A355041 Numbers k such that A152763(2^k) < A152763(2^k-1).

14, 18, 30, 42, 60, 70, 82, 88, 106, 126, 130, 166, 168, 196, 213, 240, 258

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).

Conjecture: there is no prime in this sequence. Among the primes p <= 257, the prime p for which A152763(2^p)/A152763(2^p-1) is the smallest is p = 193, where A152763(2^p)/A152763(2^p-1) = 143/140.

Links

Table of n, a(n) for n=1..17.

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.
Values of A152763(2^k)/A152763(2^k-1) for known terms:
  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

(PARI) val(n, p) = (n - vecsum(digits(n, p)))/(p-1);
q(n, p) = val(2*n, p) - 2*val(n, p);
r(n) = my(list = factor((2^(n+1)-1)/(2^n+1)), w=#list~, rat=2, ex); for(i=1, w, ex=q(2^n-1, list[i, 1]); rat*=(ex+list[i, 2]+1)/(ex+1)); rat \\ A152763(2^n)/A152763(2^n-1)
isA355041(n) = (r(n) < 1)

Crossrefs

Cf. A152763, A000108, A038003 (the odd Catalan numbers).

Sequence in context: A304144 A327933 A079349 * A212047 A244034 A154864

Adjacent sequences: A355038 A355039 A355040 * A355042 A355043 A355044

Keyword

nonn,hard,more

Author

Jianing Song, Jun 16 2022

Status

approved