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A359350 Irregular triangle T(n,k) (n >= 1, k >= 1) read by rows: row n is constructed by replacing A336811(n,k) with the largest partition into consecutive parts of A000217(A336811(n,k)).

%I A359350 #39 Sep 01 2023 14:16:44
%S A359350 1,2,1,3,2,1,1,4,3,2,1,2,1,1,5,4,3,2,1,3,2,1,2,1,1,1,6,5,4,3,2,1,4,3,
%T A359350 2,1,3,2,1,2,1,2,1,1,1,7,6,5,4,3,2,1,5,4,3,2,1,4,3,2,1,3,2,1,3,2,1,2,
%U A359350 1,2,1,1,1,1,1,8,7,6,5,4,3,2,1,6,5,4,3,2,1,5,4,3,2,1,4,3,2,1,4,3,2,1,3,2,1
%N A359350 Irregular triangle T(n,k) (n >= 1, k >= 1) read by rows: row n is constructed by replacing A336811(n,k) with the largest partition into consecutive parts of A000217(A336811(n,k)).
%C A359350 All divisors of all terms in row n are also all parts of all partitions of n.
%C A359350 The terms of row n listed in nonincreasing order give the n-th row of A176206.
%C A359350 The number of k's in row n is equal to A000041(n-k), 1 <= k <= n.
%C A359350 The number of terms >= k in row n is equal to A000070(n-k), 1 <= k <= n.
%C A359350 The number of k's in the first n rows (or in the first A014153(n-1) terms of the sequence) is equal to A000070(n-k), 1 <= k <= n.
%C A359350 The number of terms >= k in the first n rows (or in the first A014153(n-1) terms of the sequence) is equal to A014153(n-k), 1 <= k <= n.
%C A359350 Row n is constructed replacing A336811(n,k) with the largest partition into consecutive parts of A359279(n,k).
%H A359350 Paolo Xausa, <a href="/A359350/b359350.txt">Table of n, a(n) for n = 1..10980</a> (rows 1..21 of the triangle, flattened)
%e A359350 Triangle begins:
%e A359350   1;
%e A359350   2, 1;
%e A359350   3, 2, 1, 1;
%e A359350   4, 3, 2, 1, 2, 1, 1;
%e A359350   5, 4, 3, 2, 1, 3, 2, 1, 2, 1, 1, 1;
%e A359350   6, 5, 4, 3, 2, 1, 4, 3, 2, 1, 3, 2, 1, 2, 1, 2, 1, 1, 1;
%e A359350   ...
%e A359350 Or also the triangle begins:
%e A359350   [1];
%e A359350   [2, 1];
%e A359350   [3, 2, 1],          [1];
%e A359350   [4, 3, 2, 1],       [2, 1],       [1];
%e A359350   [5, 4, 3, 2, 1],    [3, 2, 1],    [2, 1],    [1],    [1];
%e A359350   [6, 5, 4, 3, 2, 1], [4, 3, 2, 1], [3, 2, 1], [2, 1], [2, 1], [1], [1];
%e A359350   ...
%e A359350 For n = 3 the third row is [3, 2, 1, 1]. The divisors of these terms are [1, 3], [1, 2], [1], [1]. These six divisors are also all parts of all partitions of 3. They are [3], [2, 1], [1, 1, 1].
%t A359350 A359350row[n_]:=Flatten[Table[ConstantArray[Range[n-m,1,-1],PartitionsP[m]-PartitionsP[m-1]],{m,0,n-1}]];Array[A359350row,10] (* _Paolo Xausa_, Sep 01 2023 *)
%Y A359350 Row sums give A014153 (convolution of A000041 and A000027).
%Y A359350 Row lengths give A000070.
%Y A359350 Row n has A000041(n-1) blocks.
%Y A359350 This triangle has the same row sums as A176206, A299779 and A359279.
%Y A359350 Cf. A000217, A336811, A336812, A338156, A359280.
%K A359350 nonn,tabf
%O A359350 1,2
%A A359350 _Omar E. Pol_, Dec 27 2022