A007805 -id:A007805 - cp's OEIS frontend

A305316 a(n) = sqrt(5b(n)^2 - 4) with b(n) = Fibonacci(6n+5) = A134497(n).

11, 199, 3571, 64079, 1149851, 20633239, 370248451, 6643838879, 119218851371, 2139295485799, 38388099893011, 688846502588399, 12360848946698171, 221806434537978679, 3980154972736918051, 71420983074726546239, 1281597540372340914251, 22997334743627409910279, 412670427844921037470771, 7405070366464951264563599, 132878596168524201724674011

Offset: 0

Wolfdieter Lang, Jul 10 2018

This sequence gives all solutions of one of two classes of positive proper solutions a(n) = a5(n) of the Pell equation a(n)^2 - 5*b(n) = -4 with b(n) = b5(n) = F(6*n+5) = A134497(n), with the Fibonacci numbers F = A000045. These solutions are obtained from the fundamental positive solution [11, 5] (to be read as a column vector) by application of positive powers of the automorphic matrix A = matrix([9, 20], [4, 9]) of determinant +1.
The other class of positive proper solutions is obtained similarly from the fundamental solution [1,1] and is given by [a1(n), b1(n)], with a1(n) = A305315(n) and b1(n) = A134493(n) = F(6*n+1).
The remaining positive solutions are improper and are obtained by application of positive powers of the same matrix A on the fundamental improper solution [4, 2]. They are given by [a3(n), b3(n)], with a3(n) = F(6*n+3) = 4* A049629(n) and b3(n) = A134495(n) = 2*A007805(n).
For the explicit form of powers of the automorphic matrix A in terms of Chebyshev polynomials S(n, 18) see a comment in A305315.
The relation to a proof using this Pell equation of the well known fact that each odd-indexed Fibonacci number appears as largest member in Markoff (Markov) triples with smallest member 1 see also A305315.

			See A305315 for the three classes of solutions of this Pell equation

Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A000045, A007805, A049629, A049660, A134493, A134495, A134497, A305315.

I:=[11, 199]; [n le 2 select I[n] else 18*Self(n-1)-Self(n-2): n in [1..25]]; // Vincenzo Librandi, Jul 22 2018

f[n_] := Sqrt[5 Fibonacci[6 n + 5]^2 - 4]; Array[f, 17, 0] (* or *)
CoefficientList[ Series[(x + 11)/(x^2 - 18x + 1), {x, 0, 18}], x] (* or *)
LinearRecurrence[{18, -1}, {11, 199}, 18] (* Robert G. Wilson v, Jul 21 2018 *)

x='x+O('x^99); Vec((11+x)/(1-18*x+x^2)) \\ Altug Alkan, Jul 11 2018

a(n) = sqrt(5*(F(6*n+5))^2 - 4), with F(6*n+5) = A134497(n), n >= 0.
a(n) = 11*S(n, 18) + S(n-1, 18), n >= 0, with the Chebyshev polynomials S(n, 18) = A049660(n+1) and S(-1, 18) = 0.
a(n) = 18*a(n-1) - a(n-2), n >= 1, with a(-1) = -1 and a(0) = 11.
G.f.: (11 + x)/(1 - 18*x + x^2).

A307937 Numbers that can be written as the sum of four or more consecutive squares in more than one way.

Original entry on oeis.org

3655, 3740, 4510, 4760, 5244, 5434, 5915, 7230, 7574, 8415, 11055, 11900, 12524, 14905, 17484, 18879, 19005, 19855, 20449, 20510, 21790, 22806, 23681, 25580, 25585, 27230, 27420, 28985, 31395, 34224, 37114, 39606, 41685, 42419, 44919, 45435, 45955, 48026, 48139, 48225, 49015, 53941, 57164, 62006

Offset: 1

Robert Israel, May 06 2019

nonn

Numbers that are in A174071 in two or more ways.
The first number with more than two representations as a sum of four or more consecutive positive squares is 147441 = 18^2 + ... + 76^2 = 29^2 + ... + 77^2 = 85^2 + ... + 101^2.
If x = 2*A049629(n) and y = A007805(n) for n >= 1 (satisfying the Pell equation x^2 - 5*y^2 = -1), then the sequence contains 5*x^2+10 = Sum_{(5*y-3)/2 <= i <= (5*y+3)/2} i^2 = Sum_{x-2 <= i <= x+2} i^2 = 25*y^2 + 5.

			a(1) = 3655 is in the sequence because 3655 = 8^2 + ... + 22^2 = 25^2 + ... + 29^2.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A007805, A049629, A174071.

