A077259 -id:A077259 - cp's OEIS frontend

A053141 a(0)=0, a(1)=2 then a(n) = a(n-2) + 2sqrt(8a(n-1)^2 + 8*a(n-1) + 1).

0, 2, 14, 84, 492, 2870, 16730, 97512, 568344, 3312554, 19306982, 112529340, 655869060, 3822685022, 22280241074, 129858761424, 756872327472, 4411375203410, 25711378892990, 149856898154532, 873430010034204, 5090723162050694, 29670908962269962, 172934730611569080

Offset: 0

Wolfdieter Lang

Solution to b(b+1) = 2a(a+1) in natural numbers including 0; a = a(n), b = b(n) = A001652(n).
The solution of a special case of a binomial problem of H. Finner and K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de).
Also the indices of triangular numbers that are half other triangular numbers [a of T(a) such that 2T(a)=T(b)]. The T(a)'s are in A075528, the T(b)'s are in A029549 and the b's are in A001652. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 30 2002
Sequences A053141 (this entry), A016278, A077259, A077288 and A077398 are part of an infinite series of sequences. Each depends upon the polynomial p(n) = 4k*n^2 + 4k*n + 1, when 4k is not a perfect square. Equivalently, they each depend on the equation k*t(x)=t(z) where t(n) is the triangular number formula n(n+1)/2. The dependencies are these: they are the sequences of positive integers n such that p(n) is a perfect square and there exists a positive integer m such that k*t(n)=t(m). A053141 is for k=2, A016278 is for k=3, A077259 is for k=5. - Robert Phillips (bobanne(AT)bellsouth.net), Oct 11 2007, Nov 27 2007
Jason Holt observes that a pair drawn from a drawer with A053141(n)+1 red socks and A001652(n) - A053141(n) blue socks will as likely as not be matching reds: (A053141+1)*A053141/((A001652+1)*A001652) = 1/2, n>0. - Bill Gosper, Feb 07 2010
The values x(n)=A001652(n), y(n)=A046090(n) and z(n)=A001653(n) form a nearly isosceles Pythagorean triple since y(n)=x(n)+1 and x(n)^2 + y(n)^2 = z(n)^2; e.g., for n=2, 20^2 + 21^2 = 29^2. In a similar fashion, if we define b(n)=A011900(n) and c(n)=A001652(n), a(n), b(n) and c(n) form a nearly isosceles anti-Pythagorean triple since b(n)=a(n)+1 and a(n)^2 + b(n)^2 = c(n)^2 + c(n) + 1; i.e., the value a(n)^2 + b(n)^2 lies almost exactly between two perfect squares; e.g., 2^2 + 3^2 = 13 = 4^2 - 3 = 3^2 + 4; 14^2 + 15^2 = 421 = 21^2 - 20 = 20^2 + 21. - Charlie Marion, Jun 12 2009
Behera & Panda call these the balancers and A001109 are the balancing numbers. - Michel Marcus, Nov 07 2017

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
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.
A. Behera and G. K. Panda, On the Square Roots of Triangular Numbers, Fib. Quart., 37 (1999), pp. 98-105.
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.
P. Catarino, H. Campos, and P. Vasco, On some identities for balancing and cobalancing numbers, Annales Mathematicae et Informaticae, 45 (2015) pp. 11-24.
Refik Keskin and Olcay Karaatli, Some New Properties of Balancing Numbers and Square Triangular Numbers, Journal of Integer Sequences, Vol. 15 (2012), Article #12.1.4.
aBa Mbirika, Janee Schrader, and Jürgen Spilker, Pell and Associated Pell Braid Sequences as GCDs of Sums of k Consecutive Pell, Balancing, and Related Numbers, J. Int. Seq. (2023) Vol. 26, Art. 23.6.4.
J. S. Myers, R. Schroeppel, S. R. Shannon, N. J. A. Sloane, and P. Zimmermann, Three Cousins of Recaman's Sequence, arXiv:2004:14000 [math.NT], April 2020.
G. K. Panda, Sequence balancing and cobalancing numbers, Fib. Q., Vol. 45, No. 3 (2007), 265-271. See p. 266.
Michael Penn, (co) balancing numbers, YouTube video, 2022.
Robert Phillips, Polynomials of the form 1+4ke+4ke^2, 2008.
Robert Phillips, A triangular number result, 2009.
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, 2021.
Burkard Polster, Nice merging together, Mathologer video (2015).
B. Polster and M. Ross, Marching in squares, arXiv preprint arXiv:1503.04658 [math.HO], 2015.
A. Tekcan, M. Tayat, and M. E. Ozbek, The diophantine equation 8x^2-y^2+8x(1+t)+(2t+1)^2=0 and t-balancing numbers, ISRN Combinatorics, Volume 2014, Article ID 897834, 5 pages.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A000129, A001108, A001109, A001652, A001653.
Cf. A011900, A029549, A053142, A075528, A103200.
Partial sums of A001542.
Cf. A000194, A001541, A002024, A016278, A046090, A055997, A077259, A077288, A077398, A084068.

