A001110 -id:A001110 - cp's OEIS frontend

A049684 a(n) = Fibonacci(2n)^2.

0, 1, 9, 64, 441, 3025, 20736, 142129, 974169, 6677056, 45765225, 313679521, 2149991424, 14736260449, 101003831721, 692290561600, 4745030099481, 32522920134769, 222915410843904, 1527884955772561, 10472279279564025, 71778070001175616, 491974210728665289

Offset: 0

Clark Kimberling

This is the r=9 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.

Apparently, this sequence consists of those nonnegative integers k for which x*(x^2-1)*y*(y^2-1) = k*(k^2-1) has a solution in nonnegative integers x, y. If k = a(n), x = A000045(2*n-1) and y = A000045(2*n+1) are a solution. See A374375 for numbers k*(k^2-1) that can be written as a product of two or more factors of the form x*(x^2-1). - Pontus von Brömssen, Jul 14 2024

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 27.
H. J. H. Tuenter, Fibonacci summation identities arising from Catalan's identity, Fib. Q., 60:4 (2022), 312-319.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059.
S. Barbero, U. Cerruti, and N. Murru, On polynomial solutions of the Diophantine equation (x + y - 1)^2 = wxy, Rendiconti Sem. Mat. Univ. Pol. Torino (2020) Vol. 78, No. 1, 5-12.
Pridon Davlianidze, Problem B-1264, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 1 (2020), p. 82; It's All About Catalan, Solution to Problem B-1264, ibid., Vol. 59, No. 1 (2021), pp. 87-88.
E. Kilic, Y. T. Ulutas, and N. Omur, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, table 1, k=2.
R. Stephan, Boring proof of a nonlinearity
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

First differences give A033890.
First differences of A103434.
Bisection of A007598 and A064841.
a(n) = A064170(n+2) - 1 = (1/5) A081070.
Cf. A000032, A000045, A001622, A001906, A056854, A065563, A092184, A374375.

Join[{a=0, b=1}, Table[c=7*b-1*a+2; a=b; b=c, {n, 60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)
Fibonacci[Range[0, 40, 2]]^2 (* Harvey P. Dale, Mar 22 2012 *)
Table[Fibonacci[n - 1] Fibonacci[n + 1] - 1, {n, 0, 40, 2}] (* Bruno Berselli, Feb 12 2015 *)
LinearRecurrence[{8, -8, 1},{0, 1, 9},21] (* Ray Chandler, Sep 23 2015 *)

numlib::fibonacci(2*n)^2 $ n = 0..35; // Zerinvary Lajos, May 13 2008

PARI
```
a(n)=fibonacci(2*n)^2
```

[fibonacci(2*n)^2 for n in range(0, 21)] # Zerinvary Lajos, May 15 2009

G.f.: (x+x^2) / ((1-x)*(1-7*x+x^2)).
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3) with n>2, a(0)=0, a(1)=1, a(2)=9.
a(n) = 7*a(n-1) - a(n-2) + 2 = A001906(n)^2.
a(n) = (A000032(4*n)-2)/5. [This is in Koshy's book (reference under A065563) on p. 88, attributed to Lucas 1876.] - Wolfdieter Lang, Aug 27 2012
a(n) = 1/5*(-2 + ( (7+sqrt(45))/2 )^n + ( (7-sqrt(45))/2 )^n). - Ralf Stephan, Apr 14 2004
a(n) = 2*(T(n, 7/2)-1)/5 with twice the Chebyshev polynomials of the first kind evaluated at x=7/2: 2*T(n, 7/2)= A056854(n). - Wolfdieter Lang, Oct 18 2004
a(n) = F(2*n-1)*F(2*n+1)-1, see A064170 - Bruno Berselli, Feb 12 2015
a(n) = Sum_{i=1..n} F(4*i-2) for n>0. - Bruno Berselli, Aug 25 2015
From Peter Bala, Nov 20 2019: (Start)
Sum_{n >= 1} 1/(a(n) + 1) = (sqrt(5) - 1)/2.
Sum_{n >= 1} 1/(a(n) + 4) = (3*sqrt(5) - 2)/16. More generally, it appears that
Sum_{n >= 1} 1/(a(n) + F(2*k+1)^2) = ((2*k+1)*F(2*k+1)*sqrt(5) - Lucas(2*k+1))/ (2*F(2*k+1)*F(4*k+2)) for k = 0,1,2,....
Sum_{n >= 2} 1/(a(n) - 1) = (8 - 3*sqrt(5))/9. (End)
E.g.f.: (1/5)*(-2*exp(x) + exp((16*x)/(1 + sqrt(5))^4) + exp((1/2)*(7 + 3*sqrt(5))*x)). - Stefano Spezia, Nov 23 2019
Product_{n>=2} (1 - 1/a(n)) = phi^2/3, where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 01 2021
a(n) = A092521(n-1)+A092521(n). - R. J. Mathar, Nov 22 2024

