A276600 -id:A276600 - cp's OEIS frontend

A106525 Values of x in x^2 - 49 = 2*y^2.

9, 11, 21, 43, 57, 119, 249, 331, 693, 1451, 1929, 4039, 8457, 11243, 23541, 49291, 65529, 137207, 287289, 381931, 799701, 1674443, 2226057, 4660999, 9759369, 12974411, 27166293, 56881771, 75620409, 158336759, 331531257, 440748043

Offset: 1

Andras Erszegi (erszegi.andras(AT)chello.hu), May 07 2005

From Wolfdieter Lang, Sep 27 2016: (Start)
These are the x members of all positive solutions (x(n), y(n)), proper and improper, of the Pell equation x^2 - 2*y^2 = 7^2.
The y(n) members are given in 2*A276600(n+2).
This sequence is composed of the two y members of the two proper classes of solutions of the Pell equation x^2 - 2*y^2 = 7^2 and of 7 times the proper solutions X of the Pell equation X^2 - 2*Y^2 = +1. See A275793, A275795 and 7*A001541. See A275793 for further information, and the Nagell reference. (End)
The sums of the consecutive integers in the following sequences will be squares: for n, i >= 1, if mod(i,3)=0 then 7*n+1, 7*n+2, ..., a(i)*n + (A001541(i/3)-1)/2; otherwise, if mod(i,3)=1 or 2 then 7*n+4, 7*n+5, ..., a(i)*n + (a(i)-1)/2.

			In the following, aa(n) denotes A001541(n):
a(9)=693; as mod(9,3)=0, a(9)=aa(3)*7=99*7=693, also 693^2-49=2*490^2
a(10)=1451; as mod(10,3)=1, a(10)=(aa(5)+aa(2)+aa(3)-aa(4))/2 =(3363+17+99-577)/2=1451, also 1451^2-49=2*1026^2.
The solutions (proper and every third pair improper) of x^2 - 2*y^2 = +49 begin [9, 4], [11, 6], [21, 14], [43, 30], [57, 40], [119, 84], [249, 176], [331, 234], [693, 490], [1451, 1026], [1929, 1364], [4039, 2856], [8457, 5980], [11243, 7950], [23541, 16646], ... - _Wolfdieter Lang_, Sep 27 2016

Colin Barker, Table of n, a(n) for n = 1..1001 (offset adapted by _Georg Fischer_, Jan 31 2019).
Index entries for linear recurrences with constant coefficients, signature (0,0,6,0,0,-1).

Cf. 14*A001109, 7*A001541, A106525, A106526.

Cf. A275793, A275794, A275795, A275796, 2*A276600.

I:=[9,11,21,43,57,119]; [n le 6 select I[n] else 6*Self(n-3)-Self(n-6): n in [1..40]]; // Vincenzo Librandi, Oct 26 2018

LinearRecurrence[{0,0,6,0,0,-1}, {9,11,21,43,57,119}, 50] (* Vincenzo Librandi, Oct 26 2018 *)

Vec((9+11*x+21*x^2-11*x^3-9*x^4-7*x^5)/(1-6*x^3+x^6) + O(x^50)) \\ Colin Barker, Sep 28 2016

def A106525_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(9+11*x+21*x^2-11*x^3-9*x^4-7*x^5)/(1-6*x^3+x^6) ).list()
a=A106525_list(41); a[1:] # G. C. Greubel, Sep 15 2021

a(3*k)   = A001541(k)*7 for k >= 2.
a(3*k+1) = (A001541(k+2) + A001541(k-1) + A001541(k) - A001541(k+1))/2;
a(3*k+2) = (A001541(k+2) + A001541(k-1) - A001541(k) + A001541(k+1))/2.
a(3*n) = A275793(n), a(3*n+1) = A275795(n), a(3*n+2) = 7*A001541(n+1), n >= 0. - Wolfdieter Lang, Sep 27 2016
From Colin Barker, Mar 29 2012: (Start)
a(n) = 6*a(n-3) - a(n-6).
G.f.: x*(9 + 11*x + 21*x^2 - 11*x^3 - 9*x^4 - 7*x^5)/(1 - 6*x^3 + x^6). (End)

Entry revised by N. J. A. Sloane, Oct 26 2018 at the suggestion of Georg Fischer.

