A053141 -id:A053141 - cp's OEIS frontend

A075528 Triangular numbers that are half other triangular numbers.

0, 3, 105, 3570, 121278, 4119885, 139954815, 4754343828, 161507735340, 5486508657735, 186379786627653, 6331426236682470, 215082112260576330, 7306460390622912753, 248204571168918457275, 8431648959352604634600, 286427860046819639119128

Offset: 0

Christian G. Bower, Sep 19 2002

This is the sequence of 1/2 the areas, x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n)=x(n)+1, z(n)) with x(0)=0, y(0)=1, z(0)=1, a(0)=0 and x(1)=3, y(1)=4, z(1)=5, a(1)=3. - George F. Johnson, Aug 24 2012

Colin Barker, Table of n, a(n) for n = 0..653
Martin V. Bonsangue, Gerald E. Gannon and Laura J. Pheifer, Misinterpretations can sometimes be a good thing, Math. Teacher, vol. 95, No. 6 (2002) pp. 446-449.
Harvey J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.
Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000129, A000217, A001652, A002315, A029546, A029549, A046090, A046729, A053141, A084159, A096979.

CoefficientList[ Series[ 3x/(1 - 35 x + 35 x^2 - x^3), {x, 0, 15}], x] (* Robert G. Wilson v, Jun 24 2011 *)

concat(0, Vec(3*x/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Jun 18 2015

a(n) = 3*A029546(n-1) = A029549(n)/2.
G.f.: 3*x/((1-x)*(1-34*x+x^2)).
From George F. Johnson, Aug 24 2012: (Start)
a(n) = ((3+2*sqrt(2))^(2*n+1) + (3-2*sqrt(2))^(2*n+1) - 6)/64.
8*a(n)+1 = A000129(2*n+1)^2.
16*a(n)+1 = A002315(n)^2.
128*a(n)^2 + 24*a(n) + 1 is a perfect square.
a(n+1) = 17*a(n) + 3/2 + 3*sqrt((8*a(n)+1)*(16*a(n)+1))/2.
a(n-1) = 17*a(n) + 3/2 - 3*sqrt((8*a(n)+1)*(16*a(n)+1))/2.
a(n-1)*a(n+1) = a(n)*(a(n)-3); a(n+1) = 34*a(n) - a(n-1) + 3.
a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2); a(n) = A096979(2*n)/2.
a(n) = A084159(n)*A046729(n)/4 = A001652(n)*A046090(n)/4.
Lim_{n->infinity} a(n)/a(n-1) =  17 + 12*sqrt(2).
Lim_{n->infinity} a(n)/a(n-2) = (17 + 12*sqrt(2))^2 = 577 + 408*sqrt(2).
Lim_{n->infinity} a(n)/a(n-r) = (17 + 12*sqrt(2))^r.
Lim_{n->infinity} a(n-r)/a(n) = (17 + 12*sqrt(2))^(-r) = (17 - 12*sqrt(2))^r.
(End)
a(n) = 34*a(n-1) - a(n-2) + 3, n >= 2. - R. J. Mathar, Nov 07 2015
a(n) = A000217(A053141(n)). - R. J. Mathar, Aug 16 2019
a(n) = (a(n-1)*(a(n-1)-3))/a(n-2) for n > 2. - Vladimir Pletser, Apr 08 2020
Sum_{n>=1} 1/a(n) = 2*(3 - 2*sqrt(2)). - Amiram Eldar, Dec 04 2024

A011900 a(n) = 6*a(n-1) - a(n-2) - 2 with a(0) = 1, a(1) = 3.

Original entry on oeis.org

1, 3, 15, 85, 493, 2871, 16731, 97513, 568345, 3312555, 19306983, 112529341, 655869061, 3822685023, 22280241075, 129858761425, 756872327473, 4411375203411, 25711378892991, 149856898154533, 873430010034205, 5090723162050695, 29670908962269963, 172934730611569081

Offset: 0

Mario Velucchi (mathchess(AT)velucchi.it)

Members of Diophantine pairs.
Solution to b*(b-1) = 2*a*(a-1) in natural numbers; a = a(n), b = b(n) = A046090(n).
Also the indices of centered octagonal numbers which are also centered square numbers. - Colin Barker, Jan 01 2015
Also positive integers y in the solutions to 4*x^2 - 8*y^2 - 4*x + 8*y = 0. - Colin Barker, Jan 01 2015
Also the number of perfect matchings on a triangular lattice of width 3 and length n. - Sergey Perepechko, Jul 11 2019

			G.f. = 1 + 3*x + 15x^2 + 85*x^3 + 493*x^4 + 2871*x^5 + 16731*x^6 + ... - _Michael Somos_, Feb 23 2019

Mario Velucchi "The Pell's equation ... an amusing application" in Mathematics and Informatics Quarterly, to appear 1997.

