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A064896
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Numbers of the form (2^(m*r)-1)/(2^r-1) for positive integers m, r.
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17
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1, 3, 5, 7, 9, 15, 17, 21, 31, 33, 63, 65, 73, 85, 127, 129, 255, 257, 273, 341, 511, 513, 585, 1023, 1025, 1057, 1365, 2047, 2049, 4095, 4097, 4161, 4369, 4681, 5461, 8191, 8193, 16383, 16385, 16513, 21845, 32767, 32769, 33825, 37449, 65535, 65537
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OFFSET
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1,2
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COMMENTS
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Binary expansion of n consists of single 1's diluted by (possibly empty) equal-sized blocks of 0's.
According to Stolarsky's Theorem 2.1, all numbers in this sequence are sturdy numbers; this sequence is a subsequence of A125121. - T. D. Noe, Jul 21 2008
These are the numbers k > 0 for which k + 2^m = k*2^n + 1 has a solution m,n > 0. For k > 1, these are numbers k such that (k - 2^x)*2^y + 1 = k has a solution in positive integers x,y. In other words, (k - 1)/(k - 2^x) = 2^y for some x,y > 0. If t = (2^m - 1)/(2^n - 1) is a term of this sequence (i.e. if and only if n|m), then t' = t + 2^m = t*2^n + 1 is also a term. Primes in this sequence (A245730) include: all Mersenne primes (A000668), all Fermat primes (A019434), and other primes (73, 262657, 4432676798593, ...). - Thomas Ordowski, Feb 14 2024
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LINKS
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FORMULA
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EXAMPLE
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73 is included because it is 1001001 in binary, whose 1's are diluted by blocks of two 0's.
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MAPLE
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f := proc(p) local m, r, t1; t1 := {}; for m from 1 to 10 do for r from 1 to 10 do t1 := {op(t1), (p^(m*r)-1)/(p^r-1)}; od: od: sort(convert(t1, list)); end; f(2); # very crude!
# Alternative:
N:= 10^6: # to get all terms <= N
A:= sort(convert({1, seq(seq((2^(m*r)-1)/(2^r-1), m=2..1/r*ilog2(N*(2^r-1)+1)), r=1..ilog2(N-1))}, list)); # Robert Israel, Jun 12 2015
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PROG
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(PARI) lista(nn) = {v = [1]; x = (2^nn-1); for (m=2, nn, r = 1; while ((y = (2^(m*r)-1)/(2^r-1)) <=x, v = Set(concat(v, y)); r++); ); v; } \\ Michel Marcus, Jun 12 2015
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CROSSREFS
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KEYWORD
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base,easy,nonn
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AUTHOR
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STATUS
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approved
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