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A061313
Minimal number of steps to get from 1 to n by (a) subtracting 1 or (b) multiplying by 2.
7
0, 1, 3, 2, 5, 4, 4, 3, 7, 6, 6, 5, 6, 5, 5, 4, 9, 8, 8, 7, 8, 7, 7, 6, 8, 7, 7, 6, 7, 6, 6, 5, 11, 10, 10, 9, 10, 9, 9, 8, 10, 9, 9, 8, 9, 8, 8, 7, 10, 9, 9, 8, 9, 8, 8, 7, 9, 8, 8, 7, 8, 7, 7, 6, 13, 12, 12, 11, 12, 11, 11, 10, 12, 11, 11, 10, 11, 10, 10, 9, 12, 11, 11, 10, 11, 10, 10, 9, 11, 10
OFFSET
1,3
COMMENTS
Also number of steps to get from n to 1 by process of adding 1 if odd, or dividing by 2 if even.
It is straightforward to prove that the number n appears F(n) times in this sequence, where F(n) is the n-th Fibonacci number (A000045). - Gary Gordon, May 31 2019
Conjecture: a(n)+2 is the sum of the terms of the Hirzebruch (negative) continued fraction for the Stern-Brocot tree fraction A007305(n)/A007306(n). - Andrey Zabolotskiy, Apr 17 2020
FORMULA
a(2n) = a(n)+1; a(2n+1) = a(n+1)+2; a(1) = 0.
Is Sum_{k=1..n} a(k) asymptotic to C*n*log(n) where 3 > C > 2? - Benoit Cloitre, Aug 31 2002
G.f.: x/(1-x) * Sum_{k>=0} (x^2^k + x^2^(k+1)/(1+x^2^k)). - Ralf Stephan, Jun 14 2003
a(n) = A080791(n-1) + A029837(n). - Ralf Stephan, Jun 14 2003
a(n) = 2*A023416(n-1) + A000120(n-1) = A023416(A062880(n)) = A023416(A000695(n)) + 1. - Ralf Stephan, Jul 16 2003
a(n) = A119477(n) - 1. - Philippe Deléham, Nov 03 2008
EXAMPLE
a(2) = 1 since 2 = 1*2, a(3) = 3 since 3 = 1*2*2-1, a(11) = 6 since 11 = (1*2*2-1)*2*2-1.
MATHEMATICA
f[n_] := Block[{c = 0, m = n}, While[m != 1, If[ EvenQ[m], While[ EvenQ[m], m = m/2; c++ ], m++; c++ ]]; Return[c]]; Table[f[n], {n, 1, 100}]
PROG
(PARI) a(n)=if(n<2, 0, s=n; c=1; while((s+s%2)/(2-s%2)>1, s=(s+s%2)/(2-s%2); c++); c)
(PARI) xpcount(n) = { p = 1; for(x=1, n, p1 = x; ct=0; while(p1>1, if(p1%2==0, p1/=2; ct++, p1 = p1*p+1; ct++) ); print1(ct, ", ") ) }
(PARI) a(n) = if(n--, 2*(logint(n, 2)+1)) - hammingweight(n); \\ Kevin Ryde, Oct 21 2021
(Haskell)
a061313 n = fst $ until ((== 1) . snd) (\(u, v) -> (u + 1, f v)) (0, n)
where f n = if r == 0 then n' else n + 1 where (n', r) = divMod n 2
-- Reinhard Zumkeller, Sep 05 2015
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Henry Bottomley, Jun 06 2001
STATUS
approved