A055792 a(n) and floor(a(n)/2) are both squares; i.e., squares which remain squares when written in base 2 and last digit is removed.

0, 1, 9, 289, 9801, 332929, 11309769, 384199201, 13051463049, 443365544449, 15061377048201, 511643454094369, 17380816062160329, 590436102659356801, 20057446674355970889, 681362750825443653409, 23146276081390728245001, 786292024016459316676609

Offset: 0

Author

Henry Bottomley, Jul 14 2000

Keywords

• nonn
• base
• easy

Comments

a(n) > 0 is a square such that a(n) - 1 is a product of powers. - Michel Lagneau, Feb 16 2012

Examples

			a(2) = 9 because 9 = 3^2 = 1001_2 and 100_2 = 4 = 2^2.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 0..654
• M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
• Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
• Giovanni Lucca, Circle chains inside the arbelos and integer sequences, Int'l J. Geom. (2023) Vol. 12, No. 1, 71-82.
• Index to sequences related to truncating digits of squares.
• Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Crossrefs

Cf. A023110, A247375.

Programs

Mathematica

LinearRecurrence[{35, -35, 1}, {0, 1, 9, 289}, 25] (* Paolo Xausa, Jul 22 2024 *)

PARI

concat(0, Vec(-x*(9*x^2-26*x+1)/((x-1)*(x^2-34*x+1)) + O(x^100))) \\ Colin Barker, Sep 15 2014

PARI

is(n)=issquare(n) && issquare(n\2) \\ Charles R Greathouse IV, May 07 2015

Formula

a(n) = 34*a(n-1) - a(n-2) - 16 = A001541(n-1)^2 = 2*A001542(n-1)^2 + 1 = 8*A001110(n-1) + 1.

From Colin Barker, Sep 15 2014: (Start)

a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3) for n > 3.

G.f.: -x*(9*x^2 - 26*x + 1) / ((x-1)*(x^2 - 34*x + 1)). (End)

a(n) = c*k^n + 1/2 + o(1) with k = 17+sqrt(288) = 33.97... and c = 17/4 - sqrt(18). - Charles R Greathouse IV, May 07 2015

a(n) = (4 + 2*(17 + 12*sqrt(2))^(1-n) + (34 - 24*sqrt(2))*(17 + 12*sqrt(2))^n)/8 for n > 0. - Colin Barker, Mar 02 2016