Colin Barker, Table of n, a(n) for n = 0..1000
H. J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
S. N. Perepechko, Number of perfect matchings on triangular lattices of fixed width, DIMA'2015 slides. [see: page 12]
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A000194, A001109, A001541, A001653, A002024, A006062, A011906, A029549, A046090, A053141, A075528, A156035.

I:=[1,3]; [n le 2 select I[n] else 6*Self(n-1) - Self(n-2) - 2: n in [1..40]]; // Vincenzo Librandi, Dec 05 2015

f:= gfun:-rectoproc({a(n)=6*a(n-1)-a(n-2)-2,a(0)=1,a(1)=3},a(n),remember):
seq(f(n),n=0..40); # Robert Israel, Dec 16 2015

a[0] = 1; a[1] = 3; a[n_]:= a[n]= 6 a[n-1] -a[n-2] -2; Table[a@ n, {n,0,40}] (* Michael De Vlieger, Dec 05 2015 *)
Table[(Fibonacci[2n + 1, 2] + 1)/2, {n, 0, 40}] (* Vladimir Reshetnikov, Sep 16 2016 *)
LinearRecurrence[{7,-7,1},{1,3,15},40] (* Harvey P. Dale, Feb 16 2017 *)
a[ n_] := (4 + ChebyshevT[n, 3] + ChebyshevT[n + 1, 3])/8; (* Michael Somos, Feb 23 2019 *)

Vec((1-4*x+x^2)/((1-x)*(1-6*x+x^2)) + O(x^50)) \\ Altug Alkan, Dec 06 2015

[(1+lucas_number1(2*n+1,2,-1))//2 for n in range(41)] # G. C. Greubel, Oct 17 2024

a(n) = (A001653(n+1) + 1)/2.
a(n) = (((1+sqrt(2))^(2*n-1) - (1-sqrt(2))^(2*n-1))/sqrt(8) + 1)/2.
a(n) = 7*(a(n-1) - a(n-2)) + a(n-3); a(1) = 1, a(2) = 3, a(3) = 15. Also a(n) = 1/2 + ( (1-sqrt(2))/(-4*sqrt(2)) )*(3-2*sqrt(2))^n + ( (1+sqrt(2))/(4*sqrt(2)) )*(3+2*sqrt(2))^n. - Antonio Alberto Olivares, Dec 23 2003
Sqrt(2) = Sum_{n>=0} 1/a(n); a(n) = a(n-1) + floor(1/(sqrt(2) - Sum_{k=0..n-1} 1/a(k))) (n>0) with a(0)=1. - Paul D. Hanna, Jan 25 2004
For n>k, a(n+k) = A001541(n)*A001653(k) - A053141(n-k-1); e.g., 493 = 99*5 - 2. For n<=k, a(n+k)=A001541(n)*A001653(k) - A053141(k-n); e.g., 85 = 3*29 - 2. - Charlie Marion, Oct 18 2004
a(n+1) = 3*a(n) - 1 + sqrt(8*a(n)^2 - 8*a(n) + 1), a(1)=1. - Richard Choulet, Sep 18 2007
a(n+1) = a(n) * (a(n) + 2) / a(n-1) for n>=1 with a(0)=1 and a(1)=3. - Paul D. Hanna, Apr 08 2012
G.f.: (1 - 4*x + x^2)/((1-x)*(1 - 6*x + x^2)). - R. J. Mathar, Oct 26 2009
Sum_{k=a(n)..A001109(n+1)} k = a(n)*A001109(n+1) = A011906(n+1). Example n=2, 3+4+5+6=18, 3*6=18. - Paul Cleary, Dec 05 2015
a(n) = (sqrt(1+8*A001109(n+1)^2)+1)/2 - A001109(n+1). - Robert Israel, Dec 16 2015
a(n) = a(-1-n) for all n in Z. - Michael Somos, Feb 23 2019
E.g.f.: (2*exp(x) + exp(3*x)*(2*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)))/4. - Stefano Spezia, Mar 16 2024
a(n) = A053141(n) + 1 = A000194(A029549(n))+1 = A002024(A075528(n))+1. - Pontus von Brömssen, Sep 11 2024

More terms and comments from Wolfdieter Lang

A077259 First member of the Diophantine pair (m,k) that satisfies 5*(m^2 + m) = k^2 + k; a(n) = m.

