A282146 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 = 5.
6, 12, 18, 24, 27, 30, 33, 39, 51, 54, 60, 81, 90, 102, 111, 120, 126, 128, 134, 135, 150, 156, 165, 177, 186, 195, 207, 216, 228, 237, 246, 252, 255, 261, 270, 282, 291, 300, 303, 312, 321, 333, 342, 354, 363, 372, 376, 378, 387, 396, 405, 408, 417, 429, 438, 447
Offset: 1
Examples
447 in base 5 is 3242. If we split the number in 32 and 42 we have 2*1 + 3*2 = 8 for the left side and 4*1 + 2*2 = 8 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,5),i=1..10^3);
Comments