A276598 -id:A276598 - cp's OEIS frontend

A276600 Values of m such that m^2 + 6 is a triangular number (A000217).

0, 2, 3, 7, 15, 20, 42, 88, 117, 245, 513, 682, 1428, 2990, 3975, 8323, 17427, 23168, 48510, 101572, 135033, 282737, 592005, 787030, 1647912, 3450458, 4587147, 9604735, 20110743, 26735852, 55980498, 117214000, 155827965, 326278253, 683173257, 908231938

Offset: 1

Colin Barker, Sep 07 2016

2*a(n+2) gives the y members of all positive solutions (x(n), y(n)), proper and improper, of the Pell equation x^2 - 2*y^2 = 7^2, n >= 0. The corresponding x members are x(n) = A106525(n). - Wolfdieter Lang, Sep 29 2016

			7 is in the sequence because 7^2 + 6 = 55, 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 (0,0,6,0,0,-1).

Cf. A000217, A230044.
Cf. A001109 (k=0), A106328 (k=1), A077241 (k=2), A276598 (k=3), A276599 (k=5), A276601 (k=9), A276602 (k=10), where k is the value added to n^2.
Cf. A275794, A275796, A106525.

I:=[0,2,3,7,15,20]; [n le 6 select I[n] else 6*Self(n-3) - Self(n-6): n in [1..41]]; // G. C. Greubel, Sep 15 2021

LinearRecurrence[{0,0,6,0,0,-1}, {0,2,3,7,15,20}, 41] (* G. C. Greubel, Sep 15 2021 *)

concat(0, Vec(x^2*(2+3*x+7*x^2+3*x^3+2*x^4)/(1-6*x^3+x^6) + O(x^40)))

def A276600_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x^2*(2+3*x+7*x^2+3*x^3+2*x^4)/(1-6*x^3+x^6) ).list()
a=A276600_list(41); a[1:] # G. C. Greubel, Sep 15 2021

a(n) = 6*a(n-3) - a(n-6) for n>6.
G.f.: x^2*(2 + 3*x + 7*x^2 + 3*x^3 + 2*x^4)/(1 - 6*x^3 + x^6).
From Wolfdieter Lang, Sep 29 2016: (Start)
Trisection:
a(2+3*n) = 15*S(n-1,6) - 2*S(n-2,6) = A275794(n),
a(3+3*n) = 20*S(n-1,6) - 3*S(n-2,6) = A275796(n),
a(4+3*n) = 7*(6*S(n-1,6) - S(n-2,6)) = 7*A001109(n+1) for n >= 0, with the Chebyshev polynomials S(n, 6) = A001109(n+1), n >= -1, with S(-2, 6) = -1.
(End)

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

A328791 Triangular numbers of the form k^2 + 3.

Original entry on oeis.org

3, 28, 903, 30628, 1040403, 35343028, 1200622503, 40785822028, 1385517326403, 47066803275628, 1598885794044903, 54315050194251028, 1845112820810490003, 62679520857362409028, 2129258596329511416903, 72332112754346025765628, 2457162575051435364614403

Offset: 1

Jon E. Schoenfield, Oct 27 2019

nonn

There exist triangular numbers of the form k^2 + j for j=0 (A001110), j=1 (A164055), j=2 (A214838), and j=3 (this sequence), but not for j=4,7,8,13,16,18,... (A328792).

Cf. A001110, A164055, A214838, A328792.
Intersection of A000217 and A117950.
Cf. A276598 (the k's).

limit = 10**7 # rough limit for k
A000217 = set(k*(k+1)//2 for k in range(14*limit//10))
A117950 = set(k**2 + 3 for k in range(limit))
print(sorted(A000217 & A117950)) # Michael S. Branicky, Mar 28 2021

a(1) = 3, a(2) = 28; for n > 2, a(n) = 34*a(n-1) - a(n-2) - 46.

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A276600 Values of m such that m^2 + 6 is a triangular number (A000217).

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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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A328791 Triangular numbers of the form k^2 + 3.

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