Original entry on oeis.org

0, 2, 6, 44, 116, 798, 2090, 14328, 37512, 257114, 673134, 4613732, 12078908, 82790070, 216747218, 1485607536, 3889371024, 26658145586, 69791931222, 478361013020, 1252365390980, 8583840088782, 22472785106426, 154030760585064, 403257766524696, 2763969850442378

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Nov 01 2002

			a(3) = (2*6) - 2 + (2*17) = 12 - 2 + 34 = 44.
G.f. = 2*x + 6*x^2 + 44*x^3 + 116*x^4 + 798*x^5 + 2090*x^6 + 14328*x^7 + ... - _Michael Somos_, Jul 15 2018

G. C. Greubel, Table of n, a(n) for n = 0..1000
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.
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Hermann Stamm-Wilbrandt, 6 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (1,18,-18,-1,1).

Cf. A007805, A077260, A077261, A077262.

Cf. A053141.

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

f := gfun:-rectoproc({a(-2) = 2, a(-1) = 0, a(0) = 0, a(1) = 2, a(n) = 18*a(n - 2) - a(n - 4) + 8}, a(n), remember): map(f, [$ (0 .. 40)])[]; # Vladimir Pletser, Jul 24 2020

LinearRecurrence[{1, 18, -18, -1, 1}, {0, 2, 6, 44, 116}, 30] (* G. C. Greubel, Jul 15 2018 *)
a[ n_] := With[{m = Max[n, -1 - n]}, SeriesCoefficient[ 2 x (x + 1)^2 / ((1 - x) (x^2 - 4 x - 1) (x^2 + 4 x - 1)), {x, 0, m}]]; (* Michael Somos, Jul 15 2018 *)

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

Let b(n) be A007805(n). Then with a(0)=0, a(1)=2, a(2*n+2) = 2*a(2*n+1) - a(2*n) + 2*b(n), a(2*n+3) = 2*a(2*n+2) - a(2*n+1) + 2*b(n+1).
a(n) = (A000045(A007310(n+1))-1)/2. - Vladeta Jovovic, Nov 02 2002 [corrected by R. J. Mathar, Sep 16 2009]
a(0)=0, a(1)=2, a(n+2) = 4 + 9*a(n) + 2*sqrt(1 +20*a(n) +20*a(n)^2). - Herbert Kociemba, May 12 2008
a(0)=0, a(1)=2, a(2)=6, a(3)=44, a(n) = 18*a(n-2) - a(n-4) + 8. - Robert Phillips, Sep 01 2008
G.f.: 2*x*(1+x)^2/((1-x)*(1+4*x-x^2)*(1-4*x-x^2)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jul 15 2018
a(2*n) = A049651(2*n); a(2*n+1) = A110679(2*n+1). See "6 interlaced bisections" link. - Hermann Stamm-Wilbrandt, Apr 18 2019
a(n) = a(n-1) + 18*a(n-2) - 18*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Jul 24 2020
From Vladimir Pletser, Feb 07 2021: (Start)
a(n) = ((5+sqrt(5))*(2+sqrt(5))^n + (5-sqrt(5))*(2-sqrt(5))^n)/20 - 1/2 for even n;
a(n) = ((5+3*sqrt(5))*(2+sqrt(5))^n + (5-3*sqrt(5))*(2-sqrt(5))^n)/20 - 1/2 for odd n. (End)

More terms from Colin Barker, Mar 23 2014

A055997 Numbers k such that k*(k - 1)/2 is a square.

Original entry on oeis.org

1, 2, 9, 50, 289, 1682, 9801, 57122, 332929, 1940450, 11309769, 65918162, 384199201, 2239277042, 13051463049, 76069501250, 443365544449, 2584123765442, 15061377048201, 87784138523762, 511643454094369, 2982076586042450, 17380816062160329, 101302819786919522

