A122458 -id:A122458 - cp's OEIS frontend

A160268 Odd numbers k for which A006694( (k-1)/2 )< A006694( (A000265(3k+1)-1)/2 ) .

11, 23, 37, 41, 43, 59, 61, 79, 83, 97, 103, 107, 113, 121, 139, 143, 147, 149, 163, 167, 169, 171, 173, 177, 181, 183, 191, 193, 199, 201, 203, 227, 237, 243, 249, 251, 263, 271, 283, 287, 289, 293, 303, 313, 317, 321, 323, 347, 351, 353, 355, 359, 363, 367, 373, 379

Offset: 1

Vladimir Shevelev, May 07 2009

nonn

Conjecture: For every k in the sequence, the number k^2 is in the sequence as well.

Composite numbers in the sequence which are not perfect squares are 143, 147, 171, 183 etc. [R. J. Mathar, May 16 2009]

A006694, A160198, A122458, A159885, A159945, A160266, A160267.

Edited and extended by R. J. Mathar, May 16 2009

A160364 Let f be defined as in A159885 and f^k be the k-th iteration of f. Then a(n) is the least k for which either {A000120(f^k(2n+1)) < A000120(2n+1)}&{A006694((f^k(2n+1)-1)/2)<=A006694(n)} or {A000120(f^k(2n+1))<=A000120(2n+1)}&{A006694((f^k(2n+1)-1)/2) < A006694(n)}.

Original entry on oeis.org

2, 1, 1, 5, 3, 1, 1, 2, 5, 1, 2, 1, 1, 1, 1, 5, 2, 5, 3, 33, 3, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 5, 7, 1, 5, 10, 1, 1, 2, 5, 5, 1, 1

Offset: 1

Vladimir Shevelev, May 11 2009

Using induction, one can prove that the Collatz (3x+1)-conjecture follows from the finiteness of a(n) for every n.

			Beginning with n=1, we have f(2n+1)=f(3)=5. Here A000120(3)=A000120(5)=2 and A006694((3-1)/2)= A006694((5-1)/2)=1. None of values did not become less than. Therefore a(1)>1. Since f(5)=1 and A000120(1)=1 and A006694(0)=0, then a(2)=2.

A000120 A006694 A160267 A006694 A122458 A160266 A159885 A159945 A160198

A160558 a(n) is the ordinal number of series in which the value of A160348(n) is defined.

Original entry on oeis.org

0, 1, 1, 2, 3, 2, 2, 4, 2, 5, 6, 4, 7, 8, 5, 8, 9, 4, 10, 11, 8, 12, 13, 8, 12, 14, 4, 15, 16, 11, 8, 17, 12, 11, 18, 8, 19, 20, 14, 21, 22, 15, 20, 23, 11, 8, 24, 17, 25

Offset: 0

Vladimir Shevelev, May 19 2009

a(n)>=a((f(2n+1)-1)/2), where f is defined as in A159885. E.g., for n=4, we have a(4)=3>a(((9*3+1)/4-1)/2)=a(3)=2.

			Put a(0)=0. According to example to A160348, in the first series we find A160348(1) and A160348(2), therefore a(1)=a(2)=1. In the second series we find A160348(3), A160348(5), A160348(6) and A160348(8), thus a(3)=a(5)=a(6)=a(8)=2.

Cf A160348 A160322 A160266 A160267 A259667 A159945 A160198 A006519 A122458.

A317083 a(n) is the first term less than the initial 2n+1 in the reduced Collatz trajectory.

Original entry on oeis.org

1, 1, 1, 5, 7, 5, 5, 5, 13, 11, 1, 5, 19, 23, 11, 23, 25, 5, 7, 19, 31, 37, 17, 23, 37, 29, 5, 47, 43, 19, 23, 61, 49, 19, 13, 61, 55, 1, 29, 19, 61, 47, 1, 37, 67, 61, 35, 91, 73, 7, 19, 61, 79, 91, 41, 61, 85, 65, 11, 101, 91, 59, 47, 77, 97, 37, 25, 43, 103

Offset: 0

Michel Lagneau, Jul 21 2018

nonn

			a(3)= 5 because, starting with 7, the iteration produces 11,17,13,5 and 5 is the first term less than 7.

Cf. A122458, A256598.

f[n_]:=NestWhileList[(3*#+1)/2^IntegerExponent[3*#+1,2]&,2*n+1,#>1&];
nextOddK[n_]:=Module[{m=3n+1},While[EvenQ[m],m=m/2];m];dt[n_]:=Module[{m=n,cnt=0},If[n>1,While[m=nextOddK[m];cnt++;m>n]];cnt];
Table[Part[f[i],dt[2i+1]+1],{i,0,70}]

cp's OEIS Frontend

A160268 Odd numbers k for which A006694( (k-1)/2 )< A006694( (A000265(3k+1)-1)/2 ) .

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A160364 Let f be defined as in A159885 and f^k be the k-th iteration of f. Then a(n) is the least k for which either {A000120(f^k(2n+1)) < A000120(2n+1)}&{A006694((f^k(2n+1)-1)/2)<=A006694(n)} or {A000120(f^k(2n+1))<=A000120(2n+1)}&{A006694((f^k(2n+1)-1)/2) < A006694(n)}.

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A160558 a(n) is the ordinal number of series in which the value of A160348(n) is defined.

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A317083 a(n) is the first term less than the initial 2n+1 in the reduced Collatz trajectory.

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Mathematica