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%I #23 Jun 13 2021 06:05:02
%S 1,2,3,4,7,8,10,15,16,31,32,36,42,63,64,127,128,136,170,255,256,292,
%T 511,512,528,682,1023,1024,2047,2048,2080,2184,2340,2730,4095,4096,
%U 8191,8192,8256,10922,16383,16384,16912,18724,32767,32768,32896,34952,43690,65535,65536,131071
%N Numbers of the form 2^(n-1)*(2^(n*m)-1)/(2^n-1), n >= 1, m >= 1.
%C In other words, numbers whose binary representation consists of one or more repeating blocks with only one 1 in each block.
%C Also, fixed points of the permutations A139706 and A139708.
%C Each a(n) is a term of A064896 multiplied by some power of 2. As such, this sequence must also be a subsequence of A125121.
%C Also the numbers that uniquely index a Haar graph (i.e., 5 and 6 are not in the sequence since H(5) is isomorphic to H(6)). - _Eric W. Weisstein_, Aug 19 2017
%C From _Gus Wiseman_, Apr 04 2020: (Start)
%C The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. This sequence lists all positive integers k such that the k-th composition in standard order is constant. For example, the sequence together with the corresponding constant compositions begins:
%C 0: () 136: (4,4)
%C 1: (1) 170: (2,2,2,2)
%C 2: (2) 255: (1,1,1,1,1,1,1,1)
%C 3: (1,1) 256: (9)
%C 4: (3) 292: (3,3,3)
%C 7: (1,1,1) 511: (1,1,1,1,1,1,1,1,1)
%C 8: (4) 512: (10)
%C 10: (2,2) 528: (5,5)
%C 15: (1,1,1,1) 682: (2,2,2,2,2)
%C 16: (5) 1023: (1,1,1,1,1,1,1,1,1,1)
%C 31: (1,1,1,1,1) 1024: (11)
%C 32: (6) 2047: (1,1,1,1,1,1,1,1,1,1,1)
%C 36: (3,3) 2048: (12)
%C 42: (2,2,2) 2080: (6,6)
%C 63: (1,1,1,1,1,1) 2184: (4,4,4)
%C 64: (7) 2340: (3,3,3,3)
%C 127: (1,1,1,1,1,1,1) 2730: (2,2,2,2,2,2)
%C 128: (8) 4095: (1,1,1,1,1,1,1,1,1,1,1,1)
%C (End)
%H Ivan Neretin, <a href="/A272919/b272919.txt">Table of n, a(n) for n = 1..1000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HaarGraph.html">Haar Graph</a>
%F From _Gus Wiseman_, Apr 04 2020: (Start)
%F A333381(a(n)) = A027750(n).
%F For n > 0, A124767(a(n)) = 1.
%F If n is a power of two, A333628(a(n)) = 0, otherwise = 1.
%F A333627(a(n)) is a power of 2.
%F (End)
%p N:= 10^6: # to get all terms <= N
%p R:= select(`<=`,{seq(seq(2^(n-1)*(2^(n*m)-1)/(2^n-1), m = 1 .. ilog2(2*N)/n), n = 1..ilog2(2*N))},N):
%p sort(convert(R,list)); # _Robert Israel_, May 10 2016
%t Flatten@Table[d = Reverse@Divisors[n]; 2^(d - 1)*(2^n - 1)/(2^d - 1), {n, 17}]
%Y Cf. A064896, A139708.
%Y Cf. A137706 (smallest number indexing a new Haar graph).
%Y Compositions in standard order are A066099.
%Y Strict compositions are ranked by A233564.
%Y Cf. A000120, A027750, A070939, A098504, A124767, A164894, A228351, A238279.
%K nonn
%O 1,2
%A _Ivan Neretin_, May 10 2016
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