A282143 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 = 2.

3, 6, 9, 12, 15, 18, 19, 24, 25, 30, 33, 36, 38, 45, 48, 50, 51, 60, 63, 66, 69, 72, 75, 76, 81, 87, 90, 96, 100, 102, 105, 117, 120, 126, 129, 131, 132, 138, 143, 144, 150, 152, 153, 162, 165, 174, 179, 180, 189, 192, 193, 195, 200, 204, 205, 210, 219, 231, 234

Offset: 1

Author

Paolo P. Lava, Giovanni Resta, Feb 07 2017

Keywords

• nonn
• base
• easy

Comments

All the palindromic numbers in base 2 with an even number of digits belong to the sequence.

Here the fulcrum is between two digits while in the sequence from A282107 to A282115 is one of the digits.

Examples

			143 in base 2 is 10001111. If we split the number in 10001 and 111 we have 1*1 + 0*2 + 0*3 + 0*4 + 1*5 = 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

Crossrefs

Cf. A282107 - A282115, A282144 - A282151.

Programs

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,2),i=1..10^3);