A200994 -id:A200994 - cp's OEIS frontend

A352182 Twice A200994.

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

A200993 Triangular numbers, T(m), that are two-thirds of another triangular number; T(m) such that 3T(m) = 2T(k) for some k.

Original entry on oeis.org

0, 10, 990, 97020, 9506980, 931587030, 91286021970, 8945098566040, 876528373449960, 85890835499530050, 8416425350580494950, 824723793521388975060, 80814515339745539060940, 7918997779501541438997070, 775980967875811315482651930, 76038215854050007375860892080

Offset: 0

Charlie Marion, Dec 15 2011

For n>1, a(n) = 98*a(n-1) - a (n-2) + 10. In general, for m>0, let b(n) be those triangular numbers such that for some triangular number c(n), (m+1)*b(n) = m*c(n). Then b(0) = 0, b(1)= A014105(m) and for n>1, b(n) = 2*A069129(m+1)*b(n-1) - b(n-2) + A014105(m).

Further, c(0) = 0, c(1) = A000384(m+1) and for n>1, c(n) = 2*A069129(m+1)*c(n-1) - c(n-2) + A000384(m+1).

			3*0 = 2*0.
3*10 = 2*15.
3*990 = 2*1485.
3*97020 = 2*145530.

Colin Barker, Table of n, a(n) for n = 0..500
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. A001652, A029549, A053141, A075528, A200994-A201008.

m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(10*x/((1-x)*(x^2-98*x+1)))); // G. C. Greubel, Jul 15 2018

LinearRecurrence[{99,-99,1},{0,10,990},20] (* Harvey P. Dale, Feb 25 2018 *)

concat(0, Vec(10*x/((1-x)*(1-98*x+x^2)) + O(x^40))) \\ Colin Barker, Mar 02 2016

G.f. 10*x / ((1-x)*(x^2-98*x+1)). - R. J. Mathar, Dec 20 2011
a(n) = 99*a(n-1)-99*a(n-2)+a(n-3) for n>2. - Colin Barker, Mar 02 2016
a(n) = (-10+(5-2*sqrt(6))*(49+20*sqrt(6))^(-n)+(5+2*sqrt(6))*(49+20*sqrt(6))^n)/96. - Colin Barker, Mar 07 2016

A278620 Expansion of x/(1 - 99x + 99x^2 - x^3).

Original entry on oeis.org

0, 1, 99, 9702, 950698, 93158703, 9128602197, 894509856604, 87652837344996, 8589083549953005, 841642535058049495, 82472379352138897506, 8081451533974553906094, 791899777950154143899707, 77598096787581131548265193, 7603821585405000737586089208, 745096917272902491151888477192

Offset: 0

Bruno Berselli, Nov 24 2016

Colin Barker, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

First differences: A173205.

Cf. A045502, A108741, A123479, A200993, A200994.

P:=proc(q) local a,b,c,n; a:=0; b:=1; print(a); print(b);for n from 1 to q do
c:=98*b-a+1; a:=b; b:=c; print(b); od; end: P(100); # Paolo P. Lava, Nov 30 2016

CoefficientList[x/(1 - 99 x + 99 x^2 - x^3) + O[x]^20, x]
LinearRecurrence[{99,-99,1},{0,1,99},20] (* Harvey P. Dale, Aug 22 2020 *)

makelist(coeff(taylor(x/((1-x)*(1-98*x+x^2)), x, 0, n), x, n), n, 0, 20);

concat(0, Vec(1/(1-99*x+99*x^2-x^3) + O(x^20)))

gf = x/((1-x)*(1-98*x+x^2)); print(taylor(gf, x, 0, 20).list())

O.g.f.: x/((1 - x)*(1 - 98*x + x^2)).
E.g.f.: ((5-2*sqrt(6))*exp((5-2*sqrt(6))^2*x) + (5+2*sqrt(6))*exp((5+2*sqrt(6))^2*x) - 10*exp(x))/960.
a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3) for n>2.
a(n) = 98*a(n-1) - a(n-2) + 1 for n>1.
a(n) = a(-n-1) = ((5+2*sqrt(6))^(2*n+1) + (5-2*sqrt(6))^(2*n+1))/960 - 1/96.
a(n) = floor((5+2*sqrt(6))^(2*n+1)/960).
a(n)*a(n-2) = a(n-1)*(a(n-1)-1) for n>1.
Lim_{i -> infinity} a(i)/a(i-1) = (5 + 2*sqrt(6))^2.
From the closed form: a(n) + a(-n) = A108741(n).
a(n) = A200993(n)/10 = A200994(n)/15.
a(n) = A123479(n)/20 for n>0.
a(n) = A045502(n)/40.

A200999 Triangular numbers, T(m), that are four-thirds of another triangular number; T(m) such that 3T(m) = 4T(k) for some k.

Original entry on oeis.org

0, 28, 5460, 1059240, 205487128, 39863443620, 7733302575180, 1500220836141328, 291035108908842480, 56459310907479299820, 10952815280942075322628, 2124789705191855133290040, 412198249991938953782945160, 79964335708730965178758071028

Offset: 0

Charlie Marion, Feb 15 2012

Numbers h such that 6*h+1 and 8*h+1 are both squares. [Bruno Berselli, Jul 07 2014]

			3*0 = 4*0.
3*28 = 4*21.
3*5640 = 4*4095.
3*1059240 = 4*794430.

Colin Barker, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (195,-195,1).

Cf. A200994, A245031.

LinearRecurrence[{195, -195, 1}, {0, 28, 5460}, 20] (* T. D. Noe, Feb 15 2012 *)

concat(0, Vec(28*x/((1-x)*(1-194*x+x^2)) + O(x^15))) \\ Colin Barker, Mar 02 2016

For n>1, a(n) = 194*a(n-1) - a (n-2) + 28.  See A200998 for generalization.
From Colin Barker, Mar 02 2016: (Start)
a(n) = ((97+56*sqrt(3))^(-n)*(-1+(97+56*sqrt(3))^n)*(-7+4*sqrt(3)+(7+4*sqrt(3))*(97+56*sqrt(3))^n))/96.
a(n) = 195*a(n-1)-195*a(n-2)+a(n-3) for n>2.
G.f.: 28*x / ((1-x)*(1-194*x+x^2)).
(End)

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

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

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A352182 Twice A200994.

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A200993 Triangular numbers, T(m), that are two-thirds of another triangular number; T(m) such that 3*T(m) = 2*T(k) for some k.

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A278620 Expansion of x/(1 - 99*x + 99*x^2 - x^3).

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A200999 Triangular numbers, T(m), that are four-thirds of another triangular number; T(m) such that 3*T(m) = 4*T(k) for some k.

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

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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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A200993 Triangular numbers, T(m), that are two-thirds of another triangular number; T(m) such that 3T(m) = 2T(k) for some k.

A278620 Expansion of x/(1 - 99x + 99x^2 - x^3).

A200999 Triangular numbers, T(m), that are four-thirds of another triangular number; T(m) such that 3T(m) = 4T(k) for some k.

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