Better description and more terms from Michael Somos

A027568 Numbers that are both triangular and tetrahedral.

Original entry on oeis.org

0, 1, 10, 120, 1540, 7140

Offset: 1

Patrick De Geest

From Anthony C Robin, Oct 27 2022: (Start)
For numbers to be triangular and tetrahedral, we look for solutions r*(r+1)*(r+2)/6 = t*(t+1)/2 = a(n). The corresponding r and t are r = A224421(n-1) and t = A102349(n).
Writing m=r+1 and s=2t+1, this problem is equivalent to solving the Diophantine equation 3 + 4*(m^3 - m) = 3*s^2. The integer solutions for this equation are m = 0, 1, 2, 4, 9, 21, 35 and the corresponding values of s are 1, 1, 3, 9, 31, 111, 239. (End)

J.-M. De Koninck, Ces nombres qui nous fascinent, Ellipses (Paris), 2008 (entry 10, page 3; entry 120, page 41).
L. J. Mordell, Diophantine Equations, Ac. Press, page 258.
P. Odifreddi, Il museo dei numeri, Rizzoli, 2014, page 224.
J. Roberts, The Lure of the Integers, page 53.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 21.

E. T. Avanesov, Solution of a problem on figurate numbers (in Russian), Acta Arith. 12 1966/1967 pages 409-420.
Patrick De Geest, Palindromic Tetrahedrals
M. Gardner, Letter to N. J. A. Sloane, circa Aug 11 1980, concerning A001110, A027568, A039596, etc.
J. Roberts, The Lure of the Integers, pp. 53. (Annotated scanned copy)
Eric Weisstein's World of Mathematics, Tetrahedral Number

Intersection of A000217 and A000292.

Cf. A102349, A102461, A102772, A224421.

{seq(binomial(i,3),i=0..100000) } intersect {seq(binomial(k,2), k= 0..100000)}; # Zerinvary Lajos, Apr 26 2008

With[{trno=Accumulate[Range[0,1000]]},Intersection[trno,Accumulate[ trno]]] (* Harvey P. Dale, May 25 2014 *)

for(n=0,1e3,if(ispolygonal(t=n*(n+1)*(n+2)/6,3),print1(t", "))) \\ Charles R Greathouse IV, Apr 07 2013

A108741 Member r=100 of the family of Chebyshev sequences S_r(n) defined in A092184.

Original entry on oeis.org

0, 1, 100, 9801, 960400, 94109401, 9221760900, 903638458801, 88547347201600, 8676736387298001, 850231618608002500, 83314021887196947001, 8163923913326692803600, 799981229484128697805801, 78389996565531285692164900, 7681419682192581869134354401, 752700738858307491889474566400

Offset: 0

Henry Bottomley, Jun 22 2005

Partial sums of A046173. [Joerg Arndt, Jun 10 2013]

Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A000217, A004189.

I:=[0,1,100]; [n le 3 select I[n] else 99*Self(n-1)-99*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Feb 02 2016

LinearRecurrence[{99, -99, 1}, {0, 1, 100}, 20] (* Vincenzo Librandi, Feb 02 2016 *)

a(n) = ((49+20*sqrt(6))^n+(49-20*sqrt(6))^n -2)/96 = 98*a(n-1)-a(n-2)+2 = 99*a(n-1)-99*a(n-2)+a(n-3) = (a(n-1)-1)^2/a(n-2) = A004189(n)^2.
G.f.: -x*(x+1)/((x-1)*(x^2-98*x+1)). [Colin Barker, Oct 24 2012]
From Wolfdieter Lang, Feb 01 2016: (Start)
a(n) = (T(n, 49) - 1)/48 = (T(2*n, 5) - 1)/48 with Chebyshev's T polynomials A053120. See the name.
a(n) = A000217((T(n, 5) - 1)/2)/3. n >= 0.
a(n) = S(n-1, 10)^2 = A004189(n)^2, with Chebyshev's S polynomials A049310. This is the triangular number = 3*square number identity. Cf. the famous triangular number = square number identity: A000217((T(n, 3) - 1)/2) = S(n-1, 6)^2. A001109 and A001110. (End)

