A089508 -id:A089508 - cp's OEIS frontend

A081016 a(n) = (Lucas(4n+3) + 1)/5, or Fibonacci(2n+1)Fibonacci(2n+2), or A081015(n)/5.

1, 6, 40, 273, 1870, 12816, 87841, 602070, 4126648, 28284465, 193864606, 1328767776, 9107509825, 62423800998, 427859097160, 2932589879121, 20100270056686, 137769300517680, 944284833567073, 6472224534451830

Offset: 0

R. K. Guy, Mar 01 2003

a(n-1) is, together with b(n) := A089508(n), n >= 1, the solution to a binomial problem; see A089508.
Numbers k such that 1 - 2*k + 5*k^2 is a square. - Artur Jasinski, Oct 26 2008
Also solution y of Diophantine equation x^2 + 4*y^2 = h^2 for which x = y-1. - Carmine Suriano, Jun 23 2010

Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 26.

G. C. Greubel, Table of n, a(n) for n = 0..1190
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081015.
Partial sums of A033889. Bisection of A001654. Equals A003482 + 1.
Cf. A145995, A178898.

List([0..30], n-> (Lucas(1,-1,4*n+3)[2] +1)/5 ); # G. C. Greubel, Jul 13 2019

[(Lucas(4*n+3) +1)/5: n in [0..30]]; // G. C. Greubel, Dec 18 2017

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 25 do printf(`%d,`,(luc(4*n+3)+1)/5) od: # James Sellers, Mar 03 2003

LinearRecurrence[{8,-8,1}, {1,6,40}, 30] (* Bruno Berselli, Aug 31 2017 *)

a(n)=([0,1,0; 0,0,1; 1,-8,8]^n*[1;6;40])[1,1] \\ Charles R Greathouse IV, Sep 28 2015

first(n) = Vec((1-2*x)/((1-x)*(1-7*x+x^2)) + O(x^n)) \\ Iain Fox, Dec 19 2017

[(lucas_number2(4*n+3,1,-1) +1)/5 for n in (0..30)] # G. C. Greubel, Jul 13 2019

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: (1 - 2*x)/((1 - x)*(1 - 7*x + x^2)).
a(n) = F(1) + F(5) + F(9) +...+ F(4*n+1) = F(2*n)*F(2*n+3) + 1, where F(j) = Fibonacci(j).
a(n) = 7*a(n-1) - a(n-2) - 1, n >= 2. - R. J. Mathar, Nov 07 2015

A081018 a(n) = (Lucas(4n+1)-1)/5, or Fibonacci(2n)*Fibonacci(2n+1), or A081017(n)/5.

Original entry on oeis.org

0, 2, 15, 104, 714, 4895, 33552, 229970, 1576239, 10803704, 74049690, 507544127, 3478759200, 23843770274, 163427632719, 1120149658760, 7677619978602, 52623190191455, 360684711361584, 2472169789339634, 16944503814015855, 116139356908771352, 796030994547383610

Offset: 0

R. K. Guy, Mar 01 2003

Another interpretation of this sequence is: nonnegative integers k such that (k + 1)^2 + (2k)^2 is a perfect square. So apart from a(0) = 0, a(n) + 1 and 2a(n) form the legs of a Pythagorean triple. - Nick Hobson, Jan 13 2007
Also solution y of Diophantine equation x^2 + 4*y^2 = k^2 for which x=y+1. - Carmine Suriano, Jun 23 2010
Also the index of the first of two consecutive heptagonal numbers whose sum is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 20 2014
Nonnegative integers k such that G(x) = k for some rational number x where G(x) = x/(1-x-x^2) is the generating function of the Fibonacci numbers. - Tom Edgar, Aug 24 2015
The integer solutions of the equation a(b+1) = (a-b)(a-b-1) or, equivalently, binomial(a, b) = binomial(a-1, b+1) are given by (a, b) = (a(n+1), A003482(n)=Fibonacci(2*n) * Fibonacci(2*n+3)) (Lind and Singmaster). - Tomohiro Yamada, May 30 2018

