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Primes p such that min(d(p-1), d(p+1)) is larger than the corresponding values of all previous primes, where d(n) is the number of divisors of n (A000005).
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%I #7 Sep 29 2018 04:35:50

%S 2,3,5,7,17,19,41,197,199,449,701,881,3079,4159,18089,40699,51679,

%T 90271,388961,403649,446081,906751,1276001,12227489,37487449,53308529,

%U 59522849,109245401,285258401,459712639,1381951999,2560742911,2969200961,8505402751

%N Primes p such that min(d(p-1), d(p+1)) is larger than the corresponding values of all previous primes, where d(n) is the number of divisors of n (A000005).

%C Problem 104 in Sierpinski's book is to prove that this sequence is infinite.

%C The corresponding values of min(d(p-1), d(p+1)) are 1, 2, 3, 4, 5, 6, 8, 9, 12, 14, 16, 18, 20, 28, 32, 36, 40, 48, 50, 56, 64, 80, 96, 128, 144, 160, 168, 192, 216, 256, 288, 320, 336, 384, ...

%D W. Sierpinski, 250 Problems in Elementary Number Theory, New York: American Elsevier, 1970, problem #104, pp. 9, 58-59.

%t s={}; f[p_] := Min[DivisorSigma[0,p-1], DivisorSigma[0,p+1]]; p=2; fm=0; Do[f1 = f[p]; If[f1>fm, AppendTo[s,p]; fm=f1]; p=NextPrime[p], {k, 1, 100}]; s

%o (PARI) f(p)=min(numdiv(p-1),numdiv(p+1));

%o fm=0;forprime(p=1, 1000, f1=f(p); if(f1>fm, print1(p,", "); fm=f1))

%Y Cf. A000005, A000040, A145339.

%K nonn

%O 1,1

%A _Amiram Eldar_, Sep 28 2018