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

A034706 Numbers which are sums of consecutive triangular numbers.

Original entry on oeis.org

0, 1, 3, 4, 6, 9, 10, 15, 16, 19, 20, 21, 25, 28, 31, 34, 35, 36, 45, 46, 49, 52, 55, 56, 64, 66, 74, 78, 80, 81, 83, 84, 85, 91, 100, 105, 109, 110, 116, 119, 120, 121, 130, 136, 144, 145, 153, 155, 161, 164, 165, 166, 169, 171, 185, 190, 196, 199, 200, 202, 210

Offset: 1

Erich Friedman

nonn

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
D. Subramaniam, E. Trevino, and P. Pollack, On sums of consecutive triangular numbers, INTEGERS 20A (2020) A15.

Complement gives A050941.

Cf. A000217 (1 consec), A001110 (2 consec), A129803 (3 consec), A131557 (5 consec), A257711 (7 consec), A034705, A269414 (subsequence of primes).

-- import Data.Set (deleteFindMin, union, fromList); import Data.List (inits)
a034706 n = a034706_list !! (n-1)
a034706_list = f 0 (tail $ inits $ a000217_list) (fromList [0]) where
   f x vss'@(vs:vss) s
     | y < x = y : f x vss' s'
     | otherwise = f w vss (union s $ fromList $ scanl1 (+) ws)
     where ws@(w:_) = reverse vs
           (y, s') = deleteFindMin s
-- Reinhard Zumkeller, May 12 2015

isA034706 := proc(n)
    local a,b;
    for a from 0 do
        if a*(a+1)/2 > n then
            return false;
        end if;
        for b from a do
            tab := (1+b-a)*(a^2+b*a+a+b^2+2*b)/6 ;
            if tab = n then
                return true;
            elif tab > n then
                break;
            end if;
        end do:
    end do:
end proc:
for n from 0 to 100 do
    if isA034706(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Dec 14 2015

M = 1000; (* to get all terms <= M *)
nmax = (Sqrt[8 M + 1] - 1)/2 // Ceiling;
Table[Sum[n(n+1)/2, {n, j, k}], {j, 0, nmax}, {k, j, nmax}] // Flatten // Union // Select[#, # <= M&]& (* Jean-François Alcover, Mar 10 2019 *)

A005214 Triangular numbers together with squares (excluding 0).

Original entry on oeis.org

1, 3, 4, 6, 9, 10, 15, 16, 21, 25, 28, 36, 45, 49, 55, 64, 66, 78, 81, 91, 100, 105, 120, 121, 136, 144, 153, 169, 171, 190, 196, 210, 225, 231, 253, 256, 276, 289, 300, 324, 325, 351, 361, 378, 400, 406, 435, 441, 465, 484, 496, 528, 529, 561, 576, 595, 625, 630, 666, 676

Offset: 1

Russ Cox, Jun 14 1998

Douglas R. Hofstadter, Fluid Concepts and Creative Analogies: Computer Models of the Fundamental Mechanisms of Thought, (together with the Fluid Analogies Research Group), NY: Basic Books, 1995, p. 15.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Douglas R. Hofstadter, Analogies and Sequences: Intertwined Patterns of Integers and Patterns of Thought Processes, DIMACS Conference on Challenges of Identifying Integer Sequences, Rutgers University, October 10 2014; Part 1, Part 2.
Eric Weisstein's World of Mathematics, Square Triangular Number.

