A005319 -id:A005319 - cp's OEIS frontend

A106328 Numbers j such that 8*(j^2) + 9 = k^2 for some positive number k.

0, 3, 18, 105, 612, 3567, 20790, 121173, 706248, 4116315, 23991642, 139833537, 815009580, 4750223943, 27686334078, 161367780525, 940520349072, 5481754313907, 31950005534370, 186218278892313, 1085359667819508, 6325939728024735, 36870278700328902, 214895732473948677

Offset: 1

Pierre CAMI, Apr 29 2005

The ratio k(n) /(2*j(n)) tends to sqrt(2) as n increases.
The squares of the numbers in this sequence are one less than a triangular number: a(n)^2 = A164080(n). For example, 18^2 is 324, and 325 is a triangular number. a(n)^2 + 1 = A164055(n). a(n)^2 = A072221(n)(A072221(n)+1)/2 - 1. - Tanya Khovanova & Alexey Radul, Aug 09 2009
For n > 0, a(n+1) is the n-th almost balancing number of first type (see Tekcan and Erdem). - Stefano Spezia, Nov 25 2022

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Tanya Khovanova, Recursive Sequences
Soumeya M. Tebtoub, Hacène Belbachir, and László Németh, Integer sequences and ellipse chains inside a hyperbola, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A001109, A005319, A072221, A103328, A164080, A164055.

a106328 n = a106328_list !! (n-1)
a106328_list = 0 : 3 : zipWith (-) (map (* 6) (tail a106328_list)) a106328_list
-- Reinhard Zumkeller, Jan 10 2012

s=0;lst={};Do[s+=n;If[Sqrt[s-1]==Floor[Sqrt[s-1]],AppendTo[lst,Sqrt[s-1]]],{n,8!}];lst (* Vladimir Joseph Stephan Orlovsky, Apr 02 2009 *)
Rest@ CoefficientList[Series[3 x^2/(1 - 6 x + x^2), {x, 0, 24}], x] (* Michael De Vlieger, Nov 02 2020 *)

concat(0, Vec(3*x^2/(1-6*x+x^2) + O(x^40))) \\ Michel Marcus, Sep 07 2016

a(n)=([0,1;-1,6]^n*[-3;0])[1,1] \\ Charles R Greathouse IV, Sep 07 2016

a(1)=0, a(2)=3 then a(n) = 6*a(n-1) - a(n-2).
a(n) = ((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1))*3/4/sqrt(2). - Max Alekseyev, Jan 11 2007
a(n) = 3*A001109(n). - M. F. Hasler, R. J. Mathar, Jun 03 2009
a(n) = (3/4)*A005319(n-1).
G.f.: 3*x^2/(1 - 6*x + x^2). - Philippe Deléham, Nov 17 2008
E.g.f.: 3 - 3*exp(3*x)*(4*cosh(2*sqrt(2)*x) - 3*sqrt(2)*sinh(2*sqrt(2)*x))/4. - Stefano Spezia, Nov 25 2022

More terms from Max Alekseyev, Jan 11 2007

A182190 a(n) = 6*a(n-1) - a(n-2) + 4 with n > 1, a(0)=0, a(1)=4.

Original entry on oeis.org

0, 4, 28, 168, 984, 5740, 33460, 195024, 1136688, 6625108, 38613964, 225058680, 1311738120, 7645370044, 44560482148, 259717522848, 1513744654944, 8822750406820, 51422757785980, 299713796309064

Offset: 0

Kenneth J Ramsey, Apr 17 2012

Also, nonnegative m such that 2m(m+2)+1 is a square. - Bruno Berselli, Oct 22 2012

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A000129, A182189, A182431.

