A117947 -id:A117947 - cp's OEIS frontend

A120880 G.f. satisfies: A(x) = A(x^3)(1 + 2x + 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

Paul D. Hanna, Jul 11 2006

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

			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 +...

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

Cf. A117947, A039966, A062756, A081603.

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

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))
```

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.

A117939 Triangle related to powers of 3 partitions of n.

Original entry on oeis.org

1, 2, 1, 1, -2, 1, 2, 0, 0, 1, 4, 2, 0, 2, 1, 2, -4, 2, 1, -2, 1, 1, 0, 0, -2, 0, 0, 1, 2, 1, 0, -4, -2, 0, 2, 1, 1, -2, 1, -2, 4, -2, 1, -2, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 4, 2, 0, 0, 0, 0, 0, 0, 0, 2, 1, 2, -4, 2, 0, 0, 0, 0, 0, 0, 1, -2, 1, 4, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 1, 8, 4, 0, 4, 2, 0, 0, 0, 0, 4, 2, 0, 2, 1

Offset: 0

Paul Barry, Apr 05 2006

			Triangle begins
  1;
  2,  1;
  1, -2, 1;
  2,  0, 0,  1;
  4,  2, 0,  2,  1;
  2, -4, 2,  1, -2,  1;
  1,  0, 0, -2,  0,  0, 1;
  2,  1, 0, -4, -2,  0, 2,  1;
  1, -2, 1, -2,  4, -2, 1, -2, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A059151, A117940, A117944, A117946.

Cf. A120854 (matrix log), A117941 (inverse), A117947 (matrix square-root).

T[n_, k_]:= Sum[JacobiSymbol[Binomial[n, j], 3]*JacobiSymbol[Binomial[n-j, k], 3], {j, 0, n}]; Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 29 2021 *)

T(n,k)=(matrix(n+1,n+1,r,c,(binomial(r-1,c-1)+1)%3-1)^2)[n+1,k+1] \\ Paul D. Hanna, Jul 08 2006

def A117939(n, k): return sum(jacobi_symbol(binomial(n, j), 3)*jacobi_symbol(binomial(n-j, k), 3) for j in (0..n))
flatten([[A117939(n, k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Oct 29 2021

Triangle T(n,k) = Sum_{j=0..n} L(C(n,j)/3)*L(C(n-j,k)/3) where L(j/p) is the Legendre symbol of j and p.
T(n, k) mod 2 = A117944(n,k).
T(n, 0) = A059151(n).
T(n, 1) = A117946(n).
Sum_{k=0..n} T(n, k) = A117940(n).
Matrix square of triangle A117947. Matrix log is the integer triangle A120854. - Paul D. Hanna, Jul 08 2006

A117963 Antidiagonal sums of a Legendre-binomial triangle for p = 3.

Original entry on oeis.org

1, 1, 2, 0, 2, 2, 1, 3, 4, -2, 2, 0, 2, 2, 4, 0, 4, 4, -1, 3, 2, 2, 4, 6, 1, 7, 8, -6, 2, -4, 4, 0, 4, -2, 2, 0, 2, 2, 4, 0, 4, 4, 2, 6, 8, -4, 4, 0, 4, 4, 8, 0, 8, 8, -5, 3, -2, 4, 2, 6, -1, 5, 4, 0, 4, 4, 2, 6, 8, 2, 10, 12, -5, 7, 2, 6, 8, 14, 1, 15, 16, -14, 2, -12, 8, -4, 4, -6, -2, -8, 8, 0, 8, -4, 4, 0, 4, 4, 8, -6, 2

Offset: 0

Paul Barry, Apr 05 2006

Diagonal sums of A117947.

			The triangle mentioned in the name starts:
