A350869 a(n) = Sum_{i=0..10^n-1} i^3.

0, 2025, 24502500, 249500250000, 2499500025000000, 24999500002500000000, 249999500000250000000000, 2499999500000025000000000000, 24999999500000002500000000000000, 249999999500000000250000000000000000, 2499999999500000000025000000000000000000

Offset

0,2

Comments

These terms k = x.y satisfy equation x.y = (x+y)^2, when x and y have the same number of digits, "." means concatenation, and y may not begin with 0. So, this is a subsequence of A350870 and A238237.

Links

Table of n, a(n) for n=0..10.

Iva Kodrnja, On Remarkable Properties of Number 2025, KoG, 29 (29), 74-80, 2025. See pp. 74, 78.

Index entries for linear recurrences with constant coefficients, signature (11100,-11100000,1000000000).

Formula

a(n) = 10^(2*n) * (10^n-1)^2 / 4 = A037182(n)^2.

a(n) = A000217(10^n-1)^2.

a(n) = A038544(n) - 10^(3*n).

From Elmo R. Oliveira, Nov 19 2025: (Start)

G.f.: 2025*x*(1 + 1000*x)/((1-100*x)*(1-1000*x)*(1-10000*x)).

E.g.f.: exp(100*x)*(1 - 2*exp(900*x) + exp(9900*x))/4.

a(n) = 11100*a(n-1) - 11100000*a(n-2) + 1000000000*a(n-3). (End)

Example

a(1) = Sum_{i=0..9} i^3 = (Sum_{i=0..9} i)^2 = 2025.

Mathematica

a[n_] := (10^n*(10^n - 1)/2)^2; Array[a, 11, 0] (* Amiram Eldar, Jan 20 2022 *)

Prog

(PARI) a(n) = my(x=10^n-1); (x*(x+1)/2)^2; \\ Michel Marcus, Jan 22 2022

Crossrefs

Cf. A000217, A037156, A037182, A038544, A238237, A350870.

Sequence in context: A360216 A392423 A250811 * A031768 A156856 A031543

Adjacent sequences: A350866 A350867 A350868 * A350870 A350871 A350872

Keyword

nonn,base,easy

Author

Bernard Schott, Jan 20 2022

Status

approved