A308702 a(n) is the least integer k such that both k^2+2*n-1 and k^2+4*n are squares.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 3, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 1

Offset

2,2

Comments

Motivated by the following original problem: the population of a city is a square number. With 100 more people, it would be a square plus one, with again 100 more people it would be a square. What is the initial population of the city? Answer: 2401. See Images des Mathématiques link.

One remarks that one can solve the problem when the population increase is even and greater than 1. So in the sequence definition, 2*n is the population increase and k^2 is the initial population.

One also notes that a(n) always exists with a(n) <= n-1.

Links

Michel Marcus, Table of n, a(n) for n = 2..1000

Ana Rechtman, Juin 1er défi, Images des Mathématiques, CNRS, 2019 (in French).

Example

For n=2, that is a population increase of 4, the population is 1 (a square), since we have 1+4-1 = 4 (a square) and 1+8 = 9 (a square).
For n=28, that is a population increase of 56, the population is 9 (a square), since we have 9+56-1 = 64 (a square) and 9+112 = 121 (a square).

Mathematica

okQ[m_, k_] := IntegerQ[Sqrt[m]] && IntegerQ[Sqrt[m+k-1]] && IntegerQ[ Sqrt[m+2k]];
findm[k_] := Module[{m = 1}, While[!okQ[m^2, k], m++]; m];
a[n_] := findm[2n];
Table[a[n], {n, 2, 72}] (* Jean-François Alcover, Jun 18 2019, from PARI *)

Prog

(PARI) isok(m, k) = issquare(m) && issquare(m+k-1) && issquare(m+2*k);
findm(k) = my(m=1); while (!isok(m^2, k), m++); m;
a(n) = findm(2*n);

Crossrefs

Cf. A000290 (squares), A308703.

Sequence in context: A318893 A338497 A137317 * A326200 A094759 A214701

Adjacent sequences: A308699 A308700 A308701 * A308703 A308704 A308705

Keyword

nonn

Author

Michel Marcus, Jun 18 2019

Status

approved