a053141 n = a053141_list !! n
a053141_list = 0 : 2 : map (+ 2)
   (zipWith (-) (map (* 6) (tail a053141_list)) a053141_list)
-- Reinhard Zumkeller, Jan 10 2012

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

A053141 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[0,2]) ;
    else
        6*procname(n-1)-procname(n-2)+2 ;
    end if;
end proc: # R. J. Mathar, Feb 05 2016

Join[{a=0,b=1}, Table[c=6*b-a+1; a=b; b=c, {n,60}]]*2 (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)
a[n_] := Floor[1/8*(2+Sqrt[2])*(3+2*Sqrt[2])^n]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Nov 28 2013 *)
Table[(Fibonacci[2n + 1, 2] - 1)/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)

concat(0,Vec(2/(1-x)/(1-6*x+x^2)+O(x^30))) \\ Charles R Greathouse IV, May 14 2012

{x=1+sqrt(2); y=1-sqrt(2); P(n) = (x^n - y^n)/(x-y)};
a(n) = round((P(2*n+1) - 1)/2);
for(n=0, 30, print1(a(n), ", ")) \\ G. C. Greubel, Jul 15 2018

[(lucas_number1(2*n+1, 2, -1)-1)/2 for n in range(30)] # G. C. Greubel, Apr 27 2020

a(n) = (A001653(n)-1)/2 = 2*A053142(n) = A011900(n)-1. [Corrected by Pontus von Brömssen, Sep 11 2024]
a(n) = 6*a(n-1) - a(n-2) + 2, a(0) = 0, a(1) = 2.
G.f.: 2*x/((1-x)*(1-6*x+x^2)).
Let c(n) = A001109(n). Then a(n+1) = a(n)+2*c(n+1), a(0)=0. This gives a generating function (same as existing g.f.) leading to a closed form: a(n) = (1/8)*(-4+(2+sqrt(2))*(3+2*sqrt(2))^n + (2-sqrt(2))*(3-2*sqrt(2))^n). - Bruce Corrigan (scentman(AT)myfamily.com), Oct 30 2002
a(n) = 2*Sum_{k = 0..n} A001109(k). - Mario Catalani (mario.catalani(AT)unito.it), Mar 22 2003
For n>=1, a(n) = 2*Sum_{k=0..n-1} (n-k)*A001653(k). - Charlie Marion, Jul 01 2003
For n and j >= 1, A001109(j+1)*A001652(n) - A001109(j)*A001652(n-1) + a(j) = A001652(n+j). - Charlie Marion, Jul 07 2003
From Antonio Alberto Olivares, Jan 13 2004: (Start)
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3).
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. (End)
a(n) = sqrt(2)*cosh((2*n+1)*log(1+sqrt(2)))/4 - 1/2 = (sqrt(1+4*A029549)-1)/2. - Bill Gosper, Feb 07 2010 [typo corrected by Vaclav Kotesovec, Feb 05 2016]
a(n+1) + A055997(n+1) = A001541(n+1) + A001109(n+1). - Creighton Dement, Sep 16 2004
From Charlie Marion, Oct 18 2004: (Start)
For n>k, a(n-k-1) = A001541(n)*A001653(k)-A011900(n+k); e.g., 2 = 99*5 - 493.
For n<=k, a(k-n) = A001541(n)*A001653(k) - A011900(n+k); e.g., 2 = 3*29 - 85 + 2. (End)
a(n) = A084068(n)*A084068(n+1). - Kenneth J Ramsey, Aug 16 2007
Let G(n,m) = (2*m+1)*a(n)+ m and H(n,m) = (2*m+1)*b(n)+m where b(n) is from the sequence A001652 and let T(a) = a*(a+1)/2. Then T(G(n,m)) + T(m) = 2*T(H(n,m)). - Kenneth J Ramsey, Aug 16 2007
Let S(n) equal the average of two adjacent terms of G(n,m) as defined immediately above and B(n) be one half the difference of the same adjacent terms. Then for T(i) = triangular number i*(i+1)/2, T(S(n)) - T(m) = B(n)^2 (setting m = 0 gives the square triangular numbers). - Kenneth J Ramsey, Aug 16 2007
a(n) = A001108(n+1) - A001109(n+1). - Dylan Hamilton, Nov 25 2010
a(n) = (a(n-1)*(a(n-1) - 2))/a(n-2) for n > 2. - Vladimir Pletser, Apr 08 2020
a(n) = (ChebyshevU(n, 3) - ChebyshevU(n-1, 3) - 1)/2 = (Pell(2*n+1) - 1)/2. - G. C. Greubel, Apr 27 2020
E.g.f.: (exp(3*x)*(2*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - 2*exp(x))/4. - Stefano Spezia, Mar 16 2024
a(n) = A000194(A029549(n)) = A002024(A075528(n)). - Pontus von Brömssen, Sep 11 2024

