A336626 Triangular numbers that are eight times another triangular number.
0, 120, 528, 139128, 609960, 160554240, 703893960, 185279454480, 812293020528, 213812329916328, 937385441796000, 246739243443988680, 1081741987539564120, 284736873122033021040, 1248329316235215199128, 328586104843582662292128, 1440570949193450800230240, 379188080252621270252095320
Offset: 1
Examples
a(2) = 120 is a term because it is triangular and 120/8 = 15 is also triangular. a(3) = 1154*a(1) - a(-1) + 648 = 0 - 120 + 648 = 528; a(4) = 1154*a(2) - a(0) + 648 = 1154*120 - 0 + 648 = 139128, etc. . From _Peter Luschny_, Oct 19 2020: (Start) Related sequences in context, as computed by the Julia function: n [A336623, A336624, A336625, A336626 ] [0] [0, 0, 0, 0 ] [1] [5, 15, 15, 120 ] [2] [11, 66, 32, 528 ] [3] [186, 17391, 527, 139128 ] [4] [390, 76245, 1104, 609960 ] [5] [6335, 20069280, 17919, 160554240 ] [6] [13265, 87986745, 37520, 703893960 ] [7] [215220, 23159931810, 608735, 185279454480 ] [8] [450636, 101536627566, 1274592, 812293020528 ] [9] [7311161, 26726541239541, 20679087, 213812329916328] (End)
Links
- Vladimir Pletser, Table of n, a(n) for n = 1..653
- Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
- Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
- Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
- Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
- V. Pletser, Recurrent relations for triangular multiples of other triangular numbers, Indian J. Pure Appl. Math. 53 (2022) 782-791
- Index entries for linear recurrences with constant coefficients, signature (1,1154,-1154,-1,1).
Crossrefs
Programs
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Julia
function omnibus() println("[A336623, A336624, A336625, A336626]") println([0, 0, 0, 0]) t, h = 1, 1 for n in 1:999999999 d, r = divrem(t, 8) if r == 0 d2 = 2*d s = isqrt(d2) d2 == s * (s + 1) && println([s, d, n, t]) end t, h = t + h + 1, h + 1 end end omnibus() # Peter Luschny, Oct 19 2020 -
Maple
f := gfun:-rectoproc({a(n) = 1154*a(n - 2) - a(n - 4) + 648, a(2) = 120, a(1) = 0, a(0) = 0, a(-1) = 120}, a(n), remember); map(f, [$ (1 .. 1000)])[]; # -
Mathematica
LinearRecurrence[{1, 1154, -1154, -1, 1}, {0, 120, 528, 139128, 609960}, 18]
Formula
a(n) = 8*A336624(n).
a(n) = 1154*a(n-2) - a(n-4) + 648, for n>=2 with a(2)=120, a(1)=0, a(0)=0, a(-1)=120.
a(n) = a(n-1) + 1154*a(n-2) - 1154*a(n-3) - a(n-4) + a(n-5), for n>=3 with a(3)=528, a(2)=120, a(1)=0, a(0)=0, a(-1)=120.
a(n) = ((10*sqrt(2))/17 + 15/17)*(17 + 12*sqrt(2))^n + (-(10*sqrt(2))/17 + 15/17)*(17 - 12*sqrt(2))^n + (-15/17 - (45*sqrt(2))/68)*(-17 - 12*sqrt(2))^n + (-15/17 + (45*sqrt(2))/68)*(-17 + 12*sqrt(2))^n - 27*(-4 + 3*sqrt(2))*sqrt(2)*(-1/(-17 + 12*sqrt(2)))^n/(1088*(-17 + 12*sqrt(2))) - 27*(4 + 3*sqrt(2))*sqrt(2)*(-1/(-17 - 12*sqrt(2)))^n/(1088*(-17 - 12*sqrt(2))) - 9/16 - 9*(-3 + 2*sqrt(2))*sqrt(2)*(-1/(17 - 12*sqrt(2)))^n/(272*(17 - 12*sqrt(2))) - 9*(3 + 2*sqrt(2))*sqrt(2)*(-1/(17 + 12*sqrt(2)))^n/(272*(17 + 12*sqrt(2))).
Let b(n) be A336625(n). Then a(n) = b(n)*(b(n)+1)/2.
G.f.: 24*x^2*(5 + 17*x + 5*x^2)/(1 - x - 1154*x^2 + 1154*x^3 + x^4 - x^5). - Stefano Spezia, Oct 05 2020
From Vladimir Pletser, Feb 21 2021: (Start)
a(n) = ((11*(1 + sqrt(2))^2 - (-1)^n*6*(4 + 3*sqrt(2)))*(1 + sqrt(2))^(4n) + (11*(1 - sqrt(2))^2 - (-1)^n*6*(4 - 3*sqrt(2)))*(1 - sqrt(2))^(4n))/32 - 9/16.
a(n) = ((1 + 2*sqrt(2))^2*(1 + sqrt(2))^(4n) + (1 - 2*sqrt(2))^2*(1 - sqrt(2))^(4n))/32 - 9/16 for even n.
a(n) = ((5 + 4*sqrt(2))^2*(1 + sqrt(2))^(4n) + (5 - 4*sqrt(2))^2*(1 - sqrt(2))^(4n))/32 - 9/16 for odd n. (End)
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