A085405 Common residues of binomial(3n+2,n+1)/(3n+2) modulo 2.

1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Author

Paul D. Hanna, Jun 29 2003

Keywords

• nonn

Comments

The positions of ones are given by A022340 and runs of zeros are given by A085407: both are related to the Fibonacci sequence.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..65539
• Index entries for characteristic functions

Crossrefs

Cf. A005940, A006013, A022340 (ones), A085407 (zeros), A085357, A277332, A323239.

Programs

PARI

A085405(n) = ((binomial((3*n)+2, n+1)/((3*n)+2))%2); \\ Antti Karttunen, Jan 12 2019

PARI

A085405(n) = if(n%2,0,while(n>0, my(nextn=(n>>1)); if(1==(nextn%2)*(n%2), return(0)); n = nextn); (1)); \\ (Much faster than above program) - Antti Karttunen, Jan 12 2019

Formula

a(n) = C(3n+2, n+1)/(3n+2) (Mod 2) = A006013(n) (Mod 2), where A006013 is the self-convolution of A001764 (ternary trees).

a(n) = A323239(A005940(1+n)). - Antti Karttunen, Jan 12 2019