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A090402
Symmetric triangle, read by rows, where the terms in each row, after being divided by their corresponding binomial coefficients, have an integer sum.
0
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.
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
KEYWORD
nonn,tabl
AUTHOR
Paul D. Hanna, Nov 28 2003
STATUS
approved