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 38s and Bunches of 1s (Posted on 2017-06-30)
Prove every number in the sequence 38, 381, 3811, 38111, 381111, ... is composite.

 See The Solution Submitted by Brian Smith Rating: 4.5000 (2 votes)

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 No proof but a pattern found | Comment 1 of 7
Number and its prime factors:

38 2 19
381  3 127
3811  37 103
38111  23 1657
381111  3 127037
3811111  17 37 73 83
38111111  233 163567
381111111  3 3 42345679
3811111111  37 113 613 1487
38111111111  31 2333 526957
381111111111  3 2399 52954163
3811111111111  37 103003003003
38111111111111  23333 1633356667
381111111111111  3 73 1740233384069
3811111111111111  37 2287 45038479669
38111111111111111  353 661 163333566667
381111111111111111  3 3 131 323249458109509
3811111111111111111  37 114346289 900798827
38111111111111111111  19 227 541 2857 5716953331
381111111111111111111  3 879449 1140233 126685261

4   word 255:kill "38and1s.txt":open "38and1s.txt" for output as #2
5   N=38
10   for I=2 to 20
20     print N;:Fctors=fnFactor(N):N=N*10+1
25     print #2,N;
30   next
40   close #2
999   end
1000   fnFactor(Num)
1005   local Fctrs,S,N,I
1010     N=abs(Num):Ct=0:Fctrs=0:S=""
1020     if N>0 then Limit=sqrt(N):else Limit=0
1030     if Limit<>int(Limit) then Limit=int(Limit+1)
1040     Dv=2:gosub *DivideIt
1050     Dv=3:gosub *DivideIt
1060     Dv=5:gosub *DivideIt
1070     Dv=7
1080     loop
1090      if Dv>Limit then goto *Afterloop
1100      gosub *DivideIt:Dv=Dv+4 '11
1110      gosub *DivideIt:Dv=Dv+2 '13
1120      gosub *DivideIt:Dv=Dv+4 '17
1130      gosub *DivideIt:Dv=Dv+2 '19
1140      gosub *DivideIt:Dv=Dv+4 '23
1150      gosub *DivideIt:Dv=Dv+6 '29
1160      gosub *DivideIt:Dv=Dv+2 '31
1170      gosub *DivideIt:Dv=Dv+6 '37
1180      if inkey=chr(27) then S\$=chr(27):end
1190    endloop
1200    *Afterloop
1210    if N>1 then S=str(S)+str(N):Ct=Ct+1
1220   print S
1221   print #2,S
1230   return(Ct)
1240
1250   *DivideIt
1255   Did=0
1260    loop
1270     Q=int(N/Dv)
1280     if Q*Dv=N and N>0 then
1290       :N=Q:Fctrs=Fctrs+1:S=str(S)+str(Dv):Ct=Ct+1
1291       :Did=Did+1
1300       :if N>0 then Limit=sqrt(N):else Limit=0:endif
1310       :if Limit<>int(Limit) then Limit=int(Limit+1):endif
1320      :else
1330      :goto *Afterdo
1340     :endif
1350    endloop
1360    *Afterdo
1370    return

Wolfram Alpha supplies:

3811111111111111111111 37 393380951 261840342653
38111111111111111111111 17 73 1372549 22374429543379
381111111111111111111111 3 55711 158823283 14357381249

Every third one seems to be a multiple of three, and the one after that is a multiple of 37. The one after than would seem to be unpredicatable, but is actually always divisible by 2333..., with one more 3 each occurrence. it's disquised by the fact that 233333 = 353*661 and 2333333 = 19*227*541 and 23333333 =  17*1372549.

 Posted by Charlie on 2017-06-30 09:04:10

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