Charlie Marion - cp's OEIS frontend

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

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

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)

A200994 Triangular numbers, T(m), that are three-halves of another triangular number; T(m) such that 2T(m) = 3T(k) for some k.

Original entry on oeis.org

0, 15, 1485, 145530, 14260470, 1397380545, 136929032955, 13417647849060, 1314792560174940, 128836253249295075, 12624638025870742425, 1237085690282083462590, 121221773009618308591410, 11878496669252312158495605, 1163971451813716973223977895

Offset: 0

Charlie Marion, Feb 15 2012

For n > 1, a(n) = 98*a(n-1) - a(n-2) + 15. 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).

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

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, A200993-A201008.

A201008 Triangular numbers, T(m), that are five-sixths of another triangular number: T(m) such that 6T(m)=5T(k) for some k.

Original entry on oeis.org

0, 55, 26565, 12804330, 6171660550, 2974727580825, 1433812522297155, 691094661019647940, 333106192798948009980, 160556493834431921162475, 77387896922003387052303025, 37300805759911798127288895630

Offset: 0

Charlie Marion, Dec 20 2011

			6*0 = 5*0;
6*55 = 5*66;
6*26565 = 5*31878;
6*12804330 = 5*15365196.

Vincenzo Librandi, Table of n, a(n) for n = 0..229
Index entries for linear recurrences with constant coefficients, signature (483,-483,1).

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

I:=[0, 55, 26565]; [n le 3 select I[n] else 483*Self(n-1)-483*Self(n-2)+Self(n-3): n in [1..15]]; // Vincenzo Librandi, Dec 22 2011

LinearRecurrence[{483,-483,1},{0,55,26565},30] (* Vincenzo Librandi, Dec 22 2011 *)

makelist(expand(((11-2*sqrt(30))^(2*n+1)+(11+2*sqrt(30))^(2*n+1)-22)/192), n, 0, 11); /* Bruno Berselli, Dec 21 2011 */

concat(0,Vec(55/(1-x)/(1-482*x+x^2)+O(x^98))) \\ Charles R Greathouse IV, Dec 23 2011

For n > 1, a(n) = 482*a(n-1) - a(n-2) + 55. See A200993 for generalization.
From Bruno Berselli, Dec 21 2011: (Start)
G.f.: 55*x/((1-x)*(1-482*x+x^2)).
a(n) = a(-n-1) = 483*a(n-1)-483*a(n-2)+a(n-3).
a(n) = ((11-2*r)^(2*n+1)+(11+2*r)^(2*n+1)-22)/192, where r=sqrt(30). (End)

a(11) corrected by Bruno Berselli, Dec 21 2011

a(6) corrected by Vincenzo Librandi, Dec 22 2011

A200998 Triangular numbers, T(m), that are three-quarters of another triangular number: T(m) such that 4T(m)=3T(k) for some k.

Original entry on oeis.org

0, 21, 4095, 794430, 154115346, 29897582715, 5799976931385, 1125165627105996, 218276331681631860, 42344483180609474865, 8214611460706556491971, 1593592278893891349967530, 309148687493954215337208870, 59973251781548223884068553271

Offset: 0

Charlie Marion, Dec 20 2011

For n>1, a(n) = 194*a(n-1) - a (n-2) + 21. See A200993 for generalization.

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

Colin Barker, Table of n, a(n) for n = 0..425
Index entries for linear recurrences with constant coefficients, signature (195,-195,1).

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

I:=[0,21]; [n le 2 select I[n] else  194*Self(n-1) - Self(n-2) + 21: n in [1..20]]; // Vincenzo Librandi, Mar 03 2016

LinearRecurrence[{195, -195, 1}, {0, 21, 4095}, 30] (* Vincenzo Librandi, Mar 03 2016 *)

concat(0, Vec(21/(1 - 195*x + 195*x^2 - x^3) + O(x^99))) \\ Charles R Greathouse IV, Dec 20 2011

G.f.: (21*x)/(1 - 195*x + 195*x^2 - x^3).
From Colin Barker, Mar 02 2016: (Start)
a(n) = 195*a(n-1)-195*a(n-2)+a(n-3) for n>2.
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))/128.
(End)

A201003 Triangular numbers, T(m), that are four-fifths of another triangular number: T(m) such that 5T(m) = 4T(k) for some k.

