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A037227
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If n = 2^m*k, k odd, then a(n) = 2*m+1.
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14
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1, 3, 1, 5, 1, 3, 1, 7, 1, 3, 1, 5, 1, 3, 1, 9, 1, 3, 1, 5, 1, 3, 1, 7, 1, 3, 1, 5, 1, 3, 1, 11, 1, 3, 1, 5, 1, 3, 1, 7, 1, 3, 1, 5, 1, 3, 1, 9, 1, 3, 1, 5, 1, 3, 1, 7, 1, 3, 1, 5, 1, 3, 1, 13, 1, 3, 1, 5, 1, 3, 1, 7, 1, 3, 1, 5, 1, 3, 1, 9, 1, 3, 1, 5, 1, 3, 1, 7, 1, 3, 1, 5, 1, 3, 1, 11, 1, 3, 1, 5, 1, 3
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OFFSET
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1,2
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COMMENTS
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Take the number of rightmost zeros in the binary expansion of n, double it, and increment it by 1. - Ralf Stephan, Aug 22 2013
Gives the maximum possible number of n X n complex Hermitian matrices with the property that all of their nonzero real linear combinations are nonsingular (see Adams et al. reference). - Nathaniel Johnston, Dec 11 2013
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LINKS
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FORMULA
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a(n) = Sum_{d divides n} (-1)^(d+1)*mu(d)*tau(n/d). Multiplicative with a(p^e) = 2*e+1 if p = 2; 1 if p > 2. - Vladeta Jovovic, Apr 27 2003
a(n*2^(k+1) + 2^k) = 2*k + 1 for n,k >= 0; thus a(2*n+1) = 1, a(4*n+2) = 3, a(8*n+4) = 5, a(16*n+8) = 7 and so on. Note the square array ( n*2^(k+1) + 2^k - 1 )n, k>=0 is the transpose of A075300.
G.f.: Sum_{n >= 0} (2*n + 1)*x^(2^n)/(1 - x^(2^(n+1))). (End)
Dirichlet g.f.: zeta(s)*(2^s+1)/(2^s-1).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 3. (End)
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MAPLE
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nmax:=102: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p):= 2*p+1: od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Feb 07 2013
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MATHEMATICA
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f[n_]:=Module[{z=Last[Split[IntegerDigits[n, 2]]]}, If[Union[z]={0}, 2* Length[ z]+1, 1]]; Array[f, 110] (* Harvey P. Dale, Jun 16 2019, after Ralf Stephan *)
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PROG
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(Haskell)
(R)
maxrow <- 6 # by choice
a <- 1
for(m in 0:maxrow){
for(k in 0:(2^m-1)) {
a[2^(m+1) +k] <- a[2^m+k]
a[2^(m+1)+2^m+k] <- a[2^m+k]
}
a[2^(m+1) ] <- a[2^(m+1)] + 2
}
a
(Python)
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CROSSREFS
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KEYWORD
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nonn,easy,nice,mult
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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