A308702 - OEIS

%I #19 Oct 19 2024 15:57:32

%S 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,

%T 3,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,

%U 50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,1

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

%C 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.

%C 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.

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

%H Michel Marcus, <a href="/A308702/b308702.txt">Table of n, a(n) for n = 2..1000</a>

%H Ana Rechtman, <a href="http://images-archive.math.cnrs.fr/Juin-2019-1er-defi.html">Juin 1er défi</a>, Images des Mathématiques, CNRS, 2019 (in French).

%e 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).

%e 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).

%t okQ[m_, k_] := IntegerQ[Sqrt[m]] && IntegerQ[Sqrt[m+k-1]] && IntegerQ[ Sqrt[m+2k]];

%t findm[k_] := Module[{m = 1}, While[!okQ[m^2, k], m++]; m];

%t a[n_] := findm[2n];

%t Table[a[n], {n, 2, 72}] (* _Jean-François Alcover_, Jun 18 2019, from PARI *)

%o (PARI) isok(m, k) = issquare(m) && issquare(m+k-1) && issquare(m+2*k);

%o findm(k) = my(m=1); while (!isok(m^2,k), m++); m;

%o a(n) = findm(2*n);

%Y Cf. A000290 (squares), A308703.

%K nonn

%O 2,2

%A _Michel Marcus_, Jun 18 2019