A053141 -id:A053141 - cp's OEIS frontend

A341893 Indices of triangular numbers that are one-tenth of other triangular numbers.

0, 1, 6, 12, 55, 246, 474, 2107, 9360, 18018, 80029, 355452, 684228, 3039013, 13497834, 25982664, 115402483, 512562258, 986657022, 4382255359, 19463867988, 37466984190, 166410301177, 739114421304, 1422758742216, 6319209189385, 28066884141582, 54027365220036, 239963538895471, 1065802482958830, 2051617119619170

Offset: 1

Vladimir Pletser, Feb 23 2021

The indices of triangular numbers that are one-tenth of other triangular numbers [t of T(t) such that T(t)=T(u)/10].
First member of the Diophantine pair (t, u) that satisfies 10*(t^2 + t) = u^2 + u; a(n) = t.
The T(t)'s are in A068085 and the u's are in A341895.
Also, nonnegative t such that 40*t^2 + 40*t + 1 is a square.
Can be defined for negative n by setting a(n) = a(-1-n) for all n in Z.

			a(4) = 12 is a term because its triangular number, (12*13) / 2 = 78 is one-tenth of 780, the triangular number of 39.
a(4) = 38 a(1) - a(-2) +18 = 0 - 6 +18 = 12 ;
a(5) = 38 a(2) - a(-1) + 18 = 38*1 - 1 +18 = 55.

Vladimir Pletser, Table of n, a(n) for n = 1..1000
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
V. Pletser, Recurrent relations for triangular multiples of other triangular numbers, Indian J. Pure Appl. Math. 53 (2022) 782-791
Index entries for linear recurrences with constant coefficients, signature (1,38,-38,-1,1).

Cf. A068085, A341895.
Cf. A336623, A336624, A336626, A336625, A053141, A001652, A075528, A029549, A061278, A001571, A076139, A076140, A077259, A077262, A077260, A077261, A077288, A077291, A077289, A077290, A077398, A077401, A077399, A077400, A000217.
Cf. A180003.

f := gfun:-rectoproc({a(-3) = 6, a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(2) = 6, a(n) = 38*a(n-3)-a(n-6)+18}, a(n), remember); map(f, [`$`(0 .. 1000)])[] ;

Rest@ CoefficientList[Series[(x^2*(1 + 5*x + 6*x^2 + 5*x^3 + x^4))/(1 - x - 38*x^3 + 38*x^4 + x^6 - x^7), {x, 0, 31}], x] (* Michael De Vlieger, May 19 2022 *)

a(n) = (-1 + sqrt(8*b(n) + 1))/2 where b(n) = A068085(n).
a(n) = 38 a(n-3) - a(n-6) + 18 for n > 3, with a(-2) = 6, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 6.
a(n) = a(n-1) + 38*(a(n-3) - a(n-4)) - (a(n-6) - a(n-7)) for n >= 4 with a(-2) = 6, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 6.
G.f.: x^2*(1 + 4*x+x^2)*(1+x+x^2)/ ((1-x)*(1-38*x^3+x^6)). - Stefano Spezia, Feb 24 2021
a(n) = A180003(n) - 1. - Hugo Pfoertner, Feb 28 2021

A111647 a(n) = A001541(n)A001653(n+1)A002315(n).

Original entry on oeis.org

1, 105, 20213, 3998709, 791704585, 156753394977, 31036379835581, 6145046450172525, 1216688160731724433, 240898110778299543129, 47696609245941810082565, 9443687732585695622131557

Offset: 0

Charlie Marion, Aug 24 2005

			a(1) = 105 = 3*5*7.

G. C. Greubel, Table of n, a(n) for n = 0..434
Index entries for linear recurrences with constant coefficients, signature (204,-1190,204,-1).

Cf. A001541, A001653, A002315, A111648, A111649.

