Numbers with Non-Decreasing Digits

[nino]

Let be K(n)=all the numbers obtained with the conditions of your problem.
***If n=1 then k(n)=1
1

***If n=2 then k(n)=3
11
12

22

***If n=3 then k(n)=10
111
112
113
122
123
133

222
223
233

333

***If n=4 then k(n)=35
1111
1112
1113
1114
1122
1123
1124
1133
1134
1144

1222
1223
1224
1233
1234
1244

1333
1334
1344

1444

2222
2223
2224
2233
2234
2244

2333
2334
2344

2444

3333
3334
3344

3444

4444

It is easy to see that k(n)= S(1<=i<=n) [S(1<=j<=i) of {T(sub_i)(sub_j)}]
where T(sub_i)(sub_j) is the j-th triangular number and S represents a sumatoria (to add up)

Note: I hope you can understand me, I don't speak and write english very well.

 

[nino]

Let be the matrix A=

[ \ t1 t2 t3 t4 ... t(n-3) t(n-2) t(n-1) tn ]
[ t1 \ t2 t3 t4 ... t(n-3) t(n-2) t(n-1) tn ]
[ t1 t2 \ t3 t4 ... t(n-3) t(n-2) t(n-1) tn ]
[ t1 t2 t3 \ t4 ... t(n-3) t(n-2) t(n-1) tn ]
[ t1 t2 t3 t4 \ ... t(n-3) t(n-2) t(n-1) tn ]
[ t1 t2 t3 t4 ... \ t(n-3) t(n-2) t(n-1) tn ]
[ t1 t2 t3 t4 ... t(n-3) \ t(n-2) t(n-1) tn ]
[ t1 t2 t3 t4 ... t(n-3) t(n-2) \ t(n-1) tn ]
[ t1 t2 t3 t4 ... t(n-3) t(n-2) t(n-1) \ tn ]
[ t1 t2 t3 t4 ... t(n-3) t(n-2) t(n-1) tn \ ]

where tn stands the n-th triangular number. Then, the number you are looking for is the sum of all the t(n's) under the line.

 

[Gokul43201]

nino, you're missing the condition that a<=i. So, the first digit can not be larger than 1...