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.

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

Offset: 0

Author

Paul D. Hanna, Jul 11 2006

Keywords

• nonn
• changed

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

Examples

			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 +...
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
• Andy Liu, Solution to Problem 2, Crux Mathematicorum, 17 (1991), 5-6.

Crossrefs

Cf. A117947, A039966, A062756, A081603.

Programs

Haskell

a120880 n = sum $ map (a039966 . (n -)) $ takeWhile (<= n) a005836_list
-- Reinhard Zumkeller, Jan 28 2012

Mathematica

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

PARI

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

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.