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.

A251623 Primes p with property that the sum of the 4th powers of the successive gaps between primes <= p is a prime number.

Original entry on oeis.org

5, 19, 29, 41, 61, 67, 83, 89, 103, 113, 167, 179, 229, 263, 281, 283, 307, 317, 359, 461, 467, 509, 563, 571, 613, 739, 743, 761, 1019, 1031, 1051, 1093, 1229, 1291, 1297, 1319, 1409, 1447, 1609, 1621, 1667, 1747, 1801, 1877, 1979, 2113, 2137, 2161
Offset: 1

Views

Author

Abhiram R Devesh, Dec 06 2014

Keywords

Examples

			a(1)=5; primes less than or equal to 5: [2, 3, 5]; 4th power of prime gaps: [1, 16]; sum of 4th power of prime gaps: 17.
a(2)=19; primes less than or equal to 13: [2, 3, 5, 7, 11, 13, 17, 19]; 4th powers of prime gaps (see A140299): [1, 16, 16, 256, 16, 256, 16]; sum of these: 577.
		

Crossrefs

Cf. A006512 (with gaps), A247177 (with squares of gaps), A247178 (with cubes of gaps).

Programs

  • Mathematica
    p = 2; q = 3; s = 0; lst = {}; While[p < 2500, s = s + (q - p)^4; If[ PrimeQ@ s, AppendTo[lst, q]]; p = q; q = NextPrime@ q]; lst (* Robert G. Wilson v, Dec 19 2014 *)
  • PARI
    p = 2; q = 3; s = 1; for (i = 1, 100, p = q; q = nextprime (q + 1); if (isprime (s = s + (q - p)^4), print1 (q ", "))) \\ Zak Seidov, Jan 19 2015
  • Python
    import sympy
    p=2
    s=0
    while 10000>p>0:
        np=sympy.nextprime(p)
        if sympy.isprime(s):
            print(p)
        d=np-p
        s+=(d**4)
        p=np
    

A252655 Smallest prime p with property that the sum of the n-th power of the successive gaps between primes <= p is also a prime number.

Original entry on oeis.org

5, 5, 7, 5, 43, 13, 7, 5, 241, 13, 43, 41, 19, 41, 7, 5, 13, 83, 43, 229, 811, 41, 31, 167, 811, 127, 367, 419, 79, 43, 43, 83, 673, 19, 109, 83, 13, 331, 523, 409, 199
Offset: 1

Views

Author

Abhiram R Devesh, Dec 19 2014

Keywords

Comments

First appearance of p, by power, beginning with 5: 1, 3, ??, 6, ??, 13, ??, 86, 23, ??, 12, 5, ... . - Robert G. Wilson v, Jan 11 2015

Examples

			n=1: p=5; primes less than or equal to 5: [2, 3, 5]; prime gaps: [1, 2]; sum of prime gaps: 3.
n=2: p=5; primes less than or equal to 5: [2, 3, 5]; squares of prime gaps: [1, 4]; sum of squares of prime gaps: 5.
n=3: p=7; primes less than or equal to 7: [2, 3, 5, 7]; cubes of prime gaps: [1, 8, 8]; sum of cubes of prime gaps: 17.
n=4: p=5; primes less than or equal to 5: [2, 3, 5]; 4th power of prime gaps: [1, 16]; sum of 4th power of prime gaps: 17.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Block[{p = 2, s = 0}, While[ !PrimeQ@ s, q = NextPrime@ p; s = s + (q - p)^n; p = q]; p]; Array[f, 60] (* Robert G. Wilson v, Jan 11 2015 *)
  • Python
    import sympy
    c=1
    while c>0:
        p=2
        d=0
        s=0
        while p>0:
            s=s+(d**c)
            sp=sympy.isprime(s)
            if sp ==True:
                print(c,p)
                p=-1
                c=c+1
            else:
                np=sympy.nextprime(p)
                d=np-p
                p=np

A252768 Primes p with property that the sum of the k-th powers of the successive gaps between primes <= p are prime numbers for k = 1 to n.

Original entry on oeis.org

5, 5, 13, 14593, 372313, 2315773, 541613713
Offset: 1

Views

Author

Abhiram R Devesh, Dec 21 2014

Keywords

Comments

This is a subsequence of A006512 (greater of twin primes), see comment by Robert G. Wilson v there. - Michel Marcus, Jan 23 2015

Examples

			n=3, p=13, List of primes [2, 3, 5, 7, 11, 13] and respective prime gaps are [1, 2, 2, 4, 2].
k=1: Sum of prime gaps = 11.
k=2: Sum of squares of prime gaps = 29.
k=3: Sum of cubes of prime gaps = 89.
		

Crossrefs

Programs

  • PARI
    a(n) = {vp = primes(200000); vdp = vector(#vp-1, k, vp[k+1] - vp[k]); vpp = vector(n, k, 1); k = 2; while (sum(m=1, n, isprime(vpp[m])) != n, for (j=1, n, vpp[j] += vdp[k]^j;); k++;); vp[k];} \\ Michel Marcus, Jan 23 2015
Showing 1-3 of 3 results.