A281857 Numbers occurring in a curious cubic identity.

153, 165033, 166500333, 166650003333, 166665000033333, 166666500000333333, 166666650000003333333, 166666665000000033333333, 166666666500000000333333333, 166666666650000000003333333333, 166666666665000000000033333333333, 166666666666500000000000333333333333

Offset: 1

Author

Wolfdieter Lang, Feb 07 2017

Keywords

• nonn
• easy

Comments

See A246057 for the van der Poorten et al. reference and a comment.

153 is the Armstrong number A005188(10). [Typo corrected by Jeremy Tan, Feb 25 2023]

Examples

			1^3 + 5^3 + 3^3 = 153, 16^3 + 50^3 + 33^3 = 165033, 166^3 + 500^3 + 333^3 = 166500333, ...
		

Links

• Colin Barker, Table of n, a(n) for n = 1..333
• Index entries for linear recurrences with constant coefficients, signature (1111,-112110,1111000,-1000000).

Crossrefs

Cf. A002277, A005188, A093143, A246057, A281858, A281859, A281860.

Programs

Mathematica

Table[FromDigits@ Join[ReplacePart[ConstantArray[6, n], 1 -> 1], ReplacePart[ConstantArray[0, n], 1 -> 5], ConstantArray[3, n]], {n, 12}] (* Michael De Vlieger, Feb 08 2017 *)

PARI

Vec(9*x*(17 - 550*x + 33500*x^2) / ((1 - x)*(1 - 10*x)*(1 - 100*x)*(1 - 1000*x)) + O(x^15)) \\ Colin Barker, Feb 08 2017

PARI

a(n) = (((10^n - 4)/6)^3) + ((10^n/2)^3) + (((10^n - 1)/3)^3) \\ Jean-Jacques Vaudroz, Aug 11 2024

Formula

a(n) = A246057(n-1)^3 + A093143(n)^3 + A002277(n)^3, n >= 1.

From Colin Barker, Feb 08 2017: (Start)

G.f.: 9*x*(17 - 550*x + 33500*x^2) / ((1 - x)*(1 - 10*x)*(1 - 100*x)*(1 - 1000*x)).

a(n) = (-2 + 2^(1+n)*5^n - 100^n + 1000^n) / 6.

a(n) = 1111*a(n-1) - 112110*a(n-2) + 1111000*a(n-3) - 1000000*a(n-4) for n>4. (End)