A348554 - OEIS

%I #9 Dec 13 2021 17:07:25

%S 2,2,3,2,4,2,5,2,3,4,6,2,7,2,4,8,2,3,6,9,2,4,5,10,2,11,2,3,4,6,8,12,2,

%T 13,2,4,7,14,2,3,5,6,10,15,2,4,8,16,2,17,2,3,4,6,9,12,18,2,19,2,4,5,8,

%U 10,20,2,3,6,7,14,21,2,4,11,22,2,23,2,3,4,6,8,12,16,24,2,5,10,25

%N Irregular triangle read by rows: row n gives the divisors d of 2*n with 1 < d < 2*n, for n >= 2.

%C This gives the rows 2*n of A137510, for n >= 2.

%C The length of row n is A069930(n) = tau(2*n) - 2 = A099777(n) - 2.

%C The sum of row n is A346880(n) = A062731(n) - (2*n + 1).

%F T(n, k) = A137510(2*n, k), for n >= 2 and k = 1, 2, ..., A069930(n).

%e The irregular triangle T(n, k) begins:

%e n, 2*n / k 1  2  3  4  5  6  7 ...

%e ----------------------------------

%e 2,   4:    2

%e 3,   6:    2  3

%e 4,   8:    2  4

%e 5,  10:    2  5

%e 6   12:    2  3  4 6

%e 7,  14:    2  7

%e 8,  16:    2  4  8

%e 9,  18:    2  3  6  9

%e 10, 20:    2  4  5 10

%e 11, 22:    2 11

%e 12, 24:    2  3  4  6  8 12

%e 13, 26:    2 13

%e 14, 28:    2  4  7 14

%e 15, 30:    2  3  5  6 10 15

%e 16, 32:    2  4  8  1

%e 17, 34:    2 17

%e 18, 36:    2  3  4  6  9 12 18

%e 19, 38:    2 19

%e 20, 40:    2  4  5  8 10 20

%e ...

%t Flatten@Table[Select[Divisors[2n],1<#<2n&],{n,2,25}] (* _Giorgos Kalogeropoulos_, Oct 22 2021 *)

%o (PARI) row(n) = select(x->((x>1) && (x<2*n)), divisors(2*n)); \\ _Michel Marcus_, Oct 23 2021

%Y Cf. A069930, A099777, A137510, A346880.

%K nonn,tabf

%O 2,1

%A _Wolfdieter Lang_, Oct 22 2021