N:= 10^5: # to get all terms <= N
R:= 'R':
dups:= NULL:
for m from 4 while m*(m+1)*(2*m+1)/6 <= N do
   for k from 1 do
       v:= m*(6*k^2 + 6*k*m + 2*m^2 - 6*k - 3*m + 1)/6;
       if v > N then break fi;
       if assigned(R[v]) then
         dups:= dups, v;
       else
         R[v]:= [k, k+m-1];
       fi;
od od:
sort(convert({dups},list));

M = 10^5;
dups = {}; Clear[rQ]; rQ[_] = False;
For[m = 4, m(m+1)(2m+1)/6 <= M, m++, For[k = 1, True, k++, v = m(6k^2 + 6k m + 2m^2 - 6k - 3m + 1)/6; If[v > M, Break[]]; If[rQ[v], AppendTo[dups, v], rQ[v] = True]]];
dups // Sort (* Jean-François Alcover, May 07 2019, after Robert Israel *)

A324178 Integers k such that floor(sqrt(k)) + floor(sqrt(k/5)) divides k.

Original entry on oeis.org

1, 2, 3, 4, 6, 12, 24, 28, 35, 40, 45, 50, 60, 66, 77, 91, 112, 128, 153, 190, 200, 220, 231, 276, 312, 338, 378, 406, 435, 450, 480, 496, 512, 561, 578, 648, 703, 722, 741, 780, 800, 840, 882, 903, 946, 968, 990, 1058, 1152, 1176, 1250, 1300, 1352, 1378

Offset: 1

Jinyuan Wang, Mar 09 2019

nonn

This sequence is infinite for the same reason that A324175 is: if x > y satisfies x^2 - 5*y^2 = -1 (x=A075796(j), y=A007805(j-1), j>0), then x < 5*y. Let k = 5*y^2 + m. By the pigeonhole principle there exists a number m belonging to [0, 2*x - 1] such that x + y | 5*y^2 + m, so such a k is a term.

Cf. A007805, A075796, A324174, A324175, A324176, A324177.

Select[Range[1378], Mod[#, Floor@ Sqrt@ # + Floor@ Sqrt[#/5]] == 0 &] (* Giovanni Resta, Apr 05 2019 *)

is(n) = n%(floor(sqrt(n)) + floor(sqrt(n/5))) == 0;

A075869 Numbers k such that 5*k^2 - 9 is a square.

Original entry on oeis.org

3, 51, 915, 16419, 294627, 5286867, 94868979, 1702354755, 30547516611, 548152944243, 9836205479763, 176503545691491, 3167227616967075, 56833593559715859, 1019837456457918387, 18300240622682815107

Offset: 1

Gregory V. Richardson, Oct 16 2002

Lim. n-> Inf. a(n)/a(n-1) = phi^6 = 9 + 4*sqrt(5).

A. H. Beiler, "The Pellian", ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

Tanya Khovanova, Recursive Sequences
J. J. O'Connor and E. F. Robertson, Pell's Equation
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. 3*A007805.

LinearRecurrence[{18, -1}, {3, 51}, 20] (* Harvey P. Dale, Dec 27 2018 *)

a(n) = 3*sqrt(5)/10*((2+sqrt(5))^(2*n-1)-(2-sqrt(5))^(2*n-1)) = 18*a(n-1) - a(n-2).

G.f.: 3*x*(1-x)/(1-18*x+x^2). [Philippe Deléham, Nov 17 2008; corrected by Georg Fischer, May 15 2019]

A269028 a(n) = 40*a(n - 1) - a(n - 2) for n>1, a(0) = 1, a(1) = 1.

Original entry on oeis.org

1, 1, 39, 1559, 62321, 2491281, 99588919, 3981065479, 159143030241, 6361740144161, 254310462736199, 10166056769303799, 406387960309415761, 16245352355607326641, 649407706263983649879, 25960062898203738668519, 1037753108221885563090881

Offset: 0

Ilya Gutkovskiy, Feb 18 2016

In general, the ordinary generating function for the recurrence relation b(n) = k*b(n - 1) - b(n - 2) with n>1 and b(0)=1, b(1)=1, is (1 - (k - 1)*x)/(1 - k*x +x^2). This recurrence gives the closed form b(n) = (2^( -n - 1)*((k - 2)*(k - sqrt(k^2 - 4))^n + sqrt(k^2 - 4)*(k - sqrt(k^2 - 4))^n - (k - 2)*(sqrt(k^2 - 4) + k)^n + sqrt(k^2 - 4)*(sqrt(k^2 - 4) + k)^n))/sqrt(k^2 - 4).

