A058842 -id:A058842 - cp's OEIS frontend

A073536 Breaking indices for A058842 (i.e., n such that A058842(n) is not equal to 3*A058842 (n-1) ).

3, 9, 12, 15, 17, 27, 34, 39, 46, 49, 52, 54, 66, 70, 73, 81, 84, 90, 95, 102, 106, 110, 116, 119, 124, 132, 140, 143, 149, 153, 158, 161, 165, 171, 177, 180, 183, 186, 189, 194, 198, 209, 215, 221, 224, 226, 233, 235, 241, 244, 248, 251, 255, 259, 262, 272

Offset: 1

Benoit Cloitre, Aug 27 2002

Cf. A058842.

It seems that a(n) = 5*n +O(log(n)) and that a(n)=5*n for infinitely many values of n.

It appears that a(n)=A077468(n+1). - Benoit Cloitre, Jun 04 2004

A058840 From Renyi's "beta expansion of 1 in base 3/2": sequence gives y(0), y(1), ...

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0

Offset: 0

Claude Lenormand (claude.lenormand(AT)free.fr), Jan 05 2001

Let r be a real number strictly between 1 and 2, x any real number between 0 and 1; define y = (y(i)) by x(0) = x; x(i+1) = r*x(i)-1 if r*x(i)>1 and r*x(i) otherwise; y(i) = integer part of x(i+1): y = (y(i)) is an infinite word on the alphabet (0,1). Here we take r = 3/2 and x = 1.

Kempner considers a "canonical" expansion of a real number in a non-integer base using the greedy algorithm. The greedy algorithm takes the largest possible integer digit in the range 0 <= digit < base at each digit position from high to low. For base 3/2, Kempner gives the present sequence of digits, except instead a(1)=0, as an example canonical 2 = 10.01000001001... Kempner notes too that a(1) omitted and the rest shifted down is a base-3/2 non-canonical 1 = .101000001001.... (canonical would be 1 = 1.000...). - Kevin Ryde, Dec 06 2019

A. Renyi (1957), Representation for real numbers and their ergodic properties, Acta. Math. Acad. Sci. Hung., 8, 477-493.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Aubrey J. Kempner, Anormal Systems of Numeration, American Mathematical Monthly, volume 43, number 10, December 1936, pages 610-617.

Cf. A058841, A058842.

import Data.Ratio ((%), numerator, denominator)
a058840 n = a058840_list !! n
a058840_list = 1 : renyi' 1 where
   renyi' x = y : renyi' r  where
      (r, y) | q > 1     = (q - 1, 1)
             | otherwise = (q, 0)
      q = 3%2 * x
-- Reinhard Zumkeller, Jul 01 2011

r = 3/2; x = 1; a[0] = a[1] = 1;
For[n = 2, n<105, n++, x = If[r x > 1, r x - 1, r x]; a[n] = Floor[r x]];
Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Dec 21 2018, a solution I owe to Benoit Cloitre *)

a_vector(len) = my(v=vector(len),c=2,d=1); for(i=1,len, if(c>=d,c-=d;v[i]=1); c*=3;d*=2); v; \\ Kevin Ryde, Dec 06 2019

More terms from Larry Reeves (larryr(AT)acm.org), Feb 22 2001

A058841 From Renyi's "beta expansion of 1 in base 3/2": sequence gives lengths of runs of 0's in A058840.

