%I #44 Apr 12 2023 09:23:31

%S 1,2,3,4,5,6,6,7,0,8,0,9,9,10,0,11,0,12,12,13,0,14,0,15,15,15,16,0,0,

%T 17,0,0,18,18,0,19,0,0,20,0,20,21,21,0,22,0,0,23,0,0,24,24,0,25,0,25,

%U 26,0,0,27,27,0,28,0,0,28,29,0,0,0,30,30,30,0,31,0,0,0,32,0,0,0

%N Irregular triangle read by rows: T(n,k) is the sum of all parts of the partition of n into consecutive parts that contains 2*k-1 parts, and the first element of the column k is in row A000384(k).

%C This triangle is formed from the odd-indexed columns of the triangle A285891.

%H Paolo Xausa, <a href="/A352499/b352499.txt">Table of n, a(n) for n = 1..10490</a> (rows 1..800 of triangle, flattened).

%F T(n,k) = n*A351824(n,k).

%F T(n,k) = n*[(2*k-1)|n], where 1 <= k <= floor((sqrt(8*n+1)+1)/4) and [] is the Iverson bracket. - _Paolo Xausa_, Apr 12 2023

%e Triangle begins:

%e 1;

%e 2;

%e 3;

%e 4;

%e 5;

%e 6, 6;

%e 7, 0;

%e 8, 0;

%e 9, 9;

%e 10, 0;

%e 11, 0;

%e 12, 12;

%e 13, 0;

%e 14, 0;

%e 15, 15, 15;

%e 16, 0, 0;

%e 17, 0, 0;

%e 18, 18, 0;

%e 19, 0, 0;

%e 20, 0, 20;

%e 21, 21, 0;

%e 22, 0, 0;

%e 23, 0, 0;

%e 24, 24, 0;

%e 25, 0, 25;

%e 26, 0, 0;

%e 27, 27, 0;

%e 28, 0, 0, 28;

%e ...

%e For n = 21 the partitions of 21 into on odd number of consecutive parts are [21] and [8, 7, 6], so T(21,1) = 1 and T(21,2) = 8 + 7 + 6 = 21. There is no partition of 21 into five consecutive parts so T(21,3) = 0.

%t A352499[rowmax_]:=Table[Boole[Divisible[n,2k-1]]n,{n,rowmax},{k,Floor[(Sqrt[8n+1]+1)/4]}];A352499[50] (* _Paolo Xausa_, Apr 12 2023 *)

%Y Row sums give A352257.

%Y Row n has A351846(n) terms.

%Y The number of nonzero terms in row n equals A082647(n).

%Y Cf. A000384, A237048, A245579, A285891, A299765, A351824, A352425.

%K nonn,tabf

%O 1,2

%A _Omar E. Pol_, Mar 19 2022