A053141 -id:A053141 - cp's OEIS frontend

A103625 a(n) = 3 + 7a(n-2) + sqrt(1 + 48a(n-2) + 48*a(n-2)^2), with a(1) = 0, a(2) = 0, a(3) = 2.

0, 0, 2, 4, 34, 62, 480, 870, 6692, 12124, 93214, 168872, 1298310, 2352090, 18083132, 32760394, 251865544, 456293432, 3508034490, 6355347660, 48860617322, 88518573814, 680540608024, 1232904685742, 9478707895020, 17172147026580, 132021369922262, 239177153686384

Offset: 1

Pierre CAMI, Mar 29 2005

Define j(n) = sqrt(48*a(n)^2 + 48*a(n) + 1), then j(n) is prime for n=3, 4, 5, 6, 7, 25, 28, 32, 35, 48, 65, 66, 88, 96, 113, 119, 151, 155, 182, 220, 231, 316, 488, 531, 599, 722, 1049, 1176, ...
For n > 1, first member of the Diophantine pair (m,k) that satisfies 12*(m^2 + m) = k^2 + k; a(n)=m. - Herbert Kociemba, May 12 2008
Former name: Define a(1)=0, a(2)=0, a(3)=2, a(4)=4, a(5)=34, a(6)=62, a(7)=480, a(8)=870 such that from i=1 to 8: 48*a(i)^2 + 48*a(i) + 1 = j(i)^2 with j(1)=1, j(2)=1, j(3)=17, j(4)=31, j(5)=239, j(6)=433, j(7)=3329, j(8)=6031. Then a(n) = a(n-8) + 28*sqrt(48*(a(n-4)^2) + 48*a(n-4) + 1). - G. C. Greubel, Mar 22 2024

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,14,-14,-1,1).

Cf. A053141, A103200, A103737, A125905.

R:=PowerSeriesRing(Integers(), 30); [0,0] cat Coefficients(R!(2*(x^2+x+1)/(1-x-14*x^2+14*x^3+x^4-x^5))); // G. C. Greubel, Jul 15 2018

a[1]=0; a[2]=0; a[3]=2; a[n_]:=a[n]= 3+7a[n-2]+Sqrt[1+48a[n-2]+48a[n-2]^2]; Table[a[n],{n,1,20}] (* Herbert Kociemba, May 12 2008 *)
Rest@CoefficientList[Series[2*x^3*(1+x+x^2)/(1-x-14*x^2+14*x^3+x^4-x^5), {x,0,30}], x] (* G. C. Greubel, Jul 15 2018 *)
LinearRecurrence[{1,14,-14,-1,1},{0,0,2,4,34},30] (* Harvey P. Dale, Jun 04 2021 *)

my(x='x+O('x^30)); concat([0,0], Vec(2*x^3*(1+x+x^2)/(1-x-14*x^2+14*x^3 + x^4-x^5))) \\ G. C. Greubel, Jul 15 2018

@CachedFunction
def b(n): return chebyshev_U(n, -2) # A125905
def A103625(n): return (1/8)*(-16*int(n==0) -4 +5*(-1)^n*(3*b(n) +11*b(n-1)) +5*b(n) +19*b(n-1))
[A103625(n) for n in range(1,41)] # G. C. Greubel, Mar 22 2024

G.f.: 2*x^3*(1+x+x^2)/((1-x)*(1-4*x+x^2)*(1+4*x+x^2)). - Ralf Stephan, May 18 2007
a(n) = (1/8)*(-16*[n=0] - 4 + 5*(-1)^n*(3*A125905(n) + 11*A125905(n-1)) + (5*A125905(n) + 19*A125905(n-1))), where A125905(n) = ChebyshevU(n, -2). - G. C. Greubel, Mar 22 2024
E.g.f.: (15*cosh(sqrt(3)*x)*(2*cosh(2*x) + sinh(2*x))/2 - sqrt(3)*(4*cosh(x) + sinh(x))*(cosh(x) + 4*sinh(x))*sinh(sqrt(3)*x) - 3*(4 + exp(x)))/6. - Stefano Spezia, Jun 02 2024

Terms a(17) onward added by G. C. Greubel, Jul 15 2018

Edited by G. C. Greubel, Mar 22 2024

A103715 Define a(1)=0, a(2)=0, a(3)=1, a(4)=3, a(5)=18, a(6)=22, a(7)=119, a(8)=285. Then a(n) = a(n-8) + 4sqrt(420a(n-4)^2 + 420*a(n-4) + 1).

