A216869 -id:A216869 - cp's OEIS frontend

A221671 Maximum number of squares in a non-constant arithmetic progression (AP) of length n.

1, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12

Offset: 1

Jonathan Sondow, Jan 24 2013

Let s(n;d,i) denote the number of squares in AP i, i+d, i+2d, ..., i+(n-1)d. Then a(n) is the maximum of s(n;d,i) over all such APs with d > 0.
González-Jiménez and Xarles (2013) compute a(n) up to a(52) = 12 using elliptic curves (see Table 2, where their Q(N) = a(N)). They do not seem to have noticed that a(n) = A193832(n) for n != 5 in the range where they computed a(n). I conjecture that this formula holds for all n != 5.
Bombieri & Zannier prove that a(n) << n^(3/5) (log n)^c for some constant c > 0. It is conjectured that a(n) ~ sqrt(8n/3). - Charles R Greathouse IV, Jan 21 2022

			The AP 1, 25, 49 = 1^2, 5^2, 7^2 shows that a(3) = 3. By Fermat and Euler, no four squares are in AP, so a(4) = 3 (see A216869). Then the AP 49, 169, 289, 409, 529 = 7^2, 13^2, 17^2, 409, 23^2 shows that a(5) = 4 (see A216870).

Andrew Granville, "Squares in arithmetic progressions and infinitely many primes", The American Mathematical Monthly 124, no. 10 (2017), pp. 951-954.

Enrico Bombieri and Umberto Zannier, A note on squares in arithmetic progressions, II., Atti della Accademia Nazionale dei Lincei. Classe di Scienze Fisiche, Matematiche e Naturali. Rendiconti Lincei. Matematica e Applicazioni 13, no. 2 (2002), pp. 69-75.
Enrique González-Jiménez and Xavier Xarles, On a conjecture of Rudin on squares in Arithmetic Progressions, arXiv 2013.

Cf. A001318, A080995, A193832, A216869, A216870, A221672.

(* note that an extension to more than 52 terms may not be correct *) row[n_] := Join[Table[2*n-1, {2*n-1}], Table[2*n, {n}]]; row[2] = {3, 3, 4, 4, 4}; Flatten[Table[row[n], {n, 1, 6}]][[1 ;; 52]] (* Jean-François Alcover, Jan 25 2013, from formula *)

a(n) = A193832(n) for n < 53 except for n = 5.
a(n) >= A193832(n) for all n. (Proof. A193832 equals the partial sums of A080995 (characteristic function of generalized pentagonal numbers A001318) and a term in the AP 1+24*k is a square if and only if k = A001318(x) = x*(3*x-1)/2 for some x. See González-Jiménez and Xarles (2013) Lemma 2.)
a(A221672(n)) = n.

A221672 Length of shortest non-constant arithmetic progression (AP) containing n squares.

Original entry on oeis.org

1, 2, 3, 5, 8, 13, 16, 23, 27, 36, 41, 52

Offset: 1

Jonathan Sondow, Jan 28 2013

Same as where records occur in A221671 (maximum number of squares in a non-constant AP of length n).
González-Jiménez and Xarles (2013) conjecture that for n >= 5 the sequence a(n)-1 equals the tail 7, 12, 15, 22, 26, 35, 40, 51, ... of A001318 (generalized pentagonal numbers k*(3*k-1)/2 for k = 0, +-1, +-2, ...). They prove it up to a(12)-1 = 51 = 6*(3*6-1)/2.
See A221671 for additional comments.
Also 8, 13, 16, 23, 27, 36, 41, 52 are where records occur for 8 <= n <= 52 in A193832 (number of squares in the arithmetic progression {24k + 1: 0 <= k <= n-1} [Granville]). - Jonathan Sondow, Dec 15 2017

			The AP 1, 25, 49 = 1^2, 5^2, 7^2 shows that a(n) = n for n = 1, 2, 3 (see A216869).
By Fermat and Euler, no four squares are in AP, so the AP 49, 169, 289, 409, 529 = 7^2, 13^2, 17^2, 409, 23^2 shows that a(4) = 5 (see Dickson and A216870).
As k*(3*k-1)/2 = 0, 1, 2, 5, 7 for k = 0, +-1, +-2, and 24*k*(3*k-1)/2 + 1 = (6*k-1)^2 is a square, the AP 24*n+1 for the 8 numbers n = 0, 1, ..., 7 contains 5 squares, so a(5) <= 8. González-Jiménez and Xarles (2013) prove a(5) > 7, so a(5) = 8.