Index entries for linear recurrences with constant coefficients, signature (40,-1).

Cf. A001519, A001835, A001653, A049685, A070997, A070998, A072256, A078922, A160682, A007805, A075839, A157014, A159664, A159668, A157877, A238379, A097315.

[n le 2 select 1 else 40*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 19 2016

Table[Cosh[n Log[20 + Sqrt[399]]] - Sqrt[19/21] Sinh[n Log[20 + Sqrt[399]]], {n, 0, 17}]
Table[(2^(-n - 2) (38 (40 - 2 Sqrt[399])^n + 2 Sqrt[399] (40 - 2 Sqrt[399])^n - 38 (40 + 2 Sqrt[399])^n + 2 Sqrt[399] (40 + 2 Sqrt[399])^n))/Sqrt[399], {n, 0, 17}]
LinearRecurrence[{40, -1}, {1, 1}, 17]

G.f.: (1 - 39*x)/(1 - 40*x + x^2).
a(n) = cosh(n*log(20 + sqrt(399))) - sqrt(19/21)*sinh(n*log(20 + sqrt(399))).
a(n) = (2^(-n - 2)*(38*(40 - 2*sqrt(399))^n + 2*sqrt(399)*(40 - 2*sqrt(399))^n - 38*(40 + 2*sqrt(399))^n + 2*sqrt(399)*(40 + 2*sqrt(399))^n))/sqrt(399).
Sum_{n>=0} 1/a(n) = 2.0262989201139499769986...

A278475 a(n) = floor(phi^7*a(n-1)) for n>0, a(0) = 1, where phi is the golden ratio (A001622).

Original entry on oeis.org

1, 29, 841, 24417, 708933, 20583473, 597629649, 17351843293, 503801085145, 14627583312497, 424703717147557, 12331035380591649, 358024729754305377, 10395048198255447581, 301814422479162285225, 8763013300093961719105, 254429200125204052139269, 7387209816931011473757905

Offset: 0

Ilya Gutkovskiy, Nov 23 2016

In general, the ordinary generating function for the recurrence relation b(n) = floor(phi^k*b(n - 1)) with n>0 and b(0) = 1, is (1 - x)/(1 - (phi^k + (-phi)^(-k))*x + x^2) if k is even, and (1 - x - x^2)/((1 - x)*(1 - (phi^k + (-phi)^(-k))*x - x^2)) if k is odd.

Index entries for linear recurrences with constant coefficients, signature (30,-28,-1).

Cf. A001622.

Cf. similar sequences with recurrence relation b(n) = floor(phi^k*b(n-1)) for n>0, b(0) = 1: A000012 (k = 1), A001519 (k = 2), A024551 (k = 3), A049685 (k = 4), A214993 (k = 5), A007805 (k = 6), this sequence (k = 7).

RecurrenceTable[{a[0] == 1, a[n] == Floor[GoldenRatio^7 a[n - 1]]}, a, {n, 17}]
LinearRecurrence[{30, -28, -1}, {1, 29, 841}, 18]

Vec( (1 - x - x^2)/((1 - x)*(1 - 29*x - x^2)) + O(x^50) ) \\ G. C. Greubel, Nov 24 2016

G.f.: (1 - x - x^2)/((1 - x)*(1 - 29*x - x^2)).
a(n) = 30*a(n-1) - 28*a(n-2) - a(n-3).
a(n) = ((-29 - 13*sqrt(5))^(-n)*(-7*(407 + 182*sqrt(5))*2^(n+3) + 13*(1885 + 843*sqrt(5))*(-29 - 13*sqrt(5))^n + 28*(25319 + 11323*sqrt(5))*(-843 - 377*sqrt(5))^n))/(377*(1885 + 843*sqrt(5))).

cp's OEIS Frontend

A305316 a(n) = sqrt(5*b(n)^2 - 4) with b(n) = Fibonacci(6*n+5) = A134497(n).

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A307937 Numbers that can be written as the sum of four or more consecutive squares in more than one way.

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A324178 Integers k such that floor(sqrt(k)) + floor(sqrt(k/5)) divides k.

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A075869 Numbers k such that 5*k^2 - 9 is a square.

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A269028 a(n) = 40*a(n - 1) - a(n - 2) for n>1, a(0) = 1, a(1) = 1.

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Magma

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A278475 a(n) = floor(phi^7*a(n-1)) for n>0, a(0) = 1, where phi is the golden ratio (A001622).

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Formula

A305316 a(n) = sqrt(5b(n)^2 - 4) with b(n) = Fibonacci(6n+5) = A134497(n).