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Triangle read by rows: T(n,k) is the total number of appearances of k as the smallest part in all partitions of n.
5

%I #48 Dec 18 2023 14:48:57

%S 1,2,1,4,0,1,7,2,0,1,12,1,0,0,1,19,4,2,0,0,1,30,3,1,0,0,0,1,45,8,1,2,

%T 0,0,0,1,67,7,4,1,0,0,0,0,1,97,15,3,1,2,0,0,0,0,1,139,15,4,1,1,0,0,0,

%U 0,0,1,195,27,8,4,1,2,0,0,0,0,0,1,272,29,8,3,1,1,0,0,0,0,0,0,1

%N Triangle read by rows: T(n,k) is the total number of appearances of k as the smallest part in all partitions of n.

%C In other words, T(n,k) is the total number of appearances of k in all partitions of n whose smallest part is k.

%C The sum of row n equals spt(n), the smallest part partition function (see A092269).

%C T(n,k) is also the sum of row k in the slice n of tetrahedron A209314.

%H Alois P. Heinz, <a href="/A220504/b220504.txt">Rows n = 1..141, flattened</a>

%e Triangle begins:

%e 1;

%e 2, 1;

%e 4, 0, 1;

%e 7, 2, 0, 1;

%e 12, 1, 0, 0, 1;

%e 19, 4, 2, 0, 0, 1;

%e 30, 3, 1, 0, 0, 0, 1;

%e 45, 8, 1, 2, 0, 0, 0, 1;

%e 67, 7, 4, 1, 0, 0, 0, 0, 1;

%e 97, 15, 3, 1, 2, 0, 0, 0, 0, 1;

%e 139, 15, 4, 1, 1, 0, 0, 0, 0, 0, 1;

%e 195, 27, 8, 4, 1, 2, 0, 0, 0, 0, 0, 1;

%e 272, 29, 8, 3, 1, 1, 0, 0, 0, 0, 0, 0, 1;

%e ...

%e The partitions of 6 with the smallest part in brackets are

%e ..........................

%e . [6]

%e ..........................

%e . [3]+[3]

%e ..........................

%e . 4 +[2]

%e . [2]+[2]+[2]

%e ..........................

%e . 5 +[1]

%e . 3 + 2 +[1]

%e . 4 +[1]+[1]

%e . 2 + 2 +[1]+[1]

%e . 3 +[1]+[1]+[1]

%e . 2 +[1]+[1]+[1]+[1]

%e . [1]+[1]+[1]+[1]+[1]+[1]

%e ..........................

%e There are 19 smallest parts of size 1. Also there are four smallest parts of size 2. Also there are two smallest parts of size 3. There are no smallest part of size 4 or 5. Finally there is only one smallest part of size 6. So row 6 gives 19, 4, 2, 0, 0, 1. The sum of row 6 is 19+4+2+0+0+1 = A092269(6) = 26.

%p b:= proc(n, i) option remember; local j, r; if n=0 or i<1 then 0

%p else `if`(irem(n, i, 'r')=0, [0$(i-1), r], []); for j from 0

%p to n/i do zip((x, y)->x+y, %, [b(n-i*j, i-1)], 0) od; %[] fi

%p end:

%p T:= n-> b(n, n):

%p seq(T(n), n=1..20); # _Alois P. Heinz_, Jan 20 2013

%t b[n_, i_] := b[n, i] = Module[{j, q, r, pc}, If [n == 0 || i<1, 0, {q, r} = QuotientRemainder[n, i]; pc = If[r == 0, Append[Array[0&, i-1], q], {}]; For[j = 0, j <= n/i, j++, pc = Plus @@ PadRight[{pc, b[n-i*j, i-1]}]]; pc]]; T[n_] := b[n, n]; Table[T[n], {n, 1, 20}] // Flatten (* _Jean-François Alcover_, Jan 30 2014, after _Alois P. Heinz_ *)

%Y Columns 1-3: A000070, A087787, A174455.

%Y Row sums give A092269.

%Y Cf. A026794, A182703, A209314.

%K nonn,tabl

%O 1,2

%A _Omar E. Pol_, Jan 19 2013