A282109 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 = 4.
17, 21, 25, 29, 34, 38, 42, 46, 51, 55, 59, 63, 66, 68, 70, 74, 78, 83, 84, 87, 91, 95, 100, 116, 129, 136, 145, 152, 161, 168, 177, 184, 197, 204, 213, 220, 229, 236, 245, 252, 257, 259, 263, 264, 267, 271, 272, 273, 280, 289, 296, 305, 312, 325, 332, 336, 341
Offset: 1
Examples
83 in base 4 is 1103. If j = 2 (digit 0) we have 1*1 + 1*2 = 3 for the left side and 3*1 = 3 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,4),i=1..10^3);
Comments