Union of A000290 and A000217.
Cf. A001110, A054686, A157259, A117704 (first differences), A193711 (partial sums), A193748, A193749 (partitions into).
Cf. A010052, A010054, A193714, A193715.
Cf. A241241 (subsequence), A242401 (complement).

import Data.List.Ordered (union)
a005214 n = a005214_list !! (n-1)
a005214_list = tail $ union a000290_list a000217_list
-- Reinhard Zumkeller, Feb 15 2015, Aug 03 2011

a := proc(n) floor(sqrt(n)): floor(sqrt(n+n)):
`if`(n+n = %*% + % or n = %% * %%, n, NULL) end: # Peter Luschny, May 01 2014

With[{upto=700},Module[{maxs=Floor[Sqrt[upto]], maxt=Floor[(Sqrt[8upto+1]- 1)/2]},Union[Join[Range[maxs]^2, Table[(n(n+1))/2,{n,maxt}]]]]] (* Harvey P. Dale, Sep 17 2011 *)

upTo(lim)=vecsort(concat(vector(sqrtint(lim\1),n,n^2), vector(floor(sqrt(2+2*lim)-1/2),n,n*(n+1)/2)),,8) \\ Charles R Greathouse IV, Aug 04 2011

isok(m) = ispolygonal(m,3) || ispolygonal(m,4); \\ Michel Marcus, Mar 13 2021

From Reinhard Zumkeller, Aug 03 2011: (Start)
A010052(a(n)) + A010054(a(n)) > 0.
A010052(a(A193714(n))) = 1.
A010054(a(A193715(n))) = 1. (End)
a(n) ~ c * n^2, where c = 3 - 2*sqrt(2) = A157259 - 4 = 0.171572... . - Amiram Eldar, Apr 04 2025

A056771 a(n) = a(-n) = 34*a(n-1) - a(n-2), and a(0)=1, a(1)=17.

Original entry on oeis.org

1, 17, 577, 19601, 665857, 22619537, 768398401, 26102926097, 886731088897, 30122754096401, 1023286908188737, 34761632124320657, 1180872205318713601, 40114893348711941777, 1362725501650887306817, 46292552162781456490001

Offset: 0

Henry Bottomley, Aug 16 2000

The sequence satisfies the Pell equation a(n)^2 - 18 * A202299(n+1)^2 = 1. - Vincenzo Librandi, Dec 19 2011
Also numbers n such that n - 1 and 2*n + 2 are squares. - Arkadiusz Wesolowski, Mar 15 2015
And they, n - 1 and 2*n + 2, are the squares of A005319 and A003499. - Michel Marcus, Mar 15 2015
This sequence {a(n)} gives all the nonnegative integer solutions of the Pell equation a(n)^2 - 32*(3*A091761(n))^2 = +1. - Wolfdieter Lang, Mar 09 2019

			G.f. = 1 + 17*x + 577*x^2 + 19601*x^3 + 665857*x^4 + 22619537*x^5 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..600 (terms 0..200 from Vincenzo Librandi)
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (34,-1).

Cf. A001075, A001541, A001091, A001079, A023038, A011943, A001081, A023039, A001085 and note relationship with square triangular number sequences A001110 and A001109. A091761.

Row 3 of array A188644.

I:=[1, 17]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 18 2011

LinearRecurrence[{34,-1},{1,17},30] (* Vincenzo Librandi, Dec 18 2011 *)
a[ n_] := ChebyshevT[ 2 n, 3]; (* Michael Somos, May 28 2014 *)

makelist(expand(((17+sqrt(288))^n+(17-sqrt(288))^n))/2, n, 0, 15); /* Vincenzo Librandi, Dec 18 2011 */

