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A080766
A unitary phi reciprocal amicable number: consider two different numbers r, s which satisfy the following equation for some integer k: uphi(r) = uphi(s) = (1/k) * r * s / (r-s); or equivalently, 1/uphi(r) = 1/uphi(s) = k * (1/s - 1/r); sequence gives r numbers.
3
2, 6, 14, 21, 240, 240, 242, 5520, 5566, 6578, 10881, 13056, 14880, 15444, 46200, 47908, 57600, 60480, 65280, 65292, 91392, 152320, 169728, 239540, 285696, 399168, 665280, 702000, 941625, 1405404, 2246400, 4285440, 4856832, 6591375, 10463040, 11713536, 13172544
OFFSET
1,1
COMMENTS
Here uphi(n) = A047994(n) is the unitary totient function: if n = Product p_i^e_i, uphi(n) = Product (p_i^e_i - 1).
Solutions sorted by increasing r, then increasing s, so repeated values are possible here; they occur at a(5) = a(6), a(59) = a(60), etc. - Michael S. Branicky, Sep 30 2025
LINKS
Michael S. Branicky, Table of n, a(n) for n = 1..66
PROG
(Python)
from math import prod
from sympy import factorint
from itertools import count, islice
def uphi(n): return prod(p_i**e_i-1 for p_i, e_i in factorint(n).items())
def agen(): # generator of terms (increasing r, then increasing s)
uphi_level_sets = dict()
for r in count(1):
v = uphi(r)
if v in uphi_level_sets:
for s in uphi_level_sets[v]:
k, rem = divmod(r*s, v*(r-s))
if rem == 0:
yield r # use s, k for A080767, A080768, resp.
else:
uphi_level_sets[v] = []
uphi_level_sets[v].append(r)
print(list(islice(agen(), 25))) # Michael S. Branicky, Sep 30 2025
CROSSREFS
KEYWORD
nonn
EXTENSIONS
Kohmoto found 2nd, 6th, 13th, 25th terms. Dean Hickerson calculated the other terms.
Offset corrected by Sean A. Irvine, Sep 29 2025
a(31) and beyond from Michael S. Branicky, Sep 29 2025
STATUS
approved