Offset: 1

Barry E. Williams, Jun 14 2000

Numbers k such that (k-th triangular number - k) is a square.
Gives solutions to A007913(2x)=A007913(x-1). - Benoit Cloitre, Apr 07 2002
Number of closed walks of length 2k on the grid graph P_2 X P_3. - Mitch Harris, Mar 06 2004
If x = A001109(n - 1), y = a(n) and z = x^2 + y, then x^4 + y^3 = z^2. - Bruno Berselli, Aug 24 2010
The product of any term a(n) with an even successor a(n + 2k) is always a square number. The product of any term a(n) with an odd successor a(n + 2k + 1) is always twice a square number. - Bradley Klee & Bill Gosper, Jul 22 2015
It appears that dividing even terms by two and taking the square root gives sequence A079496. - Bradley Klee, Jul 25 2015
The bisections of this sequence are a(2n - 1) = A055792(n) and a(2n) = A088920(n). - Bernard Schott, Apr 19 2020

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 193.
P. Tauvel, Exercices d'Algèbre Générale et d'Arithmétique, Dunod, 2004, Exercice 35 pages 346-347.

Colin Barker, Table of n, a(n) for n = 1..100
Dario Alpern, a^4+b^3=c^2.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 46, 56.
Phil Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Kenneth Ramsey, Generalized Proof re Square Triangular Numbers, message 62 in Triangular_and_Fibonacci_Numbers Yahoo group, Oct 10, 2011.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A007913, A001541.
A001109(n-1) = sqrt{[(a(n))^2 - (a(n))]/2}.
a(n) = A001108(n-1)+1.
A001110(n-1)=a(n)*(a(n)-1)/2.
Cf. A001652, A001653, A046090.
Identical to A115599, but with additional leading term.

I:=[1,2,9]; [n le 3 select I[n] else 7*Self(n-1)-7*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Mar 20 2015

A:= gfun:-rectoproc({a(n) = 6*a(n-1)-a(n-2)-2, a(1) = 1, a(2) = 2}, a(n), remember):
map(A,[$1..100]); # Robert Israel, Jul 22 2015

Table[ 1/4*(2 + (3 - 2*Sqrt[2])^k + (3 + 2*Sqrt[2])^k ) // Simplify, {k, 0, 20}] (* Jean-François Alcover, Mar 06 2013 *)
CoefficientList[Series[(1 - 5 x + 2 x^2) / ((1 - x) (1 - 6 x + x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 20 2015 *)
(1 + ChebyshevT[#, 3])/2 & /@ Range[0, 20] (* Bill Gosper, Jul 20 2015 *)
a[1]=1;a[2]=2;a[n_]:=(a[n-1]+1)^2/a[n-2];a/@Range[25] (* Bradley Klee, Jul 25 2015 *)
LinearRecurrence[{7,-7,1},{1,2,9},30] (* Harvey P. Dale, Dec 06 2015 *)

Vec((1-5*x+2*x^2)/((1-x)*(1-6*x+x^2))+O(x^66)) /* Joerg Arndt, Mar 06 2013 */

t(n)=(1+sqrt(2))^(n-1);
for(k=1,24,print1(round((1/4)*(t(k)^2 + t(k)^(-2) + 2)),", ")) \\ Hugo Pfoertner, Nov 29 2019

