A106329 -id:A106329 - cp's OEIS frontend

A164055 Triangular numbers that are one plus a perfect square.

1, 10, 325, 11026, 374545, 12723490, 432224101, 14682895930, 498786237505, 16944049179226, 575598885856165, 19553418069930370, 664240615491776401, 22564627508650467250, 766533094678624110085, 26039560591564569275626

Offset: 1

Tanya Khovanova & Alexey Radul, Aug 08 2009

a(n+1) is the second element of the n-th set of three consecutive triangular numbers whose product is a perfect square. - Arkadiusz Wesolowski, Apr 27 2012

The triangular numbers of this sequence satisfy the Diophantine equation T(k) = k*(k+1)/2 = m^2 + 1, which is equivalent to (2k+1)^2 - 2*(2m)^2 = 9. Now, with x=2k+1 and y=2m, the Pell-Fermat equation x^2 - 2*y^2 = 9 appears. The solutions x and y of this equation are respectively in A106329 and A075848. The indices k=(x-1)/2 of the triangular numbers of this sequence are in A072221, while the indices m=y/2 of the corresponding square numbers are in A106328. - Bernard Schott, Mar 09 2019

			10 is in this sequence because it is a triangular number A000217(4) and is equal to a square plus 1: 10 = 3^2 + 1.

Reinhard Zumkeller, Table of n, a(n) for n = 1..500
K. B. Subramaniam, Almost Square Triangular Numbers, The Fibonacci Quarterly, Vol. 37, No. 3 (1999), pp. 194-197.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A001110, A106329, A075848 (solutions of x^2 - 2 * y^2 = 9).

Cf. A072221 (indices of the triangular terms of this sequence), A106328 (indices of the corresponding square numbers).

a164055 n = a164055_list !! (n-1)
a164055_list = 1 : 10 : 325 : zipWith (+) a164055_list
   (map (* 35) $ tail $ zipWith (-) (tail a164055_list) a164055_list)
-- Reinhard Zumkeller, Apr 29 2012

LinearRecurrence[{35,-35,1}, {1,10,325}, 50] (* G. C. Greubel, Sep 09 2017 *)

my(x='x+O('x^50)); Vec(x*(1-25*x+10*x^2)/((1-x)*(x^2-34*x+1))) \\ G. C. Greubel, Sep 09 2017

A000217 INTERSECT A002522.
a(n) = A000217(A072221(n-1)).
From R. J. Mathar, Sep 22 2009: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
G.f.: x*(1-25*x+10*x^2)/((1-x)*(x^2-34*x+1)). (End)
A010054(a(n)) * A010052(a(n) - 1) = 1. - Reinhard Zumkeller, Apr 29 2012
a(n) = (A106329(n) - 1)*(A106329(n) + 1)/8 = (A106328(n))^2 + 1. - Bernard Schott, Mar 10 2019
a(n) = (14 + 9*(17+12*sqrt(2))^(1-n) - 9*(-17+12*sqrt(2))*(17+12*sqrt(2))^n) / 32. - Colin Barker, Mar 23 2019
a(n) = 9 * A001110(n) + 1 (Subramaniam, 1999). - Amiram Eldar, Jan 13 2022

Comment molded into formula by R. J. Mathar, Sep 22 2009

A301383 Expansion of (1 + 3x - 2x^2)/(1 - 7x + 7x^2 - x^3).

Original entry on oeis.org

1, 10, 61, 358, 2089, 12178, 70981, 413710, 2411281, 14053978, 81912589, 477421558, 2782616761, 16218279010, 94527057301, 550944064798, 3211137331489, 18715879924138, 109084142213341, 635788973355910, 3705649697922121, 21598109214176818, 125883005587138789, 733699924308655918

Offset: 0

Bruno Berselli, Mar 20 2018

y solutions to A000217(x-1) + A000217(x) = A000290(y-1) + A000290(y+2). The corresponding x values are listed in A075841.
y solutions to A000217(x-1) + A000217(x) = A000290(y-1) + A000290(y+1) are in A002315, and A075870 gives the x values.
y solutions to A000217(x-1) + A000217(x) = A000290(y-1) + A000290(y) are in A046090, and A001653 gives the x values.
Also, indices y for which 4*A000217(y) + 5 is a square. The next integers k such that k*A000217(y) + 5 is a square for infinitely many y values are 11, 20, 22, 29, 31, ...
First differences are in A106329.

Robert Israel, Table of n, a(n) for n = 0..1304
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Subsequence of A301451.

