A057526 Number of applications of f to reduce n to 1, where f(k) is the integer among k/2,(k-1)/4, (k+1)/4.

0, 1, 1, 2, 1, 2, 2, 3, 2, 2, 2, 3, 2, 3, 3, 4, 3, 3, 2, 3, 2, 3, 3, 4, 3, 3, 3, 4, 3, 4, 4, 5, 4, 4, 3, 4, 3, 3, 3, 4, 3, 3, 3, 4, 3, 4, 4, 5, 4, 4, 3, 4, 3, 4, 4, 5, 4, 4, 4, 5, 4, 5, 5, 6, 5, 5, 4, 5, 4, 4, 4, 5, 4, 4, 3, 4, 3, 4, 4, 5, 4, 4, 3, 4, 3, 4, 4, 5, 4, 4, 4

Offset: 1

Author

Clark Kimberling, Sep 03 2000

Keywords

• nonn

Comments

Also the number of zeros in the symmetric signed digit expansion of n with q=2 (i.e. the representation of n in the (-1,0,1)2 number system). - _Ralf Stephan, Jun 30 2003

Also the degree of the Stern polynomial B[n,t]. The Stern polynomials B[n,t] are defined by B[0,t]=0, B[1,t]=1, B[2n,t]=tB[n,t], B[2n+1,t]=B[n+1,t]+B[n,t] for n>=1 (see S. Klavzar et al. and A125184). - Emeric Deutsch, Dec 04 2006

In this sequence, n occurs exactly 3^n times. - T. D. Noe, Mar 01 2011

Examples

			a(34)=4, which counts these reductions: 34->17->4->2->1.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• C. Heuberger and H. Prodinger, On minimal expansions in redundant number systems: Algorithms and quantitative analysis, Computing 66(2001), 377-393.
• S. Klavzar, U. Milutinovic and C. Petr, Stern polynomials, Adv. Appl. Math. 39 (2007) 86-95.
• G. Manku and J. Sawada, A loopless Gray code for minimal signed-binary representations, 13th Annual European Symposium on Algorithms (ESA), LNCS 3669 (2005), 438-447.
• Maciej Ulas and Oliwia Ulas, On certain arithmetic properties of Stern polynomials, arXiv:1102.5109 [math.CO], 2011. See p. 10.

Crossrefs

Cf. A125184.

One less than A277329.

Programs

Maple

a:=proc(n) if n=1 then 0 elif n mod 2=0 then a(n/2)+1 elif n mod 4=1 then a((n-1)/2) else a((n+1)/2) fi end: seq(a(n),n=2..91); # Emeric Deutsch, Dec 04 2006

Mathematica

a[n_] := a[n] = Which[n == 1, 0, Mod[n, 2] == 0, a[n/2] + 1, Mod[n, 4] == 1, a[(n-1)/2], True, a[(n+1)/2]];
Array[a, 100] (* Jean-François Alcover, Apr 20 2020 *)

PARI

ep(r, n)=local(t=n/2^(r+2)); floor(t+5/6)-floor(t+4/6)-floor(t+2/6)+floor(t+1/6);
a(n)=sum(r=0, log(3*n)\log(2)-1, !ep(r, n)) ;
(Scheme, with memoization-macro definec)
(definec (A057526 n) (cond ((= 1 n) 0) ((even? n) (+ 1 (A057526 (/ n 2)))) ((= 3 (modulo n 4)) (+ 1 (A057526 (/ (+ 1 n) 4)))) (else (+ 1 (A057526 (/ (+ -1 n) 4))))))
;; Antti Karttunen, Oct 27 2016 after the first recurrence of Klavzar et al. as given by Emeric Deutsch in the Formula section.

Formula

a(1)=0, a(2n)=a(n)+1, a(4n-1)=a(n)+1, a(4n+1)=a(n)+1 for n>=1 (Klavzar et al. Proposition 12). - Emeric Deutsch, Dec 04 2006

a(1)=0, a(2n)=a(n)+1, a(4n+1)=a(2n), a(4n+3)=a(2n+2) for n>=1 (Klavzar et al. Corollary 13). - Emeric Deutsch, Dec 04 2006

a(n) = A277329(n)-1. - Antti Karttunen, Oct 27 2016

Extensions

0 prepended by T. D. Noe, Feb 28 2011