a(n) = (1 + polchebyshev(n-1, 1, 3))/2; \\ Michel Marcus, Apr 21 2020

a(n) = 6*a(n - 1) - a(n - 2) - 2; n >= 3, a(1) = 1, a(2) = 2.
G.f.: x*(1 - 5*x + 2*x^2)/((1 - x)*(1 - 6*x + x^2)).
a(n) - 1 + sqrt(2*a(n)*(a(n) - 1)) = A001652(n - 1). - Charlie Marion, Jul 21 2003; corrected by Michel Marcus, Apr 20 2020
a(n) = IF(mod(n; 2)=0; (((1 - sqrt(2))^n + (1 + sqrt(2))^n)/2)^2; 2*((((1 - sqrt(2))^(n + 1) + (1 + sqrt(2))^(n + 1)) - (((1 - sqrt(2))^n + (1 + sqrt(2))^n)))/4)^2). The odd-indexed terms are a(2n + 1) = [A001333(2n)]^2; the even-indexed terms are a(2n) = [A001333(2n - 1)]^2 + 1 = 2*[A001653(n)]^2. - Antonio Alberto Olivares, Jan 31 2004; corrected by Bernard Schott, Apr 20 2020
A053141(n + 1) + a(n + 1) = A001541(n + 1) + A001109(n + 1). - Creighton Dement, Sep 16 2004
a(n) = (1/2) + (1/4)*(3+2*sqrt(2))^(n-1) + (1/4)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Feb 21 2006; corrected by Michel Marcus, Apr 20 2020
a(n) = A001653(n)-A001652(n-1). - Charlie Marion, Apr 10 2006; corrected by Michel Marcus, Apr 20 2020
a(2k) = A001541(k)^2. - Alexander Adamchuk, Nov 24 2006
a(n) = 2*A001653(m)*A011900(n-m-1) +A002315(m)*A001652(n-m-1) - A001108(m) with mA001653(m)*A011900(m-n) - A002315(m)*A046090(m-n) - A001108(m). See Link to Generalized Proof re Square Triangular Numbers. - Kenneth J Ramsey, Oct 13 2011
a(n) = +7*a(n-1) -7*a(n-2) +1*a(n-3). - Joerg Arndt, Mar 06 2013
a(n) * a(n+2) = (A001108(n)-A001652(n)+3*A046090(n))^2. - Robert Israel, Jul 23 2015
sqrt(a(n+1)*a(n-1)) = a(n)+1 - Bradley Klee & Bill Gosper, Jul 25 2015
a(n) = 1 + sum{k=0..n-2} A002315(k). - David Pasino, Jul 09 2016; corrected by Michel Marcus, Apr 20 2020
E.g.f.: (2*exp(x) + exp((3-2*sqrt(2))*x) + exp((3+2*sqrt(2))*x))/4. - Ilya Gutkovskiy, Jul 09 2016
sqrt(a(n)*(a(n)-1)/2) = A001542(n)/2. - David Pasino, Jul 09 2016
Limit_{n -> infinity} a(n)/a(n-1) = A156035. - César Aguilera, Apr 07 2018
a(n) = (1/4)*(t^2 + t^(-2) + 2), where t = (1+sqrt(2))^(n-1). - Ridouane Oudra, Nov 29 2019
sqrt(a(n)) + sqrt(a(n) - 1) = (1 + sqrt(2))^(n - 1). - Ridouane Oudra, Nov 29 2019
sqrt(a(n)) - sqrt(a(n) - 1) = (-1 + sqrt(2))^(n - 1). - Bernard Schott, Apr 18 2020

A077288 First member of the Diophantine pair (m,k) that satisfies 6(m^2 + m) = k^2 + k: a(n) = m.

Original entry on oeis.org

0, 1, 3, 14, 34, 143, 341, 1420, 3380, 14061, 33463, 139194, 331254, 1377883, 3279081, 13639640, 32459560, 135018521, 321316523, 1336545574, 3180705674, 13230437223, 31485740221, 130967826660, 311676696540, 1296447829381, 3085281225183, 12833510467154

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Nov 03 2002

Also nonnegative m such that 24*m^2 + 24*m + 1 is a square. - Gerald McGarvey, Apr 02 2005

			a(3) = 2*3 - 1 + 9 = 14, a(4) = 2*14 - 3 + 9 = 34, etc.
G.f. = x + 3*x^2 + 14*x^3 + 34*x^4 + 143*x^5 + 341*x^6 + 1420*x^7 + 3380*x^8 + ... - _Michael Somos_, Jul 15 2018

G. C. Greubel, Table of n, a(n) for n = 0..1000
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-1,1).

The k values are in A077291
Cf. A077289, A077290, A077291.
Cf. A053141.

m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(1+x)^2/((1-x)*(1-10*x^2+x^4)))); // G. C. Greubel, Jul 15 2018

f := gfun:-rectoproc({a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(n) = 10*a(n - 2) - a(n - 4) + 4}, a(n), remember); map(f, [$ (0 .. 100)]); - Vladimir Pletser, Jul 24 2020

CoefficientList[Series[x*(1 + x)^2/((1 - x)*(1 - 10 x^2 + x^4)), {x, 0, 40}],x] (* T. D. Noe, Jun 04 2012 *)
LinearRecurrence[{1, 10, -10, -1, 1}, {0, 1, 3, 14, 34}, 50] (* G. C. Greubel, Jul 15 2018 *)
a[ n_] := With[{m = Max[n, -1 - n]}, SeriesCoefficient[ x (1 + x)^2 / ((1 - x) (1 - 10 x^2 + x^4)), {x, 0, m}]]; (* Michael Somos, Jul 15 2018 *)

my(x='x+O('x^30)); concat([0], Vec(x*(1+x)^2/((1-x)*(1-10*x^2+x^4)))) \\ G. C. Greubel, Jul 15 2018

