A056792 Minimal number of steps to get from 0 to n by (a) adding 1 or (b) multiplying by 2.

0, 1, 2, 3, 3, 4, 4, 5, 4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8, 6, 7, 7, 8, 7, 8, 8, 9, 6, 7, 7, 8, 7, 8, 8, 9, 7, 8, 8, 9, 8, 9, 9, 10, 7, 8, 8, 9, 8, 9, 9, 10, 8, 9, 9, 10, 9, 10, 10, 11, 7, 8, 8, 9, 8, 9, 9, 10, 8, 9, 9, 10, 9, 10, 10, 11, 8, 9, 9, 10, 9, 10, 10, 11, 9, 10, 10, 11, 10, 11

Offset: 0

Author

N. J. A. Sloane, Sep 01 2000

Keywords

• nonn
• easy

Comments

A stopping problem: begin with n and at each stage if even divide by 2 or if odd subtract 1. That is, iterate A029578 while nonzero.

From Peter Kagey, Jul 16 2015: (Start)

The number of appearances of n in this sequence is identically A000045(n). Proof:

By application of the formula,

"a(0) = 0, a(2*n+1) = a(2*n) + 1 and a(2*n) = a(n) + 1",

it can be seen that:

{i: a(i) = n} = {2*i: a(i) = n-1, n>0} U {2*i+1: a(i) = n-2, n>1}.

Because the two sets on the left hand side share no elements:

|{i: a(i) = n}| = |{i: a(i) = n-1, n>0}| + |{i: a(i) = n-2, n>1}|

Notice that

|{i : a(i) = 1}| = |{1}| = 1 = A000045(1) and

|{i : a(i) = 2}| = |{2}| = 1 = A000045(2).

Therefore the number of appearances of n in this sequence is A000045(n). (End)

Examples

			12 = 1100 in binary, so a(12)=2+4-1=5.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
• Hugo Pfoertner, Addition chains
• Index to sequences related to the complexity of n

Crossrefs

Equals A056791 - 1. The least inverse (indices of record values) of A056792 is A052955 prepended with 0. See also A014701, A115954, A056796, A056817.

Cf. A000120, A070939, A007088: base 2 sequences.

Analogous sequences with a different multiplier k: A061282 (k=3), A260112 (k=4).

Programs

Haskell

c i = if i `mod` 2 == 0 then i `div` 2 else i - 1
b 0 foldCount = foldCount
b sheetCount foldCount = b (c sheetCount) (foldCount + 1)
a056792 n = b n 0 -- Peter Kagey, Sep 02 2015

Maple

a:= n-> (l-> nops(l)+add(i, i=l)-1)(convert(n, base, 2)):
seq(a(n), n=0..105);  # Alois P. Heinz, Jul 16 2015

Mathematica

f[ n_Integer ] := (c = 0; k = n; While[ k != 0, If[ EvenQ[ k ], k /= 2, k-- ]; c++ ]; c); Table[ f[ n ], {n, 0, 100} ]
f[n_] := Floor@ Log2@ n + DigitCount[n, 2, 1]; Array[f, 100] (* Robert G. Wilson v, Jul 31 2012 *)

PARI

a(n)=if(n<1,0,n-valuation(n!*sum(i=1,n,1/i),2))

PARI

a(n)=if(n<1,0,1+a(if(n%2,n-1,n/2)))

PARI

a(n)=if(n<1,0,n=binary(n);sum(i=1,#n,n[i])+#n-1) \\ Charles R Greathouse IV, Apr 11 2012

Formula

a(0) = 0, a(2*n+1) = a(2*n) + 1 and a(2*n) = a(n) + 1.

a(n) = n-valuation(A000254(n), 2) for n>0. - Benoit Cloitre, Mar 09 2004

a(n) = A000120(n) + A070939(n) - 1. - Michel Marcus, Jul 17 2015

a(n) = (weight of binary expansion of n) + (length of binary expansion of n) - 1.

Extensions

More terms from James Sellers, Sep 06 2000

More terms from David W. Wilson, Sep 07 2000