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