Original entry on oeis.org

0, 0, 1, 3, 18, 22, 119, 285, 1516, 1844, 9797, 23407, 124334, 151226, 803275, 1919129, 10193912, 12398728, 65858793, 157345211, 835776490, 1016544510, 5399617791, 12900388213, 68523478308, 83344251132, 442702800109

Offset: 1

Pierre CAMI, Mar 27 2005

By construction, a(n) is an integer so 420*a(n)^2 + 420*a(n) + 1 = j(n)^2. The sequence j(n) starts: 1, 1, 29, 71, 379, 461, 2449, 5851, ...

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A103200, A053141.

m:=25; R:=PowerSeriesRing(Integers(), m); [0,0] cat Coefficients(R!(x^3*(x^2+1)*(x^4+2*x^3+14*x^2+2*x+1)/((1-x)*(x^8- 82*x^4 +1)))); // G. C. Greubel, Jul 15 2018

Rest[CoefficientList[Series[x^3*(x^2+1)*(x^4+2*x^3+14*x^2+2*x+1)/((1-x)*(x^8-82*x^4+1)), {x, 0, 30}], x]] (* G. C. Greubel, Jul 15 2018 *)

x='x+O('x^30); concat([0,0], Vec(x^3*(x^2+1)*(x^4+2*x^3+14*x^2 +2*x +1)/((1-x)*(x^8-82*x^4+1)))) \\ G. C. Greubel, Jul 15 2018

From R. J. Mathar, Nov 13 2009: (Start)
a(n) = a(n-1) + 82*a(n-4) - 82*a(n-5) - a(n-8) + a(n-9).
G.f.: x^3*(x^2+1)*(x^4+2*x^3+14*x^2+2*x+1)/((1-x)*(x^8-82*x^4+1)). (End)

Extended by R. J. Mathar, Nov 13 2009

A201004 Triangular numbers, T(m), that are five-quarters of another triangular number; T(m) such that 4T(m) = 5T(k) for some k.

Original entry on oeis.org

0, 45, 14535, 4680270, 1507032450, 485259768675, 156252138480945, 50312703331095660, 16200534220474321620, 5216521706289400466025, 1679703788890966475738475, 540859403501184915787322970, 174155048223592651917042257910, 56077384668593332732371819724095

Offset: 0

Charlie Marion, Feb 15 2012

			4*0 = 5*0.
4*45 = 5*36.
4*14535 = 5*11628.
4*4680270 = 5*3744216.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (323,-323,1).

Cf. A001652, A029549, A053141, A075528, A200993-A201008, A298271.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(45*x/((1-x)*(1-322*x+x^2)))); // G. C. Greubel, Jul 15 2018

LinearRecurrence[{323, -323, 1}, {0, 45, 14535}, 20] (* T. D. Noe, Feb 15 2012 *)
CoefficientList[Series[-45 x/((x - 1) (x^2 - 322 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 11 2014 *)

concat(0, Vec(45*x/((1-x)*(1-322*x+x^2)) + O(x^15))) \\ Colin Barker, Mar 02 2016

For n > 1, a(n) = 322*a(n-1) - a(n-2) + 45. See A200994 for generalization.
G.f.: 45*x / ((1-x)*(x^2-322*x+1)). - R. J. Mathar, Aug 10 2014
From Colin Barker, Mar 02 2016: (Start)
a(n) = (-18 + (9-4*sqrt(5))*(161+72*sqrt(5))^(-n) + (9+4*sqrt(5))*(161+72*sqrt(5))^n)/128.
a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3) for n > 2. (End)
a(n) = 45*A298271(n). - Amiram Eldar, Dec 01 2024

a(7) corrected by R. J. Mathar, Aug 10 2014

A309714 The smallest possible nonnegative difference between the sum of the first n positive integers (A000217) and the sum of any number of the directly following and consecutive integers.

Original entry on oeis.org

1, 0, 2, 5, 2, 6, 1, 6, 12, 5, 12, 3, 11, 0, 9, 19, 6, 17, 2, 14, 27, 10, 24, 5, 20, 36, 15, 32, 9, 27, 2, 21, 41, 14, 35, 6, 28, 51, 20, 44, 11, 36, 1, 27, 54, 17, 45, 6, 35, 65, 24, 55, 12, 44, 77, 32, 66, 19, 54, 5, 41, 78, 27, 65, 12, 51, 91, 36, 77, 20, 62

Offset: 1

Bob Andriesse, Aug 13 2019

nonn

a(n) = 0 if a positive integer m exists, such that m * (m + 1) = 2 * n * (n + 1). Let k = m - n, then n = (2 * k - 1 + sqrt(8 * k^2 + 1)) / 2. All k for which 8 * k^2 + 1 is a perfect square (A001109) yield a value for n for which a(n) = 0.

a(A053141(n)) = 0 for all n.

			a(2) = 1 + 2 - 3 = 0;
a(3) = 1 + 2 + 3 - 4 = 2;
a(7) = 1 + 2 + 3 + 4 + 5 + 6 + 7 - (8 + 9 + 10) = 1.
a(A053141(2)) = a(14) = 0, because A000217(20) = 2 * A000217(14).

