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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)

Sequence is believed to be infinite.

Joseph Silverman showed that the abc-conjecture implies that there are infinitely many primes which are not in the sequence. - Benoit Cloitre, Jan 09 2003

Graves and Murty (2013) improved Silverman's result by showing that for any fixed k > 1, the abc-conjecture implies that there are infinitely many primes == 1 (mod k) which are not in the sequence. - Jonathan Sondow, Jan 21 2013

The squares of these numbers are Fermat pseudoprimes to base 2 (A001567) and Catalan pseudoprimes (A163209). - T. D. Noe, May 22 2003

Primes p that divide the numerator of the harmonic number H((p-1)/2); that is, p divides A001008((p-1)/2). - T. D. Noe, Mar 31 2004

In a 1977 paper, Wells Johnson, citing a suggestion from Lawrence Washington, pointed out the repetitions in the binary representations of the numbers which are one less than the two known Wieferich primes; i.e., 1092 = 10001000100 (base 2); 3510 = 110110110110 (base 2). It is perhaps worth remarking that 1092 = 444 (base 16) and 3510 = 6666 (base 8), so that these numbers are small multiples of repunits in the respective bases. Whether this is mathematically significant does not appear to be known. - John Blythe Dobson, Sep 29 2007

A002326((a(n)^2 - 1)/2) = A002326((a(n)-1)/2). - Vladimir Shevelev, Jul 09 2008, Aug 24 2008

It is believed that p^2 does not divide 3^(p-1) - 1 if p = a(n). This is true for n = 1 and 2. See A178815, A178844, A178900, and Ostafe-Shparlinski (2010) Section 1.1. - Jonathan Sondow, Jun 29 2010

These primes also divide the numerator of the harmonic number H(floor((p-1)/4)). - H. Eskandari (hamid.r.eskandari(AT)gmail.com), Sep 28 2010

1093 and 3511 are prime numbers p satisfying congruence 429327^(p-1) == 1 (mod p^2). Why? - Arkadiusz Wesolowski, Apr 07 2011. Such bases are listed in A247208. - Max Alekseyev, Nov 25 2014. See A269798 for all such bases, prime and composite, that are not powers of 2. - Felix Fröhlich, Apr 07 2018

A196202(A049084(a(1))) = A196202(A049084(a(2))) = 1. - Reinhard Zumkeller, Sep 29 2011

If q is prime and q^2 divides a prime-exponent Mersenne number, then q must be a Wieferich prime. Neither of the two known Wieferich primes divide Mersenne numbers. See Will Edgington's Mersenne page in the links below. - Daran Gill, Apr 04 2013

There are no other terms below 4.97*10^17 as established by PrimeGrid (see link below). - Max Alekseyev, Nov 20 2015. The search was done via PrimeGrid's PRPNet and the results were not double-checked. Because of the unreliability of the testing, the search was suspended in May 2017 (cf. Goetz, 2017). - Felix Fröhlich, Apr 01 2018. On Nov 28 2020, PrimeGrid has resumed the search (cf. Reggie, 2020). - Felix Fröhlich, Nov 29 2020. As of Dec 29 2022, PrimeGrid has completed the search to 2^64 (about 1.8 * 10^19) and has no plans to continue further. - Charles R Greathouse IV, Sep 24 2024

Are there other primes q >= p such that q^2 divides 2^(p-1)-1, where p is a prime? - Thomas Ordowski, Nov 22 2014. Any such q must be a Wieferich prime. - Max Alekseyev, Nov 25 2014

Primes p such that p^2 divides 2^r - 1 for some r, 0 < r < p. - Thomas Ordowski, Nov 28 2014, corrected by Max Alekseyev, Nov 28 2014

For some reason, both p=a(1) and p=a(2) also have more bases b with 1 < b < p that make b^(p-1) == 1 (mod p^2) than any smaller prime p; in other words, a(1) and a(2) belong to A248865. - Jeppe Stig Nielsen, Jul 28 2015

Let r_1, r_2, r_3, ..., r_i be the set of roots of the polynomial X^((p-1)/2) - (p-3)! * X^((p-3)/2) - (p-5)! * X^((p-5)/2) - ... - 1. Then p is a Wieferich prime iff p divides sum{k=1, p}(r_k^((p-1)/2)) (see Example 2 in Jakubec, 1994). - Felix Fröhlich, May 27 2016

