OFFSET
1,2
COMMENTS
The number of partitions in row n equals A001227(n), the number of odd divisors of n, and equals the number of partitions of n into consecutive parts, and equals the number of subparts in the symmetric representation of sigma(n).
The sum of row n equals A245579(n), the sum of all parts of all partitions of n into consecutive parts.
The length of row n equals A000593(n), the sum of the odd divisors of n.
Row n has length 1 if and only if n is a power of 2.
Is the right border the same as A006519?
EXAMPLE
Triangle begins:
[1];
[2];
[3], [1,1,1];
[4];
[5], [1,1,1,1,1];
[6], [2,2,2];
[7], [1,1,1,1,1,1,1];
[8];
[9], [3,3,3], [1,1,1,1,1,1,1,1,1];
[10], [2,2,2,2,2];
[11], [1,1,1,1,1,1,1,1,1,1,1];
[12], [4,4,4];
[13], [1,1,1,1,1,1,1,1,1,1,1,1,1];
[14], [2,2,2,2,2,2,2];
[15], [5,5,5], [3,3,3,3,3], [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1];
[16];
...
For n = 10, in row 10 there are two partitions, equaling the number of odd divisors of 10, they are [1, 5], and equaling the number of partitions of 10 into consecutive parts, they are [10], [4, 3, 2, 1], and equaling the number of subparts in the symmetric representation of sigma(10), they are [9, 9].
The sum of row 10 is [10] + [2 + 2 + 2 + 2 + 2] = 20 equaling the sum of all parts of all partitions of 10 into consecutive parts, that is [10] + [4 + 3 + 2 + 1] = 20.
The length of row 10 is equal to 6 equaling the sum of the odd divisors of 10, that is 1 + 5 = 6.
MATHEMATICA
Table[ConstantArray[n/#, #] & /@ Select[Divisors[n], OddQ], {n, 15}] // Flatten (* Michael De Vlieger, Jul 15 2022 *)
PROG
(PARI) row(n) = my(v=[]); fordiv(n, d, if ((n/d)%2, v = concat(v, vector(n/d, k, d)))); Vecrev(v); \\ Michel Marcus, Jul 16 2022
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Omar E. Pol, Jul 13 2022
STATUS
approved