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A343034

Positive numbers m such that m^2 with last digit z deleted is still a perfect square k^2, and z divides m-k.

1, 13, 19, 487, 721, 18493, 27379, 702247, 1039681, 26666893, 39480499, 1012639687, 1499219281, 38453641213, 56930852179, 1460225726407, 2161873163521, 55450123962253, 82094249361619, 2105644484839207, 3117419602578001, 79959040299927613, 118379850648602419, 3036337886912410087

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OFFSET

1,2

COMMENTS

This sequence is the answer to the problem A1872 proposed on French mathematical site Diophante (see link).

Equivalent to the two Diophantine equations: m^2 = 10*k^2 + z and m-k = q*z for some q >= 1.

There exist solutions iff z = 1 or z = 9.

When (m, k, z) is a solution, then (19m+60k, 6m+19k, z) is another solution.

There is only one solution such that z = m-k: (13, 4, 9), see 1st example.

There exist two distinct families of solutions corresponding to z = 1 and z = 9, odd indices correspond to z = 1 and even indices to z = 9.

-> For z = 1, all solutions m of Pell equation m^2 - 10*k^2 = 1 are terms because z = 1 divides every m-k.

First few solutions (m, k) are (1, 0), (19, 6), (721, 228), (27379, 86568), ... with m = A078986(q) and corresponding k = 6*A078987(q).

-> For z = 9, solutions m must satisfy m^2 - 10*k^2 = 9 with 9 divides m-k. Among the 3 fundamental solutions (3, 0), (7, 2), (13, 4) of Pell equation m^2 -10*k^2 = 9, only (13, 4) gives solutions where 9 divides m-k.

First few solutions (m, k) are (13, 4), (487, 154), (18493, 5848), ... with m = A228209(3q).

LINKS

Table of n, a(n) for n=1..24.

Diophante, A1872, Carrément magiques (in French).

Index entries for linear recurrences with constant coefficients, signature (0,38,0,-1).

FORMULA

a(2n+1) = A078986(n) for n >= 0.

a(2n) = A228209(3n) for n >= 1.

a(n+4) = 38*a(n+2) - a(n), a(1) = 1, a(2) = 13, a(3)= 19, a(4) = 487.

G.f.: x*(1 + 13*x - 19*x^2 - 7*x^3)/(1 - 38*x^2 + x^4). - Stefano Spezia, Apr 03 2021

EXAMPLE

For m = 13, 13^2 = 169, 4^2 = 16, 13^2 - 10*4^2 = 9 and 9 = 13-4 divides 13-4.