Arthur Wieferich showed that if p is not a term of this sequence, then the First Case of Fermat's Last Theorem has no solution in x, y and z for prime exponent p (cf. Wieferich, 1909). - Felix Fröhlich, May 27 2016

Let U_n(P, Q) be a Lucas sequence of the first kind, let e be the Legendre symbol (D/p) and let p be a prime not dividing 2QD, where D = P^2 - 4*Q. Then a prime p such that U_(p-e) == 0 (mod p^2) is called a "Lucas-Wieferich prime associated to the pair (P, Q)". Wieferich primes are those Lucas-Wieferich primes that are associated to the pair (3, 2) (cf. McIntosh, Roettger, 2007, p. 2088). - Felix Fröhlich, May 27 2016

Any repeated prime factor of a term of A000215 is a term of this sequence. Thus, if there exist infinitely many Fermat numbers that are not squarefree, then this sequence is infinite, since no two Fermat numbers share a common factor. - Felix Fröhlich, May 27 2016

If the Diophantine equation p^x - 2^y = d has more than one solution in positive integers (x, y), with (p, d) not being one of the pairs (3, 1), (3, -5), (3, -13) or (5, -3), then p is a term of this sequence (cf. Scott, Styer, 2004, Corollary to Theorem 2). - Felix Fröhlich, Jun 18 2016

Odd primes p such that Chi_(D_0)(p) != 1 and Lambda_p(Q(sqrt(D_0))) != 1, where D_0 < 0 is the fundamental discriminant of the imaginary quadratic field Q(sqrt(1-p^2)) and Chi and Lambda are Iwasawa invariants (cf. Byeon, 2006, Proposition 1 (i)). - Felix Fröhlich, Jun 25 2016

If q is an odd prime, k, p are primes with p = 2*k+1, k == 3 (mod 4), p == -1 (mod q) and p =/= -1 (mod q^3) (Jakubec, 1998, Corollary 2 gives p == -5 (mod q) and p =/= -5 (mod q^3)) with the multiplicative order of q modulo k = (k-1)/2 and q dividing the class number of the real cyclotomic field Q(Zeta_p + (Zeta_p)^(-1)), then q is a term of this sequence (cf. Jakubec, 1995, Theorem 1). - Felix Fröhlich, Jun 25 2016

From Felix Fröhlich, Aug 06 2016: (Start)

Primes p such that p-1 is in A240719.

Prime terms of A077816 (cf. Agoh, Dilcher, Skula, 1997, Corollary 5.9).

p = prime(n) is in the sequence iff T(2, n) > 1, where T = A258045.

p = prime(n) is in the sequence iff an integer k exists such that T(n, k) = 2, where T = A258787. (End)

Conjecture: an integer n > 1 such that n^2 divides 2^(n-1)-1 must be a Wieferich prime. - Thomas Ordowski, Dec 21 2016

The above conjecture is equivalent to the statement that no "Wieferich pseudoprimes" (WPSPs) exist. While base-b WPSPs are known to exist for several bases b > 1 other than 2 (see for example A244752), no base-2 WPSPs are known. Since two necessary conditions for a composite to be a base-2 WPSP are that, both, it is a base-2 Fermat pseudoprime (A001567) and all its prime factors are Wieferich primes (cf. A270833), as shown in the comments in A240719, it seems that the first base-2 WPSP, if it exists, is probably very large. This appears to be supported by the guess that the properties of a composite to be a term of A001567 and of A270833 are "independent" of each other and by the observation that the scatterplot of A256517 seems to become "less dense" at the x-axis parallel line y = 2 for increasing n. It has been suggested in the literature that there could be asymptotically about log(log(x)) Wieferich primes below some number x, which is a function that grows to infinity, but does so very slowly. Considering the above constraints, the number of WPSPs may grow even more slowly, suggesting any such number, should it exist, probably lies far beyond any bound a brute-force search could reach in the forseeable future. Therefore I guess that the conjecture may be false, but a disproof or the discovery of a counterexample are probably extraordinarily difficult problems. - Felix Fröhlich, Jan 18 2019