A276598 Values of m such that m^2 + 3 is a triangular number (A000217).

Original entry on oeis.org

0, 5, 30, 175, 1020, 5945, 34650, 201955, 1177080, 6860525, 39986070, 233055895, 1358349300, 7917039905, 46143890130, 268946300875, 1567533915120, 9136257189845, 53250009223950, 310363798153855, 1808932779699180, 10543232880041225, 61450464500548170

Offset: 1

Colin Barker, Sep 07 2016

			5 is in the sequence because 5^2 + 3 = 28, which is a triangular number.

Colin Barker, Table of n, a(n) for n = 1..1000
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Soumeya M. Tebtoub, Hacène Belbachir, and László Németh, Integer sequences and ellipse chains inside a hyperbola, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000129, A000217, A230044.
Cf. A001109 (k=0), A106328 (k=1), A077241 (k=2), A276599 (k=5), A276600 (k=6), A276601 (k=9), A276602 (k=10), where k is the value added to n^2.
Cf. A328791 (the resulting triangular numbers).

[n le 2 select 5*(n-1) else 6*Self(n-1) - Self(n-2): n in [1..31]]; // G. C. Greubel, Sep 15 2021

CoefficientList[Series[5*x/(1 - 6*x + x^2), {x, 0, 20}], x] (* Wesley Ivan Hurt, Sep 07 2016 *)
LinearRecurrence[{6,-1},{0,5},30] (* Harvey P. Dale, Apr 26 2019 *)
(5/2)*Fibonacci[2*Range[30] -2, 2] (* G. C. Greubel, Sep 15 2021 *)

concat(0, Vec(5*x^2/(1-6*x+x^2) + O(x^30)))

a(n)=([0,1;-1,6]^n*[-5;0])[1,1] \\ Charles R Greathouse IV, Sep 07 2016

[(5/2)*lucas_number1(2*n-2, 2, -1) for n in (1..30)] # G. C. Greubel, Sep 15 2021

a(n) = 5*A001109(n-1).
a(n) = 5*( (3 - 2*sqrt(2))*(3 + 2*sqrt(2))^n - (3 + 2*sqrt(2))*(3 - 2*sqrt(2))^n )/(4*sqrt(2)).
a(n) = 6*a(n-1) - a(n-2) for n>2.
G.f.: 5*x^2 / (1-6*x+x^2).
a(n) = (5/2)*A000129(2*n-2). - G. C. Greubel, Sep 15 2021

A275793 The x members of the positive proper solutions (x = x1(n), y = y1(n)) of the first class for the Pell equation x^2 - 2*y^2 = +7^2.

Original entry on oeis.org

9, 43, 249, 1451, 8457, 49291, 287289, 1674443, 9759369, 56881771, 331531257, 1932305771, 11262303369, 65641514443, 382586783289, 2229879185291, 12996688328457, 75750250785451, 441504816384249, 2573278647520043, 14998167068736009, 87415723764896011

Offset: 0

Wolfdieter Lang, Sep 27 2016

This gives the (increasingly sorted) positive x members of the first class of the proper solutions (x1(n), y1(n)) to the Pell equation x^2 - 2*y^2 = +7^2. For the y1(n) solutions see 2*A275794(n). The solutions for the second class (x2(n), y2(n)) are found in A275795(n) and 2*A275796(n).
All solutions, including the improper ones, are given in A106525(n) and 2*A276600(n+2).
See also the comments on A263012 which apply here mutatis mutandis.
This is for the Pell equations x^2 - 2*y^2 = z^2, besides z^2 = 1 the first instance with proper solutions. For z^2 > 1 there seem to be always two classes of such solutions. For z^2 = 1 there is only one class of proper solutions. These z^2 values seem to appear for z from A058529 (prime factors are +1 or -1 (mod 8)).

