cp's OEIS Frontend

A244051 Triangle read by rows in which row n lists the parts of the partitions of n into equal parts, in nonincreasing order.

%I A244051 #64 Oct 16 2023 16:14:56
%S A244051 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 A244051 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 A244051 1,10,5,5,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1
%N A244051 Triangle read by rows in which row n lists the parts of the partitions of n into equal parts, in nonincreasing order.
%C A244051 Row n has length sigma(n) = A000203(n).
%C A244051 Row sums give n*A000005(n) = A038040(n).
%C A244051 Column 1 is A000027.
%C A244051 Both columns 2 and 3 are A032742, n > 1.
%C A244051 For any k > 0 and t > 0, the sequence contains exactly one run of k consecutive t's. - _Rémy Sigrist_, Feb 11 2019
%C A244051 From _Omar E. Pol_, Dec 04 2019: (Start)
%C A244051 The number of parts congruent to 0 (mod m) in row m*n equals sigma(n) = A000203(n).
%C A244051 The number of parts greater than 1 in row n equals A001065(n), the sum of aliquot parts of n.
%C A244051 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 A244051 The number of partitions in row n equals A000005(n), the number of divisors of n.
%C A244051 The number of partitions in row n with an odd number of parts equals A001227(n).
%C A244051 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 A244051 The decreasing records in row n give the n-th row of A056538.
%C A244051 Row n has n 1's which are all at the end of the row.
%C A244051 First n rows contain A000217(n) 1's.
%C A244051 The number of k's in row n is A126988(n,k).
%C A244051 The number of odd parts in row n is A002131(n).
%C A244051 The k-th block in row n has A027750(n,k) parts.
%C A244051 Right border gives A000012. (End)
%C A244051 The r-th row of the triangle begins at index k = A160664(r-1). - _Samuel Harkness_, Jun 21 2022
%H A244051 Samuel Harkness, <a href="/A244051/b244051.txt">Table of n, a(n) for n = 1..10000</a>
%H A244051 Samuel Harkness, <a href="/A244051/a244051.jpg">Log-log Scatterplot of the first 1000000 terms</a>
%e A244051 Triangle begins:
%e A244051    [1];
%e A244051    [2], [1,1];
%e A244051    [3], [1,1,1];
%e A244051    [4], [2,2], [1,1,1,1];
%e A244051    [5], [1,1,1,1,1];
%e A244051    [6], [3,3], [2,2,2], [1,1,1,1,1,1];
%e A244051    [7], [1,1,1,1,1,1,1];
%e A244051    [8], [4,4], [2,2,2,2], [1,1,1,1,1,1,1,1];
%e A244051    [9], [3,3,3], [1,1,1,1,1,1,1,1,1];
%e A244051   [10], [5,5], [2,2,2,2,2], [1,1,1,1,1,1,1,1,1,1];
%e A244051   [11], [1,1,1,1,1,1,1,1,1,1,1];
%e A244051   [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 A244051   [13], [1,1,1,1,1,1,1,1,1,1,1,1,1];
%e A244051   [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 A244051   [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 A244051   [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 A244051   ...
%e A244051 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 A244051 From _Omar E. Pol_, Dec 04 2019: (Start)
%e A244051 The structure of the above triangle is as follows:
%e A244051    1;
%e A244051    2 11;
%e A244051    3    111;
%e A244051    4 22     1111;
%e A244051    5             11111;
%e A244051    6 33 222            111111;
%e A244051    7                          1111111;
%e A244051    8 44     2222                      11111111;
%e A244051    9    333                                    111111111;
%e A244051   ... (End)
%t A244051 A244051row[n_]:=Flatten[Map[ConstantArray[#,n/#]&,Reverse[Divisors[n]]]];
%t A244051 Array[A244051row,10] (* _Paolo Xausa_, Oct 16 2023 *)
%o A244051 (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 A244051 Cf. A000005, A000012, A000041, A000203, A000217, A001065, A001227, A002131, A027750, A032742, A038040, A048050, A056538, A126988, A160664, A237593, A245579, A299765, A328365, A309400 (mirror).
%K A244051 nonn,tabf,easy
%O A244051 1,2
%A A244051 _Omar E. Pol_, Nov 08 2014