A137821 Numbers k such that Sum_{j=1..2k} Catalan(j) == 0 (mod 3).

1, 4, 6, 13, 15, 18, 19, 40, 42, 45, 46, 54, 55, 58, 60, 121, 123, 126, 127, 135, 136, 139, 141, 162, 163, 166, 168, 175, 177, 180, 181, 364, 366, 369, 370, 378, 379, 382, 384, 405, 406, 409, 411, 418, 420, 423, 424, 486, 487, 490, 492, 499, 501, 504, 505

Offset: 1

Author

M. F. Hasler, Feb 25 2008

Keywords

• nonn

Comments

It would be natural to prepend an initial term a(1)=0 (for which the sum is to be considered empty, thus zero), but we omit it to avoid confusion w.r.t. indices of A107755.

Links

• M. F. Hasler, Table of n, a(n) for n = 1..499.

Crossrefs

Cf. A107755 (twice this), A137822-A137824.

Programs

Mathematica

Flatten[Position[Accumulate[CatalanNumber[Range[1100]]],?(Mod[#,3]==0&)]]/2 (* _Harvey P. Dale, Jun 19 2025 *)

PARI

n=0; A137821=vector(499,i,{ if( bitand(i,i-1), while(n++ & s+=binomial(4*n-2,2*n-1)/(2*n)*(10*n-1)/(2*n+1),),s=Mod(0,3); n=2*n+1+log(i+.5)\log(2)%2 ); n})

Formula

a(n) = A107755(n)/2 = Sum_{k=0..n} A137822(k).

a(2^j) = 2 a(2^j-1) + 1 (resp. +2) for j even (resp. odd).

Sum_{k=1..2n} Catalan(k) = Sum_{k=1..n} Catalan(2k-1) * (10k-1)/(2k+1), thus:

{ a(m) } = { n>0 | Sum_{k=1..n} Catalan(2k-1) * (10k-1)/(2k+1) == 0 (mod 3) }.