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A342994 a(n) = (1000^n - 1)*(220/333).

%I A342994 #42 Jul 01 2025 08:54:47
%S A342994 660,660660,660660660,660660660660,660660660660660,660660660660660660,
%T A342994 660660660660660660660,660660660660660660660660,
%U A342994 660660660660660660660660660,660660660660660660660660660660,660660660660660660660660660660660,660660660660660660660660660660660660
%N A342994 a(n) = (1000^n - 1)*(220/333).
%C A342994 Why is this sequence interesting? Answer: Squares that can be expressed as the product of a number and its reversal in exactly two ways are in A325150.
%C A342994 There are only 3 ways to get such squares, according to the Diophantine equation q = m^2 = k * rev(k) = t * rev(t).
%C A342994 1) When q, m do not end with 0, then m = k is a palindrome belonging to A117281; example: for m = k = A117281(1) = 252, q = 252^2 = 252*252 = 144*441 = 63504 = A325150(1).
%C A342994 2) When q = m^2 both end with 0, there exist these 2 possibilities:
%C A342994 2.1) k and t also both end with 0, then m = r * 10^u where r belongs to A325151 and u >= 1; example: for r = A325151(1) = 403, u = 1, m = 4030, k = 16900 and t = 96100 with q = 16240900 = 4030^2 = 16900 * 961 = 96100 * 169 = A325150(4).
%C A342994 2.2) k ends with 0 but not t, then m belongs to this sequence; so another equivalent name is: numbers with trailing zeros whose square can be expressed as the product of a number ending with 0 and its reversal, and agian as the product of a number and its reversal, but this time without trailing zero (see examples).
%H A342994 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1001,-1000).
%F A342994 a(n) = (1000^n - 1)*(220/333).
%F A342994 G.f.: 660*x/(1 - 1001*x + 1000*x^2). - _Stefano Spezia_, Apr 28 2021
%F A342994 a(n) = 1001*a(n-1) - 1000*a(n-2). - _Wesley Ivan Hurt_, Apr 28 2021
%F A342994 E.g.f.: 220*exp(x)*(-1 + exp(999*x))/333. - _Elmo R. Oliveira_, Jul 01 2025
%e A342994 For a(1) = 660, we have 660^2 = 435600 = 6600 * 66 = 528 * 825 = A325150(2) (q = 435600, m = 660, k = 6600, t = 528).
%e A342994 For a(2) = 660660, we have 660660^2 = 436471635600 = 6606600 * 66066 = 528528 * 825825 (q = 436471635600, m = 660660, k = 6606600, t = 528528).
%e A342994 Generalization: for a(n) = 660...660, we have 660...660^2 = 660...6600 * 660...66 = 528...528 * 825...825.
%p A342994 E:= seq((1000^n - 1)*(220/333), n=1..11);
%t A342994 Table[(1000^n - 1)*(220/333), {n, 1, 11}] (* _Amiram Eldar_, Apr 28 2021 *)
%o A342994 (PARI) Vec(660*x/((1000*x-1)*(x-1)) + O(x^13)) \\ _Elmo R. Oliveira_, Jul 01 2025
%Y A342994 Cf. A117281, A325150, A325151.
%K A342994 nonn,base,easy
%O A342994 1,1
%A A342994 _Bernard Schott_, Apr 28 2021