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Triangle read by rows: T(n,k) = number of palindromic compositions of n in which no part exceeds k, 1 <= k <= n.
4

%I #22 Mar 09 2015 05:33:13

%S 1,1,2,1,1,2,1,3,3,4,1,2,3,3,4,1,5,6,7,7,8,1,3,6,6,7,7,8,1,8,11,14,14,

%T 15,15,16,1,5,11,12,14,14,15,15,16,1,13,20,27,28,30,30,31,31,32,1,8,

%U 20,23,28,28,30,30,31,31,32,1,21,37,52,55,60,60,62,62,63,63,64

%N Triangle read by rows: T(n,k) = number of palindromic compositions of n in which no part exceeds k, 1 <= k <= n.

%C A palindromic composition of a natural number m is an ordered partition of m into N+1 natural numbers (or parts), p_0, p_1, ..., p_N, of the form m = p_0 + p_1 + ... + p_N such that p_j = p_{N-j}, for each j in {0,...,N}. Two palindromic compositions, sum_{j=0..N} p_j and sum_{j=0..N} q_j (say), are identical if and only if p_j = q_j, j = 0,...,N; otherwise they are taken to be distinct.

%C Partial sums of rows of A233323.

%C T(n,k) is defined for n,k >= 0. T(n,k) = T(n,n) = A016116(n) for k>= 0. - _Alois P. Heinz_, Dec 11 2013

%H Alois P. Heinz, <a href="/A233324/b233324.txt">Rows n = 1..141, flattened</a>

%H V. E. Hoggatt, Jr., and Marjorie Bicknell, <a href="http://www.fq.math.ca/Scanned/13-4/hoggatt1.pdf">Palindromic compositions</a>, Fibonacci Quart., Vol. 13(4), 1975, pp. 350-356.

%e Triangle T(n,k) begins:

%e 1;

%e 1, 2;

%e 1, 1, 2;

%e 1, 3, 3, 4;

%e 1, 2, 3, 3, 4;

%e 1, 5, 6, 7, 7, 8;

%e 1, 3, 6, 6, 7, 7, 8;

%e 1, 8, 11, 14, 14, 15, 15, 16;

%e 1, 5, 11, 12, 14, 14, 15, 15, 16;

%e 1, 13, 20, 27, 28, 30, 30, 31, 31, 32;

%p T:= proc(n, k) option remember; `if`(n<=k, 1, 0)+

%p add(T(n-2*j, k), j=1..min(k, iquo(n, 2)))

%p end:

%p seq(seq(T(n, k), k=1..n), n=1..14); # _Alois P. Heinz_, Dec 11 2013

%t T[n_, k_] := T[n, k] = If[n <= k, 1, 0] + Sum[T[n-2*j, k], {j, 1, Min[k, Quotient[ n, 2]]}]; Table[Table[T[n, k], {k, 1, n}], {n, 1, 14}] // Flatten (* _Jean-François Alcover_, Mar 09 2015, after _Alois P. Heinz_ *)

%o (PARI) T(n,k)=if(n<1,return(n==0));sum(i=1,k,T(n-2*i,k))+(n<=k) \\ _Charles R Greathouse IV_, Dec 11 2013

%Y Cf. A233323.

%Y T(n,2) = A053602(n+1) = A123231(n). T(2n,3) = A001590(n+3). T(2n,4) = A001631(n+4). - _Alois P. Heinz_, Dec 11 2013

%K nonn,tabl

%O 1,3

%A _L. Edson Jeffery_, Dec 11 2013