In base 19, 1111111111111111111 equals 109912203092239643840221 decimal, which is prime. The next one occurs in base 31, where the string of 31 unit digits equals 568972471024107865287021434301977158534824481, another prime.
Prime testing was done using a probabilistic prime test:
5 point 80
10 for Nb=2 to 999
20 Sum=0
30 for I=0 to Nb-1
40 Sum=Sum+Nb^I
50 next
90 if fnPrime(Sum)>0 then print Nb,Sum
100 next
980 end
990 '
10000 fnOddfact(N)
10010 local K=0,P
10030 while N @ 2=0
10040 N=N\2
10050 K=K+1
10060 wend
10070 P=pack(N,K)
10080 return(P)
10090 '
10100 fnPrime(N)
10110 local I,X,J,Y,Q,K,T,Ans
10120 if N @ 2=0 then Ans=0:goto *EndPrime
10125 O=fnOddfact(N-1)
10130 Q=member(O,1)
10140 K=member(O,2)
10150 I=0
10160 repeat
10170 repeat
10180 X=fnLrand(N)
10190 until X>1
10200 J=0
10210 Y=modpow(X,Q,N)
10220 loop
10230 if or{and{J=0,Y=1},Y=N-1} then goto *ProbPrime
10240 if and{J>0,Y=1} then goto *NotPrime
10250 J=J+1
10260 if J=K then goto *NotPrime
10270 Y=(Y*Y) @ N
10280 endloop
10290 *ProbPrime
10300 I=I+1
10310 until I>50
10320 Ans=1
10330 goto *EndPrime
10340 *NotPrime
10350 Ans=0
10360 *EndPrime
10370 return(Ans)
10380 '
10400 fnLrand(N)
10410 local R
10415 N=int(N)
10420 R=(int(rnd*10^(alen(N)+2))) @ N
10430 return(R)
10440 '
10500 fnNxprime(X)
10510 if X @ 2=0 then X=X+1
10520 while fnPrime(X)=0
10530 X=X+2
10540 wend
10550 return(X)
10560 '
base resulting prime, in decimal notation
2 3
3 13
19 109912203092239643840221
31 568972471024107865287021434301977158534824481
Overflow in 10210
the overflow occurring at n = 487 (variable Nb).
Using 2,3,19,31 into Sloane's OEIS, shows that the next such base is 7547 (Sloane's A088790).
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Posted by Charlie
on 2013-07-24 12:40:09 |
computer solution
