A282149 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 = 8.
9, 18, 27, 36, 45, 54, 63, 66, 72, 75, 84, 93, 102, 111, 129, 132, 141, 144, 150, 159, 198, 201, 207, 216, 258, 273, 288, 330, 345, 360, 387, 402, 417, 432, 459, 474, 489, 504, 513, 515, 524, 528, 533, 542, 551, 576, 581, 585, 590, 599, 600, 642, 647, 657, 672
Offset: 1
Examples
672 in base 8 is 1240. If we split the number in 12 and 40 we have 2*1 + 1*2 = 4 for the left side and 4*1 + 0*2 = 4 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,8),i=1..10^3);
Comments