A179680 The number of exponents >1 in a recursive reduction of 2n-1 until reaching an odd part equal to 1.

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

Offset: 1

Author

Vladimir Shevelev, Jul 24 2010

Keywords

• nonn

Comments

Let N = 2n-1. Then consider the following algorithm of updating pairs (v,m) indicating highest exponent of 2 (2-adic valuation) and odd part: Initialize at step 1 by v(1) = A007814(N+1) and m(1) = A000265(N+1). Iterate over steps i>=2: v(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 v(i) which are larger than 1.

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

A179382(n) = 1 + the number of iterations taken by the algorithm when starting from N = 2n-1. - Antti Karttunen, Oct 02 2017

Examples

			For n = 9, 2*n-1 = 17, we have v_1 = v_2 = v_3 = 1, v_4 = 5. Thus a(9) = 1.
For n = 10, 2*n-1 = 19, we have v_1 = 2, v_2 = 3, v_3 = v_4 = v_5 = 1, v_6 = v_7 = 2, v_8 = 1, v_9 = 5. Thus a(10) = 5.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..8192

Crossrefs

Cf. A179676, A179460, A007814, A002326, A179382, A292239, A292265, A292266.

Programs

Maple

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

Mathematica

a7814[n_] := IntegerExponent[n, 2];
a265[n_] := n/2^IntegerExponent[n, 2];
a[n_] := Module[{l, m, k, nn}, nn = 2n-1; k = 0; l = a7814[nn+1]; m = a265[nn+1]; If[l>1, k++]; While[m != 1, l = a7814[nn+m]; If[l>1, k++]; m = a265[nn+m]]; k];
Array[a, 80] (* Jean-François Alcover, Jul 30 2018, after R. J. Mathar *)

Sage

def A179680(n):
    s, m, N = 0, 1, 2*n - 1
    while True:
        k = N + m
        v = valuation(k, 2)
        if v > 1: s += 1
        m = k >> v
        if m == 1: break
    return s
print([A179680(n) for n in (1..80)]) # Peter Luschny, Oct 07 2017

Scheme

(define (A179680 n) (let ((x (+ n n -1))) (let loop ((s (- 1 (A000035 n))) (k 1)) (let ((m (A000265 (+ x k)))) (if (= 1 m) s (loop (+ s (if (> (A007814 (+ x m)) 1) 1 0)) m)))))) ;; Antti Karttunen, Oct 02 2017