Original entry on oeis.org

0, 1, 5, 2, 2, 1, 9, 6, 4, 6, 2, 2, 1, 11, 3, 2, 7, 2, 5, 4, 6, 3, 3, 5, 2, 4, 7, 7, 2, 5, 3, 4, 2, 3, 5, 5, 2, 2, 2, 2, 4, 3, 10, 5, 5, 2, 1, 6, 1, 5, 2, 3, 2, 3, 3, 2, 9, 6, 9, 6, 8, 2, 7, 5, 3, 2, 2, 4, 3, 1, 14, 9, 3, 6, 7, 3, 2, 2, 3, 4, 3, 2, 6, 4, 2

Offset: 0

Claude Lenormand (claude.lenormand(AT)free.fr), Jan 05 2001

A. Renyi (1957), Representation for real numbers and their ergodic properties, Acta. Math. Acad. Sci. Hung., 8, 477-493.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Cf. A058840, A058842.

import Data.List (group)
a058841 n = a058841_list !! n
a058841_list =
   0 : (map length $ filter ((== 0) . head) $ group a058840_list)
-- Reinhard Zumkeller, Jul 01 2011

nmax = 500; r = 3/2; x = 1; (* b = A058840 *) b[0] = b[1] = 1;
For[n=2, n <= nmax, n++, x = If[r x > 1, r x - 1, r x]; b[n] = Floor[r x]];
Join[{0}, Length /@ Select[Split[Table[b[n], {n, 0, nmax}]], #[[1]] == 0&]] (* Jean-François Alcover, Dec 21 2018, using Benoit Cloitre's code for A058840 *)

More terms from Larry Reeves (larryr(AT)acm.org), Feb 22 2001

Data corrected for n>33 by Reinhard Zumkeller, Jul 01 2011

A073533 Let x(1)=1, x(n+1) = (4/3)x(n) - floor((4/3)x(n)); then a(n)=x(n)*3^n.

Original entry on oeis.org

1, 4, 16, 64, 13, 52, 208, 832, 3328, 13312, 53248, 212992, 851968, 3407872, 13631488, 11479231, 45916924, 183667696, 734670784, 2938683136, 1294379341, 5177517364, 20710069456, 82840277824, 331361111296, 1325444445184

Offset: 1

Benoit Cloitre, Aug 27 2002

It seems that the sequence x(n) = a(n)/3^n which satisfies 0 infinity sum(k=1,n,x(k))/n = C < 0.38 < 1/2
It appears that a(n) = 13*4^(n-5) for n > 4. - Creighton Dement, Nov 04 2004
This is not true, as a(16) mod 13 = 10. - Reinhard Zumkeller, Jun 05 2015

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Cf. A058842.

import Data.Ratio (numerator, (%))
a073533 n = a073533_list !! (n-1)
a073533_list = f 1 3 1 where
   f n p3 x = numerator(y * fromIntegral p3) : f (n + 1) (p3 * 3) y
              where y = z - fromIntegral (floor z); z = 4%3 * x
-- Reinhard Zumkeller, Jun 05 2015

cp's OEIS Frontend

A073536 Breaking indices for A058842 (i.e., n such that A058842(n) is not equal to 3*A058842 (n-1) ).

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A058840 From Renyi's "beta expansion of 1 in base 3/2": sequence gives y(0), y(1), ...

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A058841 From Renyi's "beta expansion of 1 in base 3/2": sequence gives lengths of runs of 0's in A058840.

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A073533 Let x(1)=1, x(n+1) = (4/3)x(n) - floor((4/3)x(n)); then a(n)=x(n)*3^n.

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cp's OEIS Frontend

A073536 Breaking indices for A058842 (i.e., n such that A058842(n) is not equal to 3*A058842 (n-1) ).

Views

Author

Keywords

Crossrefs

Formula

A058840 From Renyi's "beta expansion of 1 in base 3/2": sequence gives y(0), y(1), ...

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

Extensions

A058841 From Renyi's "beta expansion of 1 in base 3/2": sequence gives lengths of runs of 0's in A058840.

Views

Author

Keywords

References

Links

Crossrefs

Programs

Haskell

Mathematica

Extensions

A073533 Let x(1)=1, x(n+1) = (4/3)*x(n) - floor((4/3)*x(n)); then a(n)=x(n)*3^n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

A073533 Let x(1)=1, x(n+1) = (4/3)x(n) - floor((4/3)x(n)); then a(n)=x(n)*3^n.