A341845 a(n) = A061026(2n): smallest k such that 2n divides phi(k), phi = A000010.

3, 5, 7, 15, 11, 13, 29, 17, 19, 25, 23, 35, 53, 29, 31, 51, 103, 37, 191, 41, 43, 69, 47, 65, 101, 53, 81, 87, 59, 61, 311, 85, 67, 137, 71, 73, 149, 229, 79, 123, 83, 129, 173, 89, 181, 141, 283, 97, 197, 101, 103, 159, 107, 109, 121, 113, 229, 177, 709, 143

Offset

1,1

Comments

A061026(n) = A061026(2n) for odd n > 1 since phi(m) is even for m >= 3. In this sequence the redundant values are omitted.

We have the obvious inequality A070846(n) >= A307437(n) >= a(n). For odd p = prime(k), A307437(p) = a(p), and if A341861(k) > 0 we have A070846(p) = a(p).

The smallest n such that A070846(n) > A307437(n) > a(n) is n = 40, where A070846(40) = 241, A307437(40) = 187 and a(40) = 123.

Links

Jianing Song, Table of n, a(n) for n = 1..20000

Example

a(12) = 35 since phi(35) = 24 is divisible by 2*12, and there is no m < 12 such that phi(m) is divisible by 2*12.
a(16) = 51 since phi(51) = 32 is divisible by 2*16, and there is no m < 16 such that phi(m) is divisible by 2*16.

Prog

(PARI) a(n) = for(m=1, (2*n)^2, if(eulerphi(m)%(2*n)==0, return(m)))
(Python)
from sympy import totient as phi
def a(n):
  k = 1
  while phi(k)%(2*n) != 0: k += 1
  return k
print([a(n) for n in range(1, 61)]) # Michael S. Branicky, Feb 21 2021

Crossrefs

Cf. A061026, A000010, A070846, A307437, A341861.

Sequence in context: A227685 A051445 A002181 * A073692 A132012 A160690

Adjacent sequences: A341842 A341843 A341844 * A341846 A341847 A341848

Keyword

nonn

Author

Jianing Song, Feb 21 2021

Status

approved