A278620 -id:A278620 - cp's OEIS frontend

A045502 Numbers k such that 2k+1 and 3k+1 are squares.

0, 40, 3960, 388080, 38027920, 3726348120, 365144087880, 35780394264160, 3506113493799840, 343563341998120200, 33665701402321979800, 3298895174085555900240, 323258061358982156243760, 31675991118006165755988280, 3103923871503245261930607720

Offset: 0

Fred Schwab (fschwab(AT)nrao.edu)

Problem 1 for the 3rd grade of the 38th Mathematics Competition of the Republic of Slovenia (1998) was to prove that if k is a natural number such that 2*k+1 and 3*k+1 are perfect squares, then k is divisible by 40 (see link with solution Crux Mathematicorum and formula Mar 25 2021). - Bernard Schott, Mar 25 2021

Colin Barker, Table of n, a(n) for n = 0..500.
John Albert, Pell Equations, Putnam practice, November 17, 2004 (1-2).
R. S. Luthar, Problem E2606, Amer. Math. Monthly, 84 (1977), 823-824.
R. E. Woodrow, Problem 1 for the third grade of the 38th Mathematics competition of the Republic of Slovenia (1998), Crux Mathematicorum, The Olympiad Corner, p. 208, April 1999, Vol. 25, No. 4.
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).
Index to sequences related to Olympiads and other Mathematical Competitions.

Cf. A001078, A245031, A278620.

a:=[0,40,3960];; for n in [4..15] do a[n]:=99*a[n-1]-99*a[n-2]+a[n-3]; od; a; # Muniru A Asiru, Jul 17 2018

I:=[0,40,3960]; [n le 3 select I[n] else 99*Self(n-1) -99*Self(n-2) + Self(n-3): n in [1..15]]; // G. C. Greubel, Jan 13 2020

seq(coeff(series(40*x/((1-x)*(x^2-98*x+1)), x,n+1),x,n),n=0..15); # Muniru A Asiru, Jul 17 2018

f[0]=0; f[1]=2; f[n_]:= f[n]= 10*f[n-1] -f[n-2]; a[n_]:= f[n]*f[n+1];
CoefficientList[Series[40x/((1-x)(1-98x+x^2)), {x,0,15}], x] (* Michael De Vlieger, Jul 20 2018 *)
Table[5*(ChebyshevT[n, 49] +48*ChebyshevU[n-1, 49] -1)/12, {n,0,15}] (* G. C. Greubel, Jan 13 2020 *)
LinearRecurrence[{99,-99,1},{0,40,3960},20] (* Harvey P. Dale, Dec 02 2023 *)

concat(0, Vec(40*x/((1-x)*(1-98*x+x^2))+O(x^20))) \\ Colin Barker, Mar 23 2017

[4*chebyshev_U(n-1,5)*chebyshev_U(n,5) for n in (0..15)] # G. C. Greubel, Jan 13 2020

From Colin Barker, Mar 23 2017: (Start)
O.g.f.: 40*x / ((1 - x)*(1 - 98*x + x^2)).
a(n) = 99*a(n-1)- 99*a(n-2) + a(n-3) for n>2.
a(n) = (-10 + (5 - 2*sqrt(6))*(49 + 20*sqrt(6))^(-n) + (5 + 2*sqrt(6))*(49 + 20*sqrt(6))^n)/24. (End)
From G. C. Greubel, Jan 13 2020: (Start)
a(n) = 5*(ChebyshevT(n, 49) + 48*ChebyshevU(n-1, 48) - 1)/12.
a(n) = 4*ChebyshevU(n-1, 5)*ChebyshevU(n, 5). (End)
a(n) = 40*A278620(n). - Bernard Schott, Mar 25 2021

A278438 Numbers m such that T(m) + 2*T(m+1) is a square, where T = A000217.

Original entry on oeis.org

7, 799, 78407, 7683199, 752875207, 73774087199, 7229107670407, 708378777612799, 69413891098384007, 6801852948864019999, 666512175097575576007, 65311391306613542428799, 6399849835873029582446407, 627119972524250285537319199, 61451357457540654953074835207

Offset: 1

Bruno Berselli, Nov 23 2016

It is well known that T(m) + k*T(m+1) is always a square for k=1. For k=3, the nonnegative values of m are the terms of A278310.
Square roots of T(m) + 2*T(m+1) are listed by A168520 (after 0).
Negative values of m for which T(m) + 2*T(m+1) is a square: -1, -2, -82, -7922, -776162, ...