			The first positive proper fundamental solution (x = x1(n), y = y1(n)) of x^2 - 2*y^2  = 49 are [9, 4], [43, 30], [249, 176], [1451, 1026], [8457, 5980], [49291, 34854], [287289, 203144], [1674443, 1184010], ...
The first positive proper fundamental solution of the second class (x = x2(n), y = y2(n)) are [11, 6], [57, 40], [331, 234], [1929, 1364], [11243, 7950], [65529, 46336], [381931, 270066], [2226057, 1574060], ...

T. Nagell, Introduction to Number Theory, Wiley, 1951, Theorem 109, pp. 207-208.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000129, A001109, A058529, A106525.

Cf. A263012, A275794, A275795, A275796, 2*A276600.

I:=[9,43]; [n le 2 select I[n] else 6*Self(n-1) - Self(n-2): n in [1..31]]; // G. C. Greubel, Sep 15 2021

RecurrenceTable[{a[n]== 6a[n-1] -a[n-2], a[-1]==11, a[0]==9}, a, {n,0,25}] (* Michael De Vlieger, Sep 28 2016 *)
Table[9*Fibonacci[2*n+1, 2] - Fibonacci[2*n, 2], {n,0,30}] (* G. C. Greubel, Sep 15 2021 *)

a(n) = round((((3-2*sqrt(2))^n*(-8+9*sqrt(2))+(3+2*sqrt(2))^n*(8+9*sqrt(2)))) / (2*sqrt(2))) \\ Colin Barker, Sep 28 2016

Vec((9-11*x)/(1-6*x+x^2) + O(x^30)) \\ Colin Barker, Oct 02 2016

def P(n): return lucas_number1(n, 2, -1);
[9*P(2*n+1) - P(2*n) for n in (0..30)] # G. C. Greubel, Sep 15 2021

a(n) = 43*S(n-1, 6) - 9*S(n-2, 6), with the Chebyshev polynomials S(n, 6) = A001109(n+1), n >= -1, with S(-2, 6) = -1.
O.g.f: (9 - 11*x)/(1 - 6*x + x^2).
a(n) = 6*a(n-1) - a(n-2) for n >= 1, with a(-1) = 11 and a(0) = 9.
a(n) = (((3-2*sqrt(2))^n*(-8+9*sqrt(2))+(3+2*sqrt(2))^n*(8+9*sqrt(2)))) / (2*sqrt(2)). - Colin Barker, Sep 28 2016
a(n) = 9*A000129(2*n+1) - A000129(2*n). - G. C. Greubel, Sep 15 2021

A276599 Values of n such that n^2 + 5 is a triangular number (A000217).

Original entry on oeis.org

1, 4, 10, 25, 59, 146, 344, 851, 2005, 4960, 11686, 28909, 68111, 168494, 396980, 982055, 2313769, 5723836, 13485634, 33360961, 78600035, 194441930, 458114576, 1133290619, 2670087421, 6605301784, 15562409950, 38498520085, 90704372279, 224385818726

Offset: 1

Colin Barker, Sep 07 2016

			4 is in the sequence because 4^2 + 5 = 21, which is a triangular number.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A000129, A000217, A230044.

Cf. A001109 (k=0), A106328 (k=1), A077241 (k=2), A276598 (k=3), A276600 (k=6), A276601 (k=9), A276602 (k=10), where k is the value added to n^2.

I:=[1,4,10,25]; [n le 4 select I[n] else 6*Self(n-2) - Self(n-4): n in [1..31]]; // G. C. Greubel, Sep 15 2021

CoefficientList[Series[(1+x)*(1+3*x+x^2)/((1+2*x-x^2)*(1-2*x-x^2)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Sep 07 2016 *)
LinearRecurrence[{0,6,0,-1},{1,4,10,25},30] (* Harvey P. Dale, Feb 13 2017 *)

