OFFSET
1,1
COMMENTS
There are infinitely many such k.
k+1 is never prime.
k is never a multiple of 4 or 2p, where p is a 3 mod 4 prime.
If k is even then its continued fraction has an even length of 1's. This implies that k+1's continued fraction will have an odd length of 1's.
Given a pair of twins (k,k+1), if the minimum of the length of 1's among them is m, then k > 1/4 * F(m+1)^2 * F(m+2)^2 where F(n)=A000045(n).
Given k having a length of m1 and k+1 having a length of m2, then (m2-m1) divides m1+1 with the exception of m2-m1=2.
LINKS
Asher Shmidov, Table of n, a(n) for n = 1..2661
EXAMPLE
The second twin pair (corresponding to n = 2) is (74, 75).
For k = 74, sqrt(74) = [8; 1, 1, 1, 1, 16]. The periodic part is (1, 1, 1, 1, 16), which consists of 1's and a final partial quotient of 16.
For k = 75, sqrt(75) = [8; 1, 1, 1, 16]. The periodic part is (1, 1, 1, 16), which consists of 1's and a final partial quotient of 16.
Since 74 and 75 are consecutive, 74 belongs to this sequence.
PROG
(Python)
def twin_list(term_count=24):
f = [0, 1, 1]
def fib(n):
while len(f) <= n:
f.append(f[-1] + f[-2])
return f[n]
k_set = set()
m1 = 1
while True:
least_k = (fib(m1+1) * fib(m1+2))**2 // 4
if len(k_set) >= term_count:
if sorted(list(k_set))[term_count-1] < least_k:
break
for d in range(1, m1+2):
if (m1+1) % d != 0 and d != 2:
continue
m2 = m1 + d
F_d = fib(d)
if m1 % 3 == 2 or m2 % 3 == 2:
continue
num = fib(m1+1) * fib(m2+1)
two_q_minus1 = num // F_d
if two_q_minus1 % 2 == 1:
q = two_q_minus1 // 2 + 1
k1 = q**2 + 1 + two_q_minus1 * fib(m1) // fib(m1+1)
k2 = q**2 + 1 + two_q_minus1 * fib(m2) // fib(m2+1)
k_set.add(min(k1, k2)) # least of the pair. k2 could be least as m2>m1 doesn't necessarily imply k2>k1.
m1 += 1
k_sorted = sorted(list(k_set))
k_trunc = k_sorted[:term_count]
return k_trunc
CROSSREFS
KEYWORD
nonn
AUTHOR
Asher Shmidov, Apr 10 2026
STATUS
approved