Original entry on oeis.org

0, 36, 11628, 3744216, 1205625960, 388207814940, 125001710784756, 40250162664876528, 12960427376379457296, 4173217365031520372820, 1343763031112773180590780, 432687522800947932629858376, 139324038578874121533633806328, 44861907734874666185897455779276

Offset: 0

Charlie Marion, Dec 20 2011

Also, numbers m such that 8*m+1 and 10*m+1 are squares. Example: 8*1205625960+1 = 98209^2 and 12056259601 = 109801^2. - Bruno Berselli, Mar 03 2016

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

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:=20; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(36*x/((1-x)*(1-322*x+x^2)))); // G. C. Greubel, Jul 15 2018

triNums = Table[(n^2 + n)/2, {n, 0, 4999}]; Select[triNums, MemberQ[triNums, (5/4)#] &] (* Alonso del Arte, Dec 20 2011 *)
CoefficientList[Series[-36 x/((x - 1) (x^2 - 322 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 11 2014 *)
LinearRecurrence[{323,-323,1},{0,36,11628},20] (* Harvey P. Dale, Dec 21 2015 *)

concat(0, Vec(36*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) + 36. See A200993 for generalization.
G.f.: 36*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)/160.
a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3) for n>2. (End)
a(n) = 36*A298271(n). - Amiram Eldar, Dec 01 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

A198468 Consider triples a<=b

Original entry on oeis.org

4, 7, 11, 9, 16, 12, 22, 29, 14, 37, 17, 24, 46, 29, 56, 19, 24, 67, 22, 28, 79, 47, 92, 21, 24, 37, 54, 106, 27, 42, 121, 137, 29, 53, 78, 154, 32, 59, 87, 172, 41, 50, 191, 34, 45, 55, 72, 211, 37, 42, 79, 117, 232, 128, 254, 39, 94, 277, 42, 63, 102, 301

Offset: 1

Charlie Marion, Dec 19 2011

nonn

The definition can be generalized to define Pythagorean k-triples a<=bA198453 for more about Pythagorean k-triples.

			3*2 + 3*2 = 4*3
4*3 + 6*5 = 7*6
5*4 + 10*9 = 11*10
6*5 + 7*6 = 9*8
6*5 + 15*14 = 16*15

A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.

Ron Knott, Pythagorean Triples and Online Calculators

Cf. A103606, A198453-A198469.

A198467 Consider triples a<=b

Original entry on oeis.org

3, 6, 10, 7, 15, 10, 21, 28, 11, 36, 14, 22, 45, 27, 55, 15, 21, 66, 18, 25, 78, 45, 91, 15, 19, 34, 52, 105, 22, 39, 120, 136, 23, 50, 76, 153, 26, 56, 85, 171, 36, 46, 190, 27, 40, 51, 69, 210, 30, 36, 76, 115, 231, 126, 253, 31, 91, 276, 34, 58, 99, 300

Offset: 1

Charlie Marion, Dec 19 2011

nonn

The definition can be generalized to define Pythagorean k-triples a<=bA198453 for more about Pythagorean k-triples.

			3*2 + 3*2 = 4*3
4*3 + 6*5 = 7*6
5*4 + 10*9 = 11*10
6*5 + 7*6 = 9*8
6*5 + 15*14 = 16*15

A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.

Ron Knott, Pythagorean Triples and Online Calculators

Cf. A103606, A198453-A198469.

A198466 Consider triples a<=b

Original entry on oeis.org

3, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 10, 10, 11, 11, 12, 12, 12, 13, 13, 13, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 17, 18, 18, 18, 18, 19, 19, 19, 19, 20, 20, 20, 21, 21, 21, 21, 21, 22, 22, 22, 22, 22, 23, 23, 24, 24, 24, 25, 25, 25, 25, 26, 26, 26, 26, 27, 27, 27, 27, 28, 28, 28, 28, 28

Offset: 1

Charlie Marion, Dec 19 2011

nonn

The definition can be generalized to define Pythagorean k-triples a<=bA198453 for more about Pythagorean k-triples.

			3*2 + 3*2 = 4*3
4*3 + 6*5 = 7*6
5*4 + 10*9 = 11*10
6*5 + 7*6 = 9*8
6*5 + 15*14 = 16*15

A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.

Ron Knott, Pythagorean Triples and Online Calculators

Cf. A103606, A198453-A198469.

cp's OEIS Frontend

User: Charlie Marion

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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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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A200994 Triangular numbers, T(m), that are three-halves of another triangular number; T(m) such that 2*T(m) = 3*T(k) for some k.

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A201008 Triangular numbers, T(m), that are five-sixths of another triangular number: T(m) such that 6*T(m)=5*T(k) for some k.

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A200998 Triangular numbers, T(m), that are three-quarters of another triangular number: T(m) such that 4*T(m)=3*T(k) for some k.

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A201003 Triangular numbers, T(m), that are four-fifths of another triangular number: T(m) such that 5*T(m) = 4*T(k) for some k.

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

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

A200994 Triangular numbers, T(m), that are three-halves of another triangular number; T(m) such that 2T(m) = 3T(k) for some k.

A201008 Triangular numbers, T(m), that are five-sixths of another triangular number: T(m) such that 6T(m)=5T(k) for some k.

A200998 Triangular numbers, T(m), that are three-quarters of another triangular number: T(m) such that 4T(m)=3T(k) for some k.

A201003 Triangular numbers, T(m), that are four-fifths of another triangular number: T(m) such that 5T(m) = 4T(k) for some k.

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