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

660, 660660, 660660660, 660660660660, 660660660660660, 660660660660660660, 660660660660660660660, 660660660660660660660660, 660660660660660660660660660, 660660660660660660660660660660, 660660660660660660660660660660660, 660660660660660660660660660660660660

Offset: 1

Author

Bernard Schott, Apr 28 2021

Keywords

• nonn
• base
• easy

Comments

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.

There are only 3 ways to get such squares, according to the Diophantine equation q = m^2 = k * rev(k) = t * rev(t).

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).

2) When q = m^2 both end with 0, there exist these 2 possibilities:

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).

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).

Examples

			For a(1) = 660, we have 660^2 = 435600 = 6600 * 66 = 528 * 825 = A325150(2) (q = 435600, m = 660, k = 6600, t = 528).
For a(2) = 660660, we have 660660^2 = 436471635600 = 6606600 * 66066 = 528528 * 825825 (q = 436471635600, m = 660660, k = 6606600, t = 528528).
Generalization: for a(n) = 660...660, we have 660...660^2 = 660...6600 * 660...66 = 528...528 * 825...825.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (1001,-1000).

Crossrefs

Cf. A117281, A325150, A325151.

Programs

Maple

E:= seq((1000^n - 1)*(220/333), n=1..11);

Mathematica

Table[(1000^n - 1)*(220/333), {n, 1, 11}] (* Amiram Eldar, Apr 28 2021 *)

PARI

Vec(660*x/((1000*x-1)*(x-1)) + O(x^13)) \\ Elmo R. Oliveira, Jul 01 2025

Formula

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

G.f.: 660*x/(1 - 1001*x + 1000*x^2). - Stefano Spezia, Apr 28 2021

a(n) = 1001*a(n-1) - 1000*a(n-2). - Wesley Ivan Hurt, Apr 28 2021

E.g.f.: 220*exp(x)*(-1 + exp(999*x))/333. - Elmo R. Oliveira, Jul 01 2025