Vec(x*(1+x)*(1+3*x+x^2)/((1+2*x-x^2)*(1-2*x-x^2)) + O(x^40))

a(n)=([0,1;-1,6]^(n\2)*if(n%2,[1;10],[-1;4]))[1,1] \\ Charles R Greathouse IV, Sep 07 2016

def P(n): return lucas_number1(n, 2, -1)
[(1/4)*(P(n+2) + P(n+1) + (-1)^n*(3*P(n) - 7*P(n-1))) for n in (1..30)] # G. C. Greubel, Sep 15 2021

a(n) = 6*a(n-2) - a(n-4) for n>4.
G.f.: x*(1+x)*(1+3*x+x^2) / ((1+2*x-x^2)*(1-2*x-x^2)).
a(n) = (1/4)*(P(n+2) + P(n+1) + (-1)^n*(3*P(n) - 7*P(n-1))), where P(n) = A000129(n). - G. C. Greubel, Sep 15 2021

A276601 Values of k such that k^2 + 9 is a triangular number (A000217).

Original entry on oeis.org

1, 6, 12, 37, 71, 216, 414, 1259, 2413, 7338, 14064, 42769, 81971, 249276, 477762, 1452887, 2784601, 8468046, 16229844, 49355389, 94594463, 287664288, 551336934, 1676630339, 3213427141, 9772117746, 18729225912, 56956076137, 109161928331, 331964339076

Offset: 1

Colin Barker, Sep 07 2016

			6 is in the sequence because 6^2+9 = 45, which is a triangular number.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A000129, A000217, A230044.

Cf. A001109 (k=0), A106328 (k=1), A077241 (k=2), A276598 (k=3), A276599 (k=5), A276600 (k=6), A276602 (k=10), where k is the value added to n^2.

I:=[1,6,12,37]; [n le 2 select I[n] else 6*Self(n-2) - Self(n-4): n in [1..31]]; // G. C. Greubel, Sep 15 2021

CoefficientList[Series[(1+x)*(1+5*x+x^2)/((1+2*x-x^2)*(1-2*x-x^2)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Sep 07 2016 *)
LinearRecurrence[{0,6,0,-1}, {1,6,12,37}, 31] (* G. C. Greubel, Sep 15 2021 *)

Vec(x*(1+x)*(1+5*x+x^2) / ((1+2*x-x^2)*(1-2*x-x^2)) + O(x^40))

def P(n): return lucas_number1(n, 2, -1)
[(1/4)*(5*P(n+1) - P(n) + (-1)^n*(P(n-1) + 5*P(n-2))) for n in (1..30)] # G. C. Greubel, Sep 15 2021

a(n) = 6*a(n-2) - a(n-4) for n>4.
G.f.: x*(1+x)*(1+5*x+x^2) / ((1+2*x-x^2)*(1-2*x-x^2)).
a(n) = (1/4)*(5*P(n+1) - P(n) + (-1)^n*(P(n-1) + 5*P(n-2))), where P(n) = A000129(n). - G. C. Greubel, Sep 15 2021

A276602 Values of k such that k^2 + 10 is a triangular number (A000217).

Original entry on oeis.org

0, 9, 54, 315, 1836, 10701, 62370, 363519, 2118744, 12348945, 71974926, 419500611, 2445028740, 14250671829, 83059002234, 484103341575, 2821561047216, 16445262941721, 95850016603110, 558654836676939, 3256079003458524, 18977819184074205, 110610836100986706

Offset: 1

Colin Barker, Sep 07 2016

			9 is in the sequence because 9^2+10 = 91, which is a triangular number.

Colin Barker, Table of n, a(n) for n = 1..1000
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Soumeya M. Tebtoub, Hacène Belbachir and László Németh, Integer sequences and ellipse chains inside a hyperbola, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000129, A000217, A230044.

Cf. A001109 (k=0), A106328 (k=1), A077241 (k=2), A276598 (k=3), A276599 (k=5), A276600 (k=6), A276601 (k=9), where k is the value added to n^2.