{a(n) = polchebyshev( n, 1, 17)}; /* Michael Somos, Apr 05 2019 */

[lucas_number2(n,34,1)/2 for n in range(0,15)] # Zerinvary Lajos, Jun 27 2008

a(n) = (r^n + 1/r^n)/2 with r = 17 + sqrt(17^2-1).
a(n) = 16*A001110(n) + 1 = A001541(2n) = (4*A001109(n))^2 + 1 = 3*A001109(2n-1) - A001109(2n-2) = A001109(2n) - 3*A001109(2n-1).
a(n) = T(n, 17) = T(2*n, 3) with T(n, x) Chebyshev's polynomials of the first kind. See A053120. T(n, 3)= A001541(n).
G.f.: (1-17*x)/(1-34*x+x^2).
G.f.: (1 - 17*x / (1 - 288*x / (17 - x))). - Michael Somos, Apr 05 2019
a(n) = cosh(2n*arcsinh(sqrt(8))). - Herbert Kociemba, Apr 24 2008
a(n) = (a^n + b^n)/2 where a = 17 + 12*sqrt(2) and b = 17 - 12*sqrt(2); sqrt(a(n)-1)/4 = A001109(n). - James R. Buddenhagen, Dec 09 2011
a(-n) = a(n). - Michael Somos, May 28 2014
a(n) = sqrt(1 + 32*9*A091761(n)^2), n >= 0. See one of the Pell comments above. - Wolfdieter Lang, Mar 09 2019

More terms from James Sellers, Sep 07 2000

Chebyshev comments from Wolfdieter Lang, Nov 29 2002

A084703 Squares k such that 2*k+1 is also a square.

Original entry on oeis.org

0, 4, 144, 4900, 166464, 5654884, 192099600, 6525731524, 221682772224, 7530688524100, 255821727047184, 8690408031080164, 295218051329678400, 10028723337177985444, 340681375412721826704, 11573138040695364122500, 393146012008229658338304, 13355391270239113019379844

Offset: 0

Amarnath Murthy, Jun 08 2003

With the exception of 0, a subsequence of A075114. - R. J. Mathar, Dec 15 2008
Consequently, A014105(k) is a square if and only if k = a(n). - Bruno Berselli, Oct 14 2011
From M. F. Hasler, Jan 17 2012: (Start)
Bisection of A079291. The squares 2*k+1 are given in A055792.
A204576 is this sequence written in binary. (End)
a(n+1), n >= 0, is the perimeter squared (x(n) + y(n) + z(n))^2 of the ordered primitive Pythagorean triple (x(n), y(n) = x(n) + 1, z(n)). The first two terms are (x(0)=0, y(0)=1, z(0)=1), a(1) = 2^2, and (x(1)=3, y(1)=4, z(1)=5), a(2) = 12^2. - George F. Johnson, Nov 02 2012

D. W. Wilson, Table of n, a(n-1) for n = 1..100 (offset=1)
Emrah Kılıç, Yücel Türker Ulutaş, and Neşe Ömür, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011), Article 11.5.6, table 3, k=2.
Thomas Koshy, Products Involving Reciprocals of Gibonacci Polynomials, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 15-24.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000129, A001109, A001110, A001333, A001541, A001542, A001652, A001653.
Cf. A014105, A029549, A046090, A046176, A055792, A075114, A079291, A084702.
Cf. A089928, A204576.
Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: this sequence (k=-1), A076218 (k=3), A278310 (k=-5).

[4*Evaluate(ChebyshevU(n), 3)^2: n in [0..30]]; // G. C. Greubel, Aug 18 2022

b[n_]:= b[n]= If[n<2, n, 34*b[n-1] -b[n-2] +2]; (* b=A001110 *)
a[n_]:= 4*b[n]; Table[a[n], {n, 0, 30}]
4*ChebyshevU[Range[-1,30], 3]^2 (* G. C. Greubel, Aug 18 2022 *)

[4*chebyshev_U(n-1, 3)^2 for n in (0..30)] # G. C. Greubel, Aug 18 2022

a(n) = 4*A001110(n) = A001542(n)^2.
a(n+1) = A001652(n)*A001652(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2. - Charlie Marion, Jul 01 2003
a(n) = A001653(k+n)*A001653(k-n) - A001653(k)^2, for k >= n >= 0; e.g. 144 = 5741*5 - 169^2. - Charlie Marion, Jul 16 2003
G.f.: 4*x*(1+x)/((1-x)*(1-34*x+x^2)). - R. J. Mathar, Dec 15 2008
a(n) = A079291(2n). - M. F. Hasler, Jan 16 2012
From George F. Johnson, Nov 02 2012: (Start)
a(n) = ((17+12*sqrt(2))^n + (17-12*sqrt(2))^n - 2)/8.
a(n+1) = 17*a(n) + 4 + 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1) = 17*a(n) + 4 - 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1)*a(n+1) = (a(n) - 4)^2.
2*a(n) + 1 = (A001541(n))^2.
a(n+1) = 34*a(n) - a(n-1) + 8 for n>1, a(0)=0, a(1)=4.
a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2) for n>0, a(0)=0, a(1)=4, a(2)=144.
a(n)*a(n+1) = (4*A029549(n))^2.
a(n+1) - a(n) = 4*A046176(n).
a(n) + a(n+1) = 4*(6*A029549(n) + 1).
a(n) = (2*A001333(n)*A000129(n))^2.
Limit_{n -> infinity} a(n)/a(n-r) = (17+12*sqrt(2))^r. (End)
Empirical: a(n) = A089928(4*n-2), for n > 0. - Alex Ratushnyak, Apr 12 2013
a(n) = 4*A001109(n)^2. - G. C. Greubel, Aug 18 2022
Product_{n>=2} (1 - 4/a(n)) = sqrt(2)/3 + 1/2 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Edited and extended by Robert G. Wilson v, Jun 15 2003

