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 A215894 a(n) = floor(2^n / n^k), where k is the largest integer such that 2^n >= n^k. 2
 1, 2, 1, 1, 1, 2, 4, 6, 1, 1, 2, 3, 5, 9, 1, 1, 2, 4, 6, 10, 17, 1, 2, 3, 5, 9, 15, 26, 1, 2, 4, 6, 11, 18, 31, 1, 2, 4, 6, 11, 19, 32, 1, 2, 3, 5, 9, 16, 28, 49, 1, 2, 4, 7, 13, 22, 38, 1, 1, 3, 5, 9, 16, 27, 47, 1, 2, 3, 5, 10, 17, 30, 51, 1, 2, 3, 5, 10 (list; graph; refs; listen; history; text; internal format)
 OFFSET 2,2 COMMENTS a(n) < n. n such that a(n) = n-1: 2, 3, 996, 3389, 149462. LINKS Table of n, a(n) for n=2..79. FORMULA a(n) = modlg(2^n, n) = floor(2^n / n^floor(n*logn(2))), where logn is the logarithm base n. In the base-b representation of k, modlg(k,b) is the most significant digit: k = c0 + c1*b + c2*b^2 + ... + cn*b^n, cn = modlg(k,b), c0 = k mod b. - Alex Ratushnyak, Aug 30 2012 EXAMPLE a(2) = floor(2^2 / 2^2) = 1, a(3) = floor(2^3 / 3) = 2, a(4)..a(9) are floor(2^n / n^2), a(10)..a(15) are floor(2^n / n^3), a(16)..a(22) are floor(2^n / n^4), and so on. MATHEMATICA Table[Floor[2^n/n^Floor[n Log[n, 2]]], {n, 2, 64}] (* Alonso del Arte, Aug 26 2012 *) PROG (Python) import math def modiv(a, b): return a - b*(a//b) def modlg(a, b): return a // b**int(math.log(a, b)) for n in range(2, 100): a = 2**n print(modlg(a, n), end=', ') (Magma) [Floor(2^n div n^Floor(n *Log(n, 2))): n in [2..100]]; // Vincenzo Librandi, Jan 09 2019 CROSSREFS Cf. A215892, A000799, A060505. Sequence in context: A117211 A246576 A358273 * A061545 A287641 A265312 Adjacent sequences: A215891 A215892 A215893 * A215895 A215896 A215897 KEYWORD nonn AUTHOR Alex Ratushnyak, Aug 25 2012 STATUS approved

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Last modified September 10 01:04 EDT 2024. Contains 375769 sequences. (Running on oeis4.)