A341515 The Collatz or 3x+1 map (A006370) conjugated by unary-binary-encoding (A156552).

1, 5, 2, 15, 3, 11, 5, 45, 4, 125, 7, 33, 11, 245, 6, 135, 13, 77, 17, 375, 10, 605, 19, 99, 9, 845, 8, 735, 23, 17, 29, 405, 14, 1445, 15, 231, 31, 1805, 22, 1125, 37, 1331, 41, 1815, 12, 2645, 43, 297, 25, 275, 26, 2535, 47, 539, 21, 2205, 34, 4205, 53, 51, 59, 4805, 20, 1215, 33, 1859, 61, 4335, 38, 3125, 67, 693

Offset: 1

Author

Antti Karttunen, Feb 14 2021

Keywords

• nonn

Comments

Collatz-conjecture can be formulated via this sequence by postulating that all iterations of a(n), starting from any n > 1, will eventually reach the cycle [2, 5, 3].

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000
• Index entries for sequences related to 3x+1 (or Collatz) problem
• Index entries for sequences computed from indices in prime factorization

Crossrefs

Cf. A005940, A006370, A064989, A156552, A329603, A341510, A347115 (Möbius transform),

Sequences related to iterations of this sequence: A352890, A352891, A352892, A352893, A352894, A352896, A352897, A352898, A352899.

Cf. A341516 (a variant).

Programs

PARI

A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); (t); };
A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
A156552(n) = { my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res };
A329603(n) = A005940(2+(3*A156552(n)));
A341515(n) = if(n%2, A064989(n), A329603(n));

Formula

If n is odd, then a(n) = A064989(n), otherwise a(n) = A329603(n) = A341510(n,2*n).

a(n) = A005940(1+A006370(A156552(n))).