cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A030102 Base-3 reversal of n (written in base 10).

Original entry on oeis.org

0, 1, 2, 1, 4, 7, 2, 5, 8, 1, 10, 19, 4, 13, 22, 7, 16, 25, 2, 11, 20, 5, 14, 23, 8, 17, 26, 1, 28, 55, 10, 37, 64, 19, 46, 73, 4, 31, 58, 13, 40, 67, 22, 49, 76, 7, 34, 61, 16, 43, 70, 25, 52, 79, 2, 29, 56, 11, 38, 65, 20, 47, 74, 5, 32, 59, 14, 41, 68, 23, 50, 77, 8, 35, 62, 17, 44, 71
Offset: 0

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Examples

			a(17) = 25 because 17 in base 3 is 122, and backwards that is 221, which is 25 in base 10.
a(18) = 2 because 18 in base 3 is 200, and backwards that is 2.
		

Crossrefs

Cf. A030341.
Cf. A263273 for a bijective variant.

Programs

  • Haskell
    a030102 = foldl (\v d -> 3 * v + d) 0 . a030341_row
    -- Reinhard Zumkeller, Dec 16 2013
  • Maple
    a030102:= proc(n) option remember;
      local y;
      y:= n mod 3;
      3^ilog[3](n)*y + procname((n-y)/3)
    end proc:
    for i from 0 to 2 do a030102(i):= i od:
    seq(a030102(i),i=0..100); # Robert Israel, Dec 24 2015
    # alternative
    A030102 := proc(n)
        local r ;
        r := ListTools[Reverse](convert(n,base,3)) ;
        add(op(i,r)*3^(i-1),i=1..nops(r)) ;
    end proc: # R. J. Mathar, May 28 2016
  • Mathematica
    A030102[n_] := FromDigits[Reverse@IntegerDigits[n, 3], 3] (* JungHwan Min, Dec 23 2015 *)
    FromDigits[#,3]&/@(Reverse/@IntegerDigits[Range[0,80],3]) (* Harvey P. Dale, Feb 05 2020 *)
  • PARI
    a(n,b=3)=subst(Polrev(base(n,b)),x,b) /* where */
    base(n,b)={my(a=[n%b]);while(0M. F. Hasler, Nov 04 2011
    
  • PARI
    a(n) = fromdigits(Vecrev(digits(n, 3)), 3); \\ Michel Marcus, Oct 10 2017
    

Formula

a(n) = t(n,0) with t(n,r) = if n=0 then r else t(floor(n/3),r*3+(n mod 3)). - Reinhard Zumkeller, Mar 04 2010
G.f. G(x) satisfies: G(x) = (1+x+x^2)*G(x^3) - (1+2*x)*(x + 2*Sum_{m>=0} 3^m*x^(3^(m+1)+1)/(x^3-1). - Robert Israel, Dec 24 2015