Named after the German mathematician Arthur Josef Alwin Wieferich (1884-1954). a(1) = 1093 was found by Waldemar Meissner in 1913. a(2) = 3511 was found by N. G. W. H. Beeger in 1922. - Amiram Eldar, Jun 05 2021

From Jianing Song, Jun 21 2025: (Start)

The ring of integers of Q(2^(1/k)) is Z[2^(1/k)] if and only if k does not have a prime factor in this sequence (k is in A342390). See Theorem 5.3 of the paper of Keith Conrad. For example, we have:

(1 + 2^(364/1093) + 2^(2*364/1093) + ... + 2^(1092*364/1093))/1093 is an algebraic integer, but it is not in Z[2^(1/1093)];

(1 + 2^(1755/3511) + 2^(2*1755/3511) + ... + 2^(3510*1755/3511))/3511 is an algebraic integer, but it is not in Z[2^(1/3511)]. (End)

Note that a(1) = 7*11, a(2) = 7*11^2, and a(3) = 7*11^3.

Harmonic numbers are defined as H(n) = Sum_{k=1..n} 1/k = A001008(n)/A002805(n).

Alternating harmonic numbers are defined as H'(n) = Sum_{k=1..n} (-1)^(k+1)*1/k = A058313(n)/A058312(n).

Numbers n such that n does not divide the denominator of the n-th harmonic number are listed in A074791. Numbers n such that n does not divide the denominator of the n-th alternating harmonic number are listed in A121594.

This sequence is the intersection of A074791 and A121594.

Comments from Max Alekseyev, Mar 07 2007: (Start)

While A125581 indeed contains the geometric progression 7*11^n as a subsequence, it also contains other geometric progressions such as: 546*1093^n, 1092*1093^n, 1755*3511^n, 3510*3511^n and 4896*5557^n (see A126196 and A126197). It may also contain some "isolated" terms (i.e. not participating in the geometric progressions) but such terms are harder to find and at the moment I have no proof that they exist.

This is a sketch of my proof that geometric progression 7*11^n and the others mentioned above belong to A125581.

Lemma 1. H'(n) = H(n) - H([n/2]).

Lemma 2. For prime p and integer n >= p, valuation(H(n),p) >= valuation(H([n/p]),p) - 1.

Theorem. For an integer b > 1 and a prime number p such that p divides the numerators of both H(b) and H([b/2]), the geometric progression b*p^n belongs to A125581.

Proof. It is enough to show that valuation(H(b*p^n),p) > -n and valuation(H'(b*p^n), p) > -n. By Lemma 2 we have valuation(H(b*p^n), p) >= valuation(H(b), p) - n >= 1 - n > -n.

From this inequality and Lemma 1, we have valuation(H'(b*p^n), p) >= min{ valuation(H(b*p^n), p), valuation(H([b*p^n/2]), p) } >= min{ 1 - n, valuation(H([b*p^n/2]), p) }. It remains to show that valuation(H([b*p^n/2]), p) >= 1 - n.

Again by Lemma 2, we have valuation(H([b*p^n/2]), p) >= valuation(H([b/2]), p) - n >= 1 - n, which completes the proof.

It is easy to check that this Theorem holds for the aforementioned geometric progressions. (End)

Generalized harmonic numbers are defined as H(m,k) = Sum_{j=1..m} 1/j^k. Alternating generalized harmonic numbers are defined as H'(m,k) = Sum_{j=1..m} (-1)^(j+1)/j^k.

Sequence contains all geometric progressions of the form (p-1)*p^k for k > 0 and some primes p > 3. Note the factorization of initial terms of {a(n)} = {4*5, 6*7, 4*5^2, 10*11, 12*13, 16*17, 6*7^2, 18*19, 4*5^3, 22*23, 28*29, 30*31, 10*11^2, 36*37, 40*41, 42*43, 12*13^2, 6*7^3, 46*47, 4*5^4, 52*53, 58*59, 60*61, 66*67, 16*17^2, 70*71, 72*73, 78*79, 18*19^2, 82*83, ...}. The smallest term that does not fit this pattern is 11026 = ((149-1)/2) * 149.

