A234787 Cubes (with at least two digits) that become squares when their rightmost digit is removed.

1000, 64000, 729000, 4096000, 15625000, 46656000, 117649000, 262144000, 531441000, 1000000000, 1771561000, 2985984000, 4826809000, 7529536000, 11390625000, 16777216000, 24137569000, 34012224000, 47045881000, 64000000000

Offset: 1

Author

Reiner Moewald, Dec 30 2013

Keywords

• nonn
• base
• easy

Comments

With the help of the Nagell-Lutz theorem it is easy to prove that there are no other solutions than those of the form 1000*n^6.

Links

• Georg Fischer, Table of n, a(n) for n = 1..300 [a(1..204) from Reiner Moewald]
• Wikipedia, Nagell-Lutz theorem.
• Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Crossrefs

Cf. A226354.

Subsequence of A000578.

Programs

PARI

lista(nn) = {for (n=1, nn, if (((cb = n^3) > 10) && issquare(cb\10), print1(cb, ", ")););} \\ Michel Marcus, Jan 10 2014

PARI

Vec(1000*x*(1 + x)*(1 + 56*x + 246*x^2 + 56*x^3 + x^4) / (1 - x)^7 + O(x^40)) \\ Colin Barker, Dec 15 2019

Formula

a(n) = 1000*n^6.

From Colin Barker, Dec 15 2019: (Start)

G.f.: 1000*x*(1 + x)*(1 + 56*x + 246*x^2 + 56*x^3 + x^4) / (1 - x)^7.

a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) for n>7.

(End)