Let b(n) be A072256. Then a(2*n+2) = 2*a(2*n+1) - a(2*n) + b(n+1), a(2*n+3) = 2*a(2*n+2) - a(2*n+1) + b(n+2), with a(0)=0, a(1)=1.
G.f.: x*(1+x)^2/((1-x)*(1-10*x^2+x^4)).
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jul 15 2018
a(n) = 10*a(n-2) - a(n-4) + 4, n > 4. - Vladimir Pletser, Feb 29 2020
a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Jul 24 2020
2*a(n) + 1 = A080806(n+1). - R. J. Mathar, Oct 01 2021

A077398 First member of the Diophantine pair (m,k) that satisfies 7*(m^2+m) = k^2+k; a(n)=m.

Original entry on oeis.org

0, 2, 5, 39, 87, 629, 1394, 10032, 22224, 159890, 354197, 2548215, 5644935, 40611557, 89964770, 647236704, 1433791392, 10315175714, 22850697509, 164395574727, 364177368759, 2620014019925, 5803987202642, 41755828744080, 92499617873520, 665473245885362

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Nov 05 2002

Equivalently, m such that 28*m*(m+1)+1 is a square. - Bruno Berselli, May 19 2014

Colin Barker, Table of n, a(n) for n = 0..1000
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (1,16,-16,-1,1).

Cf. A077397, A077399, A077400. The k values are in A077401.

Cf. A053141.

I:=[0,2,5,39,87]; [n le 5 select I[n] else Self(n-1)+16*Self(n-2) -16*Self(n-3)-Self(n-4)+Self(n-5): n in [1..30]]; // G. C. Greubel, Jan 18 2018

f := gfun:-rectoproc({a(-2) = 2, a(-1) = 0, a(0) = 0, a(1) = 2, a(n) = 16*a(n - 2) - a(n - 4) + 7}, a(n), remember): map(f, [$ (0 .. 40)])[]; # Vladimir Pletser, Jul 24 2020

LinearRecurrence[{1,16,-16,-1,1},{0,2,5,39,87}, 30] (* G. C. Greubel, Jan 18 2018 *)
CoefficientList[Series[x (2+3x+2x^2)/((1-x)(1-16x^2+x^4)),{x,0,40}],x] (* Harvey P. Dale, Aug 19 2022 *)

{a(n)=if(n<0,0,polcoeff(x*(2+3*x+2*x^2)/((1-x)*(1-16*x^2+x^4)) + x*O(x^n),n))};

G.f.: x*(2+3*x+2*x^2)/((1-x)*(1-16*x^2+x^4)).
a(n) = 16*a(n-2) - a(n-4) + 7, n >= 3. [corrected by Vladimir Pletser, Feb 29 2020]
Let b(n) be A077397 then a(n+2) = 2*a(n+1) - a(n) + b(n) with a(0)=0 a(1)=2.
a(0)=0, a(1)=2; a(n+2) = (7 + 16*a(n) + 3*sqrt(1+28*a(n)+28*a(n)^2))/2. - Herbert Kociemba, May 12 2008
a(n) = a(n-1) + 16*a(n-2) - 16*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Jul 24 2020

A012132 Numbers z such that x(x+1) + y(y+1) = z*(z+1) is solvable in positive integers x,y.

Original entry on oeis.org

3, 6, 8, 10, 11, 13, 15, 16, 18, 20, 21, 23, 26, 27, 28, 31, 33, 36, 37, 38, 40, 41, 43, 44, 45, 46, 48, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 66, 67, 68, 71, 73, 74, 75, 76, 77, 78, 80, 81, 83, 86, 88, 89, 91, 92, 93

Offset: 1

Sander van Rijnswou (sander(AT)win.tue.nl)