Name corrected by Zak Seidov, Apr 11 2011

A077260 Triangular numbers that are 1/5 of a triangular number.

Original entry on oeis.org

0, 3, 21, 990, 6786, 318801, 2185095, 102652956, 703593828, 33053933055, 226555027545, 10643263790778, 72950015275686, 3427097886697485, 23489678363743371, 1103514876252799416, 7563603483110089800, 355328363055514714491, 2435456831883085172253, 114414629388999485266710

Offset: 0

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

			Since b(3)=44 -> a(3)=44*45/2=990.

Colin Barker, Table of n, a(n) for n = 0..798
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 (1,322,-322,-1,1).

Cf. A000217, A077259, A077261, A077262.

CoefficientList[Series[(-3 x (x^2 + 6 x + 1))/((x - 1) (x^2 - 18 x + 1)*(x^2 + 18 x + 1)), {x, 0, 18}], x] (* Michael De Vlieger, Apr 21 2021 *)
LinearRecurrence[{1,322,-322,-1,1},{0,3,21,990,6786},20] (* Harvey P. Dale, Dec 12 2023 *)

concat(0, Vec(-3*x*(x^2+6*x+1) / ((x-1)*(x^2-18*x+1)*(x^2+18*x+1)) + O(x^100))) \\ Colin Barker, May 15 2015

a(n) = A077261(n)/5.
a(n) = b(n)*(b(n)+1)/2 where b(n) = A077259(n).
a(n) = (A000045(A007310(n+1))^2-1)/8. - Vladeta Jovovic, Nov 02 2002. - Definition corrected by R. J. Mathar, Sep 16 2009
G.f.: (-3*x*(x^2+6*x+1))/((x-1)*(x^2-18*x+1)*(x^2+18*x+1)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
a(n) = 322*a(n-2) - a(n-4) + 24. - Vladimir Pletser, Mar 23 2020
E.g.f.: (-6*cosh(x) - (-3 + sqrt(5))*cosh((9 - 4*sqrt(5))*x) + (3 + sqrt(5))*cosh((9 + 4*sqrt(5))*x) - 6*sinh(x) + (7 - 3*sqrt(5))*sinh((9 - 4*sqrt(5))*x) + (7 + 3*sqrt(5))*sinh((9 + 4*sqrt(5))*x))/80. - Stefano Spezia, Aug 15 2024

A077262 Second member of the Diophantine pair (m,k) that satisfies 5*(m^2 + m) = k^2 + k; a(n) = k.

Original entry on oeis.org

0, 5, 14, 99, 260, 1785, 4674, 32039, 83880, 574925, 1505174, 10316619, 27009260, 185124225, 484661514, 3321919439, 8696898000, 59609425685, 156059502494, 1069647742899, 2800374146900, 19194049946505, 50250675141714, 344423251294199, 901711778403960

Offset: 0

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

The first member of the (m,k) pairs are in A077259.

			a(3) = (-1 + sqrt(8*4950 + 1))/2 = (-1 + sqrt(39601))/2 = (199 - 1)/2 = 99.

Colin Barker, Table of n, a(n) for n = 0..1000
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 (1,18,-18,-1,1).

