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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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

Original entry on oeis.org

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

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Author

Amarnath Murthy, Feb 18 2002

Keywords

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

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

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

Original entry on oeis.org

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

Views

Author

Vladimir Pletser, Feb 23 2021

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

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.

Programs

  • Maple
    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)])[] ;
  • Mathematica
    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 *)

Formula

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

A166477 Minimum positive integer solution x of equation n=x*(x+1)/(t*(t+1)); that is, ratio of product of two consecutive integers divided by product of two consecutive integers. Here n is a nonsquare integer (see A000037).

Original entry on oeis.org

3, 2, 5, 3, 6, 15, 4, 11, 8, 12, 20, 5, 51, 27, 19, 15, 6, 11, 45, 95, 12, 54, 7, 29, 24, 30, 1343, 54, 84, 14, 185, 95, 65, 15, 41, 35, 42, 560, 9, 23, 140, 287, 24, 17, 39, 105, 1539, 10, 48, 18, 87, 1770, 104, 183, 216, 27, 455, 11, 200, 119, 45, 20, 71, 63, 72, 14060, 99
Offset: 2

Views

Author

Carmine Suriano, Oct 14 2009

Keywords

Comments

From R. J. Mathar, Oct 23 2010: (Start)
Writing x = (-1 + sqrt(1 + 4*n*t*(t+1)))/2, each solution is associated with a Diophantine equation 1 + 4*n*t*(t+1) = s^2. The sequence entries are the leading column if all solutions are presented in rows for a given n:
n Seq # solutions
-- ------- ------------------------------------------------
2 A001652 3, 20, 119, 696, 4059
3 A001571 2, 9, 35, 132, 494, 1845, 6887
4 ...
5 A077262 5, 14, 99, 260, 1785, 4674
6 A077291 3, 8, 35, 84, 351, 836, 3479, 8280
7 A077401 6, 14, 104, 231, 1665, 3689
8 A336625 15, 32, 527, 1104, 17919
9 ...
10 A341895 4, 20, 39, 175, 779, 1500, 6664, 29600
11 11, 21, 230, 429, 4598, 8568
12 8, 15, 119, 216, 1664, 3015, 23183
13 12, 77, 845, 1494, 16302
14 20, 35, 615, 1064, 18444, 31899
15 5, 9, 44, 75, 350, 594, 2759, 4680, 21725, 36849
16 ...
17 51, 84, 3399, 5576
18 27, 44, 935, 1512, 31779
19 19, 285, 455, 6649
20 15, 24, 279, 440, 5015, 7904
(End) [table reformatted by Jon E. Schoenfield, Apr 01 2018]

Examples

			For n=14, x=20; corresponding value of t is 5 since 14 = 20*21/(5*6).

Crossrefs

Cf. A000037.
Cf. A166478 (associated t). - R. J. Mathar, Oct 23 2010

Extensions

Deleted an 8 between 14 and 185. - R. J. Mathar, Oct 23 2010

A341894 For square n > 0, a(n) = 0; for nonsquare n > 0, a(n) is the rank r such that t(r) + t(r-1) = u(r) - u(r-1) - 1, where u(r) and t(r) are indices of some triangular numbers in the Diophantine relation T(u(r)) = n*T(t(r)).

Original entry on oeis.org

0, 1, 1, 0, 2, 2, 2, 2, 0, 3, 2, 2, 4, 2, 2, 0, 2, 2, 3, 2, 4, 4, 2, 2, 0, 3, 2, 4, 4, 2, 4, 2, 2, 2, 2, 0, 2, 2, 2, 4, 4, 2, 4, 2, 4, 6, 2, 2, 0, 3, 3, 4, 4, 2, 4, 2, 4, 4, 2, 2, 8, 2, 2, 0, 2, 4, 4, 2, 4, 4, 4, 2, 6, 2, 2, 6, 4, 4, 2, 2, 0, 3, 2, 2, 8, 4, 2, 4, 4, 2, 6, 4, 4, 4, 2, 4, 4, 2, 2, 0, 2, 2, 4, 2, 4, 8, 2, 2, 8, 2
Offset: 1

Views

Author

Vladimir Pletser, Mar 06 2021

Keywords

Comments

Let t(i) and u(j) be the indices of triangular numbers that satisfy the Diophantine relation T(u(j)) = n*T(t(i)) for some integers i and j. The number of solutions (t(i), u(j)) of T(u(j)) = n*T(t(i)) is 0 or 1 for square n, and an infinity for nonsquare n.
For square n, a(n) is arbitrarily set to 0.
For nonsquare n, a(n) is the index r in the sequence of t(i) and u(j) such that t(r) + t(r-1) = u(r) - u(r-1) - 1.
Alternatively, for nonsquare n, a(n) is the index r such that the ratio t(i)/t(i-r) is decreasing monotonically without jumps for increasing values of i.
Alternatively, for n > 4, a(n) is the index r such that the ratio t(r)/t(r-1) varies between (s+1)/(s-1) and (s+2)/s, with s = [sqrt(n)], where [x] = floor(x).
Alternatively, for nonsquare n, a(n) is the number of fundamental solutions (X_f, Y_f) of the generalized Pell equation X^2 - n*Y^2 = 1 - n providing odd solutions, i.e., with X_f odd and Y_f odd (or Y_f even if y_f is odd, where y_f is the fundamental solution of the associated simple Pell equation x^2 - n*y^2 = 1).