nonn

Theorem (Sierpinski, 1963): n is a term iff n^2+(n+1)^2 is a composite number. - N. J. A. Sloane, Feb 29 2020
For n > 1, A047219 is a subset of this sequence. This is because n^2 + (n+1)^2 is divisible by 5 if n is (1 or 3) mod 5 (also see A027861). - Dmitry Kamenetsky, Sep 02 2008
From Hermann Stamm-Wilbrandt, Sep 10 2014: (Start)
For n > 0, A212160 is a subset of this sequence (n^2 + (n+1)^2 is divisible by 13 if n == (2 or 10) (mod 13)).
For n >= 0, A212161 is a subset of this sequence (n^2 + (n+1)^2 is divisible by 17 if n == (6 or 10) (mod 17)).
The above are for divisibility by 5, 13, 17; notation (1,3,5), (2,10,13), (6,10,17). Divisibility by p for a and p-a-1; notation (a,p-a-1,p). These are the next tuples: (8,20,29), (15,21,37), (4,36,41), (11,41,53), ... . The corresponding sequences are a subset of this sequence (8,20,37,49,66,78,... for (8,20,29)). These sequences have no entries in the OEIS yet. For any prime of the form 4*k+1 there is exactly one of these tuples/sequences.
For n > 1, A000217 (triangular numbers) is a subset of this sequence (3,6,10,15,...); z=A000217(n), y=z-1, x=n.
For n > 0, A001652 is a subset of this sequence; z=A001652(n), x=y=A053141(n).
For n > 1, A001108(=A115598) is a subset of this sequence; z=A001108(n), x=A076708(n), y=x+1.
For n > 0, A124124(2*n+1)(=A098790(2*n)) is a subset of this sequence (6,37,218,...); z=A124124(2*n+1), x=a(n)-1, y=a(n)+1, a(m) = 6*a(m-1) - a(m-2) + 2, a(0)=0, a(1)=4.
(End)

Aviezri S. Fraenkel, Diophantine equations involving generalized triangular and tetrahedral numbers, pp. 99-114 of A. O. L. Atkin and B. J. Birch, editors, Computers in Number Theory. Academic Press, NY, 1971.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
H. Finner and K. Strassburger, Increasing sample sizes do not necessarily increase the power of UMPU-tests for 2 X 2-tables, Metrika, 54, 77-91, (2001).
Heiko Harborth, Fermat-like binomial equations, Applications of Fibonacci numbers, Proc. 2nd Int. Conf., San Jose/Ca., August 1986, 1-5 (1988).
W. Sierpinski, On triangular numbers which are sums of two smaller triangular numbers, (Polish), Wiadom. Mat. (2) 7 (1963), 27-28. See MR0182602.

Complement of A027861. - Michael Somos, Jun 08 2000
Cf. A000217, A047219, A027861, A166080.
Cf. also A212160, A212161, A001652, A001108, A115598, A124124, A098790.

Select[Range[100], !PrimeQ[#^2 + (#+1)^2]& ] (* Jean-François Alcover, Jan 17 2013, after Michael Somos *)

More terms and references from Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Feb 09 2000

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

A185223 A185128(n) is the a(n)-th triangular number.

Original entry on oeis.org

6, 18, 37, 44, 86, 91, 116, 132, 247, 278, 392, 613, 637, 662, 798, 847, 912, 1164, 1235, 1362, 1430, 1638, 1735, 1991, 2056, 2090, 2167, 2364, 2537, 2736, 3139, 3478, 3751, 3867, 4298, 4422, 4553, 5202, 6068, 6391, 6500, 7241, 7859, 7957, 8378, 9309, 9793

Offset: 1

Martin Renner, Jan 20 2012

nonn

Side lengths where both triangular numbers are the same (A053141) are not included. - R. J. Mathar, Feb 11 2018

See A185128 for further information.

			A185128(2) = 171 which is the 18th triangular number, so a(2) = 18.

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966, p. 197, no. 8.

R. J. Mathar, Table of n, a(n) for n = 1..58
N. J. A. Sloane, Annotated scan of Beiler's Table 81, based on page 197 of Beiler's "Recreations in the Theory of Numbers: The Queen of Mathematics Entertains", New York, Dover, First ed., 1964.

Cf. A000217, A185128, A185129, A185233, A185243, A185253, A185257, A185258.

lista(nn) = {v = vector(nn, n, n*(n+1)/2); for (n=2, nn, for (k=1, n-1, if (ispolygonal(v[n]+v[k], 3) && ispolygonal(v[n]-v[k], 3), print1(n, ", "));););} \\ Michel Marcus, Jan 08 2015

Edited (with a simpler definition) by N. J. A. Sloane, Dec 28 2024

A336624 Triangular numbers that are one-eighth of other triangular numbers; T(t) such that 8*T(t)=T(u) for some u where T(k) is the k-th triangular number.

Original entry on oeis.org

0, 15, 66, 17391, 76245, 20069280, 87986745, 23159931810, 101536627566, 26726541239541, 117173180224500, 30842405430498585, 135217748442445515, 35592109140254127630, 156041164529401899891, 41073263105447832786516, 180071368649181350028780, 47398510031577658781511915

Offset: 0

Vladimir Pletser, Aug 07 2020

The triangular numbers T(t) that are one-eighth of other triangular numbers T(u) : T(t)=T(u)/8. The t's are in A336623, the T(u)'s are in A336626 and the u's are in A336625.

