A065363 Sum of balanced ternary digits in n. Replace 3^k with 1 in balanced ternary expansion of n.
0, 1, 0, 1, 2, -1, 0, 1, 0, 1, 2, 1, 2, 3, -2, -1, 0, -1, 0, 1, 0, 1, 2, -1, 0, 1, 0, 1, 2, 1, 2, 3, 0, 1, 2, 1, 2, 3, 2, 3, 4, -3, -2, -1, -2, -1, 0, -1, 0, 1, -2, -1, 0, -1, 0, 1, 0, 1, 2, -1, 0, 1, 0, 1, 2, 1, 2, 3, -2, -1, 0, -1, 0, 1, 0, 1, 2, -1, 0, 1, 0, 1, 2, 1, 2, 3, 0, 1, 2, 1, 2, 3, 2, 3, 4, -1, 0, 1, 0, 1, 2, 1, 2, 3, 0, 1, 2, 1, 2
Offset: 0
Examples
5 = + 1(9) - 1(3) - 1(1) -> +1 - 1 - 1 = -1 = a(5).
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
- F. T. Adams-Watters and F. Ruskey, Generating Functions for the Digital Sum and Other Digit Counting Sequences, JIS 12 (2009) 09.5.6.
- Daniel Forgues, Table of n, a(n) for n = 0..100000
Programs
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Maple
a:= proc(n) `if`(n=0, 0, (d-> `if`(d=2, a(q+1)-1, d+a(q)))(irem(n, 3, 'q'))) end: seq(a(n), n=0..120); # Alois P. Heinz, Jan 09 2020 -
Mathematica
balTernDigits[0] := {0}; balTernDigits[n_/;n > 0] := Module[{unParsed = n, currRem, currExp = 1, digitList = {}, nextDigit}, While[unParsed > 0, If[unParsed == 3^(currExp - 1), digitList = Append[digitList, 1]; unParsed = 0, currRem = Mod[unParsed, 3^currExp]/3^(currExp - 1); nextDigit = Switch[currRem, 0, 0, 2, -1, 1, 1]; digitList = Append[digitList, nextDigit]; unParsed = unParsed - nextDigit * 3^(currExp - 1)]; currExp++]; digitList = Reverse[digitList]; Return[digitList]]; balTernDigits[n_/;n < 0] := (-1)balTernDigits[Abs[n]]; Table[Plus@@balTernDigits[n], {n, 0, 108}] (* Alonso del Arte, Feb 25 2011 *) terVal[lst_List] := Reverse[lst].(3^Range[0, Length[lst] - 1]); maxDig = 4; t = Table[0, {3 * 3^maxDig/2}]; t[[1]] = 1; Do[d = IntegerDigits[Range[0, 3^dig - 1], 3, dig]/.{2 -> -1}; d = Prepend[#, 1]&/@d; t[[terVal/@d]] = Total/@d, {dig, maxDig}]; Prepend[t, 0] (* T. D. Noe, Feb 24 2011 *) Array[Total[Prepend[IntegerDigits[#, 3], 0] //. {a___, b_, 2, c___} :> {a, b + 1, -1, c}] &, 109, 0] (* Michael De Vlieger, Jun 27 2020 *) -
PARI
bt(n)=my(d=digits(n,3),c=1); while(c, if(d[1]==2, d=concat(0,d)); c=0; for(i=2,#d, if(d[i]==2,d[i]=-1;d[i-1]+=1;c=1))); d a(n)=vecsum(bt(n)) \\ Charles R Greathouse IV, May 07 2020
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Python
def a(n): s=0 x=0 while n>0: x=n%3 n=n//3 if x==2: x=-1 n+=1 s+=x return s print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 06 2017
Formula
G.f.: (1/(1-x))*Sum_{k>=0} (x^3^k - x^(2*3^k))/(x^((3^k-1)/2)*(1 + x^3^k + x^(2*3^k))). - Franklin T. Adams-Watters, May 13 2009
a(3*n - 1) = a(n) - 1, a(3*n) = a(n), a(3*n + 1) = a(n) + 1. - Thomas König, Jun 24 2020
Comments