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.

A062756 Number of 1's in ternary (base-3) expansion of n.

Original entry on oeis.org

0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 2, 1, 2, 3, 2, 1, 2, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 2, 3, 4, 3, 2, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 2, 1, 2, 3, 2, 1, 2, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 2, 3, 4, 3, 2, 3, 2, 1, 2, 1, 2, 3, 2
Offset: 0

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Author

Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 16 2001

Keywords

Comments

Fixed point of the morphism: 0 ->010; 1 ->121; 2 ->232; ...; n -> n(n+1)n, starting from a(0)=0. - Philippe Deléham, Oct 25 2011

Crossrefs

Cf. A080846, A343785 (first differences).
Cf. A081606 (indices of !=0).
Indices of terms 0..6: A005823, A023692, A023693, A023694, A023695, A023696, A023697.
Numbers of: A077267 (0's), A081603 (2's), A160384 (1's+2's).
Other bases: A000120, A160381, A268643.

Programs

  • Haskell
    a062756 0 = 0
    a062756 n = a062756 n' + m `mod` 2 where (n',m) = divMod n 3
    -- Reinhard Zumkeller, Feb 21 2013
    
  • Mathematica
    Table[Count[IntegerDigits[i, 3], 1], {i, 0, 200}]
    Nest[Join[#, # + 1, #] &, {0}, 5] (* IWABUCHI Yu(u)ki, Sep 08 2012 *)
  • PARI
    a(n)=if(n<1,0,a(n\3)+(n%3)%2) \\ Paul D. Hanna, Feb 24 2006
    
  • PARI
    a(n)=hammingweight(digits(n,3)%2); \\ Ruud H.G. van Tol, Dec 10 2023
    
  • Python
    from sympy.ntheory import digits
    def A062756(n): return digits(n,3)[1:].count(1) # Chai Wah Wu, Dec 23 2022

Formula

a(0) = 0, a(3n) = a(n), a(3n+1) = a(n)+1, a(3n+2) = a(n). - Vladeta Jovovic, Jul 18 2001
G.f.: (Sum_{k>=0} x^(3^k)/(1+x^(3^k)+x^(2*3^k)))/(1-x). In general, the generating function for the number of digits equal to d in the base b representation of n (0 < d < b) is (Sum_{k>=0} x^(d*b^k)/(Sum_{i=0..b-1} x^(i*b^k)))/(1-x). - Franklin T. Adams-Watters, Nov 03 2005 [For d=0, use the above formula with d=b: (Sum_{k>=0} x^(b^(k+1))/(Sum_{i=0..b-1} x^(i*b^k)))/(1-x), adding 1 if you consider the representation of 0 to have one zero digit.]
a(n) = a(floor(n/3)) + (n mod 3) mod 2. - Paul D. Hanna, Feb 24 2006

Extensions

More terms from Vladeta Jovovic, Jul 18 2001