A090402 Symmetric triangle, read by rows, where the terms in each row, after being divided by their corresponding binomial coefficients, have an integer sum.

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, 1, 5, 4, 3, 2, 1, 7, 1, 2, 3, 4, 5, 1, 1, 11, 9, 7, 5, 3, 8, 8, 3, 5, 7, 9, 11

Offset

0,5

Comments

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.

Links

Table of n, a(n) for n=0..103.

Formula

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.

Example

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; ...

Prog

(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)))))

Crossrefs

Sequence in context: A248473 A307116 A212626 * A026082 A117185 A129181

Adjacent sequences: A090399 A090400 A090401 * A090403 A090404 A090405

Keyword

nonn,tabl

Author

Paul D. Hanna, Nov 28 2003

Status

approved