1,1

Y. More, Problem 11165, Amer. Math. Monthly, 112 (2005), 568.

R. J. Mathar, Table of n, a(n) for n=1,...,319.

a(2^j) = 2*a(2^j-1) + 2 (resp. + 4) if j is even (resp. odd). - M. F. Hasler, Feb 25 2008

a(n) = 2 sum( i=1..n, A137822(i) ) - Maximilian F. Hasler (MHasler(AT)univ-ag.fr), Mar 16 2008

{n: A137993(n-1) = 0}. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 07 2009]

A107755 := proc(n) option remember ; local a; if n = 1 then 2; else for a from A107755(n-1)+1 do if add(A000108(k), k=1..a) mod 3 = 0 then RETURN(a) ; fi ; od: fi ; end: # - R. J. Mathar, Feb 25 2008

c:=n->binomial(2*n, n)/(n+1): s:=0: for n from 1 to 1500 do s:=s+c(n): a[n]:=s mod 3: od: A:=[seq(a[n], n=1..1500)]: p:=proc(n) if A[n]=0 then n else fi end: seq(p(n), n=1..1500); (Deutsch)

s0 = s2 = {}; s = 0; Do[s = Mod[s + (2 n)!/n!/(n + 1)!, 3]; Switch[ Mod[s, 3], 0, AppendTo[s0, n], 2, AppendTo[s2, n]], {n, 1055}]; s0 (from Robert G. Wilson v (rgwv(AT)rgwv.com), Jun 14 2005)

(PARI) n=0; s=Mod(0, 3); A107755=vector(100, i, { if( bitand(i, i-1), while(n++&s+=binomial(2*n, n)/(n+1), ), s=Mod(0, 3); n=2*n+2+(log(i+.5)\log(2)%2)*2 ); /*print1(n", "); */ n)} \\ - M. F. Hasler, Feb 25 2008

(PARI) A107755(n)=sum( i=1, n, A137822(i) )*2 /* allows computation of a(10^4) in one second */ - Maximilian F. Hasler (MHasler(AT)univ-ag.fr), Mar 16 2008

Cf. A000108, A107756, A107757, A108784.

Cf. A137821-A137824.

Sequence in context: A013654 A108978 A135957 this_sequence A027718 A115102 A047174

Adjacent sequences: A107752 A107753 A107754 this_sequence A107756 A107757 A107758

nonn,easy

N. J. A. Sloane (njas(AT)research.att.com), Jun 11 2005

More terms from Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 12 2005

Corrected & extended by M. F. Hasler (MHasler(AT)univ-ag.fr) and Richard J. Mathar (mathar(AT)strw.leidenuniv.nl), Feb 25 2008