A214527 -id:A214527 - cp's OEIS frontend

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

Paolo P. Lava and Giorgio Balzarotti, Feb 08 2008

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

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

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

Cf. A067042, A214527.
Cf. A060979.
Cf. A225693.

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

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

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 *)

A329719 Numbers whose digits can be partitioned into at least 3 segments (not beginning with 0) where each segment is the sum of the previous two segments.

Original entry on oeis.org

112, 123, 134, 145, 156, 167, 178, 189, 213, 224, 235, 246, 257, 268, 279, 314, 325, 336, 347, 358, 369, 415, 426, 437, 448, 459, 516, 527, 538, 549, 617, 628, 639, 718, 729, 819, 1123, 1235, 1347, 1459, 1910, 2134, 2246, 2358, 2810, 2911, 3145, 3257, 3369

Offset: 1

Bi Cheng Wu, Nov 19 2019

			112 -> 1+1=2;
1235 -> 1+2=3 and 2+3=5;
224610 -> 2+2=4 and 2+4=6 and 4+6=10.

Bi Cheng Wu, Table of n, a(n) for n = 1..4958 (a(n) < 1000000)

Shares subsequences with A108203, A308104, A214527.

Cf. A019523.

part[n_] := part[n] = Select[Flatten[ Permutations /@ Reverse /@ IntegerPartitions[n, {3,n}], 1], 0 <= #[[3]] - Max[#[[1]], #[[2]]] <= 1 && AllTrue[Rest@ Rest@ Differences@ #, 0 <= # <= 1 &] &]; spl[x_, L_] := Map[ FromDigits@ Take[x, #] &, Transpose[{Most@ #, Rest[#]-1}& [ FoldList[ Plus, 1, L]]]]; sumQ[w_] := AllTrue[Range[3, Length@w], w[[#]] == w[[#-1]] + w[[#-2]] &]; ok[n_] := Block[{m = IntegerLength@ n}, AnyTrue[ spl[ IntegerDigits[n], #] & /@ part[m], sumQ[#] && Total[ IntegerLength /@ #] == m &]]; Select[ Range[100, 6000], ok] (* Giovanni Resta, Dec 03 2019 *)

def isSegmentSum(digits,segment1=None,segment2=None):
    digits = str(digits)
    N = len(digits)
    if N == 0:
        return True
    else:
        if (segment1 is None) and (segment2 is None):
            for i in range(N):
                try:
                    slice1 = digits[:i+1]
                    for j in range(N-(i+1)):
                        slice2 = digits[i+1:i+1+j+1]
                        slice3 = digits[i+1+j+1:]
                        if (isSegmentSum(slice3,slice1,slice2) and
                         len(slice3)>0 and not (slice1.startswith("0") or
                         slice2.startswith("0") or
                         (slice3.startswith("0")))):
                            return True
                except:
                    return False
        else:
            sumOfDigits = str(int(segment1)+int(segment2))
            nS = len(sumOfDigits)
            try:
                if digits[:nS] == sumOfDigits:
                    return isSegmentSum(digits[nS:],segment2,digits[:nS])
                else:
                    return False
            except:
                return False
    return False
def findSegmentSum(lower,upper):
    for i in range(lower,upper+1):
        if isSegmentSum(i):
            print(str(i))
findSegmentSum(1, 5200)

cp's OEIS Frontend

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

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