Examples

			The following table gives the first values of nonsquare n and a(n) and the sequences yielding the values of t, u, T(t) and T(u) such that T(u) = n*T(t).
n       2       3       5       6       7       8      10
a(n)    1       1       2       2       2       2       3
t    A053141 A061278 A077259 A077288 A077398 A336623  A341893*
u    A001652 A001571 A077262 A077291 A077401 A336625* A341895*
T(t) A075528 A076139 A077260 A077289 A077399 A336624  A068085*
T(u) A029549 A076140 A077261 A077290 A077400 A336626*   -
With a(n) = r, the definition t(r) + t(r-1) = u(r) - u(r-1) - 1 yields:
- For n = 2, a(n) = 1: A053141(1) + A053141(0) = A001652(1) - A001652(0) - 1, i.e., 2 + 0 = 3 + 0 - 1 = 2.
- For n = 5, a(n) = 2: A077259(2) + A077259(1) = A077262(2) - A077262(1) - 1, i.e., 6 + 2 = 14 - 5 - 1 = 8.
- For n = 10, a(n) = 3: A341893(3+1*) + A341893(2+1*) = A341895(3+1*) - A341895(2+1*) - 1, i.e., 12 + 6 = 39 - 20 - 1 = 18.
Note that for those sequences marked with an *, the first term 0 appears for n = 1, contrary to all the other sequences, where the first term 0 appears for n = 0; the numbering must therefore be adapted and 1 must be added to compensate for this shift in indices.
The monotonic decrease of t(i)/t(i-r) can be seen also as:
- For n = 2, a(n) = 1: for 1 <= i <= 6, A053141(i)/A053141(i-1) decreases monotonically from 7 to 5.829.
- For n = 5, a(n) = 2: for 3 <= i <= 8, A077259(i)/A077259(i-2) decreases monotonically from 22 to 17.948, while A077259(i)/A077259(i-1) takes values alternatively varying between 3 and 2.618 and between 7.333 and 6.855.
- For n = 10, a(n) = 3: for 4 <= i <= 10, A341893(i)/A341893(i-3) decreases monotonically from 55 to 38, while A077259(i) / A077259(i-1) takes values alternatively varying between 6 and 4.44 and between 2 and 1.925.
For n > 4, the relation (s+1)/(s-1) <=  t(r)/t(r-1) <= (s+2)/s, with s = [sqrt(n)], yields:
- For n = 5, a(n) = 2: A077259(2)/A077259(1) = 6/2 = 3, and s = [sqrt(5)] = 2, (s+1)/(s-1) = 3 and (s+2)/s = 2.
- For n = 10, a(n) = 3: A077259(3+1*)/A077259(2+1*) = 12/6 = 2, and s = [sqrt(10)] = 3, (s+1)/(s-1) = 2 and (s+2)/s = 5/3 = 1.667.
Finally, the number of fundamental solutions of the generalized Pell equation is as follows.
- For n = 2, X^2 - 2*Y^2 = -1 has a single fundamental solution, (X_f, Y_f) = (1, 1), and the rank a(n) is 1.
- For n = 5, X^2 - 5*Y^2 = -4 has two fundamental solutions, (X_f, Y_f) = (1, 1) and (-1, 1), and the rank a(n) is 2.
- For n = 10, X^2 - 10*Y^2 = -9 has three fundamental solutions, (X_f, Y_f) = (1, 1), (-1, 1), and (9, 3), and the rank a(n) is 3.

References

  • J. S. Chahal and H. D'Souza, "Some remarks on triangular numbers", in A.D. Pollington and W. Mean, eds., Number Theory with an Emphasis on the Markov Spectrum, Lecture Notes in Pure Math, Dekker, New York, 1993, 61-67.

Links

Crossrefs

Cf. A068085, A053141, A061278, A077259, A077288, A077398, A336623, A341893, A001652, A001571, A077262, A077291, A077401, A336625, A341895, A075528, A076139, A077260, A077289, A077399, A336624, A068085, A029549, A076140, A077261, A077290, A077400, A336626, A000217, A341895, A341896.
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