cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A343020 a(n) is the smallest prime p such that tau(p+1) = 2^n.

Original entry on oeis.org

2, 5, 23, 167, 839, 7559, 128519, 1081079, 20540519, 397837439, 8031343319, 188972783999, 3212537327999, 125568306863999, 2888071057871999, 190487121512687999, 4381203794791823999, 215961289494494543999, 13283916764437951631999, 540119185025730854543999, 26465840066260811872655999, 1356699703068812438127791999
Offset: 1

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Author

Jaroslav Krizek, Apr 02 2021

Keywords

Comments

tau(m) = the number of divisors of m (A000005).
Sequences of primes p such that tau(p+1) = 2^n for 2 <= n <= 5:
n = 2: 5, 7, 13, 37, 61, 73, 157, 193, 277, 313, 397, 421, ...
n = 3: 23, 29, 41, 53, 101, 103, 109, 113, 127, 137, 151, ...
n = 4: 167, 263, 269, 311, 383, 389, 439, 461, 509, 569, ...
n = 5: 839, 1319, 1511, 1559, 1847, 1889, 2039, 2309, 2687, ...
Conjecture: a(n) is also the smallest number m such that tau(m+1) = tau(m)^n.

Examples

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

Crossrefs

Programs

  • Magma
    Ax:=func; [Ax(n): n in [1..7]]
    
  • Mathematica
    Do[p = 1; While[DivisorSigma[0, Prime[p] + 1] != 2^n, p++]; Print[n, " ", Prime[p]], {n, 1, 9}] (* Vaclav Kotesovec, Apr 03 2021 *)
  • PARI
    a(n) = my(t=2^n); forprime(p=2, oo, if(numdiv(p+1)==t, return(p))); \\ Jinyuan Wang, Apr 02 2021
    
  • Python
    from sympy import isprime,nextprime
    primes=[2]
    def solve(v,k,i,j):
        global record,stack,primes
        if k==0:
            if isprime(v-1):
                record=v
            return True
        sizeok=False
        cnt=True
        while cnt:
            if i>=len(primes):
                primes.append(nextprime(primes[-1]))
            if jBert Dobbelaere, Apr 11 2021

Extensions

a(11) from Jinyuan Wang, Apr 02 2021
More terms from David A. Corneth, Apr 09 2021

A343018 a(n) is the smallest number m such that tau(m+1) = tau(m) + n.

Original entry on oeis.org

2, 1, 5, 49, 11, 35, 23, 399, 47, 1849, 59, 143, 119, 1599, 167, 575, 179, 1295, 239, 4355, 629, 2303, 359, 899, 959, 9215, 1007, 39999, 719, 20735, 839, 5183, 1799, 46655, 1259, 36863, 1679, 7055, 3023, 986049, 2879, 3599, 6479, 82943, 2519, 193599, 3359, 207935
Offset: 0

Views

Author

Jaroslav Krizek, Apr 02 2021

Keywords

Comments

tau(m) = the number of divisors of m (A000005).
Sequences of numbers m such that tau(m+1) = tau(m) + n for 0 <= n <= 5:
n = 0: 2, 14, 21, 26, 33, 34, 38, 44, 57, 75, 85, 86, 93, ... (A005237).
n = 1: 1, 3, 9, 15, 25, 63, 121, 195, 255, 361, 483, 729, ... (A055927).
n = 2: 5, 7, 13, 27, 37, 51, 61, 62, 73, 74, 91, 115, 123, ... (A230115).
n = 3: 49, 99, 1023, 1681, 1935, 2499, 8649, 9603, 20449, ... (A230653).
n = 4: 11, 17, 19, 31, 39, 43, 55, 65, 67, 69, 77, 87, 97, ... (A230654).
n = 5: 35, 169, 289, 529, 961, 1369, 2809, 3135, 4489, ... (A228453).

Examples

			For n = 3; a(3) = 49 because 49 is the smallest number such that tau(50) = 6 = tau(49) + 3 = 3 + 3.
		

Crossrefs

Programs

  • Magma
    Ax:=func; [Ax(n): n in [0..50]];
    
  • Maple
    N:= 60: # for a(0)..a(N)
    V:= Array(0..N): count:=0: t:= numtheory:-tau(1):
    for m from 1 while count < N+1 do
      s:= numtheory:-tau(m+1); v:= s - t;
      if v >= 0 and v <= N and V[v] = 0 then count:= count+1; V[v]:= m; fi;
      t:= s;
    od:
    convert(V, list); # Robert Israel, Jan 03 2025
  • Mathematica
    d = Differences @ Table[DivisorSigma[0, n], {n, 1, 10^6}]; a[n_] := If[(p = Position[d, n]) != {}, p[[1, 1]], 0]; s = {}; n = 0; While[(a1 = a[n]) > 0, AppendTo[s, a1]; n++]; s (* Amiram Eldar, Apr 03 2021 *)
  • PARI
    a(n) = my(m=1); while (numdiv(m+1) != numdiv(m) + n, m++); m; \\ Michel Marcus, Apr 03 2021
    
  • Python
    from itertools import count, pairwise
    from sympy import divisor_count
    def A343018(n): return next(m+1 for m, t in enumerate(pairwise(map(divisor_count,count(1)))) if t[1] == t[0]+n) # Chai Wah Wu, Jul 25 2022

Formula

a(n) = A086550(n) - 1.
Showing 1-2 of 2 results.