A096259 -id:A096259 - cp's OEIS frontend

A137604 Consider the sequence: b(0) = n, and for k >= 1, b(k) = b(k-1)/2 if b(k-1) is even, otherwise b(k) = n - (b(k-1)+1)/2. Then a(n) = (Sum_{0<=k<=m} b(k)) - 1 for n > 1, where m is the smallest index such that b(m) = 1; a(1) = 1.

1, 2, 3, 6, 7, 15, 21, 14, 15, 45, 30, 66, 63, 61, 105, 30, 31, 102, 171, 114, 93, 63, 134, 276, 258, 88, 351, 270, 105, 435, 465, 62, 63, 561, 374, 630, 126, 374, 570, 780, 547, 861, 126, 602, 204, 246, 196, 846, 537, 361, 1275, 1326, 264, 1431, 483, 990, 315

Offset: 1

Yasutoshi Kohmoto, Apr 23 2008

nonn

			a(6) = 6 + 3 + 4 + 2 + 1 - 1 = 15.

Cf. A096259, A137605.

f[1] = 1; f[n_] := Block[{lst = {n}, a}, While[a = lst[[ -1]]; a != 1, If[EvenQ@ a, AppendTo[lst, a/2], AppendTo[lst, lst[[1]] - (a + 1)/2]]]; Plus @@ lst - 1]; Array[f, 58] (* Robert G. Wilson v, May 15 2008 *)

More terms from Robert G. Wilson v, May 15 2008

A137607 a(0)=361, a(n+1)=a(n)/2 if a(n) is even, else a(n+1)=a(0)-(a(n)+1)/2.

Original entry on oeis.org

361, 180, 90, 45, 338, 169, 276, 138, 69, 326, 163, 279, 221, 250, 125, 298, 149, 286, 143, 289, 216, 108, 54, 27, 347, 187, 267, 227, 247, 237, 242, 121, 300, 150, 75, 323, 199, 261, 230, 115, 303, 209, 256, 128, 64, 32, 16, 8, 4, 2, 1, 360, 180, 90, 45, 338, 169, 276, 138, 69, 326, 163, 279, 221, 250, 125, 298, 149, 286, 143, 289, 216, 108, 54, 27, 347, 187, 267, 227, 247, 237, 242, 121, 300, 150, 75, 323, 199, 261, 230, 115, 303, 209, 256, 128, 64, 32, 16, 8, 4, 2, 1, 360

Offset: 0

Yasutoshi Kohmoto, Apr 23 2008

For n>=1, periodic with period 51. - Vladeta Jovovic, Apr 28 2008

The rule is the following: b(0)=361. If b(n-1) is divisible by 2 then b(n) = b(n-1)/2. If b(n-1) is not divisible by two then b(n) = b(0)-(b(n-1)+1)/2. It is periodic.

Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).

Cf. A096259.

NestList[If[Mod[#,2]==0,#/2,361-(#+1)/2]&,361,110] (* Harvey P. Dale, Mar 31 2024 *)

A137607(n) = if( !n,361, for( i=0,n%51, n=if( i, if( n%2, 360-n\2, n\2), 360)); n) \\ M. F. Hasler, Apr 28 2008

Clarified & extended by Vladeta Jovovic and M. F. Hasler, Apr 28 2008

The initial term, 361, seems anomalous. This sequence violates the rule that entries in the OEIS should not depend on an arbitrary but large parameter. - N. J. A. Sloane, May 17 2008

A137606 Numbers m such that all numbers {1...m} appear in the sequence {b(0) = m, b(n+1) = b(n)/2 if even, m-(b(n)+1)/2 otherwise}.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 10, 12, 15, 19, 24, 27, 30, 31, 34, 36, 40, 42, 51, 52, 54, 66, 70, 75, 82, 84, 87, 90, 91, 96, 99, 100, 106, 114, 120, 132, 135, 136, 147, 156, 159, 174, 175, 180, 184, 187, 190, 192, 195, 210, 211, 222, 231, 232, 234, 240, 244, 246, 252, 255, 262

Offset: 1

Yasutoshi Kohmoto, Apr 23 2008

nonn

Lemma: A sequence {b(n)} defined as above with m>1 cannot have values outside [1,m]. (For m=1, b=(1,0,0,0....).)
Corollary: Such a sequence {b(n)} is periodic with period <= m (except maybe for some initial terms).
Lemma 2: For any m>1, b(1) = floor( m/2 ) and if b(n)=m-1, then b(n+1)= [ m/2 ].
Proposition: As soon as there is a term b(n)=2^k, the (b-)sequence continues b(n+1)=2^(k-1),...,b(n+k)=1, b(n+k+1)=m-1 and then starts over with b(n+k+2)=b(1).
Corollary 2: Numbers m=2^k, k>2 cannot appear in the present sequence.
Proposition: For any b(0)=m>1, sooner or later the value 1 is reached.
Generate a sequence b(n) by the following rule. If b(n-1) is divisible by 2 then b(n) = b(n-1)/2. If b(n-1) is not divisible by 2 then b(n) = b(0)-(b(n-1)+1)/2. When b(n)=1 it ends. Sequence gives all m such that all numbers k with 1<=k<=m-2 appear in b(n), b(0)=m.
Sequence contains 1 and numbers m>1 such that 2m-1 is prime and -2 or 2 is a primitive root modulo 2m-1. - Max Alekseyev, May 16 2008

			6->3->4->2->1. 1,2,3,4=6-2 appear in b(n), b(0)=6. So 6 is a term of A137606.

Robert G. Wilson v, Table of n, a(n) for n = 1..2354.

Cf. A137607, A096259, A105874, A001122.

f[n_] := Block[{lst = {n}, a}, While[a = Last@ lst; a != 1, AppendTo[lst, If[ EvenQ@ a, a/2, lst[[1]] - (a + 1)/2]]]; Length@ lst - 1]; t = Array[f, 262]; Select[ Range @ 262, t[[ # ]] == # - 2 &] (* Robert G. Wilson v *)

b137606(n)= n=[n]; for( i=1,n[1]-1, n=concat( n, if( n[i]%2, n[1]-(n[i]+1)/2, n[i]/2 )); n[i]>1 || break); n
A137606(Nmax) = for( n=1,Nmax, n==#b137606(n) && print1(n","))

forprime(p=3,10^3, if(znorder(Mod(-2,p))==p-1||znorder(Mod(2,p))==p-1, print1((p+1)/2,", ") )) \\ Max Alekseyev, May 16 2008

This sequence as a set is the union of { 1 }, { (A105874(n)+1)/2 } and { (A001122(n)+1)/2 }. - Max Alekseyev, May 16 2008

Edited & extended by M. F. Hasler, Apr 28 2008

cp's OEIS Frontend

A137604 Consider the sequence: b(0) = n, and for k >= 1, b(k) = b(k-1)/2 if b(k-1) is even, otherwise b(k) = n - (b(k-1)+1)/2. Then a(n) = (Sum_{0<=k<=m} b(k)) - 1 for n > 1, where m is the smallest index such that b(m) = 1; a(1) = 1.

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Extensions

A137607 a(0)=361, a(n+1)=a(n)/2 if a(n) is even, else a(n+1)=a(0)-(a(n)+1)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Extensions

A137606 Numbers m such that all numbers {1...m} appear in the sequence {b(0) = m, b(n+1) = b(n)/2 if even, m-(b(n)+1)/2 otherwise}.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

Extensions