[n le 2 select 9*(n-1) else 6*Self(n-1) - Self(n-2): n in [1..31]]; // G. C. Greubel, Sep 15 2021

CoefficientList[Series[9*x/(1 - 6*x + x^2), {x, 0, 20}], x] (* Wesley Ivan Hurt, Sep 07 2016 *)
(9/2)*Fibonacci[2*(Range[30] -1), 2] (* G. C. Greubel, Sep 15 2021 *)

concat(0, Vec(9*x^2/(1-6*x+x^2) + O(x^30)))

[(9/2)*lucas_number1(2*n-2, 2, -1) for n in (1..30)] # G. C. Greubel, Sep 15 2021

a(n) = (9/(4*sqrt(2))*( (3 - 2*sqrt(2))*(3 + 2*sqrt(2))^n - (3 + 2*sqrt(2))*(3 - 2*sqrt(2))^n) ).
a(n) = 9*A001109(n-1).
a(n) = 6*a(n-1) - a(n-2) for n>2.
G.f.: 9*x^2 / (1-6*x+x^2).
a(n) = (9/2)*A000129(2*n-2). - G. C. Greubel, Sep 15 2021

A106526 Values of y in x^2 - 49 = 2*y^2.

Original entry on oeis.org

4, 6, 14, 30, 40, 84, 176, 234, 490, 1026, 1364, 2856, 5980, 7950, 16646, 34854, 46336, 97020, 203144, 270066, 565474, 1184010, 1574060, 3295824, 6900916, 9174294, 19209470, 40221486, 53471704, 111960996, 234428000, 311655930, 652556506

Offset: 1

Andras Erszegi (erszegi.andras(AT)chello.hu), May 07 2005

nonn

The expression 2*n^2 + c with c = 49 yields more squares than any other value of c in the range 1 < c < 100 and n < 5*10^4. - K. D. Bajpai, Nov 04 2013

			a(12) = 2856; as 12 mod 3 = 0, a(12) = 14*A001109(12/3) = 204*14 = 2856; also 2*2856^2 = 4039^2 - 49, i.e., A106525(12)^2 - 49;
a(13) = 5980; as 13 mod 3 = 1, a(13) = A001109(4+2) - A001109(4+1) + A001109(4) + A001109(4-1) = 6930 - 1189 + 204 + 35 = 5980; also 2*5980^2 = 8457^2 - 49, i.e., A106525(13)^2 - 49;
a(14) = 7950; as 14 mod 3 = 2, a(14) = A001109(4+2) + A001109(4+1) - A001109(4) + A001109(4-1) = 6930 + 1189 - 204 + 35 = 7950; also 2*7950^2 = 11243^2 - 49, i.e., A106525(14)^2 - 49.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,6,0,0,-1).

Cf. A001109, A106525, A276600.

I:=[4,6,14,30,40,84]; [n le 6 select I[n] else 6*Self(n-3) - Self(n-6): n in [1..41]]; // G. C. Greubel, Aug 12 2021

LinearRecurrence[{0,0,6,0,0,-1}, {4,6,14,30,40,84}, 40] (* T. D. Noe, Nov 04 2013 *)

def A106526_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (2*x)*(2 +3*x +7*x^2 +3*x^3 +2*x^4)/(1 -6*x^3 +x^6) ).list()
a=A106526_list(41); a[1:] # G. C. Greubel, Aug 12 2021

a(n) = 6*a(n-3) - a(n-6), with initial terms 4, 6, 14, 30, 40, 84. - T. D. Noe, Nov 04 2013
From G. C. Greubel, Aug 12 2021: (Start)
a(n) = 2*A276600(n+1).
G.f.: (2*x)*(2 + 3*x + 7*x^2 + 3*x^3 + 2*x^4)/(1 - 6*x^3 + x^6). (End)

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A106525 Values of x in x^2 - 49 = 2*y^2.

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A276598 Values of m such that m^2 + 3 is a triangular number (A000217).

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A275793 The x members of the positive proper solutions (x = x1(n), y = y1(n)) of the first class for the Pell equation x^2 - 2*y^2 = +7^2.

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A276599 Values of n such that n^2 + 5 is a triangular number (A000217).

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A276601 Values of k such that k^2 + 9 is a triangular number (A000217).

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A276602 Values of k such that k^2 + 10 is a triangular number (A000217).

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