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.

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

1,1

Links

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

Example

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

Mathematica

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 *)

Prog

(Python)
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)

Crossrefs

Shares subsequences with A108203, A308104, A214527.

Cf. A019523.

Sequence in context: A036301 A117723 A359142 * A157662 A340125 A095615

Adjacent sequences: A329716 A329717 A329718 * A329720 A329721 A329722

Keyword

nonn,base

Author

Bi Cheng Wu, Nov 19 2019

Status

approved