OFFSET
1,1
COMMENTS
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)
REFERENCES
Alfred Brauer, Ueber Sequenzen von Potenzresten, S.-B. Deutsch. Akad. Wiss. Berlin 1928, 9-16.
LINKS
D. H. Lehmer and Emma Lehmer, On Runs of Residues, Proc. Amer. Math. Soc, Vol. 13, No. 1 (Feb., 1962), pp. 102-106.
Don Reble, Comments on A097160
EXAMPLE
Only the first three primes have no consecutive quadratic residues, so a(1) is the third prime, 5.
53 has three consecutive quadratic resides, but not four; and each larger prime has four consecutives.
MATHEMATICA
f[l_, a_] := Module[{A = Split[l], B}, B = Last[ Sort[ Cases[A, x : {a ..} :> { Length[x], Position[A, x][[1, 1]]}]]]; {First[B], Length[ Flatten[ Take[A, Last[B] - 1]]] + 1}]; g[n_] := g[n] = f[ JacobiSymbol[ Range[ Prime[n] - 1], Prime[n]], 1][[1]]; g[1] = 1; a = Table[0, {30}]; Do[ a[[ g[n]]] = n, {n, 2556}]; Prime[a]
CROSSREFS
KEYWORD
nonn,hard
AUTHOR
Robert G. Wilson v, Jul 28 2004
EXTENSIONS
The old values of a(7) and a(8) were unproved, while a(9) and a(10) were wrong (and are still unknown), according to email message from Don Reble received by N. J. A. Sloane, Dec 11 2015, see LINKS.
STATUS
approved