A282145 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 = 4.
5, 10, 15, 18, 20, 23, 33, 40, 53, 60, 65, 67, 72, 80, 85, 92, 98, 105, 118, 125, 130, 132, 137, 150, 157, 160, 163, 170, 183, 190, 193, 195, 202, 212, 215, 222, 235, 240, 255, 260, 261, 268, 274, 281, 288, 294, 301, 307, 314, 320, 321, 326, 333, 339, 340, 346
Offset: 1
Examples
222 in base 4 is 3132. If we split the number in 31 and 32 we have 1*1 + 3*2 = 7 for the left side and 3*1 + 2*2 = 7 for the right one.
Links
- Paolo P. Lava, Table of n, a(n) for n = 1..10000
Programs
-
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,4),i=1..10^3);
Comments