			G.f. = 2*x + 15*x^2 + 104*x^3 + 714*x^4 + 4895*x^5 + 33552*x^6 + ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 28.
Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Colin Barker, Table of n, a(n) for n = 0..1000
Pridon Davlianidze, Problem B-1279, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 4 (2020), p. 368; An Unusual Generalization, Solution to Problem B-1279 by J. N. Senadheera, ibid., Vol. 59, No. 4 (2021), pp. 370-371.
Dae S. Hong, When is the Generating Function of the Fibonacci Numbers an Integer?, The College Mathematics Journal, Vol. 46, No. 2 (March 2015), pp. 110-112
D. A. Lind, The quadratic field Q(sqrt(5)) and a certain diophantine equation, Fibonacci Quart., Vol. 6, No. 3 (1968), pp. 86-93.
David Singmaster, Repeated binomial coefficients and Fibonacci numbers, Fibonacci Quart., Vol. 13, No. 4 (1973), pp. 295-298.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081017.
Partial sums of A033891. Bisection of A001654 and A059840.
Equals A089508 + 1.
Cf. A001622, A081007, A081016, A178898.

List([0..30], n-> (Lucas(1,-1, 4*n+1)[2] -1)/5); # G. C. Greubel, Jul 14 2019

[(Lucas(4*n+1)-1)/5: n in [0..30]]; // Vincenzo Librandi, Aug 24 2015

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 25 do printf(`%d,`,(luc(4*n+1)-1)/5) od: # James Sellers, Mar 03 2003

(LucasL[4*Range[0,30]+1]-1)/5 (* or *) LinearRecurrence[{8,-8,1}, {0,2,15}, 30] (* G. C. Greubel, Aug 24 2015, modified Jul 14 2019 *)

concat(0, Vec(x*(2-x)/((1-x)*(1-7*x+x^2)) + O(x^30))) \\ Colin Barker, Dec 20 2014

[(lucas_number2(4*n+1,1,-1) -1)/5 for n in (0..30)] # G. C. Greubel, Jul 14 2019

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
a(n) = Fibonacci(3) + Fibonacci(7) + Fibonacci(11) + ... + Fibonacci(4n+3).
G.f.: x*(2-x)/((1-x)*(1-7*x+x^2)). - Colin Barker, Mar 30 2012
E.g.f.: (1/5)^(3/2)*((1+phi^2)*exp(phi^4*x) - (1 + (1/phi^2))*exp(x/phi^4) - sqrt(5)*exp(x)), where 2*phi = 1 + sqrt(5). - G. C. Greubel, Aug 24 2015
From - Michael Somos, Aug 27 2015: (Start)
a(n) = -A081016(-1-n) for all n in Z.
0 = a(n) - 7*a(n+1) + a(n+2) - 1 for all n in Z.
0 = a(n)*a(n+2) - a(n+1)^2 + a(n+1) + 2 for all n in Z.
0 = a(n)*(a(n) -7*a(n+1) -1) + a(n+1)*(a(n+1) - 1) - 2 for all n in Z. (End)
a(n) = (k(n) + sqrt(k(n)*(4 + 5*k(n))))/2, where k(n) = A049684(n). - Stefano Spezia, Mar 11 2021
Product_{n>=1} (1 + 1/a(n)) = phi (A001622) (Davlianidze, 2020). - Amiram Eldar, Nov 30 2021

More terms from James Sellers, Mar 03 2003

A090162 Values of binomial(Fibonacci(2k)Fibonacci(2k+1),Fibonacci(2k-1)Fibonacci(2k)-1).

Original entry on oeis.org

1, 3003, 61218182743304701891431482520

Offset: 1

Eric W. Weisstein, Nov 23 2003 and Wolfdieter Lang, Dec 01 2003

These numbers are known to occur at least six times in Pascal's triangle.
The next term is approximately 3.537 * 10^204 and is in the b-file.
The numbers of digits in a(n), n >= 1, are given in A100022.

Hugo Pfoertner, Table of n, a(n) for n = 1..5
A. I. Shirshov, On the equation C(n, m) = C(n+1, m-1), chapter 10 in: Kvant Selecta: Algebra and Analysis, I, ed. S. Tabachnikov, Am. Math. Soc., 1999, pp. 83-86
D. Singmaster, Repeated binomial coefficients and Fibonacci numbers, Fibonacci Quarterly, 13 (1975), 295-298.
Eric Weisstein's World of Mathematics, Pascal's Triangle

Subsequence of A003015.

