%I #64 Oct 16 2023 16:14:56

%S 1,2,1,1,3,1,1,1,4,2,2,1,1,1,1,5,1,1,1,1,1,6,3,3,2,2,2,1,1,1,1,1,1,7,

%T 1,1,1,1,1,1,1,8,4,4,2,2,2,2,1,1,1,1,1,1,1,1,9,3,3,3,1,1,1,1,1,1,1,1,

%U 1,10,5,5,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1

%N Triangle read by rows in which row n lists the parts of the partitions of n into equal parts, in nonincreasing order.

%C Row n has length sigma(n) = A000203(n).

%C Row sums give n*A000005(n) = A038040(n).

%C Column 1 is A000027.

%C Both columns 2 and 3 are A032742, n > 1.

%C For any k > 0 and t > 0, the sequence contains exactly one run of k consecutive t's. - _Rémy Sigrist_, Feb 11 2019

%C From _Omar E. Pol_, Dec 04 2019: (Start)

%C The number of parts congruent to 0 (mod m) in row m*n equals sigma(n) = A000203(n).

%C The number of parts greater than 1 in row n equals A001065(n), the sum of aliquot parts of n.

%C The number of parts greater than 1 and less than n in row n equals A048050(n), the sum of divisors of n except for 1 and n.

%C The number of partitions in row n equals A000005(n), the number of divisors of n.

%C The number of partitions in row n with an odd number of parts equals A001227(n).

%C The sum of odd parts in row n equals the sum of parts of the partitions in row n that have an odd number of parts, and equals the sum of all parts in the partitions of n into consecutive parts, and equals A245579(n) = n*A001227(n).

%C The decreasing records in row n give the n-th row of A056538.

%C Row n has n 1's which are all at the end of the row.

%C First n rows contain A000217(n) 1's.

%C The number of k's in row n is A126988(n,k).

%C The number of odd parts in row n is A002131(n).

%C The k-th block in row n has A027750(n,k) parts.

%C Right border gives A000012. (End)

%C The r-th row of the triangle begins at index k = A160664(r-1). - _Samuel Harkness_, Jun 21 2022

%H Samuel Harkness, <a href= "/A244051/b244051.txt" >Table of n, a(n) for n = 1..10000</a>

%H Samuel Harkness, <a href= "/A244051/a244051.jpg" >Log-log Scatterplot of the first 1000000 terms</a>

%e Triangle begins:

%e [1];

%e [2], [1,1];

%e [3], [1,1,1];

%e [4], [2,2], [1,1,1,1];

%e [5], [1,1,1,1,1];

%e [6], [3,3], [2,2,2], [1,1,1,1,1,1];

%e [7], [1,1,1,1,1,1,1];

%e [8], [4,4], [2,2,2,2], [1,1,1,1,1,1,1,1];

%e [9], [3,3,3], [1,1,1,1,1,1,1,1,1];

%e [10], [5,5], [2,2,2,2,2], [1,1,1,1,1,1,1,1,1,1];

%e [11], [1,1,1,1,1,1,1,1,1,1,1];

%e [12], [6,6], [4,4,4], [3,3,3,3], [2,2,2,2,2,2], [1,1,1,1,1,1,1,1,1,1,1,1];

%e [13], [1,1,1,1,1,1,1,1,1,1,1,1,1];

%e [14], [7,7], [2,2,2,2,2,2,2], [1,1,1,1,1,1,1,1,1,1,1,1,1,1];

%e [15], [5,5,5], [3,3,3,3,3], [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1];

%e [16], [8,8], [4,4,4,4], [2,2,2,2,2,2,2,2], [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1];

%e...

%e For n = 6 the 11 partitions of 6 are [6], [3, 3], [4, 2], [2, 2, 2], [5, 1], [3, 2], [4, 1, 1], [2, 2, 1, 1], [3, 1, 1, 1], [2, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1]. There are only four partitions of 6 that contain equal parts so the 6th row of triangle is [6], [3, 3], [2, 2, 2], [1, 1, 1, 1, 1, 1]. The number of parts equals sigma(6) = A000203(6) = 12. The row sum is A038040(6) = 6*A000005(6) = 6*4 = 24.

%e From _Omar E. Pol_, Dec 04 2019: (Start)

%e The structure of the above triangle is as follows:

%e 1;

%e 2 11;

%e 3 111;

%e 4 22 1111;

%e 5 11111;

%e 6 33 222 111111;

%e 7 1111111;

%e 8 44 2222 11111111;

%e 9 333 111111111;

%e... (End)

%t A244051row[n_]:=Flatten[Map[ConstantArray[#,n/#]&,Reverse[Divisors[n]]]];

%t Array[A244051row,10] (* _Paolo Xausa_, Oct 16 2023 *)

%o (PARI) tabf(nn) = {for (n=1, nn, d = Vecrev(divisors(n)); for (i=1, #d, for (j=1, n/d[i], print1(d[i], "," ));); print(););} \\ _Michel Marcus_, Nov 08 2014

%Y Cf. A000005, A000012, A000041, A000203, A000217, A001065, A001227, A002131, A027750, A032742, A038040, A048050, A056538, A126988, A160664, A237593, A245579, A299765, A328365, A309400 (mirror).

%K nonn,tabf,easy

%O 1,2

%A _Omar E. Pol_, Nov 08 2014