Colin Barker, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Subsequence of A056220.
Cf. A000217, A001078, A001079, A122652, A168520, A278620.
Cf. A278310: numbers m such that T(m) + 3*T(m+1) is a square.

Iv:=[7, 799]; [n le 2 select Iv[n] else 98*Self(n-1)-Self(n-2)+112: n in [1..20]];

P:=proc(q) local n; for n from 1 to q do if type(sqrt((3*n^2+7*n+4)/2),integer) then print(n); fi; od; end: P(10^9); #  Paolo P. Lava, Nov 25 2016

Table[((5 + 2 Sqrt[6])^(2 n) + (5 - 2 Sqrt[6])^(2 n))/12 - 7/6, {n, 1, 20}]
RecurrenceTable[{a[1] == 7, a[2] == 799, a[n] == 98 a[n - 1] - a[n - 2] + 112}, a, {n, 1, 20}]
LinearRecurrence[{99,-99,1},{7,799,78407},20] (* Harvey P. Dale, Oct 18 2024 *)

Vec(x*(7 + 106*x - x^2)/((1 - x)*(1 - 98*x + x^2)) + O(x^20)) \\ Colin Barker, Nov 27 2016

def A278438():
    a, b = 7, 799
    yield a
    while True:
        yield b
        a, b = b, 98*b - a + 112
a = A278438(); print([next(a) for  in range(15)]) # _Peter Luschny, Nov 24 2016

O.g.f.: x*(7 + 106*x - x^2)/((1 - x)*(1 - 98*x + x^2)).
E.g.f.: (exp((5-2*sqrt(6))^2*x) + exp((5+2*sqrt(6))^2*x) - 14*exp(x))/12 + 1.
a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3) for n>3.
a(n) = 98*a(n-1) - a(n-2) + 112 for n>2.
a(n) = a(-n) = ((5 + 2*sqrt(6))^(2*n) + (5 - 2*sqrt(6))^(2*n))/12 - 7/6.
a(n) = A001079(2*n)/6 - 7/6.
a(n) = 2*A001078(n)^2 - 1 = A122652(n)^2/2 - 1.
a(n) = -A278620(n+1) + 106*A278620(n) + 7*A278620(n-1).
Lim_{n -> infinity} a(n)/a(n-1) = (5 + 2*sqrt(6))^2.

A352181 a(n) = A200993(n)/2.

Original entry on oeis.org

0, 5, 495, 48510, 4753490, 465793515, 45643010985, 4472549283020, 438264186724980, 42945417749765025, 4208212675290247475, 412361896760694487530, 40407257669872769530470, 3959498889750770719498535, 387990483937905657741325965, 38019107927025003687930446040

Offset: 0

N. J. A. Sloane, Mar 08 2022

Halves of triangular numbers which are also thirds of triangular numbers.

Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A200993, A200994, A278620.

a(n) = A200994(n)/3.
From Chai Wah Wu, Apr 22 2024: (Start)
a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3) for n > 2.
G.f.: -5*x/((x - 1)*(x^2 - 98*x + 1)). (End)
a(n) = 5*A278620(n). - Hugo Pfoertner, Apr 22 2024

A352182 Twice A200994.

Original entry on oeis.org

0, 30, 2970, 291060, 28520940, 2794761090, 273858065910, 26835295698120, 2629585120349880, 257672506498590150, 25249276051741484850, 2474171380564166925180, 242443546019236617182820, 23756993338504624316991210, 2327942903627433946447955790, 228114647562150022127582676240

Offset: 0

N. J. A. Sloane, Mar 08 2022

Also 3 times A200993 and 6 times A352181.

Numbers that both doubles and triples of triangular numbers.

Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A200993, A200994, A278620, A352181.

From Chai Wah Wu, Apr 22 2024: (Start)
a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3) for n > 2.
G.f.: -30*x/((x - 1)*(x^2 - 98*x + 1)). (End)
a(n) = 30*A278620. - Hugo Pfoertner, Apr 22 2024

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A045502 Numbers k such that 2*k+1 and 3*k+1 are squares.

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A278438 Numbers m such that T(m) + 2*T(m+1) is a square, where T = A000217.

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A352181 a(n) = A200993(n)/2.

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A352182 Twice A200994.

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A045502 Numbers k such that 2k+1 and 3k+1 are squares.