Cf. A077259, A077260, A077261.

f := gfun:-rectoproc({a(-2) = -6, a(-1) = -1, a(0) = 0, a(1) = 5, a(n) = 18*a(n - 2) - a(n - 4) + 8}, a(n), remember); map(f, [$ (0 .. 40)])[]; #Vladimir Pletser, Jul 26 2020

CoefficientList[Series[(x (x^3 + 5 x^2 - 9 x - 5))/((x - 1) (x^2 - 4 x - 1) (x^2 + 4 x - 1)), {x, 0, 24}], x] (* Michael De Vlieger, Apr 21 2021 *)

concat(0, Vec(x*(x^3+5*x^2-9*x-5)/((x-1)*(x^2-4*x-1)*(x^2+4*x-1)) + O(x^100))) \\ Colin Barker, May 15 2015

a(n) = (-1 + sqrt(8*b(n) + 1))/2 where b(n) are the entries in A077261.
a(n) = (sqrt(5*A000045(A007310(n+1))^2 - 4) - 1)/2. - Vladeta Jovovic, Nov 02 2002. - Definition corrected by R. J. Mathar, Sep 16 2009
G.f.: (x*(x^3+5*x^2-9*x-5))/((x-1)*(x^2-4*x-1)*(x^2+4*x-1)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
a(n) = 18*a(n-2) - a(n-4) + 8. - Vladimir Pletser, Mar 23 2020 ; a(-2) = -6, a(-1) = -1, a(0) = 0, a(1) = 5. [Edited by Vladimir Pletser, Jul 26 2020]
From Vladimir Pletser, Jul 26 2020: (Start)
Can be defined for negative n by setting a(-n) = - a(n-1) - 1 for all n in Z.
a(n) = a(n-1) + 18*a(n-2) - 18*a(n-3) - a(n-4) + a(n-5). (End)

A077261 Triangular numbers that are 5 times another triangular number.

Original entry on oeis.org

0, 15, 105, 4950, 33930, 1594005, 10925475, 513264780, 3517969140, 165269665275, 1132775137725, 53216318953890, 364750076378430, 17135489433487425, 117448391818716855, 5517574381263997080, 37818017415550449000, 1776641815277573572455, 12177284159415425861265

Offset: 0

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

			a(3)=5*990=4950.

Colin Barker, Table of n, a(n) for n = 0..797
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 (1,322,-322,-1,1).

Subsequence of A000217.

Cf. A077259, A077260, A077262.

CoefficientList[Series[(-15 x (x^2 + 6 x + 1))/((x - 1) (x^2 - 18 x + 1) (x^2 + 18 x + 1)), {x, 0, 18}], x] (* Michael De Vlieger, Apr 21 2021 *)

a(n) = 5*A077260(n).
G.f.: (-15*x*(x^2+6*x+1))/((x-1)*(x^2-18*x+1)*(x^2+18*x+1)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
a(n) = 322*a(n-2) - a(n-4) + 120. - Vladimir Pletser, Feb 09 2021
E.g.f.: (-6*cosh(x) - (-3 + sqrt(5))*cosh((9 - 4*sqrt(5))*x) + (3 + sqrt(5))*cosh((9 + 4*sqrt(5))*x) - 6*sinh(x) + (7 - 3*sqrt(5))*sinh((9 - 4*sqrt(5))*x) + (7 + 3*sqrt(5))*sinh((9 + 4*sqrt(5))*x))/16. - Stefano Spezia, Aug 15 2024

A110679 a(n+3) = 3a(n+2) + 5a(n+1) + a(n), a(0) = 1, a(1) = 2, a(2) = 11.

Original entry on oeis.org

1, 2, 11, 44, 189, 798, 3383, 14328, 60697, 257114, 1089155, 4613732, 19544085, 82790070, 350704367, 1485607536, 6293134513, 26658145586, 112925716859, 478361013020, 2026369768941, 8583840088782, 36361730124071, 154030760585064, 652484772464329

Offset: 0

Creighton Dement, Aug 02 2005

2tesseq[A*B*cyc(A)] (see program code) gives an alternative formula for A110528.

a(n) is the number of tilings of a 2 X n rectangle by using 1 X 1 squares, dominoes and right trominoes. - Roberto Tauraso, Mar 21 2017

Colin Barker, Table of n, a(n) for n = 0..1000
Robert Munafo, Sequences Related to Floretions
Hermann Stamm-Wilbrandt, 6 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (3,5,1).

Cf. A110528, A110680, A001076, A110526, A110526, A033887.

