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A062756
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Number of 1's in ternary (base-3) expansion of n.
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56
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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
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OFFSET
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0,5
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COMMENTS
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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
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LINKS
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FORMULA
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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
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MATHEMATICA
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Table[Count[IntegerDigits[i, 3], 1], {i, 0, 200}]
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PROG
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(PARI) a(n)=if(n<1, 0, a(n\3)+(n%3)%2) \\ Paul D. Hanna, Feb 24 2006
(Haskell)
a062756 0 = 0
a062756 n = a062756 n' + m `mod` 2 where (n', m) = divMod n 3
(Python)
from sympy.ntheory import digits
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 16 2001
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EXTENSIONS
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STATUS
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approved
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