A180003 -id:A180003 - cp's OEIS frontend

A341893 Indices of triangular numbers that are one-tenth of other triangular numbers.

0, 1, 6, 12, 55, 246, 474, 2107, 9360, 18018, 80029, 355452, 684228, 3039013, 13497834, 25982664, 115402483, 512562258, 986657022, 4382255359, 19463867988, 37466984190, 166410301177, 739114421304, 1422758742216, 6319209189385, 28066884141582, 54027365220036, 239963538895471, 1065802482958830, 2051617119619170

Offset: 1

Vladimir Pletser, Feb 23 2021

The indices of triangular numbers that are one-tenth of other triangular numbers [t of T(t) such that T(t)=T(u)/10].
First member of the Diophantine pair (t, u) that satisfies 10*(t^2 + t) = u^2 + u; a(n) = t.
The T(t)'s are in A068085 and the u's are in A341895.
Also, nonnegative t such that 40*t^2 + 40*t + 1 is a square.
Can be defined for negative n by setting a(n) = a(-1-n) for all n in Z.

			a(4) = 12 is a term because its triangular number, (12*13) / 2 = 78 is one-tenth of 780, the triangular number of 39.
a(4) = 38 a(1) - a(-2) +18 = 0 - 6 +18 = 12 ;
a(5) = 38 a(2) - a(-1) + 18 = 38*1 - 1 +18 = 55.

Vladimir Pletser, 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.
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 (1,38,-38,-1,1).

Cf. A068085, A341895.
Cf. A336623, A336624, A336626, A336625, A053141, A001652, A075528, A029549, A061278, A001571, A076139, A076140, A077259, A077262, A077260, A077261, A077288, A077291, A077289, A077290, A077398, A077401, A077399, A077400, A000217.
Cf. A180003.

f := gfun:-rectoproc({a(-3) = 6, a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(2) = 6, a(n) = 38*a(n-3)-a(n-6)+18}, a(n), remember); map(f, [`$`(0 .. 1000)])[] ;

Rest@ CoefficientList[Series[(x^2*(1 + 5*x + 6*x^2 + 5*x^3 + x^4))/(1 - x - 38*x^3 + 38*x^4 + x^6 - x^7), {x, 0, 31}], x] (* Michael De Vlieger, May 19 2022 *)

a(n) = (-1 + sqrt(8*b(n) + 1))/2 where b(n) = A068085(n).
a(n) = 38 a(n-3) - a(n-6) + 18 for n > 3, with a(-2) = 6, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 6.
a(n) = a(n-1) + 38*(a(n-3) - a(n-4)) - (a(n-6) - a(n-7)) for n >= 4 with a(-2) = 6, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 6.
G.f.: x^2*(1 + 4*x+x^2)*(1+x+x^2)/ ((1-x)*(1-38*x^3+x^6)). - Stefano Spezia, Feb 24 2021
a(n) = A180003(n) - 1. - Hugo Pfoertner, Feb 28 2021

A180002 Place a(n) red and b(n) blue balls in an urn; draw 5 balls without replacement; Probability(5 red balls) = Probability(3 red and 2 blue balls).

Original entry on oeis.org

4, 8, 24, 43, 179, 783, 1504, 6668, 29604, 56983, 253079, 1124043, 2163724, 9610208, 42683904, 82164403, 364934699, 1620864183, 3120083464, 13857908228, 61550154924, 118481007103, 526235577839, 2337285022803, 4499158186324, 19983094049528

Offset: 1

Paul Weisenhorn, Aug 05 2010

nonn

This is equivalent to the Pell equation A(n)^2 - 10*B(n)^2 = -9 with a(n) = (A(n)+7)/2, b(n) = (B(n)+1)/2, and the 3 fundamental solutions (1,1), (9,3), (41,13), and the solution (19,6) for the unit form.

			For n=3: a(3)=24 and b(3)=7 since binomial(24,5) = binomial(24,3)*binomial(7,2) = 42504.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,38,-38,0,-1,1).

Cf. A180003 (b(n)).

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( x*(4+4*x+ 16*x^2-133*x^3-16*x^4-4*x^5+3*x^6)/((1-x)*(1-38*x^3+x^6)) )); // G. C. Greubel, Mar 20 2019

Rest[CoefficientList[Series[x*(4+4*x+16*x^2-133*x^3-16*x^4-4*x^5 +3*x^6 )/((1-x)*(1-38*x^3+x^6)), {x,0,30}], x]] (* G. C. Greubel, Mar 20 2019 *)
LinearRecurrence[{1,0,38,-38,0,-1,1},{4,8,24,43,179,783,1504},30] (* Harvey P. Dale, May 04 2024 *)

my(x='x+O('x^30)); Vec(x*(4+4*x+16*x^2-133*x^3-16*x^4-4*x^5+3*x^6) /((1-x)*(1-38*x^3+x^6))) \\ G. C. Greubel, Mar 20 2019

a=(x*(4+4*x+16*x^2-133*x^3-16*x^4-4*x^5+3*x^6)/((1-x)*(1-38*x^3 +x^6))).series(x, 30).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Mar 20 2019

G.f.: x*(4 +4*x +16*x^2 -133*x^3 -16*x^4 -4*x^5 +3*x^6)/((1-x)*(1 -38*x^3 +x^6)).
a(n+9) = 39*a(n+6) - 39*a(n+3) + a(n).
Let r = sqrt(10) then:
a(3*n+1) = (14 + (1+r)*(19+6*r)^n + (1-r)*(19-6*r)^n)/4.
a(3*n+2) = (14 + 3*(3+r)*(19+6*r)^n + 3*(3-r)*(19-6*r)^n)/4.
a(3*n+3) = (14 + (41+13*r)*(19+6*r)^n + (41-13*r)*(19-6*r)^n)/4.
a(n) = a(n-1) + 38*a(n-3) - 38*a(n-4) - a(n-6) + a(n-7). - G. C. Greubel, Mar 20 2019

cp's OEIS Frontend

A341893 Indices of triangular numbers that are one-tenth of other triangular numbers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula