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

%I A068085 #56 May 31 2022 14:30:27
%S A068085 0,1,21,78,1540,30381,112575,2220778,43809480,162333171,3202360435,
%T A068085 63173239878,234084320106,4617801526591,91095768094695,
%U A068085 337549427259780,6658866598983886,131360034419310411,486746040024282753,9602081017933237120,189421078536877518066,701887452165588470145
%N A068085 Numbers k such that k and 10*k are both triangular numbers.
%C A068085 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).
%C A068085 Conjecture: satisfies a linear recurrence having signature (1, 0, 1442, -1442, 0, -1, 1). - _Harvey P. Dale_, Sep 03 2020
%C A068085 This conjecture is true because of the connection between (generalized) Pell equations and continued fractions of quadratic irrationals. - _Georg Fischer_, Feb 23 2021
%C A068085 From _Vladimir Pletser_, Feb 26 2021: (Start)
%C A068085 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.
%C A068085 Can be defined for negative n by setting a(n) = a(1-n) for all n in Z. (End)
%H A068085 Georg Fischer, <a href="/A068085/b068085.txt">Table of n, a(n) for n = 1..1000</a>
%H A068085 Vladimir Pletser, <a href="https://www.researchgate.net/profile/Vladimir-Pletser/publication/359808848_USING_PELL_EQUATION_SOLUTIONS_TO_FIND_ALL_TRIANGULAR_NUMBERS_MULTIPLE_OF_OTHER_TRIANGULAR_NUMBERS/">Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers</a>, 2022.
%H A068085 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,1442,-1442,0,-1,1).
%F A068085 a(n) = (99 + 1442*a(n-3) + 57*sqrt((1 + 8*a(n-3))*(1 + 88*a(n-3))))/2.
%F A068085 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
%F A068085 From __Vladimir Pletser_, Feb 26 2021: (Start)
%F A068085 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.
%F A068085 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.
%F A068085 a(n) = b(n)*(b(n)+1)/2 where b(n) is A341893(n). (End)
%e A068085 21 and 210 are both triangular numbers.
%p A068085 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
%t A068085 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
%Y A068085 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).
%K A068085 nonn,easy
%O A068085 1,3
%A A068085 _Amarnath Murthy_, Feb 18 2002
%E A068085 Edited by _Dean Hickerson_, Feb 20 2002
%E A068085 More terms from _Georg Fischer_, Feb 23 2021