A152763 Number of divisors of Catalan number A000108(n).

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

Offset: 0

Author

Omar E. Pol, Dec 14 2008

Keywords

• easy
• nonn

Comments

From Jianing Song, Jun 16 2022: (Start)

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

• Amiram Eldar, Table of n, a(n) for n = 0..10000 (terms 0..1000 from G. C. Greubel)

Crossrefs

Cf. A000005, A000108, A152761, A152762.

Programs

Maple

A000108 := proc(n) binomial(2*n,n)/(n+1) ; end: A152763 := proc(n) numtheory[tau](A000108(n)) ; end: for n from 0 to 80 do printf("%d,",A152763(n)) ; od: # R. J. Mathar, Dec 15 2008

Mathematica

DivisorSigma[0, CatalanNumber@Range[0, 40]] (* Vladimir Reshetnikov, Nov 13 2015 *)

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

Formula

a(n) = A000005(A000108(n)).

Extensions

Extended by R. J. Mathar, Dec 15 2008