A282108 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 = 3.
10, 13, 16, 20, 23, 26, 29, 30, 32, 35, 39, 48, 55, 60, 64, 69, 73, 78, 82, 87, 90, 91, 96, 100, 105, 112, 117, 121, 130, 137, 142, 144, 146, 151, 155, 160, 164, 165, 169, 173, 178, 180, 182, 187, 192, 194, 203, 207, 212, 219, 224, 233, 234, 242, 246, 247, 256
Offset: 1
Examples
35 in base 3 is 1022. If j = 2 (second 2 from the right) we have 0*1 + 1*2 = 2 for the left side and 2*1 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),j=1..k)=add(a[j]*(j-k),j=k+1..nops(a)) then RETURN(n); break: fi: od: end: seq(P(i,3),i=1..10^3);
Comments