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a(n) is the smallest prime p such that tau(p+1) = 2^n.
4

%I #30 Sep 08 2022 08:46:26

%S 2,5,23,167,839,7559,128519,1081079,20540519,397837439,8031343319,

%T 188972783999,3212537327999,125568306863999,2888071057871999,

%U 190487121512687999,4381203794791823999,215961289494494543999,13283916764437951631999,540119185025730854543999,26465840066260811872655999,1356699703068812438127791999

%N a(n) is the smallest prime p such that tau(p+1) = 2^n.

%C tau(m) = the number of divisors of m (A000005).

%C Sequences of primes p such that tau(p+1) = 2^n for 2 <= n <= 5:

%C n = 2: 5, 7, 13, 37, 61, 73, 157, 193, 277, 313, 397, 421, ...

%C n = 3: 23, 29, 41, 53, 101, 103, 109, 113, 127, 137, 151, ...

%C n = 4: 167, 263, 269, 311, 383, 389, 439, 461, 509, 569, ...

%C n = 5: 839, 1319, 1511, 1559, 1847, 1889, 2039, 2309, 2687, ...

%C Conjecture: a(n) is also the smallest number m such that tau(m+1) = tau(m)^n.

%H David A. Corneth, <a href="/A343020/b343020.txt">Table of n, a(n) for n = 1..31</a>

%e a(4) = 167 because 167 is the smallest prime p such that tau(p+1) = 16 = 2^4.

%t Do[p = 1; While[DivisorSigma[0, Prime[p] + 1] != 2^n, p++]; Print[n, " ", Prime[p]], {n, 1, 9}] (* _Vaclav Kotesovec_, Apr 03 2021 *)

%o (Magma) Ax:=func<n|exists(r){m:m in[1..10^6] | IsPrime(m) and #Divisors(m + 1) eq 2 ^ n} select r else 0>; [Ax(n): n in [1..7]]

%o (PARI) a(n) = my(t=2^n); forprime(p=2, oo, if(numdiv(p+1)==t, return(p))); \\ _Jinyuan Wang_, Apr 02 2021

%o (Python)

%o from sympy import isprime,nextprime

%o primes=[2]

%o def solve(v,k,i,j):

%o global record,stack,primes

%o if k==0:

%o if isprime(v-1):

%o record=v

%o return True

%o sizeok=False

%o cnt=True

%o while cnt:

%o if i>=len(primes):

%o primes.append(nextprime(primes[-1]))

%o if j<len(stack) and stack[j]<primes[i]:

%o f=stack[j] ; j+=1

%o else:

%o f=primes[i] ; i+=1

%o if record==None or v * f**k < record:

%o stack.append(f**2)

%o ok=solve(v*f,k-1,i,j)

%o stack.pop()

%o sizeok|=ok

%o cnt&=ok

%o else:

%o cnt=False

%o return sizeok

%o def a343020(n):

%o global record,stack

%o record,stack = None,[]

%o solve(1,n,0,0)

%o return record-1

%o # _Bert Dobbelaere_, Apr 11 2021

%Y Cf. A037992, A080371, A080372, A340799, A343018, A343019.

%K nonn

%O 1,1

%A _Jaroslav Krizek_, Apr 02 2021

%E a(11) from _Jinyuan Wang_, Apr 02 2021

%E More terms from _David A. Corneth_, Apr 09 2021