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 A355041 Numbers k such that A152763(2^k) < A152763(2^k-1). 0
 14, 18, 30, 42, 60, 70, 82, 88, 106, 126, 130, 166, 168, 196, 213, 240, 258 (list; graph; refs; listen; history; text; internal format)
 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 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

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Last modified August 8 12:30 EDT 2022. Contains 356009 sequences. (Running on oeis4.)