A036353 Square pentagonal numbers.

Original entry on oeis.org

0, 1, 9801, 94109401, 903638458801, 8676736387298001, 83314021887196947001, 799981229484128697805801, 7681419682192581869134354401, 73756990988431941623299373152801, 708214619789503821274338711878841001, 6800276705461824703444258688161258139001

Offset: 0

Jean-Francois Chariot (jeanfrancois.chariot(AT)afoc.alcatel.fr)

Lim_{n -> oo} a(n)/a(n-1) = (sqrt(2) + sqrt(3))^8 = 4801 + 1960*sqrt(6). - Ant King, Nov 06 2011

Pentagonal numbers (A000326) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 11 2015

Colin Barker, Table of n, a(n) for n = 0..252
Muniru A. Asiru, All square chiliagonal numbers, International Journal of Mathematical Education in Science and Technology, Volume 47, 2016 - Issue 7.
Byungchan Kim, Eunmi Kim, and Jeremy Lovejoy, On weighted overpartitions related to some q-series in Ramanujan's lost notebook, Int'l J. Number Theory (2021). Also at Université de Paris (France, 2020).
Eric Weisstein's World of Mathematics, Pentagonal Square Number
Index entries for linear recurrences with constant coefficients, signature (9603,-9603,1).

Cf. A000326, A001078, A001079, A001110, A046172, A046173, A248205.

Table[Floor[1/96 ( Sqrt[2] + Sqrt[3] ) ^ ( 8*n - 4 ) ] , {n, 0, 9}] (* Ant King, Nov 06 2011 *)
LinearRecurrence[{9603,-9603,1},{0,1,9801,94109401},20] (* Harvey P. Dale, Apr 14 2019 *)

for(n=0,10^9,g=(n*(3*n-1)/2); if(issquare(g),print(g)))

concat(0, Vec(x*(1+198*x+x^2)/((1-x)*(1-9602*x+x^2)) + O(x^20))) \\ Colin Barker, Jun 24 2015

a(n) = 9602*a(n-1) - a(n-2) + 200; g.f.: x*(1+198*x+x^2)/((1-x)*(1-9602*x+x^2)). - Warut Roonguthai, Jan 05 2001
a(n+1) = 4801*a(n)+100+980*(24*a(n)^2+a(n))^(1/2). - Richard Choulet, Sep 21 2007
From Ant King, Nov 06 2011: (Start)
a(n) = floor(1/96*(sqrt(2) + sqrt(3))^(8*n-4)).
a(n) = 9603*a(n-1) - 9603*a(n-2) + a(n-3).
(End)

More terms from Eric W. Weisstein

A054493 A Pellian-related recursive sequence.

Original entry on oeis.org

1, 7, 36, 175, 841, 4032, 19321, 92575, 443556, 2125207, 10182481, 48787200, 233753521, 1119980407, 5366148516, 25710762175, 123187662361, 590227549632, 2827950085801, 13549522879375, 64919664311076, 311048798676007, 1490324329068961, 7140572846668800

Offset: 0

Barry E. Williams, May 06 2000

This is the r=7 member in the r-family of sequences S_r(n+1) defined in A092184 where more information can be found.

Working with an offset of 1, this sequence is a divisibility sequence, i.e., a(n) divides a(m) whenever n divides m. Case P1 = 7, P2 = 10, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 25 2014

			G.f. = 1 + 7*x + 36*x^2 + 175*x^3 + 841*x^4 + 4032*x^5 + 19321*x^6 + ...

