A089893 a(n) = (A001317(2n)-1)/4.

0, 1, 4, 21, 64, 321, 1092, 5461, 16384, 81921, 278532, 1392661, 4210752, 21053761, 71582788, 357913941, 1073741824, 5368709121, 18253611012, 91268055061, 275951648832, 1379758244161, 4691178030148, 23455890150741

Offset: 0

Author

Ralf Stephan, Jan 10 2004

Keywords

• nonn

Comments

a(n) = (A038192(n) - 3)/12 = (A038183(n) - 1)/4, reflecting the fact that, if you look at each row of the Pascal triangle's parity as a binary number (cf. A001317), then the numbers in odd rows are thrice the numbers in even rows.

Conjecture: a(2^k) = 2^(2^(k+1)-2). [This conjecture is true. - Vladimir Shevelev, Nov 28 2010]

Conjectures: lim(n->inf, a(2n+1)/a(2n)) = 5, lim(n->inf, a(4n+2)/a(4n+1)) = 17/5, lim(n->inf, a(8n+4)/a(8n+3)) = 257/85 etc. [This follows from the formula, for n>=0, t>=1: ( 4*a(2^t*n+2^(t-1))+1 )/( 4*a(2^t*n+2^(t-1)-1)+1 ) = 3*F_t/(F_t-2), where F_t= A000215(t) - Vladimir Shevelev, Nov 28 2010]

Links

• Vladimir Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2, arXiv:1011.6083 [math.NT], 2010-2012.

Crossrefs

Cf. A000215, A001317, A038183, A038192.

Programs

Mathematica

a1317[n_] := Sum[2^k Mod[Binomial[n, k], 2] , {k, 0, n}];
a[n_] := (a1317[2n] - 1)/4;
Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Jan 18 2019 *)

PARI

a(n)=(sum(k=0,2*n+1,(binomial(2*n+1,k)%2)*2^k)-3)/12

Python

def A089893(n): return sum((bool(~(m:=n<<1)&m-k)^1)<>2 # Chai Wah Wu, May 02 2023