| Displaying 1-6 of 6 results found.
page
1
11, 1093, 1093, 3511, 3511, 5557, 104891, 1006003
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). 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.
MATHEMATICA
f[n_] := GCD @@ Numerator@ HarmonicNumber@ {n, Floor[n/2]}; f@ Select[ Range[5000], f[#] > 1 &] (* Giovanni Resta, May 13 2016 *)
5, 13, 17, 23, 41, 67, 73, 79, 107, 113, 139, 149, 157, 179, 191, 193, 223, 239, 241, 251, 263, 277, 281, 293, 307, 311, 317, 331, 337, 349, 431, 443, 449, 461, 467, 479, 487, 491, 499, 503, 541, 547, 557, 563, 569, 593, 619, 653, 683, 691, 709, 757, 769, 787
COMMENTS
For p = prime(n), Boyd defines J_p to be the set of numbers k such that p divides A001008(k), the numerator of the harmonic number H(k). For harmonic primes, J_p contains only the three numbers p-1, (p-1)p and (p-1)(p+1). It has been conjectured that there are an infinite number of these primes and that their density in the primes is 1/e. Prime p= A000040(n) is in this sequence iff neither H(k) == 0 (mod p), nor H(k) == - A177783(n) (mod p) have solutions for 1 <= k <= p-2. - Max Alekseyev, May 13 2010
PROG
(PARI) is(p)=my(K=-Mod((binomial(2*p-1, p)-1)/2/p^3, p), H=Mod(0, p)); for(k=1, p-2, H+=1/k; if(H==0||H==K, return(0))); 1 \\ Charles R Greathouse IV, Mar 16 2014
Least k such that prime(n) appears in the factorization of A001008(k) (the numerator of the k-th harmonic number).
+10
9
2, 4, 6, 3, 12, 16, 18, 22, 13, 30, 17, 40, 13, 46, 22, 58, 10, 66, 70, 72, 78, 82, 88, 11, 100, 102, 106, 25, 112, 126, 130, 5, 138, 148, 150, 156, 162, 166, 71, 178, 180, 190, 192, 196, 38, 210, 222, 22, 228, 232, 238, 240, 250, 66, 262, 33, 58, 276, 280, 282
COMMENTS
a(n)<=n for n =2,5,14,18,25,29,33,46,49,...
For p = prime(n), Boyd defines J_p to be the set of numbers k such that p divides A001008(k). This sequence gives the smallest elements of J_p. The largest elements of J_p are given by A177734. The sizes of J_p are given by A092103.
MATHEMATICA
p = Prime[n]; k = 1; sum = 1/k;
While[! Divisible[Numerator[sum], p],
k++; sum += 1/k];
Return[k]];
PROG
(PARI) a(n)=if(n<0, 0, s=1; while(numerator(sum(k=1, s, 1/k))%prime(n)>0, s++); s)
Number of values of k for which prime(n) divides A001008(k), the numerator of the k-th harmonic number.
+10
7
3, 3, 13, 638, 3, 3, 25, 3, 18, 26, 15, 3, 27, 24, 17, 23, 13, 3, 45, 3, 3
COMMENTS
For p = prime(n), Boyd defines J_p to be the set of numbers k such that p divides A001008(k). This sequence gives the size of J_p. A072984 and A177734 give the smallest and largest elements of J_p, respectively. A092101 gives primes prime(n) such that a(n) = 3 (i.e., a( A000720( A092101(m))) = 3 for all m). A092102 gives primes prime(n) such that a(n) > 3.
Boyd gives bounds: a(23) > 5870; a(31) > 2713; a(78) > 7718; and the following values: a(24)-a(30) = [7, 74, 44, 63, 3, 1273, 3]; a(32)-a(77) = [7, 38, 3, 3, 7, 3, 74, 526, 288, 3, 19, 3, 3, 41, 11, 59, 3, 31, 65, 176, 3, 3, 3, 20, 3, 106, 55, 3, 3, 89, 3, 3, 3, 79, 3, 3, 3, 47, 3, 21, 253, 29, 7, 79, 41, 19]; a(79)-a(99) = [13, 9, 703, 23, 3, 205, 105, 3, 3, 323, 3, 7, 3, 3, 3, 3, 3, 3, 13, 1763, 7, 3, 3].
