A282147 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 = 6.
7, 14, 21, 28, 35, 38, 42, 45, 52, 59, 73, 76, 83, 84, 115, 126, 146, 157, 168, 188, 199, 210, 217, 219, 226, 228, 233, 252, 257, 259, 270, 290, 301, 312, 332, 343, 354, 363, 374, 385, 405, 416, 427, 434, 438, 445, 456, 476, 487, 498, 504, 507, 518, 529, 549, 560
Offset: 1
Examples
580 in base 6 is 2404. If we split the number in 24 and 04 we have 4*1 + 2*2 = 8 for the left side and 0*1 + 4*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,6),i=1..10^3);
Comments