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

0,1

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

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

Prog

(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

Crossrefs

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

Sequence in context: A284957 A357385 A316533 * A036988 A108357 A309848

Adjacent sequences: A085402 A085403 A085404 * A085406 A085407 A085408

Keyword

nonn

Author

Paul D. Hanna, Jun 29 2003

Status

approved