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1)))); // G. C. Greubel, Jul 15 2018

CoefficientList[Series[(1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1)), {x, 0, 30}], x] (* G. C. Greubel, Jul 15 2018 *)

x='x+O('x^30); Vec((1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1))) \\ G. C. Greubel, Jul 15 2018

2*a(n) = A001109(3*n+1) + A001109(n+1).
a(n) = sqrt(A011900(2*n)*A046090(2*n)*A001109(2*n+1)).
a(n) = A001541(3*n) + 2*A001109(n)*A001541(n-1)*A001541(n).
For n>0, a(n) = A001652(3*n) - A053141(2*n)*A002315(n-1) - A001652(n-1).
G.f.: (1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
2*a(n) = A001109(n+1) + A097731(n) + 6*A097731(n-1). - R. J. Mathar, Jan 31 2024

A120744 Least k>0 such that a centered polygonal number nk(k+1)/2+1 is a perfect square; or -1 if no such number exists.

Original entry on oeis.org

2, -1, 1, 3, 2, 7, 15, 1, 16, 8, 14, 4, 5, 15, 1, 2, 5, -1, 6, 3, 2, 39, 6, 1, 21, 7, 110, 3, 15, 7, 15, -1, 2, 8, 1, 4, 989, 8, 14, 2, 45, 15, 9, 4, 5, 335, 9, 1, 29, -1, 30, 15, 10, 415, 6, 2, 10, 32, 54, 3, 77, 55, 1, 5, 2, 7, 47750, 11, 15, 23, 47, -1, 48, 24, 16, 12, 5, 8, 2639, 1, 6720, 704, 38, 4, 2, 39, 505, 3, 13, 56, 9, 20, 13, 1631, 41

Offset: 1

Alexander Adamchuk, Apr 26 2007

sign

			a(5) = 2 because A129556(2) = 2>1 and A129556(1) = 0<1.

Max Alekseyev, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Centered Polygonal Number

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

a(n) = -1 for n in A166259.

a(n) = 1 for n = k^2-1.

Edited and b-file provided by Max Alekseyev, Jan 20 2010

A309332 Number of ways the n-th triangular number T(n) = A000217(n) can be written as the sum of two positive triangular numbers.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 2, 0, 3, 0, 0, 1, 1, 3, 0, 0, 1, 0, 1, 0, 0, 3, 1, 1, 0, 1, 3, 0, 1, 1, 1, 2, 0, 1, 2, 0, 1, 1, 2, 1, 1, 1, 1, 2, 1, 0, 3, 1, 1, 1, 0, 3, 1, 1, 0, 0, 2, 0, 1, 1, 1, 1, 1, 5, 0, 1, 1, 0, 1, 0, 0, 3, 0, 3, 1, 0, 3, 1, 3, 1, 3, 3, 0, 1, 0, 0, 3, 0, 2, 0, 1

Offset: 1

Alois P. Heinz, Aug 01 2019

nonn

The order doesn't matter. 21 = 6+15 = 15+6 are not counted as distinct solutions. - N. J. A. Sloane, Feb 22 2020

			a(3) = 1: 2*3/2 + 2*3/2 = 3*4/2.
a(21) = 2: 6*7/2 + 20*21/2 = 12*13/2 + 17*18/2 = 21*22/2.
a(23) = 3: 9*10/2 + 21*22/2 = 11*12/2 + 20*21/2 = 14*15/2 + 18*19/2 = 23*24/2.

Alois P. Heinz, Table of n, a(n) for n = 1..20000

Cf. A000217, A001652, A012132, A027861, A046080 (the same for squares), A053141, A062301 (the same for primes), A108769, A309507.

a:= proc(n) local h, j, r, w; h, r:= n*(n+1), 0;
      for j from n-1 by -1 do w:= j*(j+1);
        if 2*w

a[n_] := Module[{h = n(n+1), j, r = 0, w}, For[j = n-1, True, j--, w = j(j+1); If[2w < h, Break[]]; If[ IntegerQ[Sqrt[4(h-w)+1]], r++]]; r];
Table[a[n], {n, 1, 120}] (* Jean-François Alcover, Nov 16 2022, after Alois P. Heinz *)

a(n) > 0 <=> n in { A012132 }.
a(n) = 0 <=> n in { A027861 }.
a(n) = 1 <=> n in { A108769 }.

A016278 Expansion of g.f. 1/((1-2x)(1-3x)(1-9x)).