Cf. A081016, A089508, A062527, A206351.

a := proc(n) local a,b,s,p; s:= 1+sqrt(5); p:=16^n;
a := 4-2*p*s^(-4*n-1)+(s+2)*s^(4*n-1)/p:
b := 1+p*((s-2)^(1-4*n)/2-s^(-1-4*n)*(2+s)):
GAMMA(a/5)/(GAMMA(b/5)*GAMMA(1+(a-b)/5)) end:
digits := [1, 4, 29, 205, 1412]: A := n -> round(evalf(a(n),digits[n]+10)):
A(4); # Peter Luschny, Jul 15 2017

Table[Binomial[Fibonacci[2k]Fibonacci[2k+1],Fibonacci[2k-1] Fibonacci[2k]-1], {k,4}] (* Harvey P. Dale, Aug 18 2011 *)

A090162(n)=binomial(fibonacci(2*n+1)*fibonacci(2*n),fibonacci(2*n-1)*fibonacci(2*n)-1) \\ M. F. Hasler, Feb 17 2023

def A090162(n): return C(A000045(2*n+1)*A000045(2*n),A000045(2*n-1)*A000045(2*n)-1) # See A007318 for C(.,.). - M. F. Hasler, Feb 17 2023

a(n) = binomial(A089508(n), A081016(n-1)).
a(n) = binomial(A089508(n)+1, A081016(n-1)-1).
a(n) = Gamma(x)/(Gamma(y)*Gamma(1+x-y)) with x = A206351(n+1) and y = A081016(n-1). - Peter Luschny, Jul 15 2017

A206351 a(n) = 7*a(n-1) - a(n-2) - 4 with a(1)=1, a(2)=3.

Original entry on oeis.org

1, 3, 16, 105, 715, 4896, 33553, 229971, 1576240, 10803705, 74049691, 507544128, 3478759201, 23843770275, 163427632720, 1120149658761, 7677619978603, 52623190191456, 360684711361585, 2472169789339635

Offset: 1

James R. Buddenhagen, Feb 06 2012

A Pell sequence related to Heronian triangles (rational triangles), see A206334. The connection is this: consider the problem of finding triangles with area a positive integer n, and with sides (a, b, n) where a, b are rational. Note that n is both the area and one side. For many values of n this is not possible, and the sequence of such numbers n is quite erratic (see A206334). Nonetheless, each term in this sequence is such a value of n. For example, for n = 105 you can take the other two sides, a and b, to be 10817/104, and 233/104 and the area will equal n, i.e., 105.

			G.f. = x + 3*x^2 + 16*x^3 + 105*x^4 + 715*x^5 + 4896*x^6 + 33553*x^7 + ... - _Michael Somos_, Jun 26 2018

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Pridon Davlianidze, Problem B-1279, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 4 (2020), p. 368; An Unusual Generalization, Solution to Problem B-1279 by J. N. Senadheera, ibid., Vol. 59, No. 4 (2021), pp. 370-371.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A003482, A089508, A094214.

Subsequence of A206334.

a206351 n = a206351_list !! (n-1)
a206351_list = 1 : 3 : map (subtract 4)
               (zipWith (-) (map (* 7) (tail a206351_list)) a206351_list)
-- Reinhard Zumkeller, Feb 08 2012

[Fibonacci(2*n)*Fibonacci(2*n-3): n in [1..30]]; // G. C. Greubel, Aug 12 2018

genZ := proc(n)
local start;
option remember;
    start := [1, 3];
    if n < 3 then start[n]
    else 7*genZ(n - 1) - genZ(n - 2) - 4
    end if
end proc:
seq(genZ(n),n=1..20);

LinearRecurrence[{8, -8, 1}, {1, 3, 16}, 50] (* Charles R Greathouse IV, Feb 07 2012 *)
RecurrenceTable[{a[1] == 1, a[2] == 3, a[n] == 7 a[n - 1] - a[n - 2] - 4}, a, {n, 20}] (* Bruno Berselli, Feb 07 2012 *)
a[ n_] := Fibonacci[2 n] Fibonacci[2 n - 3]; (* Michael Somos, Jun 26 2018 *)
nxt[{a_,b_}]:={b,7b-a-4}; NestList[nxt,{1,3},20][[;;,1]] (* Harvey P. Dale, Aug 29 2024 *)

