cp's OEIS Frontend

Rabung and Jordan (1970) incorrectly computed a(8) as 399: their placement of residues supporting a(8)=399 fails since 80 and 81 fall into the same 8th-power residue class. - Max Alekseyev, Aug 10 2005

Don Reble pointed out that for even n, the n-th residue class placement of prime factors q of n must obey the quadratic reciprocity law: q must be in an even class whenever n*(q-1) is a multiple of 8. - Max Alekseyev, Sep 04 2017

The most likely continuation (after 2153) is 4481, 9857, 25793, 60961, 132113, 324673. - Don Reble, Aug 02 2014 (see LINKS). - N. J. A. Sloane, Dec 11 2015

From David L. Harden: (Start)

"Proof that A097160(2)=17:

"Since the quadratic residues modulo 17 are 1,2,4,8,9,13,15 and 16, there are no 3 consecutive integers among these. Thus A097160(2)>=17.

"To show it is 17, we show there will be a run of 3 when p>17:

"Case 1. (2/p)=(3/p)=1. 1,2,3 works. (Also, 2,3,4 or 48,49,50 will work)

"Case 2. (2/p)=1 and (3/p)=-1. In this case, 8,9,10 works unless (5/p)=-1. 49,50,51 will work unless (17/p)=1. But then 15,16,17 works.

"Case 3. (2/p)=-1 and (3/p)=1. In this case, 3,4,5 works unless (5/p)=-1. But then 49/120, 169/120, 289/120 will work since 120 is invertible modulo p.

"Case 4. (2/p)=(3/p)=-1. Then 1/24, 25/24, 49/24 works. QED

Here the quadratic character of only 2 primes was used; for higher-index terms, more primes should be needed (how many?) and this promises to make computation (via these ideas) exponentially harder.

"One can attempt to carry out this kind of reasoning while eschewing fractions; then the chase for a run of quadratic residues is longer but one can obtain a universal upper bound on the onset of such a run of quadratic residues.

"Note that if the quadratic character of -1 is known, then a run of consecutive quadratic residues can include both negative and positive fractions (like -7/6, -1/6, 5/6, 11/6).

"Otherwise the help from knowing (-1/p) seems to be rather limited:

"If (-1/p)=-1, then a run of k quadratic nonresidues mod p can be turned into a run of k quadratic residues mod p by multiplying them by -1 and reversing their order. This allows the computation in this case to be that much easier. However, this also seems to make it harder for a prime p in A097160 to have (-1/p)=-1, as evidenced by the fact that all the terms included there are congruent to 1 mod 4.

"Problem: Are all terms in A097160 congruent to 1 mod 4?

"Also, beyond A097160(3)=53, all listed terms are congruent to 1 mod 8. Does this hold up (if so, why?), or is it just a result of how little computation has been done?" (End)

A000236(n) is the maximum length of a run of consecutive residues modulo prime p, starting with 1, where no two adjacent elements belong to the same n-th power residue class (in other words, there is no n-th power residue modulo p in the sequence of ratios 1/2, 2/3, ..., (A000236(n)-1)/A000236(n)). a(n) equals the smallest p admitting a run of maximum length A000236(n).

For all n, a(n) exists and equals 1, 3, 4, 6 or 9. Proof: a(4)=1 by inspection. For n > 4 (p > 7), if 2 is a quadratic residue of p, then a(n)=1; otherwise if 5 is a quadratic residue of p, then a(n)=4 or 3; otherwise 2*5=10 is a quadratic residue of p and (9, 10) are consecutive residues. However, a(n)=8 or 7 is impossible as 8 cannot be a quadratic residue (since 2 is not), leaving 9 and 6 as the other possible values.

The constant 0.133141413191633911131141331131441391... = sum(a(n)/10^(n-3)) is conjectured to be irrational.