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.

A126197 GCDs arising in A126196.

Original entry on oeis.org

11, 1093, 1093, 3511, 3511, 5557, 104891, 1006003
Offset: 1

Views

Author

Max Alekseyev and Tanya Khovanova, Mar 07 2007

Keywords

Comments

All terms are primes. Note a connection to the Wieferich primes A001220: a(2) = a(3) = A001220(1), a(3) = a(4) = A001220(2).
From John Blythe Dobson, Jan 14 2017: (Start)
All Wieferich primes p will belong to this sequence twice, because if H([p/k]) denotes the harmonic number with index floor(p/k), then p divides all of H([p/4]), H([p/2]), and H(p-1). The first two of these elements gives one solution, and the second and third another. This property of the Wieferich primes predates their name, and was apparently first proved by Glaisher in "On the residues of r^(p-1) to modulus p^2, p^3, etc.," pp. 21-22, 23 (see References).
Note also a connection to the Mirimanoff primes A014127: a(1) = A014127(1), a(8) = A014127(2). All Mirimanoff primes p will belong to this sequence, because p divides both H([p/3]) and H([2p/3]). This property of the Mirimanoff primes likewise predates their name, and was apparently first proved by Glaisher in "A general congruence theorem relating to the Bernoullian function," p. 50 (see Links).
The Wieferich primes and Mirimanoff primes would seem to be the only cases for which the value of n in A126196(n) is predictable from knowledge of p. It is not obvious that all members of the present sequence are prime; however, by definition all their divisors must be non-harmonic primes A092102. Furthermore, it is clear from the cited literature under that entry that H([n/2]) == H(n) == 0 (mod p) is only possible when n < p. Thus, all divisors of the present sequence must belong to the harmonic irregular primes A092194.
One possible reason for interest in this sequence is a 1995 result of Dilcher and Skula (see Links) which among other things shows that if a prime p were an exception to the first case of Fermat's Last Theorem, then p would divide both H([p/k]) and H([2p/k]) for every value of k from 2 to 46. To date, the only values for which such coincidences have been found have k = 2, 3, or 4. For k = 6 to hold, p would have to be simultaneously a Wieferich prime and a Mirimanoff prime, while for k = 5 to hold, p would have to be simultaneously a Wall-Sun-Sun prime and a member of A123692. The sparse numerical results for the present sequence suggest that even the more relaxed condition H([n/2]) == H(n) == 0 (mod p) is rarely satisfied. (End)

References

  • J. W. L. Glaisher, On the residues of r^(p-1) to modulus p^2, p^3, etc., Quarterly Journal of Pure and Applied Mathematics 32 (1900-1901), 1-27.

Crossrefs

Programs

  • Mathematica
    f[n_] := GCD @@ Numerator@ HarmonicNumber@ {n, Floor[n/2]}; f@ Select[ Range[5000], f[#] > 1 &] (* Giovanni Resta, May 13 2016 *)

Extensions

a(8) from Giovanni Resta, May 13 2016

A093569 For p = prime(n), the number of integers k < p-1 such that p divides A001008(k), the numerator of the harmonic number H(k).

Original entry on oeis.org

0, 0, 0, 0, 2, 0, 0, 0, 0, 2, 0, 2, 0, 2, 0, 2, 0, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 6, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 4, 0, 0, 0, 0, 0, 2, 0, 2, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 2, 0, 0, 0, 2, 0, 0, 2, 2, 2, 0, 2, 0, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

T. D. Noe, Apr 01 2004

Keywords

Comments

It is well-known that prime p >= 3 divides the numerator of H(p-1). For primes p in A092194, there are integers k < p-1 for which p divides the numerator of H(k). Interestingly, if p divides A001008(k) for k < p-1, then p divides A001008(p-k-1). Hence the terms of this sequence are usually even. The only exceptions are the two known Wieferich primes 1093 and 3511, A001220, which have 3 values of k < p-1 for which p divides A001008(k), one being k = (p-1)/2.

Examples

			a(5) = 2 because 11 = prime(5) and there are 2 values, k = 3 and 7, such that 11 divides A001008(k).
		

Crossrefs

Programs

  • Mathematica
    len=500; Table[p=Prime[i]; cnt=0; k=1; While[k
    				

A093690 Primes p that divide A007406(k), the numerator of the k-th generalized harmonic number H(k,2) = Sum 1/i^2 for i=1..k, for some k < (p-1)/2.

Original entry on oeis.org

37, 41, 43, 59, 97, 107, 127, 137, 149, 157, 163, 167, 181, 211, 241, 269, 307, 311, 373, 383, 419, 421, 433, 457, 467, 479, 487, 491, 499, 547, 563, 569, 571, 577, 601, 617, 619, 643, 653, 659, 677, 709, 727, 739, 787, 797, 811, 821, 859, 863, 883, 911, 929
Offset: 1

Views

Author

T. D. Noe, Apr 09 2004

Keywords

Comments

Because these primes are analogous to the irregular primes A000928 that divide the numerators of Bernoulli numbers, they might be called H2-irregular primes. Also see A092194. The density of these primes is about 0.4 - close to the density of irregular primes.

Crossrefs

Cf. A092194 (primes p that divide A001008(k) for some k < p-1), A093689 (least k such that prime(n) divides A007406(k)).

Programs

  • Mathematica
    nn=1000; t=Numerator[HarmonicNumber[Range[nn], 2]]; lst = {}; Do[p=Prime[n]; i=1; While[i<(p-1)/2 && Mod[t[[i]], p]>0, i++ ]; If[i<(p-1)/2, AppendTo[lst, p]], {n, 3, PrimePi[nn]}]; lst
Showing 1-3 of 3 results.