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A006047
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Number of entries in n-th row of Pascal's triangle not divisible by 3.
(Formerly M0422)
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29
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1, 2, 3, 2, 4, 6, 3, 6, 9, 2, 4, 6, 4, 8, 12, 6, 12, 18, 3, 6, 9, 6, 12, 18, 9, 18, 27, 2, 4, 6, 4, 8, 12, 6, 12, 18, 4, 8, 12, 8, 16, 24, 12, 24, 36, 6, 12, 18, 12, 24, 36, 18, 36, 54, 3, 6, 9, 6, 12, 18, 9, 18, 27, 6, 12, 18, 12, 24, 36, 18, 36, 54, 9, 18, 27, 18, 36, 54, 27, 54
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OFFSET
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0,2
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COMMENTS
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Fixed point of the morphism a -> a, 2a, 3a, starting from a(1) = 1. - Robert G. Wilson v, Jan 24 2006
This is a particular case of the number of entries in n-th row of Pascal's triangle not divisible by a prime p, which is given by a simple recursion using ⊗, the Kronecker (or tensor) product of vectors. Let v_0=(1,2,...,p). Then v_{n+1)=v_0 ⊗ v_n, where the vector v_n contains the values for the first p^n rows of Pascal's triangle (rows 0 through p^n-1). - William B. Everett (bill(AT)chgnet.ru), Mar 29 2008
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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Write n in base 3; if the representation contains r 1's and s 2's then a(n) = 3^s * 2^r. Also a(n) = Sum_{k=0..n} (C(n, k)^2 mod 3). - Avi Peretz (njk(AT)netvision.net.il), Apr 21 2001
a(n) = b(n+1), with b(1)=1, b(2)=2, b(3n)=3b(n), b(3n+1)=b(n+1), b(3n+2)=2b(n+1). - Ralf Stephan, Sep 15 2003
G.f.: Product_{n>=0} (1+2*x^(3^n)+3*x^(2*3^n)) (Northshield). - Johannes W. Meijer, Jun 05 2011
G.f. g(x) satisfies g(x) = (1 + 2*x + 3*x^2)*g(x^3). - Robert Israel, Oct 15 2015
a(3^k) = 2 for k>=0;
a(2*3^k) = 3 for k>=0;
a(n) = Product_{b_j != 0} a(b_j*3^j) where n = Sum_{j>=0} b_j*3^j is the ternary representation of n. (End)
a(n) = Sum_{k = 0..n} mod(C(n,k)^2, 3). - Peter Bala, Dec 17 2020
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EXAMPLE
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15 in base 3 is 120, here r=1 and s=1 so a(15) = 3*2 = 6.
William B. Everett's comment with p=3, n=2: v_0 = (1,2,3), v_1 = (1,2,3) => v_2 = (1*1,1*2,1*3,2*1,2*2,2*3,3*1,3*2,3*3) = (1,2,3,2,4,6,3,6,9), the first 3^2 values of the present sequence. - Wolfdieter Lang, Mar 19 2014
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MAPLE
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p:=proc(n) local ct, k: ct:=0: for k from 0 to n do if binomial(n, k) mod 3 = 0 then else ct:=ct+1 fi od: end: seq(p(n), n=0..82); # Emeric Deutsch
f:= proc(n) option remember; ((n mod 3)+1)*procname(ceil((n+1)/3)-1) end proc:
f(0):= 1: f(1):= 2:
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MATHEMATICA
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Nest[Flatten[ # /. a_Integer -> {a, 2a, 3a}] &, {1}, 4] (* Robert G. Wilson v, Jan 24 2006 *)
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PROG
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(PARI) b(n)=if(n<3, n, if(n%3==0, 3*b(n/3), if(n%3==1, 1*b((n+2)/3), 2*b((n+1)/3)))) \\ Ralf Stephan
(PARI) A006047(n) = b(1+n); \\ (The above PARI-program by Ralf Stephan is for offset-1-version of this sequence.) - Antti Karttunen, May 28 2017
(PARI) A006047(n) = { my(m=1, d); while(n, d = (n%3); m *= (1+d); n \= 3); m; }; \\ Antti Karttunen, May 28 2017
(PARI) a(n) = prod(i=1, #d=digits(n, 3), (1+d[i])) \\ David A. Corneth, May 28 2017
(PARI) upto(n) = my(res = [1], v); while(#res < n, v = concat(2*res, 3*res); res = concat(res, v)); res \\ David A. Corneth, May 29 2017
(Haskell)
a006047 = sum . map signum . a083093_row
(Scheme) (define (A006047 n) (if (zero? n) 1 (let ((d (mod n 3))) (* (+ 1 d) (A006047 (/ (- n d) 3)))))) ;; For R6RS standard. Use modulo instead of mod in older Schemes like MIT/GNU Scheme. - Antti Karttunen, May 28 2017
(Python)
from sympy.ntheory.factor_ import digits
from sympy import prod
def a(n):
d=digits(n, 3)
return n + 1 if n<3 else prod(1 + d[i] for i in range(1, len(d)))
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CROSSREFS
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Cf. A001316, A003586, A038148, A053735, A083093, A089898, A206424, A227428, A286586, A286587, A286633.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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