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%I #16 May 26 2024 12:18:41
%S 1,0,1,0,0,1,0,0,2,1,0,0,0,3,1,0,0,0,3,4,1,0,0,0,6,6,5,1,0,0,0,0,16,
%T 10,6,1,0,0,0,0,12,30,15,7,1,0,0,0,0,12,35,50,21,8,1,0,0,0,0,24,50,75,
%U 77,28,9,1,0,0,0,0,0,90,126,140,112,36,10,1
%N Triangle read by rows: T(n,k) is the number of complete compositions of n with k parts.
%C A composition (ordered partition) is complete if the set of parts both covers an interval (is gap-free) and contains 1.
%H Alois P. Heinz, <a href="/A371417/b371417.txt">Rows n = 0..200, flattened</a>
%F T(n,k) = k!*[z^n*t^k] Sum_{i>0} z^(i*(i+1)/2)*t^i * Product_{j=1..i} Sum_{k>=0} (z^(j*k)*t^k)/(k+1)!.
%e The triangle begins:
%e k=0 1 2 3 4 5 6 7 8 9 10
%e n=0: 1;
%e n=1: 0, 1;
%e n=2: 0, 0, 1;
%e n=3: 0, 0, 2, 1;
%e n=4: 0, 0, 0, 3, 1;
%e n=5: 0, 0, 0, 3, 4, 1;
%e n=6: 0, 0, 0, 6, 6, 5, 1;
%e n=7: 0, 0, 0, 0, 16, 10, 6, 1;
%e n=8: 0, 0, 0, 0, 12, 30, 15, 7, 1;
%e n=9: 0, 0, 0, 0, 12, 35, 50, 21, 8, 1;
%e n=10: 0, 0, 0, 0, 24, 50, 75, 77, 28, 9, 1;
%e ...
%e For n = 5 there are a total of 8 complete compositions:
%e T(5,3) = 3: (221), (212), (122)
%e T(5,4) = 4: (2111), (1211), (1121), (1112)
%e T(5,5) = 1: (11111)
%p b:= proc(n, i, t) option remember; `if`(n=0,
%p `if`(i=0, t!, 0), `if`(i<1 or n<i, 0, add(
%p expand(x^j*b(n-i*j, i-1, t+j)/j!), j=1..n/i)))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(add(b(n, i, 0), i=0..n)):
%p seq(T(n), n=0..12); # _Alois P. Heinz_, Apr 03 2024
%o (PARI)
%o G(N)={ my(z='z+O('z^N)); Vec(sum(i=1,N,z^(i*(i+1)/2)*t^i*prod(j=1,i,sum(k=0,N, (z^(j*k)*t^k)/(k+1)!))))}
%o my(v=G(10)); for(n=0, #v, if(n<1,print([1]), my(p=v[n], r=vector(n+1)); for(k=0, n, r[k+1] =k!*polcoeff(p, k)); print(r)))
%Y A107428 counts gap-free compositions.
%Y A251729 counts gap-free but not complete compositions.
%Y Cf. A107429 (row sums give complete compositions of n), A000670 (column sums), A152947 (number of nonzero terms per column).
%Y Cf. A066099, A124774, A373118.
%K nonn,easy,tabl,changed
%O 0,9
%A _John Tyler Rascoe_, Mar 23 2024
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