[(Fibonacci(3*n+2) +(-1)^n)/2: n in [0..30]]; // G. C. Greubel, Apr 19 2019

seriestolist(series((-1+x)/((x+1)*(x^2+4*x-1)), x=0,25)); -or- Floretion Algebra Multiplication Program, FAMP Code: -1jesseq[A*B*cyc(A)] with A = - 'j + 'k - 'ii' - 'ij' - 'ik' and B = - .5'i - .5i' - .5'ii' + .5'jj' - .5'kk' + .5'jk' + .5'kj' - .5e

a[n_] := (Fibonacci[3*n+2] + (-1)^n)/2; a /@ Range[0, 22] (* Giovanni Resta, Mar 21 2017 *)

Vec((1 - x) / ((1 + x)*(1 - 4*x - x^2)) + O(x^30)) \\ Colin Barker, Mar 21 2017

{a(n) = -(-1)^n * (fibonacci(-2 - 3*n)\2)}; /* Michael Somos, Mar 26 2017 */

[(fibonacci(3*n+2) +(-1)^n)/2 for n in (0..30)] # G. C. Greubel, Apr 19 2019

Program "FAMP" finds: 2*(-1^(n+1)) = A110528(n) - A001076(n+1) - 2*a(n). Program "Superseeker" finds: a(n) = A110526(n+1) - A110526(n); a(n) + a(n+1) = A033887(n+1).
a(n) = (-1)^n*Sum_{k=0..n} (-1)^k*Fibonacci(3*k+1). - Gary Detlefs, Jan 22 2013
a(n) = (Fibonacci(3*n+2)+(-1)^n)/2. - Roberto Tauraso, Mar 21 2017
From Colin Barker, Mar 21 2017: (Start)
G.f.: (1 - x) / ((1 + x)*(1 - 4*x - x^2)).
a(n) = 3*a(n-1) + 5*a(n-2) + a(n-3) for n>2.
(End)
a(n) = -(-1)^n * A049651(-1 - n) for all n in Z. - Michael Somos, Mar 26 2017
a(2*n) = A254627(2*n+1); a(2*n+1) = A077259(2*n+1). See "6 interlaced bisections" link. - Hermann Stamm-Wilbrandt, Apr 18 2019
2*a(n) = A015448(n+1)+(-1)^n. - R. J. Mathar, Oct 03 2021

Typo in program code fixed by Creighton Dement, Dec 11 2009

A049651 a(n) = (F(3*n+1) - 1)/2, where F=A000045 (the Fibonacci sequence).

Original entry on oeis.org

0, 1, 6, 27, 116, 493, 2090, 8855, 37512, 158905, 673134, 2851443, 12078908, 51167077, 216747218, 918155951, 3889371024, 16475640049, 69791931222, 295643364939, 1252365390980, 5305104928861, 22472785106426, 95196245354567, 403257766524696, 1708227311453353, 7236167012338110

Offset: 0

Clark Kimberling

This is the sequence A(0,1;4,1;2) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010

For n>0, a(n) is the least number whose greedy Fibonacci-union-Lucas representation (as at A214973), has n terms. - Clark Kimberling, Oct 23 2012

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 24.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Hans Koch, Golden mean renormalization for the almost Mathieu operator and related skew products, arXiv:1907.06804 [math-ph], 2019.
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
Hermann Stamm-Wilbrandt, 6 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Cf. A000045, A033887, A077259, A214973.

Pairwise sums of A049652.

[(Fibonacci(3*n+1) - 1)/2: n in [0..30]]; // G. C. Greubel, Dec 05 2017

(Fibonacci[Range[1,5!,3]]-1)/2 (* Vladimir Joseph Stephan Orlovsky, May 18 2010 *)
LinearRecurrence[{5, -3, -1}, {0, 1, 6}, 50] (* G. C. Greubel, Dec 05 2017 *)

vector(30,n,n--;(fibonacci(3*n+1) -1)/2) \\ G. C. Greubel, Dec 05 2017

[(fibonacci(3*n+1)-1)/2 for n in (0..30)] # G. C. Greubel, Apr 19 2019

From Ralf Stephan, Jan 23 2003: (Start)
a(n) = 4*a(n-1) + a(n-2) + 2, a(0)=0, a(1)=1.
G.f.: x*(1+x)/((1-x)*(1-4*x-x^2)).
a(n) is asymptotic to -1/2+(sqrt(5)+5)/20*(sqrt(5)+2)^n. (End)
a(n+1) = F(2) + F(5) + F(8) + ... + F(3n+2).
a(n) = 5*a(n-1) - 3*a(n-2) - a(n-3), a(0)=0, a(1)=1, a(2)= 6. Observation by G. Detlefs. See the W. Lang link. - Wolfdieter Lang, Oct 18 2010
a(2*n) = A077259(2*n); a(2*n+1) = A294262(2*n+1). See "6 interlaced bisections" link. - Hermann Stamm-Wilbrandt, Apr 18 2019
E.g.f.: exp(x)*(exp(x)*(5*cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x)) - 5)/10. - Stefano Spezia, May 24 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

A336625 Indices of triangular numbers that are eight times other triangular numbers.

