0,5

Row sums, after dividing terms by binomial coefficients, start with {1,2} and after that repeat {3,4}. These coefficients are not minimal, but do obey a definite rule.

T(0, k)=T(k, k)=1 (k>=0); for n>1: T(2n, n)=(n+1), T(2n+1, n)=T(2n+1, n+1)=(n+2); for 0<k<n: T(2n, k)=T(2n, 2n-k)=n-k, T(2n+1, k)=T(2n+1, 2n+1-k)=2n+1-2k. For n>0, sum_{k=0..2n}T(2n, k)/C(2n, k)=3, and sum_{k=0..2n+1}T(2n+1, k)/C(2n+1, k)=4.

Sum of T(n,k)/C(n,k) as k=0..n equals 3 (n even) or 4 (n odd):

(n=6) 1/1 + 2/6 + 1/15 + 4/20 + 1/15 + 2/6 + 1/1 = 3;

(n=7) 1/1 + 5/7 + 3/21 + 5/35 + 5/35 + 3/21 + 5/7 + 1/1 = 4.

Triangle begins:

1;

1,1;

1,2,1;

1,3,3,1;

1,1,3,1,1;

1,3,4,4,3,1;

1,2,1,4,1,2,1;

1,5,3,5,5,3,5,1;

1,3,2,1,5,1,2,3,1;

1,7,5,3,6,6,3,5,7,1;

1,4,3,2,1,6,1,2,3,4,1;

1,9,7,5,3,7,7,3,5,7,9,1; ...

(PARI) {T(n, k)=if(n<k|k<0, 0, if(n==k|k==0, 1, if(k==n\2, (n+3)\2, if(k>n\2, T(n, n-k), if(n%2==0, n\2-k, n-2*k)))))}

Sequence in context: A048805 A129571 A034931 this_sequence A026082 A117185 A129181

Adjacent sequences: A090399 A090400 A090401 this_sequence A090403 A090404 A090405

nonn,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Nov 28 2003