cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A064896 Numbers of the form (2^(m*r)-1)/(2^r-1) for positive integers m, r.

Original entry on oeis.org

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
Offset: 1

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Author

Marc LeBrun, Oct 11 2001

Keywords

Comments

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

Examples

			73 is included because it is 1001001 in binary, whose 1's are diluted by blocks of two 0's.
		

Crossrefs

Cf. A076270 (k=3), A076275 (k=4), A076284 (k=5), A076285 (k=6), A076286 (k=7), A076287 (k=8), A076288 (k=9), A076289 (k=10).
Primes in this sequence: A245730.

Programs

  • Maple
    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
  • 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

Formula

A064894(a(n)) = A056538(n).