Original entry on oeis.org

0, 15, 32, 527, 1104, 17919, 37520, 608735, 1274592, 20679087, 43298624, 702480239, 1470878640, 23863649055, 49966575152, 810661587647, 1697392676544, 27538630330959, 57661384427360, 935502769664975, 1958789677853712, 31779555538278207, 66541187662598864, 1079569385531794079, 2260441590850507680

Offset: 1

Vladimir Pletser, Aug 13 2020

Second member of the Diophantine pair (b(n), a(n)) that satisfies a(n)^2 + a(n) = 8*(b(n)^2 + b(n)) or T(a(n)) = 8*T(b(n)) where T(x) is the triangular number of x. The T(a)'s are in A336626, the T(b)'s are in A336624 and the b's are in A336623.

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

			a(3) = 34*a(1) - a(-1) + 16 = 0 - (-16) + 16 = 32,
a(4) = 34*a(2) - a(0) + 16 = 34*15 - (-1) + 16 = 527, etc.

Vladimir Pletser, Table of n, a(n) for n = 1..1000
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, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A336623, A336624, A336626, A166477 (at n=8).

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

f := gfun:-rectoproc({a(n) = 34*a(n - 2) - a(n - 4) + 16, a(2) = 15, a(1) = 0, a(0) = -1, a(-1) = -16}, a(n), remember); map(f, [$ (0 .. 1000)]); #

LinearRecurrence[{1, 34, -34, -1, 1}, {0, 15, 32, 527, 1104, 17919}, 29] (* Amiram Eldar, Aug 18 2020 *)
FullSimplify[Table[((Sqrt[2] + 1)^(2*n + 1) * (3 - Sqrt[2]*(-1)^n) - (Sqrt[2] - 1)^(2*n + 1) * (3 + Sqrt[2]*(-1)^n) - 2)/4, {n, 0, 20}]] (* Vaclav Kotesovec, Sep 08 2020 *)

concat(0, Vec(x*(15 + 17*x - 15*x^2 - x^3) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)) + O(x^22))) \\ Colin Barker, Aug 14 2020

a(n) = 34*a(n-2) - a(n-4) + 16, for n>=2 with a(2)=15, a(1)=0, a(0)=-1, a(-1)=-16.
a(n) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5), for n>=3 with a(3)=32, a(2)=15, a(1)=0, a(0)=-1, a(-1)=-16.
a(n) = (-1 + sqrt(8*b(n) + 1))/2, where b(n) is A336626(n).
G.f.: x^2*(15 + 17*x - 15*x^2 - x^3) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)). - Colin Barker, Aug 14 2020
a(n) = ((sqrt(2) + 1)^(2*n+1) * (3 - sqrt(2)*(-1)^n) - (sqrt(2) - 1)^(2*n+1) * (3 + sqrt(2)*(-1)^n) - 2)/4. - Vaclav Kotesovec, Sep 08 2020
From Vladimir Pletser, Feb 21 2021: (Start)
a(n) = ((3 - sqrt(2))*(1 + sqrt(2))^(2*n+1) + (3 + sqrt(2))*(1 - sqrt(2))^(2*n+1))/4 - 1/2 for even n.
a(n) = ((3 + sqrt(2))*(1 + sqrt(2))^(2*n+1) + (3 - sqrt(2))*(1 - sqrt(2))^(2*n+1))/4 - 1/2 for odd n. (End)

A336623 First member of the Diophantine pair (m, k) that satisfies 8*(m^2 + m) = k^2 + k; a(n) = m.

Original entry on oeis.org

0, 5, 11, 186, 390, 6335, 13265, 215220, 450636, 7311161, 15308375, 248364270, 520034130, 8437074035, 17665852061, 286612152936, 600118935960, 9736376125805, 20386377970595, 330750176124450, 692536732064286, 11235769612105511, 23525862512215145, 381685416635462940

Offset: 0

Vladimir Pletser, Aug 07 2020

The indices of triangular numbers that are one-eighth of other triangular numbers [m of T(m) such that T(m)=T(k)/8]. The T(m)'s are in A336624, the T(k)'s are in A336626 and the k's are in A336625.
Also, nonnegative m such that 32*m^2 + 32*m + 1 is a square.
Can be defined for negative n by setting a(n) = a(-1-n) for all n in Z.

			a(2) = 34 a(0) - a(-2)+16=0 -5 +16 = 11 ; a(3) = 34 a(1) - a(-1)+16 = 34*5 -0 +16 = 186, etc.

Vladimir Pletser, 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, 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 (1,34,-34,-1,1).

Cf. A336624, A336626, A336625.