Vec((1-5*x)/(1-8*x+8*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Feb 07 2012

{a(n) = fibonacci(2*n) * fibonacci(2*n - 3)}; /* Michael Somos, Jun 26 2018 */

From Bruno Berselli, Feb 07 2012: (Start)
G.f.: x*(1-5*x)/(1-8*x+8*x^2-x^3).
a(n) = A081018(n-1) + 1. (End)
a(n) = -A003482(-n) = Fibonacci(2*n)*Fibonacci(2*n-3). - Michael Somos, Jun 26 2018
a(n) = A089508(n-1) + 2 for n>1. - Bruno Berselli, Jun 20 2019 [Formula found by Umberto Cerruti]
Product_{n>=2} (1 - 1/a(n)) = 1/phi (A094214) (Davlianidze, 2020). - Amiram Eldar, Nov 30 2021
a(n) = (Fibonacci(2*n-2) + 1/Lucas(2*n-2))*(Fibonacci(2*n-1) + 1/Lucas(2*n-1)). - Peter Bala, Sep 03 2022

A260259 a(n) = F(n)*F(n+1) - (-1)^n, where F = A000045.

Original entry on oeis.org

-1, 2, 1, 7, 14, 41, 103, 274, 713, 1871, 4894, 12817, 33551, 87842, 229969, 602071, 1576238, 4126649, 10803703, 28284466, 74049689, 193864607, 507544126, 1328767777, 3478759199, 9107509826, 23843770273, 62423800999, 163427632718, 427859097161, 1120149658759

Offset: 0

Bruno Berselli, Oct 31 2015

Primes in sequence for n = 1, 3, 5, 6, 9, 24, 42, 48, 53, 71, 86, 102, 138, 182, 302, 438, 506, 926, ...

Bruno Berselli, Table of n, a(n) for n = 0..500
A. Bremner, R. Høibakk, D. Lukkassen, Crossed ladders and Euler’s quartic, Annales Mathematicae et Informaticae, 36 (2009) pp. 29-41. See p. 33.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

First bisection of A111569.
Cf. A226205: numbers of the form F(n)*F(n+1)+(-1)^n.
Cf. A000045, A001654, A003482, A059929, A089508 (first bisection, without -1), A206351.

[Fibonacci(n)*Fibonacci(n+1)-(-1)^n: n in [0..30]];

with(combinat): A260259:=n->fibonacci(n)*fibonacci(n+1)-(-1)^n: seq(A260259(n), n=0..50); # Wesley Ivan Hurt, Feb 04 2017

Table[Fibonacci[n] Fibonacci[n + 1] - (-1)^n, {n, 0, 30}]

makelist(fib(n)*fib(n+1)-(-1)^n,n,0,30);

for(n=0, 30, print1(fibonacci(n)*fibonacci(n+1)-(-1)^n", "));

a(n) = round((2^(-1-n)*(-3*(-1)^n*2^(2+n)-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/5) \\ Colin Barker, Sep 29 2016

Vec(-(1-4*x+x^2)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 29 2016

[fibonacci(n)*fibonacci(n+1)-(-1)^n for n in (0..30)]

G.f.: (-1 + 4*x - x^2)/((1 + x)*(1 - 3*x + x^2)).
a(n) = -a(-n-1) = 2*a(n-1) + 2*a(n-2) - a(n-3) for all n in Z.
a(n) = F(n+2)^2 - 2*F(n+1)^2.
a(n) = A059929(n) - A059929(n-1) with A059929(-1)=1.
a(n) = -A001654(n+1) + 4*A001654(n) - A001654(n-1).
a(n) = A206351((n+2)/2)-2 for even n; a(n) = A003482((n-1)/2)+2 for odd n.
Sum_{i>=0} 1/a(i) = .754301907697893871765121109686...
a(n) = (2^(-1-n)*(-3*(-1)^n*2^(2+n)-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Sep 29 2016

A354337 a(n) is the integer w such that (L(2n)^2, -L(2n + 1)^2, w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).

