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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.
1
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
OFFSET
1,1
COMMENTS
u(3)=7, sum(k>=1,u(k)) = 28 is an integer, hence 7 is in the sequence.
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
PROG
(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
CROSSREFS
Sequence in context: A344343 A226965 A128918 * A286054 A340376 A369792
KEYWORD
nonn,more
AUTHOR
Benoit Cloitre, Dec 04 2002
EXTENSIONS
More terms from Colin Barker, Aug 14 2013
STATUS
approved