{1},
{1, 1},
{1,-1, 1},
{1, 0, 0, 1},
{1, 1, 0, 1, 1},
{1,-1, 1, 1,-1, 1},
{1, 0, 0,-1, 0, 0, 1}.

Antti Karttunen, Table of n, a(n) for n = 0..16384 (terms 0..5000 from Alois P. Heinz)

Cf. A117947, A117964 [= a(n) mod 2].

a[n_] := Sum[JacobiSymbol[Binomial[n - k, k], 3], {k, 0, n/2}];
a /@ Range[0, 100] (* Jean-François Alcover, Oct 14 2019 *)

{a(n)=local(A=1+x+x*O(x^n)); for(i=1,#binary(n), A=subst(A,x,x^3+x*O(x^n)) *(1-4*x^3-x^6)/(1-x-x^2+x*O(x^n))); polcoeff(A,n,x)} \\ Paul D. Hanna, Jul 11 2006

A117963list(upto_n) = { my(A=1+x+x*O(x^upto_n)); for(i=1, #binary(upto_n), print1(i,", "); A=subst(A, x, x^3+x*O(x^upto_n)) *(1-4*x^3-x^6)/(1-x-x^2+x*O(x^upto_n))); print(); vector(upto_n,n,polcoeff(A,n-1)); }; \\ Antti Karttunen, Jan 01 2023, after Paul D. Hanna's program above.

A117963(n) = sum(k=0,n\2,kronecker(binomial(n-k,k),3)); \\ Antti Karttunen, Jan 01 2023

a(n) = a(3n+2)/a(2).
a(n) = Sum_{k=0..floor(n/2)} L(C(n-k,k)/3) where L(j/p) is the Legendre symbol of j and p.
From Paul D. Hanna, Jul 11 2006: (Start)
G.f. satisfies: A(x) = A(x^3)*(1 - 4*x^3 - x^6)/(1 - x - x^2).
a(n) == Fibonacci(n+1) (mod 3);
a(n) == a(n-1) + a(n-2) (mod 3). (End)

A120854 Matrix log of A117939, read by rows, consisting only of 0's, 3's and signed 2's.

Original entry on oeis.org

0, 2, 0, 3, -2, 0, 2, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 2, 3, -2, 0, 3, 0, 0, -2, 0, 0, 0, 0, 3, 0, 0, -2, 0, 2, 0, 0, 0, 3, 0, 0, -2, 3, -2, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 3, -2, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 3, -2, 0, 0, 0, 0, 0

Offset: 0

Paul D. Hanna, Jul 08 2006

The number of nonzero elements in row n equals A053735(n), the sum of ternary digits of n. Row sums are A120855(n) = 2*A062756 + A081603(n), where A062756(n) = number of 1's in ternary expansion of n and A081603(n) = number of 2's in ternary expansion of n. Triangle A117939 is related to partitions of n into powers of 3 and is the matrix square of A117947, where A117947(n,k) = balanced ternary digits of C(n,k) mod 3, also A117947(n,k) = L(C(n,k)/3) where L(j/p) is the Legendre symbol of j and p.

			Triangle begins:
0;
2, 0;
3,-2, 0;
2, 0, 0, 0;
0, 2, 0, 2, 0;
0, 0, 2, 3,-2, 0;
3, 0, 0,-2, 0, 0, 0;
0, 3, 0, 0,-2, 0, 2, 0;
0, 0, 3, 0, 0,-2, 3,-2, 0;
2, 0, 0, 0, 0, 0, 0, 0, 0, 0; ...
Matrix exponentiation gives A117939:
1;
2, 1;
1,-2, 1;
2, 0, 0, 1;
4, 2, 0, 2, 1;
2,-4, 2, 1,-2, 1;
1, 0, 0,-2, 0, 0, 1;
2, 1, 0,-4,-2, 0, 2, 1;
1,-2, 1,-2, 4,-2, 1,-2, 1; ...
and A117939 is the matrix square of A117947:
1;
1, 1;
1,-1, 1;
1, 0, 0, 1;
1, 1, 0, 1, 1;
1,-1, 1, 1,-1, 1;
1, 0, 0,-1, 0, 0, 1;
1, 1, 0,-1,-1, 0, 1, 1;
1,-1, 1,-1, 1,-1, 1,-1, 1; ...

