A230044 -id:A230044 - cp's OEIS frontend

A064784 Difference between n-th triangular number t(n) and the largest square <= t(n).

0, 2, 2, 1, 6, 5, 3, 0, 9, 6, 2, 14, 10, 5, 20, 15, 9, 2, 21, 14, 6, 28, 20, 11, 1, 27, 17, 6, 35, 24, 12, 44, 32, 19, 5, 41, 27, 12, 51, 36, 20, 3, 46, 29, 11, 57, 39, 20, 0, 50, 30, 9, 62, 41, 19, 75, 53, 30, 6, 66, 42, 17, 80, 55, 29, 2, 69, 42, 14, 84, 56, 27, 100, 71, 41, 10, 87, 56

Offset: 1

Jonathan Ayres (jonathan.ayres(AT)btinternet.com), Oct 20 2001

The second differences of a(n) - (a(n)-a(n-1))-(a(n-1)-a(n-2)) - give 2, -2, -1, 6, -6, -1, -1, 12, -12, -1, 16, -16, -1 ... 82k+2, 82k-2, -1, 82k+6, 82k-6, -1, -1, 82k+12, 82k-12, -1, 82k+16, -82k-16, -1, 82k+20, -82k-20, -1, -1, 82k+26, -82k-26, -1, 82k+30, -82k-30, -1, -1, 82k+36, -82k-36, -1, 82k+40, -82k-40, -1, 82k+44, -82k-44, -1, -1, 82k+50, -82k-50, -1, 82k+54, -82k-54, -1, -1, 82k+60, -82k-60, -1, 82k+64, -82k-64, -1, -1, 82k+70, -82k-70, -1, 82k+74, -82k-74, -1, 82k+78, -82k-78, -1, -1, ...

			n = 5: A000217(5) = 28, largest square below that is 25, so a(5) = 28 - 25 = 3.

Harry J. Smith, Table of n, a(n) for n = 1..1000
J. C. Lagarias, E. M. Rains and N. J. A. Sloane, The EKG sequence, Exper. Math. 11 (2002), 437-446.
J. C. Lagarias, E. M. Rains and N. J. A. Sloane, The EKG sequence, arXiv:math/0204011 [math.NT], 2002.
Index entries for sequences related to EKG sequence

Cf. A001108, A076816, A128549, A230038. Unique values are in A230044.

seq(n*(n+1)/2-floor(sqrt(n*(n+1)/2))^2,n=0..100);

f[n_]:=n*(n+1)/2-Floor[Sqrt[n*(n+1)/2]]^2; lst={}; Do[AppendTo[lst,f[n]],{n,0,6!}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 17 2010 *)
#-Floor[Sqrt[#]]^2&/@Accumulate[Range[100]] (* Harvey P. Dale, Oct 15 2014 *)

{ default(realprecision, 100); for (n=1, 1000, t=n*(n + 1)/2; a=t - floor(sqrt(t))^2; write("b064784.txt", n, " ", a) ) } \\ Harry J. Smith, Sep 25 2009

from math import isqrt
def A064784(n): return (m:=n*(n+1)>>1)-isqrt(m)**2 # Chai Wah Wu, Jun 01 2024

a(n) = n*(n+1)/2 - floor(sqrt(n*(n+1)/2))^2.
a(n) = A053186(A000217(n)). - R. J. Mathar, Sep 10 2016
a(A001108(n)) = 0. - Hugo Pfoertner, Jun 01 2024

Definition corrected by Harry J. Smith, Sep 25 2009

Terms corrected by Harry J. Smith, Sep 25 2009

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

Original entry on oeis.org

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)

A175035 Offsets i such that i + n*(n+1)/2 is a perfect square for some positive integer n.

Original entry on oeis.org

1, 3, 4, 6, 8, 9, 10, 13, 15, 16, 19, 21, 22, 24, 25, 26, 28, 30, 33, 34, 35, 36, 39, 43, 45, 46, 48, 49, 53, 54, 55, 58, 60, 61, 63, 64, 66, 71, 72, 75, 76, 78, 79, 80, 81, 85, 89, 90, 91, 93, 94, 97, 99, 100, 103, 105, 106, 108, 111, 114, 115, 116, 118, 120