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 181-193.
Peter Bala, Linear divisibility sequences and Chebyshev polynomials
S. Barbero, U. Cerruti, and N. Murru, On polynomial solutions of the Diophantine equation (x + y - 1)^2 = wxy, Rendiconti Sem. Mat. Univ. Pol. Torino (2020) Vol. 78, No. 1, 5-12.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
R. Stephan, Boring proof of a nonlinearity
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,-6,1)

Cf. A004254, A100047, A030221 (first differences).

A054493 := proc(n)
    option remember;
    if n <= 1 then
        6*n+1 ;
    else
        5*procname(n-1)-procname(n-2)+2 ;
    end if ;
end proc:
seq(A054493(n),n=0..10) ; # R. J. Mathar, Apr 16 2018

LinearRecurrence[{6,-6,1},{1,7,36},30] (* Harvey P. Dale, Apr 15 2015 *)
a[ n_] := ChebyshevU[n, Sqrt[7]/2]^2; (* Michael Somos, Jan 22 2017 *)

{a(n) = simplify(polchebyshev(n, 2, quadgen(28)/2)^2)}; /* Michael Somos, Jan 22 2017 */

a(n) = 5*a(n-1) - a(n-2) + 2, a(0)=1, a(1)=7.
A004254 = sqrt{21*(A054493)^2+28*(A054493)}/7. - James Sellers, May 10 2000
a(n) = (1/3)*(-2 + ((5+sqrt(21))/2)^n + ((5-sqrt(21))/2)^n). - Ralf Stephan, Apr 14 2004
G.f.: (1+x)/((1-x)*(1 - 5*x + x^2)) = (1+x)/(1 - 6*x + 6*x^2 - x^3). From the R. Stephan link.
a(n) = 6*a(n-1) - 6*a(n-2) + a(n-3), n>=2, a(-1):=0, a(0)=1, a(1)=7.
a(n) = (2*T(n, 5/2)-2)/3, with twice the Chebyshev polynomials of the first kind, 2*T(n, x=5/2)=A003501(n).
a(n) = b(n) + b(n-1), n>=1, with b(n)=A089817(n) the partial sums of S(n, 5)= U(n, 5/2)=A004254(n+1), with S(n, x)=U(n, x/2) Chebyshev's polynomials of the second kind.
From Peter Bala, Mar 25 2014: (Start)
The following formulas assume an offset of 1.
Let {u(n)} be the Lucas sequence in the quadratic integer ring Z[sqrt(7)] defined by the recurrence u(0) = 0, u(1) = 1 and u(n) = sqrt(7)*u(n-1) - u(n-2) for n >= 2. Then a(n) = u(n)^2.
Equivalently, a(n) = U(n-1,sqrt(7)/2)^2, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = 1/3*( ((sqrt(7) + sqrt(3))/2)^n - ((sqrt(7) - sqrt(3))/2)^n )^2.
a(n) = bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -5/2; 1, 7/2] and T(n,x) denotes the Chebyshev polynomial of the first kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
a(2*n - 1) = 7 * A004254(n)^2, a(2*n) = A030221(n)^2 for all n in Z. - Michael Somos, Jan 22 2017
a(n) = a(-2-n) for all n in Z. - Michael Somos, Jan 22 2017
0 = 1 + a(n)*(-2 + a(n) - 5*a(n+1)) + a(n+1)*(-2 + a(n+1)) for all n in Z. - Michael Somos, Jan 22 2017

Chebyshev comments from Wolfdieter Lang, Sep 10 2004

cp's OEIS Frontend

A049684 a(n) = Fibonacci(2n)^2.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

MuPAD

PARI

Sage

Formula

Extensions

A027568 Numbers that are both triangular and tetrahedral.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

A108741 Member r=100 of the family of Chebyshev sequences S_r(n) defined in A092184.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Formula

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

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

PARI

Formula

A034706 Numbers which are sums of consecutive triangular numbers.

Views

Author

Keywords

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

A005214 Triangular numbers together with squares (excluding 0).

Views

Author

Keywords

References

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

PARI

Formula

A056771 a(n) = a(-n) = 34*a(n-1) - a(n-2), and a(0)=1, a(1)=17.