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A120880 G.f. satisfies: A(x) = A(x^3)*(1 + 2*x + x^2); thus a(n) = 2^A062756(n), where A062756(n) is the number of 1's in the ternary expansion of n. 4
1, 2, 1, 2, 4, 2, 1, 2, 1, 2, 4, 2, 4, 8, 4, 2, 4, 2, 1, 2, 1, 2, 4, 2, 1, 2, 1, 2, 4, 2, 4, 8, 4, 2, 4, 2, 4, 8, 4, 8, 16, 8, 4, 8, 4, 2, 4, 2, 4, 8, 4, 2, 4, 2, 1, 2, 1, 2, 4, 2, 1, 2, 1, 2, 4, 2, 4, 8, 4, 2, 4, 2, 1, 2, 1, 2, 4, 2, 1, 2, 1, 2, 4, 2, 4, 8, 4, 2, 4, 2, 4, 8, 4, 8, 16, 8, 4, 8, 4, 2, 4, 2, 4, 8, 4 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

More generally, if g.f. of {a(n)} satisfies: A(x) = A(x^3)*(1 + b*x + c*x^2), then a(n) = b^A062756(n)*c^A081603(n), where A062756(n) is the number of 1's and A081603(n) is the number of 2's, in the ternary expansion of n. This sequence is not the same as A059151.

a(n) is the number of entries in the n-th row of Pascal's triangle that are congruent to 1 mod 3 minus the number of entries that are congruent to 2 mod 3. - N. Sato, Jun 22 2007 (see Liu (1991))

This sequence pertains to genotype Punnett square mathematics. Start with X = 1. Each hybrid cross involves the equation X:2X:X. Therefore, the ratio in the first (mono) hybrid cross is X=1:2X=2(1) or 2:X=1; or 1:2:1. When you move up to the next hybridization level, replace the previous cross ratio with X. X now represents 3 numbers—1:2:1. Therefore, the ratio in the second (di) hybrid cross is X = (1:2:1):2X = [2(1):2(2):2(1)] or (2:4:2):X = (1:2:1). Put it together and you get 1:2:1:2:4:2:1:2:1. Each time you move up a hybridization level, replace the previous ratio with X, and use the same equation—X:2X:X to get its ratio. - John Michael Feuk, Dec 10 2011

Also number of ways to write n as sum of two nonnegative numbers having in ternary representation no 3; see also A205565. [Reinhard Zumkeller, Jan 28 2012]

LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Andy Liu, Solution to Problem 2, Crux Mathematicorum, 17 (1991), 5-6.

FORMULA

a((3^n+1)/2) = 2^n; a(n) = a(floor(n/3))*2^[[n (mod 3)] (mod 2)], with a(0)=1. G.f.: A(x) = prod_{n>=0} (1 + x^(3^n))^2. Self-convolution of A039966. Row sums of triangle A117947(n,k) = balanced ternary of C(n,k) mod 3.

EXAMPLE

Records are 2^n at positions: 0,1,4,13,40,121,...,(3^n-1)/2,... (n>=0).

A(x) = 1 + 2*x + x^2 + 2*x^3 + 4*x^4 + 2*x^5 + x^6 + 2*x^7 + x^8 +...

MATHEMATICA

Nest[ Join[#, 2 #, #] &, {1}, 5] (* Robert G. Wilson v, Jul 27 2014 *)

PROG

(PARI) a(n)=if(n==0, 1, a(n\3)*2^((n%3)%2))

(Haskell)

a120880 n = sum $ map (a039966 . (n -)) $ takeWhile (<= n) a005836_list

-- Reinhard Zumkeller, Jan 28 2012

CROSSREFS

Cf. A117947, A039966, A062756, A081603.

Sequence in context: A305350 A009205 A086754 * A059151 A290091 A059149

Adjacent sequences:  A120877 A120878 A120879 * A120881 A120882 A120883

KEYWORD

nonn

AUTHOR

Paul D. Hanna, Jul 11 2006

STATUS

approved

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Last modified October 21 14:20 EDT 2019. Contains 328301 sequences. (Running on oeis4.)