Generalized harmonic numbers are defined as H(m,k) = Sum_{j=1..m} 1/j^k. Alternating generalized harmonic numbers are defined as H'(m,k) = Sum_{j=1..m} (-1)^(j+1)/j^k.

Note that {a(n)} contains the following geometric progressions: ((16843-1)/3)*16843^m found by Max Alekseyev, ((16843-1)/2)*16843^m found by Max Alekseyev, ((16843-1)*2/3)*16843^m, (16843-1)*16843^m, 20826*21647^m found by Max Alekseyev, ((2124679-1)/3)*2124679^m, ((2124679-1)/2)*2124679^m, ((2124679-1)*2/3)*2124679^m, (2124679-1)*2124679^m. Here {16843, 2124679} = A088164 are the only two currently known Wolstenholme Primes: primes p such that {2p-1} choose {p-1} == 1 mod p^4. See more details in Comments at A128672 and A125581.

Generalized harmonic numbers are defined as H(m,k) = Sum_{j=1..m} 1/j^k. Alternating generalized harmonic numbers are defined as H'(m,k) = Sum_{j=1..m} (-1)^(j+1)/j^k.

Sequence contains all terms of geometric progressions of the form (p-1)*p^k, k > 0, for some primes p >= 5, such as 4*5^k, 7*23^k, 15*23^k, 27*113^k, etc. Note the factorization of initial terms of {a(n)} = {4*5, 4*5^2, 10*11, 12*13, 7*23, 16*17, 18*19, 15*23, 4*5^3, 22*23, 28*29, 30*31, 10*11^2, 36*37, 40*41, 42*43, 12*13^2, 46*47, 4*5^4, 52*53, 27*113, 58*59, 60*61, 7*23^2, ...}. See more details in Comments at A128672 and A125581.

Generalized harmonic numbers are defined as H(m,k) = Sum_{j=1..m} 1/j^k. Alternating generalized harmonic numbers are defined as H'(m,k) = Sum_{j=1..m} (-1)^(j+1)/j^k.

Sequence contains all terms of geometric progressions 37^k*(37-1)/3, 37^k*(37-1)/2, 37^k*(37-1)*2/3, 37^k*(37-1) for k > 0. Note the factorization of initial terms of {a(n)} = {37*12, 37*18, 37*24, 37*36, ...}. See more details in Comments at A128672 and A125581.

Generalized harmonic numbers are defined as H(m,k) = Sum_{j=1..m}1/j^k. Alternating generalized harmonic numbers are defined as H'(m,k) = Sum_{j=1..m} (-1)^(j+1)/j^k.

Some apparent periodicity in {a(n)} (not without exceptions): a(n) = 20 for n = 2 + 4m, a(n) = 42 for n = 4 + 12m and 8 + 12m, a(n) = 76 for n = 9 + 18m, a(n) = 77 for n = 1 + 10m, a(n) = 104 for n = 7 + 12m, a(n) = 110 for n = 12m, a(n) = 136 for n = 25 + 32m, etc.

See more details in Comments at A128672 and A125581.

Generalized harmonic numbers are defined as H(m,k) = Sum_{j=1..m} 1/j^k. Alternating generalized harmonic numbers are defined as H'(m,k) = Sum_{j=1..m} (-1)^(j+1)/j^k.

Sequence contains geometric progressions of the form (p-1)*p^k for k > 0 and some prime p > 5. Note the factorization of initial terms of {a(n)} = {6*7, 10*11, 12*13, 16*17, 6*7^2, 18*19, 22*23, 28*29, 30*31, 10*11*2, 36*37, 40*41, 42*43, 12*13^2, 6*7^3, ...}. See more details in Comments at A128672 and A125581.

All listed terms are composite.

The ratio of A117731(n) and A082687(n) when they are different is listed in A125741(n) = A117731[ a(n) ] / A082687[ a(n) ] = {7, 13, 7, 7, 37, 19, 119, 41, 31, 37, 37, 43, 13, 7, 13, 49, 7, 7, 61, 71, 103, 67, 73, 139, ...}.

It appears that all (or almost all) members of geometric progressions 2*7^k, 4*13^k, 15*7^k, 3^37^k, 6*19^k, 17*7^k, 4*41^k, 10*31^k, 12*37^k, 55*13^k, 107*7^k, etc. for k>0 are in the sequence.