Original entry on oeis.org

1, 14, 145, 1370, 12541, 113534, 1023865, 9221090, 83008981, 747138854, 6724424785, 60520350410, 544684739821, 4902167424974, 44119521140905, 397075733249330, 3573681728253061, 32163135941435894, 289468224634660225

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Counting families of generalized balancing numbers, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.
Index entries for linear recurrences with constant coefficients, signature (14,-51,54).

Cf. A053141, A053142.

m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/((1-2*x)*(1-3*x)*(1-9*x)))); // Vincenzo Librandi, Jun 24 2013

CoefficientList[Series[1 / ((1 - 2 x) (1 - 3 x) (1 - 9 x)), {x, 0, 20}], x] (* Vincenzo Librandi, Jun 24 2013 *)

Vec(1/((1-2*x)*(1-3*x)*(1-9*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

a(n) = 14*a(n-1) - 51*a(n-2) + 54*a(n-3); a(n) = (4/7)*2^(n-1) + (-3/2)*3^(n-1) + (27/14)*9^(n-1). - Antonio Alberto Olivares, Apr 21 2008, Apr 22 2008

E.g.f.: exp(2*x)*(8 - 21*exp(x) + 27*exp(7*x))/14. - Stefano Spezia, Feb 12 2025

A103737 Define a(1)=0, a(2)=0, a(3)=3, a(4)=7 such that from i=1 to 4: 30a(i)^2 + 30a(i) + 1 = j(i)^2, j(1)=1, j(2)=1, j(3)=19, j(4)=41 Then a(n) = a(n-4) + 4sqrt(30(a(n-2)^2) + 30*a(n-2) + 1).

Original entry on oeis.org

0, 0, 3, 7, 76, 164, 1679, 3611, 36872, 79288, 809515, 1740735, 17772468, 38216892, 390184791, 839030899, 8566292944, 18420462896, 188068259987, 404411152823, 4128935426780, 8878624899220, 90648511129183, 194925336630027, 1990138309415256, 4279478780961384, 43692394296006459

Offset: 1

Pierre CAMI, Mar 27 2005

nonn

By construction and recurrence, 30*a(n)^2 + 30*a(n) + 1 = j(n)^2.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,22,-22,-1,1).

Cf. A053141, A103200.

m:=25; R:=PowerSeriesRing(Integers(), m); [0,0] cat Coefficients(R!(x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)))); // G. C. Greubel, Jul 15 2018

Rest[CoefficientList[Series[x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)), {x, 0, 50}], x]] (* G. C. Greubel, Jul 15 2018 *)
LinearRecurrence[{1,22,-22,-1,1},{0,0,3,7,76},40] (* Harvey P. Dale, Mar 05 2025 *)

x='x+O('x^30); concat([0,0], Vec(x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)))) \\ G. C. Greubel, Jul 15 2018

G.f.: x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009

Terms a(19) onward added by G. C. Greubel, Jul 15 2018

A166259 Positive integers n such that a centered polygonal number nk(k+1)/2+1 is not a square for any k > 0.

Original entry on oeis.org

2, 18, 32, 50, 72, 98, 128, 162, 200, 242, 338, 392, 450, 512, 578, 648, 722, 882, 968, 1058, 1152, 1250, 1352, 1458, 1682, 1800, 1922, 2048, 2178, 2312, 2401, 2450, 2662, 2738, 2809, 2888, 3042, 3174, 3200, 3362, 3528, 3698, 3750, 4050, 4225, 4232, 4418, 4489, 4608, 4802

Offset: 1

Alexander Adamchuk, Oct 10 2009

nonn

Positive integers n such that A120744(n) = -1.

Eric Weisstein's World of Mathematics, Centered Polygonal Number.

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

Edited and extended by Max Alekseyev, Jan 20 2010

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

A222393 Nonnegative integers m such that 18m(m+1)+1 is a square.

