A347287 a(n) = Sum_{k = 1..m} 2^(e_k-1) where e_k = floor(log_p_k(p_(k-1)^e_(k-1))) such that e_k > 0.
1, 3, 5, 11, 23, 39, 75, 151, 279, 559, 1071, 2127, 4255, 8351, 16687, 33327, 66095, 132191, 263263, 526511, 1052847, 2101423, 4202847, 8405695, 16794303, 33587903, 67175807, 134284671, 268568959, 537004415, 1074006399, 2148012799, 4295496447, 8590992639, 17181985279
Offset: 1
Examples
a(1) = 1 since we can find no nonzero exponent e such that 3^e < 2^1; 2^(1 - 1) = 2^0 = 1. a(2) = 3 since 3^1 < 2^2 yet 3^2 > 2^2. (We assume hereinafter that the powers listed are the largest possible smaller than the immediately previous term.) 2^(2-1) + 2^(1-1) = 2^1+2^0 = 2+1 = 3. a(3) = 5 since 2^3 > 3^1, hence 2^(3-1) + 2^(1-1) = 2^2 + 2^0 = 4+1 = 5. a(4) = 11 since 2^4 > 3^2 > 5^1, thus 2^(4-1) + 2^(2-1) + 2(1-1) = 8+2+1 = 11, etc. n Row n of A347285 (reversed) a(n) ---------------------------------------------------- 1: 1 -> 1 2: 1 2 -> 3 3: 1 3 -> 5 4: 1 2 4 -> 11 5: 1 2 3 5 -> 23 6: 1 2 3 6 -> 39 7: 1 2 4 7 -> 75 8: 1 2 3 5 8 -> 151 9: 1 2 3 5 9 -> 279 10: 1 2 3 4 6 10 -> 559 11: 1 2 3 4 6 11 -> 1071 12: 1 2 3 4 7 12 -> 2127 ...
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..3322
- Michael De Vlieger, Little-endian bitmap of a(n) for n=1..512, black = 1 and white = 0.
- Michael De Vlieger, Little-endian bitmap of a(n) for n=1..10000, black = 1 and white = 0.
Programs
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Mathematica
Array[Total[2^(-1 + NestWhile[Block[{p = Prime[#2]}, Append[#1, {p^#, #} &@ Floor@ Log[p, #1[[-1, 1]]]]] & @@ {#, Length@ # + 1} &, {{2^#, #}}, #[[-1, -1]] > 1 &][[All, -1]])] &, 35] (* Generate 10000 terms from 10000 X 10000 bitmap *) MapIndexed[FromDigits[Reverse@ #1[[1 ;; First[#2]]], 2] &, ImageData@ Import["https://oeis.org/A347287/a347287_1.png"] /. {0. -> 1, 1. -> 0}]
Formula
a(n) = row sum of 2^(m-1) where m are terms in row n of A347285.
Comments