

A056792


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


15



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
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OFFSET

0,3


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(2n+1) = a(2n) + 1 and a(2n) = a(n) + 1",
it can be seen that:
{i: a(i) = n} = {2*i: a(i) = n1, n>0} U {2*i+1: a(i)=n2, n>1}.
Because the two sets on the left hand side share no elements:
{i: a(i) = n} = {i: a(i) = n1, n>0} + {i: a(i) = n2, 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)


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


FORMULA

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


EXAMPLE

12 = 1100 in binary, so a(12)=2+41=5.


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 *)


PROG

(PARI) a(n)=if(n<1, 0, nvaluation(n!*sum(i=1, n, 1/i), 2))
(PARI) a(n)=if(n<1, 0, 1+a(if(n%2, n1, n/2)))
(PARI) a(n)=n=binary(n); sum(i=1, #n, n[i])+#n1 \\ Charles R Greathouse IV, Apr 11 2012
(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


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).
Sequence in context: A265690 A061339 A073933 * A292127 A227861 A294991
Adjacent sequences: A056789 A056790 A056791 * A056793 A056794 A056795


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane, Sep 01 2000


EXTENSIONS

More terms from James A. Sellers, Sep 06 2000
More terms from David W. Wilson, Sep 07 2000


STATUS

approved



