OFFSET
0,1
COMMENTS
This is an old school trick which says that a square of an integer that ends with 5 is easy to compute. Remove the 5, multiply the remaining number by (itself + 1), and concatenate 25 at the end. So, a(n)\100 = A002378(n). - Michel Marcus, Dec 23 2013
REFERENCES
Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See p. 136.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
G.f.: -25*(x^2 + 6*x + 1)/(x - 1)^3. - Colin Barker, Nov 14 2012
a(n) = A017329(n)^2. - Michel Marcus, Dec 23 2013
a(0) = 25, a(n) = a(n - 1) + 200*n. - Alonso del Arte, Feb 25 2020
From Amiram Eldar, Apr 18 2023: (Start)
a(n) = 5^2 * A016754(n).
Sum_{n>=0} 1/a(n) = Pi^2/200.
Sum_{n>=0} (-1)^n/a(n) = G/25, where G is Catalan's constant (A006752). (End)
E.g.f.: 25*exp(x)*(1 + 8*x + 4*x^2). - Stefano Spezia, Aug 04 2024
EXAMPLE
5^2 = 25;
15^2 = (1 * 2) concatenate 25 = 225;
25^2 = (2 * 3) concatenate 25 = 625;
35^2 = (3 * 4) concatenate 25 = 1225;
45^2 = (4 * 5) concatenate 25 = 2025;
55^2 = (5 * 6) concatenate 25 = 3025;
65^2 = (6 * 7) concatenate 25 = 4225, etc.
MATHEMATICA
25Range[1, 89, 2]^2 (* Alonso del Arte, Feb 25 2020 *)
PROG
(Magma) [(10*n+5)^2: n in [0..35]]; // Vincenzo Librandi, Aug 02 2011
(PARI) a(n)=(10*n+5)^2 \\ Charles R Greathouse IV, Jun 17 2017
(Scala) (5 to 445 by 10).map(n => n * n) // Alonso del Arte, Feb 25 2020
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
More terms from Michel Marcus, Dec 23 2013
STATUS
approved