio, Jr., <a href="/A060802/a060802_3.rtf">Proof for sequence generation - Schoenfield and Consiglio</a>
%F A060802 After the 8th term:
%F A060802 If 2^x <= n <= (2^x)+2 then a(n) = 2 ^ floor(base2log(x)-2)+2 (see A052548)
%F A060802 If (2^x)+2 < n and n+1 < (2^x + 2^x+1)/2 then a(n) and a(n+1) = a(n-1)+1
%F A060802 If (2^x+2^x+1)/2 <= n then a(n) = a(n-1)+1. - _David Consiglio, Jr._, Jan 08 2015
%e A060802 a(20)=7 because every number from 1 to 20 can be obtained from {1,2,4,6,7}.
%K A060802 nonn
%O A060802 1,2
%A A060802 _Sen-Peng Eu_, Apr 27 2001
%E A060802 a(32)-a(1024) from _David Consiglio, Jr._, Jan 08 2015