Eswarathasan and Levine conjectured that for any prime number p the set J_p is finite.
I proved that if J_p(x) is the number of integers in J_p that are less than x > 1, then J_p(x) < 129 p^(2/3) x^0.765 for any prime p. In particular, J_p has asymptotic density zero. (End)
Bing-Ling Wu and Yong-Gao Chen improved Sanna's (see previous comment) result showing that J_p(x) <= 3 x^(2/3 + 1/(25 log p)) for any prime p and any x > 1. - Carlo Sanna, Jan 12 2017
EXAMPLE
a(2) = 3 because 3 divides A001008(k) for k = 2, 7, and 22. a(4) = 13 because 7 divides A001008(k) for only the 13 values k = 6, 42, 48, 295, 299, 337, 341, 2096, 2390, 14675, 16731, 16735, and 102728. This is the 4th row in A229493.
CROSSREFS
Cf. A092193 (number of generations for each prime). Cf. A229493 (terms for each prime).
Number of generations for which prime(n) divides A001008(k) for some k.
+10
4
4, 3, 7, 30, 3, 3, 7, 3, 5, 7, 4, 3, 5, 5, 6, 6, 4, 3, 8, 3, 3
COMMENTS
For any prime p, generation m consists of the numbers p^(m-1) <= k < p^m. The zeroth generation consists of just the number 0. When there is a k in generation m such that p divides A001008(k), then that k may generate solutions in generation m+1. It is conjectured that for all primes there are solutions for only a finite number of generations. The number of generations is unknown for p=83. Boyd's table 3 states incorrectly that harmonic primes have 2 generations; harmonic primes have 3 generations.
EXAMPLE
a(4)=7 because the fourth prime, 7, divides A001008(k) for k = 6, 42, 48, 295, 299, 337, 341, 2096, 2390, 14675, 16731, 16735 and 102728. These values of k fall into 6 generations; adding the zeroth generation makes a total of 7 generations.
AUTHOR
T. D. Noe, Feb 24 2004; corrected Jul 28 2004
Primes p that divide A001008(k), the numerator of the k-th harmonic number H(k), for some k < p-1.
+10
3
11, 29, 37, 43, 53, 61, 97, 109, 137, 173, 199, 227, 257, 269, 271, 313, 347, 353, 379, 397, 401, 409, 421, 433, 439, 509, 521, 577, 599, 601, 617, 641, 643, 647, 659, 677, 733, 761, 773, 809, 821, 827, 839, 863, 911, 919, 929, 937, 941, 947, 953, 971, 1009
COMMENTS
These primes are a subset of the non-harmonic primes A092102. Because these primes are analogous to the irregular primes A000928 that divide the numerators of Bernoulli numbers, they might be called H-irregular primes. The density of these primes is about 0.4 -- very close to the density of irregular primes. These primes are called Harmonic irregular primes in the Wikipedia entry for "Regular prime" (see links). It may be noted that if p is known to be of this type and H(k) is the smallest Harmonic number divisible by p, then not only does k < p-1 hold, but k <= (p-1)/2. This is because, by symmetry, H(p-1-n) == H(n) (mod p), so that any eligible k lying between (p+1)/2 and p-1 would have a counterpart in the range between 1 and (p-1)/2. Furthermore, the minimal k cannot be exactly equal to (p-1)/2, because then p would be a Wieferich prime ( A001220) and would also divide H(Int(p/4)). Thus k <= (p-3)/2, and this inequality is sharp because exact equality holds for p = 29, 37, 3373 (see A072984). - John Blythe Dobson, Apr 09 2015
MATHEMATICA
n=2; Table[cnt=0; While[cnt==0, p=Prime[n]; k=1; h=0; While[cnt==0 && k<=(p-1)/2, h=h+1/k; If[Mod[Numerator[h], p]==0, cnt++ ]; k++ ]; n++ ]; p, {100}]
Search completed in 0.009 seconds
|