A078113 Let u(1)=u(2)=1, u(3)=n, u(k) = (1/2)*abs(2*u(k-1) -u(k-2)-u(k-3)); sequence gives values of n such that Sum_{k>=1} u(k) is an integer.

2, 6, 7, 15, 17, 33, 37, 69, 77, 141, 157, 285, 317, 573, 637, 1149, 1277, 2301, 2557, 4605, 5117, 9213, 10237, 18429, 20477, 36861, 40957, 73725, 81917, 147453, 163837, 294909, 327677, 589821, 655357, 1179645, 1310717, 2359293, 2621437, 4718589, 5242877

Offset: 1

Author

Benoit Cloitre, Dec 04 2002

Keywords

• nonn
• more

Comments

u(3)=7, Sum_{k>=1} u(k) = 28 is an integer, hence 7 is in the sequence.

Crossrefs

Cf. A154117, A133257.

Programs

PARI

A078113(maxn, maxk) = {
  u=vector(maxk);
  u[1]=1; u[2]=1;
  for(n=1, maxn,
    u[3]=n;
    for(k=4, maxk, u[k]=abs(2*u[k-1]-u[k-2]-u[k-3])/2);
    s=sum(i=1, maxk, u[i]);
    if(ceil(s)-s < 1E-11, print1(n, ", ")) \\ Arbitrary 1E-11
  )
}
A078113(1000000, 200) \\ Colin Barker, Aug 14 2013

Formula

Conjecture: a(n) = -3+2^(1/2*(-5+n))*(10-10*(-1)^n+9*sqrt(2)+9*(-1)^n*sqrt(2)). a(n) = a(n-1)+2*a(n-2)-2*a(n-3). G.f.: x*(3*x^2-4*x-2) / ((x-1)*(2*x^2-1)). - Colin Barker, Aug 14 2013

Conjecture: a(n) = 2*a(n-2) + 3, n odd>2 = A154117((n+1)/2). - Bill McEachen, Jun 21 2025

Extensions

a(11)-a(33) from Colin Barker, Aug 14 2013

a(34)-a(41) from Bill McEachen, Jun 21 2025