OFFSET
1,2
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
LINKS
T. D. Noe, Table of n, a(n) for n=1..1000
T. Chinburg and M. Henriksen, Sums of k-th powers in the ring of polynomials with integer coefficients, Acta Arithmetica, 29 (1976), 227-250.
K. B. Stolarsky, Integers whose multiples have anomalous digital frequencies, Acta Arithmetica, 38 (1980), 117-128.
EXAMPLE
73 is included because it is 1001001 in binary, whose 1's are diluted by blocks of two 0's.
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
PROG
(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
CROSSREFS
KEYWORD
base,easy,nonn
AUTHOR
Marc LeBrun, Oct 11 2001
STATUS
approved