Can be defined for negative n by setting a(n) = a(-1-n) for all n in Z.

			a(1)= 15 is a term because it is triangular and 8*15 = 120 is also triangular.
a(2) = 1154*a(0) - a(-2) + 81 = 0 - 15 + 81 = 66;
a(3) = 1154*a(1) - a(-1) + 81 = 1154*15 - 0 + 81 = 17391, etc.

Vladimir Pletser, Table of n, a(n) for n = 0..650
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (1,1154,-1154,-1,1).

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

f := gfun:-rectoproc({a(n) = 1154*a(n - 2) - a(n - 4) + 81, a(1) = 15, a(0) = 0, a(-1) = 0, a(-2) = 15}, a(n), remember): map(f, [$ (0 .. 40)])[]; #

LinearRecurrence[{1, 1154, -1154, -1, 1}, {0, 15, 66, 17391, 76245}, 18] (* Amiram Eldar, Aug 08 2020 *)
FullSimplify[Table[((Sqrt[2] + 1)^(4*n + 2)*(11 - 6*(-1)^n*Sqrt[2]) + (Sqrt[2] - 1)^(4*n + 2)*(11 + 6*(-1)^n*Sqrt[2]) - 18)/256, {n, 0, 17}]] (* Vaclav Kotesovec, Sep 08 2020 *)
Select[Accumulate[Range[0, 10^6]]/8, OddQ[Sqrt[8 # + 1]] &] (* The program generates the first 8 terms of the sequence. *) (* Harvey P. Dale, Jan 15 2024 *)

concat(0, Vec(3*x*(5 + 17*x + 5*x^2) / ((1 - x)*(1 - 34*x + x^2)*(1 + 34*x + x^2)) + O(x^40))) \\ Colin Barker, Aug 08 2020

a(n) = 1154*a(n-2) - a(n-4) + 81, for n>=2 with a(1)=15, a(0)=0, a(-1)=0, a(-2)=15.
a(n) = a(n-1) + 1154*a(n-2) - 1154*a(n-3) - a(n-4) + a(n-5), for n>=3 with a(2)=66, a(1)=15, a(0)=0, a(-1)=0, a(-2)=15.
a(n) = b(n)*(b(n)+1)/2 where b(n) is A336623(n).
G.f.: 3*x*(5 + 17*x + 5*x^2) / ((1 - x)*(1 - 34*x + x^2)*(1 + 34*x + x^2)). - Colin Barker, Aug 08 2020
a(n) = ((sqrt(2) + 1)^(4*n + 2) * (11 - 6*(-1)^n*sqrt(2)) + (sqrt(2) - 1)^(4*n + 2) * (11 + 6*(-1)^n*sqrt(2)) - 18)/256. - Vaclav Kotesovec, Sep 08 2020
From Vladimir Pletser, Feb 21 2021: (Start)
a(n) = ((11 - 6*sqrt(2))*(1 + sqrt(2))^(4n + 2) + (11 + 6*sqrt(2))*(1 - sqrt(2) )^(4n + 2) - 18) / 256 for even n.
a(n) = ((11 + 6*sqrt(2))*(1 + sqrt(2) )^(4n + 2) + (11 - 6*sqrt(2))*(1 - sqrt(2) )^(4n + 2) - 18) / 256 for odd n. (End)
128*a(n) = -9+33*A077420(n)-24*(-1)^n*A046176(n+1). - R. J. Mathar, May 05 2023

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A075528 Triangular numbers that are half other triangular numbers.

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A011900 a(n) = 6*a(n-1) - a(n-2) - 2 with a(0) = 1, a(1) = 3.

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A077259 First member of the Diophantine pair (m,k) that satisfies 5*(m^2 + m) = k^2 + k; a(n) = m.

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A055997 Numbers k such that k*(k - 1)/2 is a square.

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A077288 First member of the Diophantine pair (m,k) that satisfies 6(m^2 + m) = k^2 + k: a(n) = m.

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A077398 First member of the Diophantine pair (m,k) that satisfies 7*(m^2+m) = k^2+k; a(n)=m.

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A012132 Numbers z such that x(x+1) + y(y+1) = z*(z+1) is solvable in positive integers x,y.

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