A097160 -id:A097160 - cp's OEIS frontend

A097159 Smallest prime p such that there are n consecutive quadratic residues mod p.

2, 7, 11, 19, 43, 67, 83, 131, 283, 277, 467, 479, 1907, 1607, 2543, 1559, 5443, 5711, 6389, 14969, 25703, 10559, 20747, 52057, 136223, 90313, 162263, 18191, 167107, 31391, 376589, 607153, 671947

Offset: 1

Robert G. Wilson v, Jul 28 2004

nonn

Additional terms less than 10^6: a(35)=298483, a(36)=422231, a(40)=701399 and a(42)=366791. - T. D. Noe, Apr 03 2007

			a(22)=10559, a(23)=20747 & a(28)=18191.

Cf. A002307, A097160, A097161.

Cf. A129201.

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[b = g[n]; If[ a[[b]] < 31 && a[[b]] == 0, a[[b]] = n; Print[b, " = ", Prime[n]]], {n, 2555}]

More terms from T. D. Noe, Apr 03 2007

A097161 Number of primes modulo which the largest number of consecutive quadratic residues is n.

Original entry on oeis.org

3, 3, 3, 11, 12, 34, 54, 97, 180, 351

Offset: 1

Robert G. Wilson v, Jul 29 2004

nonn

I am not sure how many of these entries have been proved to be correct. See comments in A097160. - N. J. A. Sloane, Jan 01 2007

			7, 13 & 17 have exactly two consecutive quadratic residues.
11, 41 & 53 have exactly three consecutive quadratic residues.
19, 23, 29, 31, 37, 47, 73, 89, 97, 149 & 193 have exactly four consecutive quadratic residues.
43, 59, 61, 109, 113, 139, 229, 313, 317, 337, 397 & 457 have exactly five consecutive quadratic residues.
67, 71, 79, 107, 137, 157, 163, 167, 173, 179, 181, 191, 199, 233, 239, 241, 257, 281, 349, 353, 373, 409, 431, 577, 601, 617, 641, 769, 797, 857, 881, 1049, 1373 & 2153 have exactly six consecutive quadratic residues.

Cf. A097159, A097160.

A125607 Lesser of the smallest pair of consecutive positive reduced quadratic residues modulo p = prime(n) > 5.

Original entry on oeis.org

1, 3, 3, 1, 4, 1, 4, 1, 3, 1, 9, 1, 6, 3, 3, 9, 1, 1, 1, 3, 1, 1, 4, 1, 3, 3, 1, 1, 3, 1, 4, 4, 1, 3, 9, 1, 9, 3, 3, 1, 1, 6, 1, 4, 1, 3, 3, 1, 1, 1, 3, 1, 1, 4, 1, 3, 1, 6, 9, 6, 1, 1, 6, 4, 1, 3, 3, 1, 1, 1, 3, 4, 1, 4, 3, 1, 1, 3, 3, 1, 1, 1, 3, 1, 1, 4, 1, 3, 1, 1, 3, 4, 1, 4, 1, 6, 3, 9, 6, 3, 1, 4, 1, 3, 1

Offset: 4

Nick Hobson, Nov 30 2006

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.

			The quadratic residues of 13=prime(6) are 1, 3, 4, 9, 10 and 12. The least consecutive pair of residues is (3, 4); hence a(6)=3.

Nick Hobson, Table of n, a(n) for n = 4..1003
D. H. Lehmer and Emma Lehmer, On Runs of Residues, Proc. American Mathematical Society, Vol. 13, No. 1 (Feb., 1962), pp. 102-106.

Cf. A000445, A002307, A097159, A097160, A097161.

vector(108, m, p=prime(m+3); if(p%8==1||p%8==7, 1, if(p%12==1||p%12==11, 3, if(p%10==1||p%10==9, 4, if((p%24==1||p%24==5||p%24==19||p%24==23) && (p%28==1||p%28==3||p%28==9||p%28==19||p%28==25||p%28==27), 6, 9)))))

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A097159 Smallest prime p such that there are n consecutive quadratic residues mod p.

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A097161 Number of primes modulo which the largest number of consecutive quadratic residues is n.

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A125607 Lesser of the smallest pair of consecutive positive reduced quadratic residues modulo p = prime(n) > 5.

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