OFFSET
1,1
COMMENTS
Note that Catalan(2^k-1) is odd and that Catalan(2^k)/Catalan(2^k-1) = 2 * (2^(k+1)-1)/(2^k+1). Suppose that (2^(k+1)-1)/(2^k+1) = Product_{i=1..r} (p_i)^(e_i), let r_i be the (p_i)-adic valuation of binomial(2*(2^k-1),2^k-1), then A152763(2^k)/A152763(2^k-1) = 2 * Product_{i=1..r} (e_i+r_i+1)/(e_i+1).
EXAMPLE
14 is a term since A152763(2^14) = 4.457... * 10^721 < A152763(2^14-1) = 4.754... * 10^721. Note that Catalan(2^14)/Catalan(2^14-1) = 2 * 32767/16385, 32767/16385 = (7*31*151)/(5*29*113). We have v(N,5) = v(N,31) = v(N,113) = v(N,151) = 1, v(N,7) = 3, v(N,29) = 2 for N = binomial(2*(2^14-1),2^14-1), so A152763(2^14)/A152763(2^14-1) = 2 * ((3+1+1)/(3+1)) * ((1+1+1)/(1+1)) * ((1+1+1)/(1+1)) * ((1-1+1)/(1+1)) * ((2-1+1)/(2+1)) * ((1-1+1)/(1+1)) = 15/16 < 1.
18 is a term since A152763(2^18) = 1.178... * 10^8888 < A152763(2^18-1) = 2.121... * 10^8888. Note that Catalan(2^18)/Catalan(2^18-1) = 2 * 524287/262145, 524287/262145 = 524287/(5*13*37*109). We have v(N,5) = 5, q(N,13) = 2, v(N,37) = v(N,109) = 1, v(N,524287) = 0 for N = binomial(2*(2^18-1),2^18-1), so A152763(2^18)/A152763(2^18-1) = 2 * ((5-1+1)/(5+1)) * ((2-1+1)/(2+1)) * ((1-1+1)/(1+1)) * ((1-1+1)/(1+1)) * ((0+1+1)/(0+1)) = 5/9 < 1.
k = 14: 15/16
k = 18: 5/9
k = 30: 9/11
k = 42: 432/455
k = 60: 64/81
k = 70: 104/105
k = 82: 160/243
k = 88: 16/21
k = 106: 38/45
k = 126: 2275/2673
k = 130: 3773/6400
k = 166: 216/287
k = 168: 27/35
k = 196: 605/897
k = 213: 1683/1840
k = 240: 320/343
k = 258: 732875/810432
PROG
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Jianing Song, Jun 16 2022
EXTENSIONS
a(18)-a(20) from Amiram Eldar, Jul 24 2024
STATUS
approved