A383963 Irregular triangle read by rows: T(n,k) is the sum of the k-th pair of conjugate divisors of n. If n is a square then the central term in the row n is equal to 2*sqrt(n), with n >= 1, 1 <= k <= A000005(n).

2, 3, 3, 4, 4, 5, 4, 5, 6, 6, 7, 5, 5, 7, 8, 8, 9, 6, 6, 9, 10, 6, 10, 11, 7, 7, 11, 12, 12, 13, 8, 7, 7, 8, 13, 14, 14, 15, 9, 9, 15, 16, 8, 8, 16, 17, 10, 8, 10, 17, 18, 18, 19, 11, 9, 9, 11, 19, 20, 20, 21, 12, 9, 9, 12, 21, 22, 10, 10, 22, 23, 13, 13, 23, 24, 24, 25, 14, 11, 10, 10, 11, 14, 25

Offset: 1

Author

Omar E. Pol, Jun 17 2025

Keywords

• nonn
• tabf

Comments

Row n is a palindromic composition of A074400(n) = 2*A000203(n).

Shares infinitely many rows with the virtual sequence 2*A237270.

Examples

			Triangle begins:
   n |   Row n
   1 |   2;
   2 |   3,  3;
   3 |   4,  4;
   4 |   5,  4,  5;
   5 |   6,  6;
   6 |   7,  5,  5,  7;
   7 |   8,  8;
   8 |   9,  6,  6,  9;
   9 |  10,  6, 10;
  10 |  11,  7,  7, 11;
  11 |  12, 12;
  12 |  13,  8,  7,  7,  8, 13;
  13 |  14, 14;
  14 |  15,  9,  9, 15;
  15 |  16,  8,  8, 16;
  16 |  17, 10,  8, 10, 17;
  ...
For n = 8 the divisors of 8 are [1, 2, 4, 8] and the sums of the conjugate divisors are respectively [1 + 8 = 9], [2 + 4 = 6], [4 + 2 = 6], [8 + 1 = 9], so the 8th row is [9, 6, 6, 9].
For n = 9 the divisors of 9 are [1, 3, 9] and the sums of the conjugate divisors are respectively [1 + 9 = 10], [3 + 3 = 6], [9 + 1 = 10], so the 9th row is [10, 6, 10]. Since 9 is a square then the central term in the row is equal to 2*sqrt(9) = 2*3 = 6. Also in this case the 9th row is the same as the 9th row of the virtual sequence 2*A237270 because the 9th row of A237270 is [5, 3, 5].
		

Crossrefs

Row lengths give A000005.

Row sums give A074400 = 2*A000203.

Column 1 gives A000027 except the 1, the same for the right border.

Cf. A027750, A056538, A237270, A272025.

Programs

Mathematica

row[n_] := Module[{d = Divisors[n]}, d + Reverse[d]]; Array[row, 24] // Flatten (* Amiram Eldar, Jun 18 2025 *)

PARI

row(n) = my(d=divisors(n)); vector(#d, k, d[k]+n/d[k]); \\ Michel Marcus, Jun 18 2025

Formula

T(n,k) = A027750(n,k) + A056538(n,k).