Original entry on oeis.org

0, 4, 12, 152, 424, 5180, 14420, 175984, 489872, 5978292, 16641244, 203085960, 565312440, 6898944364, 19203981732, 234361022432, 652370066464, 7961375818340, 22161378278060, 270452416801144, 752834491387592, 9187420795420572, 25574211328900084

Offset: 1

Bruno Berselli, Feb 19 2013

a(n+2)/a(n) tends to A156164.

a(n) is congruent to {0,2,4} (mod 5, 6 and 10).

Bruno Berselli, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. nonnegative integers n such that k*n*(n+1)+1 is a square: A001652 (k=2), A001921 (k=3), A001477 (k=4), A053606 (k=5), A105038 (k=6), A105040 (k=7), A053141 (k=8), A222390 (k=10), A105838 (k=11), A061278 (k=12), A104240 (k=13); A105063 (k=17), this sequence (k=18), A101180 (k=19), A077259 (k=20) [incomplete list].

m:=22; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(4*(1+x)^2/((1-x)*(1-6*x+x^2)*(1+6*x+x^2))));

I:=[0,4,12,152,424]; [n le 5 select I[n] else Self(n-1)+34*Self(n-2)-34*Self(n-3)-Self(n-4)+Self(n-5): n in [1..25]]; // Vincenzo Librandi, Aug 18 2013

LinearRecurrence[{1, 34, -34, -1, 1}, {0, 4, 12, 152, 424}, 23]
CoefficientList[Series[4 x (1 + x)^2 / ((1 - x) (1 - 6 x + x^2) (1 + 6 x + x^2)), {x, 0, 25}], x] (* Vincenzo Librandi, Aug 18 2013 *)

makelist(expand(-1/2+((3+sqrt(2)*(-1)^n)*(3-2*sqrt(2))^(2*floor(n/2))+(3-sqrt(2)*(-1)^n)*(3+2*sqrt(2))^(2*floor(n/2)))/12), n, 1, 23);

x='x+O('x^30); concat([0], Vec(4*x*(1+x)^2/((1-x)*(1-6*x+x^2)*(1+6*x+x^2)))) \\ G. C. Greubel, Jul 15 2018

G.f.: 4*x*(1+x)^2/((1-x)*(1-6*x+x^2)*(1+6*x+x^2)).
a(n) = a(-n+1) = a(n-1)+34*a(n-2)-34*a(n-3)-a(n-4)+a(n-5).
a(n) = -1/2+((3+t*(-1)^n)*(3-2*t)^(2*floor(n/2))+(3-t*(-1)^n)*(3+2*t)^(2*floor(n/2)))/12, where t=sqrt(2).

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A341893 Indices of triangular numbers that are one-tenth of other triangular numbers.

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A111647 a(n) = A001541(n)*A001653(n+1)*A002315(n).

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A120744 Least k>0 such that a centered polygonal number nk(k+1)/2+1 is a perfect square; or -1 if no such number exists.

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A309332 Number of ways the n-th triangular number T(n) = A000217(n) can be written as the sum of two positive triangular numbers.

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A016278 Expansion of g.f. 1/((1-2x)(1-3x)(1-9x)).

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A103737 Define a(1)=0, a(2)=0, a(3)=3, a(4)=7 such that from i=1 to 4: 30*a(i)^2 + 30*a(i) + 1 = j(i)^2, j(1)=1, j(2)=1, j(3)=19, j(4)=41 Then a(n) = a(n-4) + 4*sqrt(30*(a(n-2)^2) + 30*a(n-2) + 1).

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A166259 Positive integers n such that a centered polygonal number n*k*(k+1)/2+1 is not a square for any k > 0.

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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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A111647 a(n) = A001541(n)A001653(n+1)A002315(n).

A103737 Define a(1)=0, a(2)=0, a(3)=3, a(4)=7 such that from i=1 to 4: 30a(i)^2 + 30a(i) + 1 = j(i)^2, j(1)=1, j(2)=1, j(3)=19, j(4)=41 Then a(n) = a(n-4) + 4sqrt(30(a(n-2)^2) + 30*a(n-2) + 1).

A166259 Positive integers n such that a centered polygonal number nk(k+1)/2+1 is not a square for any k > 0.

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.

A222393 Nonnegative integers m such that 18m(m+1)+1 is a square.