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

1,2

Comments

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

References

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

Links

Table of n, a(n) for n=1..52.

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.

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.

Example

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).

Mathematica

(* 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 *)

Crossrefs

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

Sequence in context: A257569 A039836 A083398 * A301640 A061420 A003057

Adjacent sequences: A221668 A221669 A221670 * A221672 A221673 A221674

Keyword

nonn,hard,more

Author

Jonathan Sondow, Jan 24 2013

Status

approved