Cf. A000217, A000290, A001652, A002315, A033539, A046090, A053142, A075841, A077444, A106329, A241976.

using Nemo
function A301383List(len)
    R, x = PowerSeriesRing(ZZ, len+2, "x")
    f = divexact(1+3*x-2*x^2, 1-7*x+7*x^2-x^3)
    [coeff(f, k) for k in 0:len]
end
A301383List(23) |> println # Peter Luschny, Mar 21 2018

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+3*x-2*x^2)/(1-7*x+7*x^2-x^3)));

f:= gfun:-rectoproc({a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3), a(0)=1,a(1)=10,a(2)=61},a(n),remember):
map(f, [$0..50]); # Robert Israel, Mar 21 2018

CoefficientList[Series[(1 + 3 x - 2 x^2)/(1 - 7 x + 7 x^2 - x^3), {x, 0, 30}], x]

makelist(coeff(taylor((1+3*x-2*x^2)/(1-7*x+7*x^2-x^3), x, 0, n), x, n), n, 0, 30);

Vec((1+3*x-2*x^2)/(1-7*x+7*x^2-x^3)+O(x^30))

m=30; L. = PowerSeriesRing(ZZ, m); f=(1+3*x-2*x^2)/(1-7*x+7*x^2-x^3); print(f.coefficients())

O.g.f.: (1 + 3*x - 2*x^2)/((1 - x)*(1 - 6*x + x^2)).
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3) = 6*a(n-1) - a(n-2) + 2.
a(n) = (3/4)*((1 + sqrt(2))^(2*n + 1) + (1 - sqrt(2))^(2*n + 1)) - 1/2.
a(n) = A033539(2*n+2) = A241976(n+1) + 1 = 3*A001652(n) + 1 = 3*A046090(n) - 2.
a(n) = A053142(n+1) + 3*A053142(n) - 2*A053142(n-1), n>0.
2*a(n) = 3*A002315(n)   - 1.
4*a(n) = 3*A077444(n+1) - 2.
E.g.f.: (3*exp(3*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - cosh(x) - sinh(x))/2. - Stefano Spezia, Mar 06 2020
Let T(n) be the n-th triangular number, A000217(n).  Then T(a(n)-3) + 2*T(a(n)-2) + 3*T(a(n)-1) + 4*T(a(n)) + 3*T(a(n)+1) + 2*T(a(n)+2) + T(a(n)+3) = (A001653(n) + A001653(n+2))^2. - Charlie Marion, Mar 16 2021

A358682 Numbers k such that 8k^2 + 8k - 7 is a square.

Original entry on oeis.org

1, 7, 43, 253, 1477, 8611, 50191, 292537, 1705033, 9937663, 57920947, 337588021, 1967607181, 11468055067, 66840723223, 389576284273, 2270616982417, 13234125610231, 77134136678971, 449570694463597, 2620290030102613, 15272169486152083, 89012726886809887, 518804191834707241

Offset: 1

Stefano Spezia, Nov 26 2022

a(n) is the n-th almost cobalancing number of second type (see Tekcan and Erdem).

			a(2) = 7 is a term since 8*7^2 + 8*7 - 7 = 441 = 21^2.

Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A002315, A006452, A077443, A077446, A106328, A106329, A216134.

Mathematica
```
LinearRecurrence[{7,-7,1},{1,7,43},24]
```

a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3) for n > 3.
a(n) = (3*(3 - 2*sqrt(2))^n*(2 + sqrt(2)) + 3*(2 - sqrt(2))*(3 + 2*sqrt(2))^n - 4)/8.
O.g.f.: x*(1 + x^2)/((1 - x)*(1 - 6*x + x^2)).
E.g.f.: (3*(2 + sqrt(2))*(cosh(3*x - 2*sqrt(2)*x) + sinh(3*x - 2*sqrt(2)*x)) + 3*(2 - sqrt(2))*(cosh(3*x + 2*sqrt(2)*x) + sinh(3*x + 2*sqrt(2)*x)) - 4*(cosh(x) + sinh(x)) - 8)/8.
a(n) = 3*A011900(n) - 2 = 6*A053142(n) + 1. - Hugo Pfoertner, Nov 26 2022

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A164055 Triangular numbers that are one plus a perfect square.

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A301383 Expansion of (1 + 3*x - 2*x^2)/(1 - 7*x + 7*x^2 - x^3).

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A358682 Numbers k such that 8*k^2 + 8*k - 7 is a square.

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A301383 Expansion of (1 + 3x - 2x^2)/(1 - 7x + 7x^2 - x^3).

A358682 Numbers k such that 8k^2 + 8k - 7 is a square.