A282144 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 = 3.
4, 8, 11, 12, 19, 24, 28, 33, 36, 40, 47, 52, 56, 57, 61, 68, 72, 80, 84, 85, 92, 97, 99, 104, 108, 109, 113, 120, 125, 141, 145, 156, 168, 170, 171, 172, 183, 193, 204, 208, 216, 218, 229, 240, 244, 245, 250, 252, 255, 257, 269, 276, 278, 280, 291, 297, 301, 312
Offset: 1
Examples
229 in base 3 is 22111. If we split the number in 22 and 111 we have 2*1 + 2*2 = 6 for the left side and 1*1 + 1*2 + 1*3 = 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,3),i=1..10^3);
Comments