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
Examples
112 -> 1+1=2; 1235 -> 1+2=3 and 2+3=5; 224610 -> 2+2=4 and 2+4=6 and 4+6=10.
Links
- Bi Cheng Wu, Table of n, a(n) for n = 1..4958 (a(n) < 1000000)
Programs
-
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 *) -
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)