A135499 Numbers for which Sum_digits(odd positions) = Sum_digits(even positions).

11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132, 143, 154, 165, 176, 187, 198, 220, 231, 242, 253, 264, 275, 286, 297, 330, 341, 352, 363, 374, 385, 396, 440, 451, 462, 473, 484, 495, 550, 561, 572, 583, 594, 660, 671, 682, 693, 770, 781, 792, 880, 891, 990

Offset: 1

Author

Paolo P. Lava and Giorgio Balzarotti, Feb 08 2008

Keywords

• easy
• nonn
• base

Comments

Conjecture: this is a subsequence of A008593 (verified for the first 50 thousand terms). - R. J. Mathar, Feb 10 2008

Subsequence of A008593. - Zak Seidov Feb 11 2008

If k is present, so is 10*k. - Robert G. Wilson v, Jul 13 2014

As Seidov said, a subsequence of multiples of 11. That follows trivially from the divisibility rule for 11. - Jens Kruse Andersen, Jul 13 2014

A225693(a(n)) = 0. - Reinhard Zumkeller, Aug 08 2014

Examples

			594, 1023, and 1397 are terms:
   594 -> 4 + 5 = 9;
  1023 -> 3 + 0 = 2 + 1;
  1397 -> 7 + 3 = 9 + 1.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000 (first 119 terms from _Paolo P. Lava_ and _Giorgio Balzarotti_)

Crossrefs

Cf. A067042, A214527.

Cf. A060979.

Cf. A225693.

Programs

Haskell

a135499 n = a135499_list !! (n-1)
a135499_list = filter ((== 0) . a225693) [1..]
-- Reinhard Zumkeller, Aug 08 2014, Jul 05 2014

Maple

P:=proc(n) local i,k,w,x; for i from 1 by 1 to n do w:=0; k:=i; while k>0 do w:=w+k-(trunc(k/10)*10); k:=trunc(k/10); od; x:=0; k:=i; while k>0 do x:=x+(k-(trunc(k/10)*10)); k:=trunc(k/100); od; if w=2*x then print(i); fi; od; end: P(3000);
# Alternative:
filter:= proc(n)
local L,d;
L:= convert(n,base,10);
d:= nops(L);
add(L[2*i],i=1..floor(d/2)) = add(L[2*i-1],i=1..floor((d+1)/2))
end proc:
select(filter,[ 11*j $ j= 1 .. 10^4 ]); # Robert Israel, May 28 2014

Mathematica

dQ[n_]:=Module[{p=Transpose[Partition[IntegerDigits[n],2,2,1,0]]},Total[First[p]]== Total[Last[p]]]; Select[Range[1000],dQ] (* Harvey P. Dale, May 26 2011 *)