A065120 Highest power of 2 dividing A057335(n).

0, 1, 2, 1, 3, 2, 1, 1, 4, 3, 2, 2, 1, 1, 1, 1, 5, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 6, 5, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 6, 5, 5, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

Author

Alford Arnold, Nov 12 2001

Keywords

• easy
• nonn

Comments

a(n) appears on row 1 of the array illustrated in A066099.

Except for initial zero, ordinal transform of A062050. After initial zero, n-th chunk consists of n, one n-1, two (n-2)'s, ..., 2^(k-1) (n-k)'s, ..., 2^(n-1) 1's. - Franklin T. Adams-Watters, Sep 11 2006

Zero together with a triangle read by rows in which row j lists the first 2^(j-1) terms of A001511 in nonincreasing order, j >= 1, see example. Also row j lists the first parts, in nonincreasing order, of the compositions of j. - Omar E. Pol, Sep 11 2013

The n-th row represents the frequency distribution of 1, 2, 3, ..., 2^(n-1) in the first 2^n - 1 terms of A003602. - Gary W. Adamson, Jun 10 2021

Examples

			A057335(7)= 30 and 30 = 2*3*5 so a(7) = 1; A057335(9)= 24 and 24 = 8*3 so a(9) = 3
From _Omar E. Pol_, Aug 30 2013: (Start)
Written as an irregular triangle with row lengths A011782:
  0;
  1;
  2,1;
  3,2,1,1;
  4,3,2,2,1,1,1,1;
  5,4,3,3,2,2,2,2,1,1,1,1,1,1,1,1;
  6,5,4,4,3,3,3,3,2,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
  ...
Column 1 is A001477. Row sums give A000225. Row lengths is A011782.
(End)
		

Crossrefs

Cf. A066099, A062050, A011782.

Cf. A003602.

Programs

Mathematica

nmax = 105;
A062050 = Flatten[Table[Range[2^n], {n, 0, Log[2, nmax] // Ceiling}]];
Module[{b}, b[_] = 0;
a[n_] := If[n == 0, 0, With[{t = A062050[[n]]}, b[t] = b[t] + 1]]];
a /@ Range[0, nmax] (* Jean-François Alcover, Jan 12 2022 *)

PARI

lista(nn) = {my(v = vector(nn)); v[1] = 1; for (i=2, nn, v[i] = mg(i-1)*v[(i+1)\2];); for (i=1, nn, print1(valuation(v[i], 2), ", "););} \\ Michel Marcus, Feb 09 2014

PARI

my(L(n)=if(n,logint(n,2),-1)); a(n) = my(p=L(n)); p - L(n-1<Kevin Ryde, Aug 06 2021

Formula

From Daniel Starodubtsev, Aug 05 2021: (Start)

a(n) = A001511(A059894(n) - 2^A000523(n) + 1) for n > 0 with a(0) = 0.

a(2n+1) = a(n), a(2n) = a(n) + A036987(n-1) for n > 1 with a(0) = 0, a(1) = 1. (End)

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 29 2003