Cf. A117939, A117947; A120855 (row sums), A062756, A081603, A053735.

/* Generated as the Matrix LOG of A117939: */ T(n,k)=local(M=matrix(n+1,n+1,r,c,(binomial(r-1,c-1)+1)%3-1)^2, L=sum(i=1,#M,-(M^0-M)^i/i));return(L[n+1,k+1])

/* Generated as the Ternary Fractal: */ T(n,k)=local(r=n,c=k,s=floor(log(n+1)/log(3))+1,u=vector(s),v=vector(s),d,e); if(n<=k,0,if(n<3&k<3,[0,0,0;2,0,0;3,-2,0][n+1,k+1], for(i=1,#u,u[i]=r%3;r=r\3);for(i=1,#v,v[i]=c%3;c=c\3); d=0;for(i=1,#v,if(u[i]!=v[i],d+=1;e=i));if(d==1,T(u[e],v[e]),0)))

Ternary fractal, T(3*n,3*k) = T(n,k), defined by: T(n,k) = 0 if n<=k or when more than 1 digit differs between the ternary expansions of n and k; else T(n,k) = T(m,j) where the only ternary digits of n, k, that differ are m, j, respectively and T(1,0)=2, T(2,1)=-2, T(2,0)=3.

A120855 Row sums of triangle A120854, which is the matrix log of triangle A117939.

Original entry on oeis.org

0, 2, 1, 2, 4, 3, 1, 3, 2, 2, 4, 3, 4, 6, 5, 3, 5, 4, 1, 3, 2, 3, 5, 4, 2, 4, 3, 2, 4, 3, 4, 6, 5, 3, 5, 4, 4, 6, 5, 6, 8, 7, 5, 7, 6, 3, 5, 4, 5, 7, 6, 4, 6, 5, 1, 3, 2, 3, 5, 4, 2, 4, 3, 3, 5, 4, 5, 7, 6, 4, 6, 5, 2, 4, 3, 4, 6, 5, 3, 5, 4, 2, 4, 3, 4, 6, 5, 3, 5, 4, 4, 6, 5, 6, 8, 7, 5, 7, 6, 3, 5, 4, 5, 7, 6

Offset: 0

Paul D. Hanna, Jul 08 2006

nonn

Triangle A117939 is related to powers of 3 partitions of n and is the matrix square of A117947(n,k) = balanced ternary digits of C(n,k) mod 3, also A117947(n,k) = L(C(n,k)/3) where L(j/p) is the Legendre symbol of j and p.

Cf. A120854, A117947; A062756, A081603, A053735.

f[n_] := DigitCount[n, 3] /. {a_, b_, c_} -> 2a + b + 0c; Array[f, 105, 0] (* Robert G. Wilson v, Jul 31 2012 *)

{a(n)=local(M=matrix(n+1,n+1,r,c,(binomial(r-1,c-1)+1)%3-1)^2, L=sum(i=1,#M,-(M^0-M)^i/i));return(sum(k=0,n,L[n+1,k+1]))}

a(n) = 2*A062756 + A081603(n), where A062756(n) = number of 1's in ternary expansion of n and A081603(n) = number of 2's in ternary expansion of n.

cp's OEIS Frontend

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.

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A117939 Triangle related to powers of 3 partitions of n.

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A117963 Antidiagonal sums of a Legendre-binomial triangle for p = 3.

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A120854 Matrix log of A117939, read by rows, consisting only of 0's, 3's and signed 2's.

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A120855 Row sums of triangle A120854, which is the matrix log of triangle A117939.

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A120880 G.f. satisfies: A(x) = A(x^3)(1 + 2x + x^2); thus a(n) = 2^A062756(n), where A062756(n) is the number of 1's in the ternary expansion of n.