Original entry on oeis.org

19, 149, 1039, 7139, 48949, 335519, 2299699, 15762389, 108037039, 740496899, 5075441269, 34787591999, 238437702739, 1634276327189, 11201496587599, 76776199786019, 526231901914549, 3606847113615839, 24721697893396339, 169445038140158549, 1161393569087713519

Offset: 1

XU Pingya, Jun 20 2022

Subsequence of A017377.

			2*(L(4)^2)^3 + 2*(-L(5)^2)^3 + (149)^3 = 2*(49)^3 + 2*(-121)^3 + (149)^3 = 125, a(2) = 149.

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000032, A002878, A005248, A017377, A081017, A089508, A092521, A288913.

Cf. A337929, A354336.

LinearRecurrence[{7,-1},{19,149},21]-1 + LucasL[2*Range[21]-3]^2

a(n) = (125 - 2*A005248(n)^6 + 2*A002878(n)^6)^(1/3).
a(n) = Lucas(4*n+2) + Lucas(4n-1) - 3 = 2*A056914(n)-3 = 15*A092521(n) + A288913(n-1).
a(n) = 2*A081017(n) - 1.
a(n) = 10*A089508(n) + 9.
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: x*(19 - 3*x - x^2)/((1 - x)*(1 - 7*x + x^2)). - Stefano Spezia, Jun 22 2022

A292116 Numbers for which there exists a nontrivial bisection of binomial coefficients as given by Theorem 12 of Ionascu et al. (2016).

Original entry on oeis.org

13, 14, 33, 34, 61, 62, 97, 98, 103, 141, 142, 193, 194, 253, 254, 321, 322, 397, 398, 481, 482, 573, 574, 673, 674, 713, 781, 782, 897, 898, 1021, 1022, 1153, 1154, 1293, 1294, 1441, 1442, 1597, 1598, 1761, 1762, 1933, 1934, 2113, 2114, 2301, 2302, 2497, 2498, 2701, 2702, 2913, 2914, 3133, 3134

Offset: 1

N. J. A. Sloane, Sep 10 2017

nonn

It would be nice to have a more precise definition.
From Ray Chandler, Sep 11 2017: (Start)
The sequence is the union of three types of numbers:
(1) A060626 beginning with the 2nd term.
(2) A089508 beginning with the 3rd term and omitting even values (every third term).
(3) A082109 beginning with the 2nd term.
Note that there appear to be other solutions that are not covered by Theorem 12.
(End)

Ray Chandler, Table of n, a(n) for n = 1..10000
Eugen J. Ionascu, Thor Martinsen, and Pantelimon Stanica, Bisecting binomial coefficients, arXiv:1610.02063 [math.CO], 2016. See p. 18.

Cf. A060626, A082109, A089508.

lim=3000; a0={};
k=4; While[c=k^2-3;c<=lim,a0=Join[a0,{c,c+1}];k+=2];
k=2; While[c=Fibonacci[2k]*Fibonacci[2k+1]-1;c<=lim,If[OddQ[c],AppendTo[a0,c]];k++];
a0=Sort[a0] (* Ray Chandler, Sep 11 2017 *)

cp's OEIS Frontend

A081016 a(n) = (Lucas(4*n+3) + 1)/5, or Fibonacci(2*n+1)*Fibonacci(2*n+2), or A081015(n)/5.

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A081018 a(n) = (Lucas(4n+1)-1)/5, or Fibonacci(2n)*Fibonacci(2n+1), or A081017(n)/5.

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A090162 Values of binomial(Fibonacci(2k)*Fibonacci(2k+1),Fibonacci(2k-1)*Fibonacci(2k)-1).

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A206351 a(n) = 7*a(n-1) - a(n-2) - 4 with a(1)=1, a(2)=3.

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A260259 a(n) = F(n)*F(n+1) - (-1)^n, where F = A000045.

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A081016 a(n) = (Lucas(4n+3) + 1)/5, or Fibonacci(2n+1)Fibonacci(2n+2), or A081015(n)/5.

A090162 Values of binomial(Fibonacci(2k)Fibonacci(2k+1),Fibonacci(2k-1)Fibonacci(2k)-1).

A354337 a(n) is the integer w such that (L(2n)^2, -L(2n + 1)^2, w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).