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A179680 The number of exponents >1 in a recursive reduction of 2n-1 until reaching an odd part equal to 1. 3
0, 1, 1, 1, 1, 3, 3, 1, 1, 5, 1, 3, 5, 5, 7, 1, 1, 3, 9, 3, 3, 3, 3, 6, 5, 2, 13, 5, 3, 15, 15, 1, 1, 17, 5, 9, 1, 5, 7, 10, 13, 21, 1, 7, 2, 3, 2, 9, 11, 9, 25, 13, 2, 27, 9, 9, 5, 11, 2, 6, 27, 5, 25, 1, 1, 33, 3, 9, 15, 35, 11, 15, 3, 11, 37, 3, 6, 5, 13, 13 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,6

COMMENTS

Let N=2n-1. Then consider the following algorithm of updating pairs (l,m) indicating highest exponents of 2 and odd part: Initialize at step 1 by l(1)= A007814(N+1) and m(1) = A000265(N+1). Iterate over steps i>=2: l(i) = A007814(N+m(i-1)), m(i)=A000265(N+m(i-1)) using the previous odd part m(i-1) until some m(k)=1. a(n) is defined as the count of the l(i) which are larger than 1.

This is an algorithm to compute A002326 because the sum l(1)+l(2)+ ... +l(k) of the exponents is A002326(n-1).

LINKS

Table of n, a(n) for n=1..80.

EXAMPLE

For n=9, 2*n-1=17, we have l_1 =l_2 =l_3 =1, l_4=5. Thus a(9)=1.

For n=10, 2*n-1=19, we have l_1=2, l_2=3, l_3 =l_4 =l_5 =1, l_6 =l_7 =2, l_8=1, l_9 = 5. Thus a(10)=5.

MAPLE

A007814 := proc(n) padic[ordp](n, 2); end proc:

A000265 := proc(n) n/2^A007814(n) ; end proc:

A179680 := proc(n) local l, m, a , N ; N := 2*n-1 ; a := 0 ; l := A007814(N+1) ; m := A000265(N+1) ; if l > 1 then a := a+1 ; end if; while m <> 1 do l := A007814(N+m) ; if l > 1 then a := a+1 ; end if; m := A000265(N+m) ; end do: a ; end proc:

seq(A179680(n), n=1..80) ; # R. J. Mathar, Apr 05 2011

CROSSREFS

Cf. A179676, A179460, A007814, A002326, A179382

Sequence in context: A133333 A171876 A133332 * A123562 A046218 A046221

Adjacent sequences:  A179677 A179678 A179679 * A179681 A179682 A179683

KEYWORD

nonn

AUTHOR

Vladimir Shevelev, Jul 24 2010

STATUS

approved

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Last modified March 28 11:45 EDT 2017. Contains 284186 sequences.