A282110 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+1..k}{(i-j)*d_i} = Sum_{i=1..j-1}{(j-i)*d_i}. Case x = 5.
26, 31, 36, 41, 46, 52, 57, 62, 67, 72, 78, 83, 88, 93, 98, 104, 109, 114, 119, 124, 127, 130, 132, 137, 142, 147, 153, 155, 158, 163, 168, 173, 179, 180, 184, 189, 194, 199, 205, 230, 251, 254, 259, 260, 264, 269, 274, 276, 285, 301, 310, 326, 335, 351, 360, 381
Offset: 1
Examples
137 in base 5 is 1022. If j=2 (the second 2 from right) we have 0*1 + 1*2 = 2 for the left side and 2*1 = 2 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),j=1..k)=add(a[j]*(j-k),j=k+1..nops(a)) then RETURN(n); break: fi: od: end: seq(P(i,5),i=1..10^3);
Comments