A282150 Numbers n with k digits in base x (MSD(n)=d_k, LSD(n)=d_1) such that, chosen one of their digits in position d_k < j < d_1, is Sum_{i=j..k}{(i-j+1)*d_i} = Sum_{i=1..j-1}{(j-i)*d_i}. Case x = 9.
10, 20, 30, 40, 50, 60, 70, 80, 83, 90, 93, 103, 113, 123, 133, 143, 163, 166, 176, 180, 186, 196, 206, 249, 253, 259, 269, 270, 326, 332, 343, 360, 416, 433, 450, 489, 506, 523, 540, 579, 596, 613, 630, 652, 669, 686, 703, 720, 730, 732, 742, 747, 752, 762, 772
Offset: 1
Examples
762 in base 9 is 1036. If we split the number in 103 and 6 we have 3*1 + 0*2 + 1*3 = 6 for the left side and 6*1 = 6 for the right one.
Links
- Paolo P. Lava, Table of n, a(n) for n = 1..10000
Programs
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Maple
P:=proc(n,h) local a,j,k: a:=convert(n, base, h): for k from 1 to nops(a)-1 do if add(a[j]*(k-j+1),j=1..k)=add(a[j]*(j-k),j=k+1..nops(a)) then RETURN(n); break: fi: od: end: seq(P(i,9),i=1..10^3);
Comments