A068085 Numbers k such that k and 10*k are both triangular numbers.

0, 1, 21, 78, 1540, 30381, 112575, 2220778, 43809480, 162333171, 3202360435, 63173239878, 234084320106, 4617801526591, 91095768094695, 337549427259780, 6658866598983886, 131360034419310411, 486746040024282753, 9602081017933237120, 189421078536877518066, 701887452165588470145

Offset: 1

Author

Amarnath Murthy, Feb 18 2002

Keywords

• nonn
• easy

Comments

Let y=sqrt(8*k+1) and x=sqrt(80*k+1), which must be integers if k and 10*k are triangular. These quantities satisfy the Pell-like equation x^2 - 10*y^2 = -9. All solutions x+y*sqrt(10) are obtained from 1+sqrt(10), 9+3*sqrt(10) and 41+13*sqrt(10) by multiplying by powers of the fundamental unit 19+6*sqrt(10).

Conjecture: satisfies a linear recurrence having signature (1, 0, 1442, -1442, 0, -1, 1). - Harvey P. Dale, Sep 03 2020

This conjecture is true because of the connection between (generalized) Pell equations and continued fractions of quadratic irrationals. - Georg Fischer, Feb 23 2021

From Vladimir Pletser, Feb 26 2021: (Start)

The triangular numbers T(t) that are one-tenth of other triangular numbers T(u) : T(t)=T(u)/10. The t's are in A341893, and the u's are in A341895.

Can be defined for negative n by setting a(n) = a(1-n) for all n in Z. (End)

Examples

			21 and 210 are both triangular numbers.
		

Links

• Georg Fischer, Table of n, a(n) for n = 1..1000
• Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
• Index entries for linear recurrences with constant coefficients, signature (1,0,1442,-1442,0,-1,1).

Crossrefs

Cf. for k and m*k both triangular: A075528 (m=2), A076139 (m=3), 0 (m=4), A077260 (m=5), A077289 (m=6), A077399 (m=7), A336624 (m=8), 0 (m=9), this sequence (m=10).

Programs

Maple

f := gfun:-rectoproc({a(-3) = 21, a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(2) = 21, a(n) = 1442*a(n-3)-a(n-6)+99}, a(n), remember); map(f, [`$`(0 .. 1000)])[] ; # Vladimir Pletser, Feb 26 2021

Mathematica

a[0]=0; a[1]=1; a[2]=21; a[n_] := a[n]=(99+1442a[n-3]+57Sqrt[(1+8a[n-3])(1+80a[n-3])])/2

Formula

a(n) = (99 + 1442*a(n-3) + 57*sqrt((1 + 8*a(n-3))*(1 + 88*a(n-3))))/2.

G.f.: -x^2*(x^4+20*x^3+57*x^2+20*x+1) / ((x-1)*(x^6-1442*x^3+1)). - Colin Barker, Jun 24 2014

From _Vladimir Pletser, Feb 26 2021: (Start)

a(n) = 1442 *a(n-3) - a(n-6) + 99, for n > 3, with a(-2) = 21, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 21.

a(n) = a(n - 1) + 1442 ( a(n - 3) - a(n - 4) ) - ( a(n - 6) - a(n - 7) ) for n >= 4 with a(-2) = 21, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 21.

a(n) = b(n)*(b(n)+1)/2 where b(n) is A341893(n). (End)

Extensions

Edited by Dean Hickerson, Feb 20 2002

More terms from Georg Fischer, Feb 23 2021