A100472 Inverse modulo 2 modulo transform of 9^n.

1, 8, 80, 640, 6560, 52480, 524800, 4198400, 43046720, 344373760, 3443737600, 27549900800, 282386483200, 2259091865600, 22590918656000, 180727349248000, 1853020188851840, 14824161510814720, 148241615108147200

Offset: 0

Author

Paul Barry, Dec 06 2004

Keywords

• easy
• nonn

Comments

9^n may be retrieved as 9^n = Sum_{k=0..n} (binomial(n,k) mod 2)*a(k).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A001019, A010060, A047999, A106400.

Programs

Magma

A106400:= func< n | 1 - 2*(&+Intseq(n, 2) mod(2)) >;
A100472:= func< n | (&+[A106400(n-j)*(Binomial(n,j) mod 2)*9^j: j in [0..n]]) >;
[A100472(n): n in [0..30]]; // G. C. Greubel, Apr 06 2023

Mathematica

A100472[n_]:= A100472[n]= Sum[(-1)^ThueMorse[n-j]*Mod[Binomial[n, j], 2]*9^j, {j,0,n}];
Table[A100472[n], {n,0,30}] (* G. C. Greubel, Apr 06 2023 *)

SageMath

@CachedFunction
def A010060(n): return (bin(n).count('1')%2)
def A100472(n): return sum((-1)^A010060(n-k)*(binomial(n, k)%2)*9^k for k in range(n+1))
[A100472(n) for n in range(31)] # G. C. Greubel, Apr 06 2023

Formula

a(n) = Sum_{k=0..n} (-1)^A010060(n-k) * (binomial(n, k) mod 2) * 9^k.

a(n) = Sum_{k=0..n} A106400(n-k) * (binomial(n, k) mod 2) * 9^k. - G. C. Greubel, Apr 06 2023