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Number of divisors of Catalan number A000108(n).
7

%I #29 Jun 16 2022 13:17:27

%S 1,1,2,2,4,8,12,8,16,16,24,32,48,72,192,96,192,256,576,512,768,768,

%T 1024,1152,1152,1728,1536,1536,4096,4096,5120,2048,6144,12288,12288,

%U 8192,12288,12288,24576,24576,36864,98304,131072,147456,196608,196608,368640

%N Number of divisors of Catalan number A000108(n).

%C From _Jianing Song_, Jun 16 2022: (Start)

%C 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.

%C 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)

%H Amiram Eldar, <a href="/A152763/b152763.txt">Table of n, a(n) for n = 0..10000</a> (terms 0..1000 from G. C. Greubel)

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

%p 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

%t DivisorSigma[0, CatalanNumber@Range[0, 40]] (* _Vladimir Reshetnikov_, Nov 13 2015 *)

%o (PARI) vector(100, n, n--; numdiv(binomial(2*n, n)/(n+1))) \\ _Altug Alkan_, Nov 13 2015

%o (PARI) val(n,p) = (n - vecsum(digits(n,p)))/(p-1); \\ p-adic valuation of n!

%o 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

%Y Cf. A000005, A000108, A152761, A152762.

%K easy,nonn

%O 0,3

%A _Omar E. Pol_, Dec 14 2008

%E Extended by _R. J. Mathar_, Dec 15 2008