L. E. Dickson, History of the Theory of Numbers, Vol. II, Chelsea, New York, 1952, pp. 435-440.

Enrique González-Jiménez and Xavier Xarles, On a conjecture of Rudin on squares in Arithmetic Progressions, arXiv 2013.

Cf. A001318, A080995, A193832, A216869, A216870, A221671.

A221671(a(n)) = n.

a(n) <= A001318(n)+1. (Proof. As 24*k*(3*k-1)/2 + 1 = (6*k-1)^2, a term in the AP 24*m+1 is a square when m is in A001318. Thus the AP 24*m+1 for m = 0, 1, ..., A001318(n) contains n squares and has length A001318(n)+1.)

A216870 A maximal length five arithmetic progression of squares in a quadratic number field.

Original entry on oeis.org

49, 169, 289, 409, 529

Offset: 1

Jonathan Sondow, Nov 20 2012

Bremner (2102): "Xarles (2011) investigated arithmetic progressions (APs) in number fields, and proved the existence of an upper bound K(d) for the maximal length of an AP of squares in a number field of degree d. He shows that K(2) = 5."
Euler showed that K(1) = 3. See A216869 for the smallest non-constant example. Another example is a(1), a(2), a(3) = 49, 169, 289 = 7^2, 13^2, 17^2.
It is known that K(3) >= 4.

			a(n) = 7^2, 13^2, 17^2, sqrt(409)^2, 23^2 for n = 1, 2, 3, 4, 5.

A. Bremner, Arithmetic progressions of squares in cubic fields, Abstract 2012.
X. Xarles, Squares in arithmetic progression over number fields, arXiv:0909.1642 [math.AG], 2009.
X. Xarles, Squares in arithmetic progression over number fields, J. Number Theory, 132 (2012), 379-389.

Cf. A216869, A221671, A221672.

NestList[120+#&,49,4] (* Harvey P. Dale, Apr 20 2013 *)

a(n+1) - a(n) = 120 for n = 1, 2, 3, 4.

A295784 Length of the longest arithmetic progression in squares mod n with slope coprime to n.

Original entry on oeis.org

2, 2, 3, 2, 3, 2, 2, 3, 3, 2, 4, 3, 2, 2, 5, 2, 4, 2, 2, 3, 5, 2, 3, 3, 2, 2, 4, 2, 4, 2, 2, 5, 3, 2, 4, 4, 2, 2, 5, 2, 5, 2, 2, 5, 5, 2, 3, 3, 2, 2, 6, 2, 3, 2, 2, 4, 5, 2, 5, 4, 2, 2, 3, 2, 6, 2, 2, 3, 7, 2, 9, 4, 2, 2, 3, 2, 6, 2, 2, 5, 7, 2, 3, 5, 2, 2, 5, 2, 3, 2, 2, 5, 3, 2, 9, 3, 2, 2, 7, 2, 7, 2, 2, 6, 6

Offset: 3

Tom Hejda, Nov 27 2017

nonn

The sequence reaches 2 infinitely many times as a(4*n)=2. (If we had a(4*n)>=3, it would imply a(4)>=3, but a(4)=2. This comes from the fact that a(m*n)<=a(m) for m,n>=3.)

			For n=17 we have residues {0,1,2,4,8,9,13,15,16} and the following arithmetic progressions of length 5: (15, 16, 0, 1, 2), (13, 15, 0, 2, 4), (9, 13, 0, 4, 8)

Tom Hejda, Table of n, a(n) for n = 3..20000
MathOverflow user Seva, Consecutive non-quadratic residues (Proves that a(n) is unbounded.)

Bounded by A000224.

Cf. A216869.

def a(n) :
    if n in [1,2] : return Infinity
    R = quadratic_residues(n)
    return max( next( m for m in itertools.count() if (a+(b-a)*m)%n not in R ) \
      for a,b in zip(R,R[1:]+R[:1]) if gcd(b-a,n) == 1 )

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A221671 Maximum number of squares in a non-constant arithmetic progression (AP) of length n.

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A221672 Length of shortest non-constant arithmetic progression (AP) containing n squares.

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A216870 A maximal length five arithmetic progression of squares in a quadratic number field.

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A295784 Length of the longest arithmetic progression in squares mod n with slope coprime to n.

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