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A317489
Irregular triangle read by rows. For n >= 3 and 1 <= k <= floor(n/3), T(n,k) is the number of palindromic compositions of n into k parts of size at least 3.
0
1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 2, 1, 1, 2, 1, 1, 0, 3, 0, 1, 1, 3, 2, 1, 0, 4, 0, 1, 1, 1, 4, 3, 1, 1, 0, 5, 0, 3, 1, 1, 5, 4, 3, 1, 1, 0, 6, 0, 6, 0, 1, 1, 6, 5, 6, 3, 1, 0, 7, 0, 10, 0, 1, 1, 1, 7, 6, 10, 6, 1, 1, 0, 8, 0, 15, 0, 4, 1, 1, 8, 7, 15, 10, 4, 1, 1, 0, 9, 0, 21, 0, 10, 0, 1, 1, 9, 8, 21, 15, 10, 4, 1, 0, 10, 0, 28, 0, 20, 0, 1, 1, 1, 10, 9, 28, 21, 20, 10, 1, 1, 0, 11, 0, 36, 0, 35, 0, 5
OFFSET
3,18
FORMULA
T(n,k) = 0 if n is odd and k is even;
T(n,k) = binomial((n-1)/2-k,(k-1)/2) if n is odd and k is odd;
T(n,k) = binomial((n-2)/2-k,(k-1)/2) if n is even and k is odd;
T(n,k) = binomial((n-2)/2-k,(k-2)/2) if n is even and k is even.
EXAMPLE
For n=24 and k=3, T(24,3) = 8 = binomial((24-2)/2-3, (3-1)/2) = binomial(8,1).
The first entries of the irregular triangle formed by the values of T(n,k) are:
1;
1;
1;
1, 1;
1, 0;
1, 1;
1, 0, 1;
1, 1, 1;
1, 0, 2;
1, 1, 2, 1;
1, 0, 3, 0;
1, 1, 3, 2;
1, 0, 4, 0, 1;
1, 1, 4, 3, 1;
1, 0, 5, 0, 3;
1, 1, 5, 4, 3, 1;
1, 0, 6, 0, 6, 0;
1, 1, 6, 5, 6, 3;
1, 0, 7, 0, 10, 0, 1;
1, 1, 7, 6, 10, 6, 1;
1, 0, 8, 0, 15, 0, 4;
1, 1, 8, 7, 15, 10, 4, 1;
1, 0, 9, 0, 21, 0, 10, 0;
1, 1, 9, 8, 21, 15, 10, 4;
1, 0, 10, 0, 28, 0, 20, 0, 1;
1, 1, 10, 9, 28, 21, 20, 10, 1;
1, 0, 11, 0, 36, 0, 35, 0, 5;
MATHEMATICA
T[n_, k_] := If[Mod[n, 2] == 1 && Mod[k, 2] == 0, 0, Binomial[Quotient[n-1, 2] - k, Quotient[k-1, 2]]];
Table[T[n, k], {n, 3, 30}, {k, 1, Quotient[n, 3]}] // Flatten (* Jean-François Alcover, Sep 13 2018, from PARI *)
PROG
(PARI) T(n, k)=if(n%2==1&&k%2==0, 0, binomial((n-1)\2-k, (k-1)\2)); \\ Andrew Howroyd, Sep 07 2018
CROSSREFS
Row sums of the triangle equal A226916(n+4).
Sequence in context: A143158 A336708 A308424 * A345647 A091950 A014750
KEYWORD
nonn,tabf
AUTHOR
STATUS
approved