A203602 Inverse permutation to A092401.
1, 3, 2, 5, 7, 4, 9, 11, 13, 15, 17, 6, 19, 21, 8, 23, 25, 27, 29, 31, 10, 33, 35, 12, 37, 39, 14, 41, 43, 16, 45, 47, 18, 49, 51, 53, 55, 57, 20, 59, 61, 22, 63, 65, 67, 69, 71, 24, 73, 75, 26, 77, 79, 28, 81, 83, 30, 85, 87, 32, 89, 91, 93, 95, 97, 34, 99
Offset: 1
Keywords
Links
Programs
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Haskell
import Data.List (elemIndex) import Data.Maybe (mapMaybe) a203602 n = a203602_list !! (n-1) a203602_list = map (+ 1) $ mapMaybe (`elemIndex` a092401_list) [1..]
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Mathematica
div[n_, m_]=Floor[n/m // Chop]-Ceiling[n/m // Chop](*n not divisible by m=>-1, else 0*); ind[m_]:=Sum[(-1)^(n) div[m, 3^n], {n, 1, Floor[Log[m]/Log[3] // FullSimplify]}] + Mod[Floor[Log[3 m]/Log[3] // FullSimplify], 2];(* returns 0 or 1 depending on if we have an 'n' term (=>1) or a '3n' term (=>0) *) f[m_] := (2* Sum[(-1)^(n) Floor[m/(3^(n)) // FullSimplify], {n, 0, Floor[Log[m]/Log[3] // FullSimplify]}] - 1)* ind[m] + (1 - ind[m]) (2* Sum[(-1)^(n) Floor[m/(3^(n + 1)) // FullSimplify], {n, 0, -1 + Floor[Log[m]/Log[3] // FullSimplify]}]); Table[f[k], {k, 1, 50}] (* Daniel Hoying, Aug 06 2020 *)
Formula
a(n) = -2*(Sum_{k=0..-1+floor(log(n)/log(3))} (-1)^k*floor(n/3^(k+1)))*(-1 + (floor(log(3*n)/log(3)) mod 2)+Sum_{k=1..floor(log(n)/log(3))} (-1)^k*(-ceiling(n/3^k) + floor(n/3^k))) + (-1 + 2*Sum_{k=0..floor(log(n)/log(3))} (-1)^k*floor(n/3^k))*((floor(log(3*n)/log(3)) mod 2)+Sum_{k=1..floor(log(n)/log(3))} (-1)^k*(-ceiling(n/3^k) + floor(n/3^k))). - Daniel Hoying, Aug 06 2020
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