Cf. A000217, A001109, A001652, A053141, A309655.

a(n) = {my(t=n*(n+1)/2, k = n+1); while(t >= k, t -= k; k++); t;} \\ Michel Marcus, Aug 16 2019

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

Vladimir Pletser, Mar 06 2021

nonn

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).

			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.

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.

Vladimir Pletser, Table of n, a(n) for n = 1..256
T. Breiteig, Quotients of triangular numbers, The Mathematical Gazette, 99, 2015, 243-255.
Keith Matthews, The Diophantine Equation x^2 - Dy^2 = N, D > 0, in integers, Expositiones Mathematicae, 18, 2000, 323-331.
Keith Matthews, Quadratic Diophantine equations BCMATH programs, 2020.
Vladimir Pletser, Searching for Multiple of Triangular Numbers being Triangular Numbers, ResearchGate, DOI: 10.13140/RG.2.2.35428.91527, 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv: 2102.13494 [math.NT], 2021.
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv: 2101.00998 [math.NT], 2021.

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.

A145856 Least number k>1 such that centered n-gonal number n*k(k-1)/2+1 is a perfect square, or 0 if no such k exists.

Original entry on oeis.org

3, 0, 2, 4, 3, 8, 16, 2, 17, 9, 15, 5, 6, 16, 2, 3, 6, 0, 7, 4, 3, 40, 7, 2, 22, 8, 111, 4, 16, 8, 16, 0, 3, 9, 2, 5, 990, 9, 15, 3, 46, 16, 10, 5, 6, 336, 10, 2, 30, 0, 31, 16, 11, 416, 7, 3, 11, 33, 55, 4, 78, 56, 2, 6, 3, 8, 47751, 12, 16, 24, 48, 0, 49, 25, 17, 13, 6, 9, 2640, 2, 6721

Offset: 1

Alexander Adamchuk, Oct 22 2008

nonn

Jonathan Vos Post, When Centered Polygonal Numbers are Perfect Squares, submitted to Mathematics Magazine, 4 May 2004, manuscript no. 04-1165, unpublished, available upon request. - Jonathan Vos Post, Oct 25 2008

Eric Weisstein's World of Mathematics, Centered Polygonal Numbers
Index entries for sequences related to centered polygonal numbers

Cf. A120744, A001542, A166259, A006451, A001921, A129444, A001570, A001652, A129556, A053606, A105038, A105040, A053141, A061278.

a(n) = 0 for n in A166259.

a(n) = A120744(n) + 1. - Alexander Adamchuk, Oct 10 2009

Edited by Max Alekseyev, Jan 23 2010

cp's OEIS Frontend

A103625 a(n) = 3 + 7*a(n-2) + sqrt(1 + 48*a(n-2) + 48*a(n-2)^2), with a(1) = 0, a(2) = 0, a(3) = 2.

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A103715 Define a(1)=0, a(2)=0, a(3)=1, a(4)=3, a(5)=18, a(6)=22, a(7)=119, a(8)=285. Then a(n) = a(n-8) + 4*sqrt(420*a(n-4)^2 + 420*a(n-4) + 1).

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A201004 Triangular numbers, T(m), that are five-quarters of another triangular number; T(m) such that 4*T(m) = 5*T(k) for some k.

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A309714 The smallest possible nonnegative difference between the sum of the first n positive integers (A000217) and the sum of any number of the directly following and consecutive integers.

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PARI

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)).

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A145856 Least number k>1 such that centered n-gonal number n*k(k-1)/2+1 is a perfect square, or 0 if no such k exists.

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A103625 a(n) = 3 + 7a(n-2) + sqrt(1 + 48a(n-2) + 48*a(n-2)^2), with a(1) = 0, a(2) = 0, a(3) = 2.

A103715 Define a(1)=0, a(2)=0, a(3)=1, a(4)=3, a(5)=18, a(6)=22, a(7)=119, a(8)=285. Then a(n) = a(n-8) + 4sqrt(420a(n-4)^2 + 420*a(n-4) + 1).

A201004 Triangular numbers, T(m), that are five-quarters of another triangular number; T(m) such that 4T(m) = 5T(k) for some k.