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A351725 Table T(n,k) read by rows: number of partitions of n into k parts of size 1, 5, 10 or 25.

%I A351725 #29 May 16 2022 04:58:50
%S A351725 1,0,1,0,0,1,0,0,0,1,0,0,0,0,1,0,1,0,0,0,1,0,0,1,0,0,0,1,0,0,0,1,0,0,
%T A351725 0,1,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,1,1,0,0,0,1,0,0,0,1,0,0,
%U A351725 1,1,0,0,0,1,0,0,0,1,0,0,0,1,1,0,0,0,1,0,0,0,1,0,0,0,0,1,1,0,0,0,1,0,0,0,1,0,0,0,0,0,1,1,0,0,0,1,0,0,0,1,0,0,1,1,0,0,1,1,0,0,0,1,0,0,0,1
%N A351725 Table T(n,k) read by rows: number of partitions of n into k parts of size 1, 5, 10 or 25.
%C A351725 Multiset transform of the binary sequence b(n)=1,1,0,0,0,1,0,0,0,0,1,0,... with g.f. 1 + x + x^5 + x^10 + x^25, where b(.) is the Inverse Euler Transform of A001299.
%H A351725 Alois P. Heinz, <a href="/A351725/b351725.txt">Rows n = 0..360, flattened</a>
%H A351725 <a href="/index/Mag#change">Index entries for sequences related to making change</a>.
%F A351725 T(n,0) = 0 if k>0.
%F A351725 T(n,n) = 1.
%F A351725 Sum_{k=0..n} k * T(n,k) = A351740(n). - _Alois P. Heinz_, Feb 17 2022
%e A351725 T(30,6)=2 counts the partitions 5+5+5+5+5+5 = 1+1+1+1+1+25.
%e A351725 The triangle starts at row n=0 and has columns k=0..n:
%e A351725 1
%e A351725 0 1
%e A351725 0 0 1
%e A351725 0 0 0 1
%e A351725 0 0 0 0 1
%e A351725 0 1 0 0 0 1
%e A351725 0 0 1 0 0 0 1
%e A351725 0 0 0 1 0 0 0 1
%e A351725 0 0 0 0 1 0 0 0 1
%e A351725 0 0 0 0 0 1 0 0 0 1
%e A351725 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 0 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 0 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 1 1 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 0 1 1 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 0 0 1 1 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 0 0 0 1 1 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 0 0 0 0 1 1 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 0 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%e A351725 0 0 1 1 1 1 2 0 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1
%p A351725 b:= proc(n, i) option remember; `if`(n=0 or i=1, x^n, b(n, i-1)+
%p A351725      (p-> `if`(p>n, 0, expand(x*b(n-p, i))))([1, 5, 10, 25][i]))
%p A351725     end:
%p A351725 T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n, 4)):
%p A351725 seq(T(n), n=0..15);  # _Alois P. Heinz_, Feb 17 2022
%t A351725 b[n_, i_] := b[n, i] = If[n == 0 || i == 1, x^n, b[n, i - 1] +
%t A351725      Function[p, If[p > n, 0, Expand[x*b[n-p, i]]]][{1, 5, 10, 25}[[i]]]];
%t A351725 T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, n}]][b[n, 4]];
%t A351725 Table[T[n], {n, 0, 15}] // Flatten (* _Jean-François Alcover_, May 16 2022, after _Alois P. Heinz_ *)
%Y A351725 Cf. A001299 (row sums), A351740.
%Y A351725 Column k=0 gives A000007.
%Y A351725 Main diagonal gives A000012.
%Y A351725 T(2n,n) gives A351742.
%K A351725 nonn,easy,look,tabl
%O A351725 0,472
%A A351725 _R. J. Mathar_, Feb 17 2022