Offset: 1

Ctibor O. Zizka, Nov 10 2009

The ansatz n*(n+1)/2+i=s^2 can be transformed into (2*n+1)^2-2*(2*s)^2 =1-8*i.
A necessary condition for solutions to this Diophantine equation is that D=2 is a quadratic residue of the squarefree part of 8*i-1 (see A057126).
A sufficient condition is then available by a sequence of tests on the continued fractions of a quadratic surd that originates from a solution of this congruence.
See Mollin and Matthews for details. - R. J. Mathar, Nov 16 2009

R. A. Mollin, Simple continued fraction solutions for Diophantine Equations, Exposit. Mathem. 19 (2001) 55-73.
Keith Matthews, The Diophantine equation x^2-Dy^2=N,D >0, Exposit. Mathem. 18 (4) (2000) 323-331 [MR1788328].

Cf. A001108, A006451, A154138 to A154154.

Cf. A230044.

Take[Rest[Ceiling[Sqrt[#]]^2-#&/@Accumulate[Range[1000]]//Union],70] (* Harvey P. Dale, Sep 07 2019 *)

is(n)=#bnfisintnorm(bnfinit(z^2-8),-8*n+1) /* Ralf Stephan, Oct 14 2013 */

Extended by R. J. Mathar, Nov 26 2009

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

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

A328792 Numbers that are not the difference between any triangular number and the largest square that does not exceed it.

Original entry on oeis.org

4, 7, 8, 13, 16, 18, 22, 23, 25, 26, 31, 33, 34, 37, 38, 40, 43, 47, 48, 49, 52, 58, 59, 60, 61, 63, 64, 67, 68, 70, 73, 76, 79, 81, 83, 85, 86, 88, 92, 93, 94, 97, 98, 99, 102, 103, 106, 108, 112, 113, 114, 115, 118, 121, 123, 124, 125, 130, 133, 134, 138

Offset: 1

Jon E. Schoenfield, Oct 27 2019

nonn

			For any triangular number t, let f(t) = t - floor(sqrt(t))^2.
0 is not a term: for each term t in A001110, f(t) = 0.
1 is not a term: for each term t > 1 in A164055, f(t) = 1.
2 is not a term: for each term t in A214838, f(t) = 2.
3 is not a term: for each term t > 3 in A328791, f(t) = 3.
4 is a term, however: there exists no triangular number t such that f(t) = 4.

The complement of A230044.

Cf. A001110, A064784, A164055, A214838, A328791.

A335761 Nonnegative numbers that are the difference between a positive tetrahedral number and a positive cubic number.

Original entry on oeis.org

0, 2, 3, 4, 8, 9, 12, 19, 20, 21, 27, 29, 34, 40, 43, 47, 48, 53, 55, 56, 57, 70, 76, 83, 87, 93, 95, 101, 103, 112, 119, 136, 138, 140, 144, 148, 152, 156, 157, 161, 164, 168, 174, 181, 193, 209, 212, 217, 219, 222, 239, 240, 253, 259, 275, 278, 279, 281, 285

Offset: 1

Ya-Ping Lu, Jun 21 2020

nonn

The sequence is the difference between the tetrahedral number (A000292) and the cubic number (A000578) such that terms are of the form A000292(i)-A000578(j), where A000292(i) >= A000578(j) >= 0.

It appears that sequence terms are more scarce than prime numbers. The number of terms in this sequence (n) and the number of prime numbers up to a(n) are shown in the figure attached in the LINKS section. It can be seen that, for a(n) > 304, n is less than pi(a(n)), where pi is the prime counting function.

			a(1)=0 because t(1)-c(1)=1-1=0;
a(2)=2 because t(3)-c(2)=10-8=2;
a(7)=12 because t(4)-c(2)=20-8=12, and t(39)-c(22)=10660-10648=12;
a(19)=55 because t(6)-c(1)=56-1=55, and t(4669)-c(2570)=16974593055-16974593000=55.

Ya-Ping Lu, The number of terms in a(n) and the number of prime numbers up to a(n)

Cf. A000292, A000578, A230044, A328792.

The difference between the i-th tetrahedral number, t(i), and j-th cubic number, c(j) is d = i*(i+1)*(i+2)/6 - j^3, where i, j >=1 and t(i) >= c(j).

cp's OEIS Frontend

A064784 Difference between n-th triangular number t(n) and the largest square <= t(n).

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

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A175035 Offsets i such that i + n*(n+1)/2 is a perfect square for some positive integer n.

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A276598 Values of m such that m^2 + 3 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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