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.

Showing 1-3 of 3 results.

A103794 Smallest number b such that b^prime(n) - (b-1)^prime(n) is prime.

Original entry on oeis.org

2, 2, 2, 2, 6, 2, 2, 2, 6, 3, 2, 40, 7, 5, 13, 3, 3, 2, 7, 18, 47, 8, 6, 2, 26, 3, 42, 2, 13, 8, 2, 8, 328, 8, 9, 45, 27, 13, 76, 15, 52, 111, 5, 15, 50, 287, 16, 5, 40, 23, 110, 368, 23, 68, 28, 96, 81, 150, 3, 143, 4, 12, 403, 4, 45, 11, 83, 21, 96, 5, 109, 350, 128, 304, 38, 4, 163
Offset: 1

Views

Author

Lei Zhou, Feb 24 2005

Keywords

Comments

Conjecture: sequence is defined for all positive indices.
For p=prime(n), Eisenstein's irreducibility criterion can be used to show that the polynomial (x+1)^p-x^p is irreducible, which is a necessary (but not sufficient) condition for a(n) to exist. - T. D. Noe, Dec 05 2005

Examples

			2^prime(1)-1^prime(1)=3 is prime, so a(1)=2;
2^prime(5)-1^prime(5)=2047 has a factor of 23;
...
6^prime(5)-5^prime(5)=313968931 is prime, so a(5)=6;
		

Crossrefs

Programs

  • Maple
    f:= proc(n) local p,b;
      p:= ithprime(n);
      for b from 2 do
        if isprime(b^p - (b-1)^p) then return b fi
      od
    end proc:
    map(f, [$1..80]); # Robert Israel, Jun 04 2024
  • Mathematica
    Do[p=Prime[k]; n=2; nm1=n-1; cp=n^p-nm1^p; While[ !PrimeQ[cp], n=n+1; nm1=n-1; cp=n^p-nm1^p]; Print[n], {k, 1, 200}]

Formula

a(n) = A222119(n) + 1. - Ray Chandler, Feb 26 2017

A222120 Number of digits in the smallest prime of the form (k+1)^p - k^p, where p = prime(n).

Original entry on oeis.org

1, 1, 2, 3, 9, 4, 6, 6, 18, 14, 10, 60, 35, 31, 53, 26, 29, 19, 57, 90, 122, 72, 65, 27, 138, 49, 168, 33, 122, 103, 39, 119, 345, 126, 143, 250, 225, 182, 315, 204, 308, 371, 134, 227, 335, 489, 255, 156, 364, 312, 476, 613, 329, 460, 372, 522, 514, 590, 133
Offset: 1

Views

Author

Vladimir Pletser, Feb 07 2013

Keywords

Comments

The smallest primes of the form (k+1)^p - k^p are in A121620. The values of k are in A222119. For the primes p = 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, ... (A000043), k = 1 and Mersenne primes 2^p - 1 (A000668) are obtained.

Examples

			a(5) = 9 because the 5th prime is 11, and the smallest prime of the form (k+1)^11 - k^11 is 6^11 - 5^11 = 313968931, which has 9 digits
		

Crossrefs

Cf. A222119 (values of k).

Programs

  • Mathematica
    Table[p = Prime[n]; k = 1; While[q = (k + 1)^p - k^p; ! PrimeQ[q], k++]; Length[IntegerDigits[q]], {n, 60}] (* T. D. Noe, Feb 12 2013 *)
  • PARI
    a222120(n) = {local(p,k); p=prime(n); while(!isprime((k+1)^p - k^p), k=k+1); ceil(log((k+1)^p - k^p)/log(10))} \\ Michael B. Porter, Feb 12 2013

A301510 Smallest positive number b such that ((b+1)^prime(n) + b^prime(n))/(2*b + 1) is prime, or 0 if no such b exists.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 3, 16, 1, 11, 6, 37, 1, 9, 120, 9, 1, 2, 67, 16, 1, 26, 103, 12, 60, 1, 239, 4, 40, 2, 44, 174, 33, 1, 3, 260, 114, 1, 161, 70, 1, 3, 2, 3, 50, 45, 472, 228, 183, 66, 37, 7, 122, 235, 68, 102, 294, 8, 13, 1, 40, 62, 143, 1, 61, 7
Offset: 2

Views

Author

Tim Johannes Ohrtmann, Mar 22 2018

Keywords

Comments

Conjecture: a(n) > 0 for every n > 1.
Records: 1, 4, 16, 37, 120, 239, 260, 472, 917, 1539, 6633, 7050, 12818, ..., which occur at n = 2, 10, 13, 17, 20, 32, 41, 52, 72, 128, 171, 290, 309, ... - Robert G. Wilson v, Jun 16 2018

Examples

			a(10) = 4 because (5^29 + 4^29)/9 = 2149818248341 is prime and (2^29 + 1^29)/3, (3^29 + 2^29)/5 and (4^29 + 3^29)/7 are all composite.
		

Crossrefs

Numbers n such that ((b+1)^n + b^n)/(2*b + 1) is prime for b = 1 to 18: A000978, A057469, A128066, A128335, A128336, A187805, A181141, A187819, A217095, A185239, A213216, A225097, A224984, A221637, A227170, A228573, A227171, A225818.

Programs

  • Mathematica
    Table[p = Prime[n]; k = 1; While[q = ((b+1)^n+b^n)/(2*b+1); ! PrimeQ[q], k++]; k, {n, 200}]
    f[n_] := Block[{b = 1, p = Prime@ n}, While[! PrimeQ[((b +1)^p + b^p)/(2b +1)], b++]; b]; Array[f, 70, 2] (* Robert G. Wilson v, Jun 13 2018 *)
  • PARI
    for(n=2, 200, b=0; until(isprime((((b+1)^prime(n)+b^prime(n))/(2*b+1))), b++); print1(b,", ")) \\ corrected by Eric Chen, Jun 06 2018

Formula

a(n) = A250201(2*prime(n)) - 1 for n >= 2. - Eric Chen, Jun 06 2018
Showing 1-3 of 3 results.