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Michael De Vlieger, <a href="/A175020/b175020.txt">Table of n, a(n) for n = 1..14167</a> (Terms n <= 10^8).
__Leroy Quet__, , Nov 03 2009
__Leroy Quet_, _, Nov 03 2009
_Leroy Quet, _, Nov 03 2009
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Equivalently, numbers in whose binary representation the runs of 1's are in increasing order of length, the runs of 0's are in decreasing order of length, and all runs of 0's are at least as long as any run of 1's. The position of [1^m] in the partitions of m will be P(m-1). It is the last partition in the list with a part of size 1; anything with a part of size 2 or more will start 100... in the binary representation, while this partition starts 101...; and any partition that does not have a part of size 1 will start 11.... Removing one part of size 1 from the partitions of size m that have such a part gives each partition of m-1 uniquely. This relationship is expressed by the second formula of A002865. - _Franklin T. Adams-Watters, _, Nov 03 2009
Contribution from R. J. Mathar, Feb 27 2010: (Start)
for n from 1 to 300 do if isA175020(n) then printf("%d, ", n) ; end if; end do; (End)
# R. J. Mathar, Feb 27 2010
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