[math-fun] a(n) is the n-th integer with digit sum = n

Hello Math-Fun, Could someone help me compute S with this rule: a(n) is the n-th integer of N with digit sum = n S starts like this, I think: S = 1,11,21,31,41,51,61,71,81,109,... 1 is the 1st integer with digit sum = 1 [1,10,100,1000,10000,...] 11 is the 2nd integer with digit sum = 2 [2,11,20,101,200,1001,2000,...] 21 is the 3rd integer with digit sum = 3 [3,12,21,30,102,120,201,210,300,...] 31 is the 4th integer with digit sum = 4 [4,13,22,31,40,103,112,121,130,...] ... 109 is the 10th integer with digit sum = 10 [19,28,37,46,55,64,73,82,91,109,127,...] Best, É.