|
| |
|
|
A152763
|
|
Number of divisors of Catalan number A000108(n).
|
|
7
|
|
|
|
1, 1, 2, 2, 4, 8, 12, 8, 16, 16, 24, 32, 48, 72, 192, 96, 192, 256, 576, 512, 768, 768, 1024, 1152, 1152, 1728, 1536, 1536, 4096, 4096, 5120, 2048, 6144, 12288, 12288, 8192, 12288, 12288, 24576, 24576, 36864, 98304, 131072, 147456, 196608, 196608, 368640
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
Conjecture: a(2^k-1) < a(2^k-2) for all k >= 3. Checked up to k = 263. Note that Catalan(2^k-1) is odd and Catalan(2^k-2)/Catalan(2^k-1) = 2^(k-1)/(2^(k+1)-3). Suppose that 2^(k+1)-3 = 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 a(2^k-2)/a(2^k-1) = k * Product_{i=1..r} (e_i-r_i+1)/(e_i+1). This seems unlikely to be less than 1. Actually, it seems that a(2^k-2)/a(2^k-1) tends to infinity as n goes to infinity.
Conjecture: a(2^k-1) != a(2^k) for all k. Checked up to k = 265. Note 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 a(2^k)/a(2^k-1) = 2 * Product_{i=1..r} (e_i+r_i+1)/(e_i+1). This seems unlikely to be equal to 1. Among the numbers k <= 265, the number k for which a(2^k)/a(2^k-1) is closest to 1 is k = 70, where a(2^k)/a(2^k-1) = 104/105. (End)
|
|
|
LINKS
|
|
|
|
FORMULA
|
|
|
|
MAPLE
|
|
|
|
MATHEMATICA
|
|
|
|
PROG
|
(PARI) vector(100, n, n--; numdiv(binomial(2*n, n)/(n+1))) \\ Altug Alkan, Nov 13 2015
(PARI) val(n, p) = (n - vecsum(digits(n, p)))/(p-1); \\ p-adic valuation of n!
a(n) = my(r=1); forprime(p=2, 2*n, r*=val(2*n, p)-val(n, p)-val(n+1, p)+1); r \\ Jianing Song, Jun 16 2022
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
easy,nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