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

f := gfun:-rectoproc({a(n) = 34*a(n - 2) - a(n - 4) + 16, a(1) = 5, a(0) = 0, a(-1) = 0,  a(-2) = 5}, a(n), remember); map(f, [$ (0 .. 50)]); #

LinearRecurrence[{1, 34, -34, -1, 1}, {0, 5, 11, 186, 390}, 24] (* Amiram Eldar, Aug 08 2020 *)
FullSimplify[Table[((3*Sqrt[2] - 2*(-1)^n)*(1 + Sqrt[2])^(2*n + 1) + (3*Sqrt[2] + 2*(-1)^n)*(Sqrt[2] - 1)^(2*n + 1) - 8)/16, {n, 0, 20}]] (* Vaclav Kotesovec, Sep 08 2020 *)

concat(0, Vec(x*(5 + 6*x + 5*x^2) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)) + O(x^22))) \\ Colin Barker, Aug 08 2020

a(n) = 34 a(n-2) - a(n-4) + 16 for n>=2, with a(1)=5, a(0)=0, a(-1)=0, a(-2)=5.
a(n) = a(n-1) + 34 a(n-2) - 34 a(n-3) - a(n-4)+ a(n-5) for n>=3 with a(2)=11, a(1)=5, a(0)=0, a(-1)=0, a(-2)=5.
a(n) = (C+((-1)^n)*D)*A^n + (E+((-1)^n)*F)*B^n -1/2 with A = (sqrt(2) + 1)^2 ; B = (sqrt(2) - 1)^2 ; C = 3*(2 + sqrt(2))/16 ; D = -(1 + sqrt(2))/8 ; E = 3*(2 - sqrt(2))/16 ; F = (sqrt(2) - 1)/8 and n>=0.
a(n) = (-1 + sqrt(8*b(n) + 1))/2 where b(n) = A336624(n).
G.f.: x*(5 + 6*x + 5*x^2) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)). - Colin Barker, Aug 08 2020
a(n) = ((3*sqrt(2) - 2*(-1)^n) * (1 + sqrt(2))^(2*n + 1) + (3*sqrt(2) + 2*(-1)^n) * (sqrt(2) - 1)^(2*n + 1) - 8)/16. - Vaclav Kotesovec, Sep 08 2020
Comment from _Vladimir Pletser, Feb 21 2021: (Start)
a(n) = ((4 + sqrt(2))(1 + sqrt(2))^(2n) + (4 - sqrt(2))(1 - sqrt(2))^(2n))/16  - 1/2 for even n.
a(n) = ((8 + 5 sqrt(2))(1 + sqrt(2))^(2n) + (8 - 5 sqrt(2))(1 - sqrt(2))^(2n))/16  - 1/2 for odd n. (End)

A336626 Triangular numbers that are eight times another triangular number.

Original entry on oeis.org

0, 120, 528, 139128, 609960, 160554240, 703893960, 185279454480, 812293020528, 213812329916328, 937385441796000, 246739243443988680, 1081741987539564120, 284736873122033021040, 1248329316235215199128, 328586104843582662292128, 1440570949193450800230240, 379188080252621270252095320

Offset: 1

Vladimir Pletser, Oct 04 2020

The triangular numbers T(t) that are eight times another triangular number T(u) : T(t) = 8*T(u). The t's are in A336625, the T(u)'s are in A336624 and the u's are in A336623.

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

			a(2) = 120 is a term because it is triangular and 120/8 = 15 is also triangular.
a(3) = 1154*a(1) - a(-1) + 648 = 0 - 120 + 648 = 528;
a(4) = 1154*a(2) - a(0) + 648 = 1154*120 - 0 + 648 = 139128, etc.
.
From _Peter Luschny_, Oct 19 2020: (Start)
Related sequences in context, as computed by the Julia function:
n   [A336623, A336624,        A336625,  A336626        ]
[0] [0,       0,              0,        0              ]
[1] [5,       15,             15,       120            ]
[2] [11,      66,             32,       528            ]
[3] [186,     17391,          527,      139128         ]
[4] [390,     76245,          1104,     609960         ]
[5] [6335,    20069280,       17919,    160554240      ]
[6] [13265,   87986745,       37520,    703893960      ]
[7] [215220,  23159931810,    608735,   185279454480   ]
[8] [450636,  101536627566,   1274592,  812293020528   ]
[9] [7311161, 26726541239541, 20679087, 213812329916328] (End)

Vladimir Pletser, Table of n, a(n) for n = 1..653
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.
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 (1,1154,-1154,-1,1).

