OFFSET
1,1
COMMENTS
This sequence was first suggested by Stefan Steinerberger, who conjectures that there are infinitely many terms.
For every fixed pair of integers a,b, there exists only a finite number of suitable values of m. - Max Alekseyev
For all known entries (up to a(20)), there is exactly one pair (a,b) which satisfies the required conditions. In every case b-a is either 1 or 2. See sequence A140602 for values with b-a = 1 and A140603 for b-a = 2. Related open questions: (1) Must the pair (a,b) for a given a(n) be unique? (2) Does every solution have b-a <= 2?
EXAMPLE
C(19,3) + C(19,5) divides C(19,8)
C(34,6) + C(34,7) divides C(34,13)
C(41,5) + C(41,7) divides C(41,12)
C(89,7) + C(89,8) divides C(89,15)
C(104,3) + C(104,4) divides C(104,7)
C(359,5) + C(359,6) divides C(359,11)
C(398,20) + C(398,21) divides C(398,41)
C(495,12) + C(495,14) divides C(495,26)
C(527,7) + C(527,9) divides C(527,16)
C(1845,15) + C(1845,17) divides C(1845,32)
C(2309,5) + C(2309,6) divides C(2309,11)
C(2729,19) + C(2729,20) divides C(2729,39)
C(3539,35) + C(3539,36) divides C(3539,71)
C(4619,11) + C(4619,12) divides C(4619,23)
C(8644,18) + C(8644,19) divides C(8644,37)
C(12923,34) + C(12923,36) divides C(12923,70)
C(14135,30) + C(14135,31) divides C(14135,61)
C(15774,24) + C(15774,26) divides C(15774,50)
C(36098,34) + C(36098,36) divides C(36098,70), and C(36569,47) + C(36569,48) divides C(36569,95). - Robin Visser, Sep 30 2023
PROG
(Sage)
for m in range(2, 100000):
for a in range(0, m//2):
for b in range(a+1, m-a+1):
if (binomial(m, a+b)%(binomial(m, a)+binomial(m, b)) == 0):
print(m) # Robin Visser, Sep 30 2023
CROSSREFS
KEYWORD
hard,more,nonn
AUTHOR
Andrew V. Sutherland, May 18 2008
EXTENSIONS
Edited by Max Alekseyev, Jun 16 2010
a(19)-a(20) from Robin Visser, Sep 30 2023
STATUS
approved