| Summary: | Let <inline-formula> <math display="inline"> <semantics> <mrow> <mi>q</mi> <mo>≥</mo> <mn>2</mn> </mrow> </semantics> </math> </inline-formula> be a positive integer and let <inline-formula> <math display="inline"> <semantics> <mrow> <mo>(</mo> <msub> <mi>a</mi> <mi>j</mi> </msub> <mo>)</mo> </mrow> </semantics> </math> </inline-formula>, <inline-formula> <math display="inline"> <semantics> <mrow> <mo>(</mo> <msub> <mi>b</mi> <mi>j</mi> </msub> <mo>)</mo> </mrow> </semantics> </math> </inline-formula>, and <inline-formula> <math display="inline"> <semantics> <mrow> <mo>(</mo> <msub> <mi>c</mi> <mi>j</mi> </msub> <mo>)</mo> </mrow> </semantics> </math> </inline-formula> (with <i>j</i> a non-negative integer) be three given <inline-formula> <math display="inline"> <semantics> <mi mathvariant="double-struck">C</mi> </semantics> </math> </inline-formula>-valued and <i>q</i>-periodic sequences. Let <inline-formula> <math display="inline"> <semantics> <mrow> <mi>A</mi> <mrow> <mo>(</mo> <mi>q</mi> <mo>)</mo> </mrow> <mo>:</mo> <mo>=</mo> <msub> <mi>A</mi> <mrow> <mi>q</mi> <mo>−</mo> <mn>1</mn> </mrow> </msub> <mo>⋯</mo> <msub> <mi>A</mi> <mn>0</mn> </msub> </mrow> </semantics> </math> </inline-formula>, where <inline-formula> <math display="inline"> <semantics> <msub> <mi>A</mi> <mi>j</mi> </msub> </semantics> </math> </inline-formula> is as is given below. Assuming that the “monodromy matrix„ <inline-formula> <math display="inline"> <semantics> <mrow> <mi>A</mi> <mo>(</mo> <mi>q</mi> <mo>)</mo> </mrow> </semantics> </math> </inline-formula> has at least one multiple eigenvalue, we prove that the linear scalar recurrence <inline-formula> <math display="inline"> <semantics> <mrow> <msub> <mi>x</mi> <mrow> <mi>n</mi> <mo>+</mo> <mn>3</mn> </mrow> </msub> <mo>=</mo> <msub> <mi>a</mi> <mi>n</mi> </msub> <msub> <mi>x</mi> <mrow> <mi>n</mi> <mo>+</mo> <mn>2</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>b</mi> <mi>n</mi> </msub> <msub> <mi>x</mi> <mrow> <mi>n</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>c</mi> <mi>n</mi> </msub> <msub> <mi>x</mi> <mi>n</mi> </msub> <mo>,</mo> <mi>n</mi> <mo>∈</mo> <msub> <mi mathvariant="double-struck">Z</mi> <mo>+</mo> </msub> </mrow> </semantics> </math> </inline-formula> is Hyers-Ulam stable if and only if the spectrum of <inline-formula> <math display="inline"> <semantics> <mrow> <mi>A</mi> <mo>(</mo> <mi>q</mi> <mo>)</mo> </mrow> </semantics> </math> </inline-formula> does not intersect the unit circle <inline-formula> <math display="inline"> <semantics> <mrow> <mi mathvariant="sans-serif">Γ</mi> <mo>:</mo> <mo>=</mo> <mo>{</mo> <mi>w</mi> <mo>∈</mo> <mi mathvariant="double-struck">C</mi> <mo>:</mo> <mo>|</mo> <mi>w</mi> <mo>|</mo> <mo>=</mo> <mn>1</mn> <mo>}</mo> </mrow> </semantics> </math> </inline-formula>. Connecting this result with a recently obtained one it follows that the above linear recurrence is Hyers-Ulam stable if and only if the spectrum of <inline-formula> <math display="inline"> <semantics> <mrow> <mi>A</mi> <mo>(</mo> <mi>q</mi> <mo>)</mo> </mrow> </semantics> </math> </inline-formula> does not intersect the unit circle.
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