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

function omnibus()
    println("[A336623, A336624, A336625, A336626]")
    println([0, 0, 0, 0])
    t, h = 1, 1
    for n in 1:999999999
        d, r = divrem(t, 8)
        if r == 0
            d2 = 2*d
            s = isqrt(d2)
            d2 == s * (s + 1) && println([s, d, n, t])
        end
        t, h = t + h + 1, h + 1
    end
end
omnibus() # Peter Luschny, Oct 19 2020

f := gfun:-rectoproc({a(n) = 1154*a(n - 2) - a(n - 4) + 648, a(2) = 120, a(1) = 0, a(0) = 0, a(-1) = 120}, a(n), remember); map(f, [$ (1 .. 1000)])[]; #

LinearRecurrence[{1, 1154, -1154, -1, 1}, {0, 120, 528, 139128, 609960}, 18]

a(n) = 8*A336624(n).
a(n) = 1154*a(n-2) - a(n-4) + 648, for n>=2 with a(2)=120, a(1)=0, a(0)=0, a(-1)=120.
a(n) = a(n-1) + 1154*a(n-2) - 1154*a(n-3) - a(n-4) + a(n-5), for n>=3 with a(3)=528, a(2)=120, a(1)=0, a(0)=0, a(-1)=120.
a(n) = ((10*sqrt(2))/17 + 15/17)*(17 + 12*sqrt(2))^n + (-(10*sqrt(2))/17 + 15/17)*(17 - 12*sqrt(2))^n + (-15/17 - (45*sqrt(2))/68)*(-17 - 12*sqrt(2))^n + (-15/17 + (45*sqrt(2))/68)*(-17 + 12*sqrt(2))^n - 27*(-4 + 3*sqrt(2))*sqrt(2)*(-1/(-17 + 12*sqrt(2)))^n/(1088*(-17 + 12*sqrt(2))) - 27*(4 + 3*sqrt(2))*sqrt(2)*(-1/(-17 - 12*sqrt(2)))^n/(1088*(-17 - 12*sqrt(2))) - 9/16 - 9*(-3 + 2*sqrt(2))*sqrt(2)*(-1/(17 - 12*sqrt(2)))^n/(272*(17 - 12*sqrt(2))) - 9*(3 + 2*sqrt(2))*sqrt(2)*(-1/(17 + 12*sqrt(2)))^n/(272*(17 + 12*sqrt(2))).
Let b(n) be A336625(n). Then a(n) = b(n)*(b(n)+1)/2.
G.f.: 24*x^2*(5 + 17*x + 5*x^2)/(1 - x - 1154*x^2 + 1154*x^3 + x^4 - x^5). - Stefano Spezia, Oct 05 2020
From Vladimir Pletser, Feb 21 2021: (Start)
a(n) = ((11*(1 + sqrt(2))^2 - (-1)^n*6*(4 + 3*sqrt(2)))*(1 + sqrt(2))^(4n) + (11*(1 - sqrt(2))^2 - (-1)^n*6*(4 - 3*sqrt(2)))*(1 - sqrt(2))^(4n))/32 - 9/16.
a(n) = ((1 + 2*sqrt(2))^2*(1 + sqrt(2))^(4n) + (1 - 2*sqrt(2))^2*(1 - sqrt(2))^(4n))/32 - 9/16 for even n.
a(n) = ((5 + 4*sqrt(2))^2*(1 + sqrt(2))^(4n) + (5 - 4*sqrt(2))^2*(1 - sqrt(2))^(4n))/32 - 9/16 for odd n. (End)

cp's OEIS Frontend

A053141 a(0)=0, a(1)=2 then a(n) = a(n-2) + 2*sqrt(8*a(n-1)^2 + 8*a(n-1) + 1).

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A077260 Triangular numbers that are 1/5 of a triangular number.

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A077262 Second member of the Diophantine pair (m,k) that satisfies 5*(m^2 + m) = k^2 + k; a(n) = k.

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A077261 Triangular numbers that are 5 times another triangular number.

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A110679 a(n+3) = 3*a(n+2) + 5*a(n+1) + a(n), a(0) = 1, a(1) = 2, a(2) = 11.

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A049651 a(n) = (F(3*n+1) - 1)/2, where F=A000045 (the Fibonacci sequence).

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A053141 a(0)=0, a(1)=2 then a(n) = a(n-2) + 2sqrt(8a(n-1)^2 + 8*a(n-1) + 1).

A110679 a(n+3) = 3a(n+2) + 5a(n+1) + a(n), a(0) = 1, a(1) = 2, a(2) = 11.