I:=[0,4]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2)+4: n in [1..20]]; // Bruno Berselli, Jun 07 2012

m = 4;n = 0; c = 0;
list3 = Reap[While[c < 22, t = 6 n - m + 4; Sow[t];m = n; n = t;c++]][[2,1]]
Table[Fibonacci[2*n+1, 2] -1, {n,0,40}] (* G. C. Greubel, May 24 2021 *)

[lucas_number1(2*n+1,2,-1) -1 for n in (0..40)] # G. C. Greubel, May 24 2021

G.f.: 4*x/((1-x)*(1-6*x+x^2)). - Bruno Berselli, Jun 07 2012
a(n) = 4*A053142(n). - Bruno Berselli, Jun 07 2012
a(n) = A001653(n+1) - 1. - Kiran S. Kedlaya, Mar 14 2021
a(n) = A000129(2*n+1) - 1. - G. C. Greubel, May 24 2021
a(n) = A143608(n)*A143608(n+1). - R. J. Mathar, Jan 31 2024
a(n)-a(n-1) = A005319(n). - R. J. Mathar, Jan 31 2024

A046727 Related to Pythagorean triples: alternate terms of A001652 and A046090.

Original entry on oeis.org

0, 3, 21, 119, 697, 4059, 23661, 137903, 803761, 4684659, 27304197, 159140519, 927538921, 5406093003, 31509019101, 183648021599, 1070379110497, 6238626641379, 36361380737781, 211929657785303, 1235216565974041, 7199369738058939, 41961001862379597, 244566641436218639

Offset: 0

N. J. A. Sloane

For a triple (a,b,c) there exist k,m such that (a,b,c) = (k^2 - m^2, 2*k*m, k^2 + m^2). Here k = A001333(n) and m = A001333(n+1), so this sequence is identical to the Pell oblongs A084159 for n > 0. - Lambert Klasen (Lambert.Klasen(AT)gmx.de), Nov 10 2004

a(n), for n >= 1, gives the odd length (in some unit) catheti (legs) of the (primitive) Pythagorean triples which have absolute length difference of the catheti equal to one. See a W. Lang comment on A001653 on how to generate all such Pythagorean triples. - Wolfdieter Lang, Mar 08 2012

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
A. Bogomolny, The Trinary Tree(s) underlying Primitive Pythagorean Triples.
Dan Romik, The dynamics of Pythagorean Triples, Trans. Amer. Math. Soc. 360 (2008), 6045-6064.
P. E. Trier, "Almost Isosceles" Right-Angled Triangles, Eureka, No. 4, May 1940, pp. 9 - 11.
Index entries for linear recurrences with constant coefficients, signature (5,5,-1).

Cf. A001109, A001333, A001652, A005319, A046090, A046729.

Essentially the same as A084159.

a046727 n = a046727_list !! n
a046727_list = 0 : f (tail a001652_list) (tail a046090_list) where
   f (x::xs) (:y:ys) = x : y : f xs ys
-- Reinhard Zumkeller, Jan 10 2012

I:=[0,3,21,119]; [n le 4 select I[n] else 5*Self(n-1)+5*Self(n-2)-Self(n-3): n in [1..30]]; // Vincenzo Librandi, Nov 04 2016

RecurrenceTable[{a[n+2]==6a[n+1] -a[n] -4*(-1)^n, a[0]==3, a[1]==21}, a, {n, 30}] (* Ron Knott, Jul 01 2013 *)
LinearRecurrence[{5,5,-1}, {0,3,21,119}, 30] (* Vincenzo Librandi, Nov 04 2016 *)

concat(0, Vec(x*(3+6*x-x^2)/((1+x)*(1-6*x+x^2)) + O(x^30))) \\ Colin Barker, Nov 03 2016

[(lucas_number2(2*n+1,2,-1) +2*(-1)^n)/4 -int(n==0) for n in range(41)] # G. C. Greubel, Feb 11 2023

Values of x obtained by repeatedly multiplying the triple (x, y, z) = (3, 4, 5) by the matrix A = ([1 2 2], [2 1 2], [2 2 3]), the Across matrix of "The Trinary Tree(s) underlying Primitive Pythagorean Triples" generating matrices. - Vim Wenders, Jan 14 2004
For n > 0, a(n) = A001333(n)*A001333(n+1). - Lambert Klasen (Lambert.Klasen(AT)gmx.de), Nov 10 2004
G.f.: x*(3+6*x-x^2)/((1+x)*(1-6*x+x^2)). - R. J. Mathar, Jul 08 2009
a(n) + a(n+1) = A005319(n+1), n > 0. - R. J. Mathar, Jul 13 2009
a(n) = 6*a(n-1) - a(n-2) - 4*(-1)^n. - Ron Knott, Jul 01 2013
From Colin Barker, Nov 03 2016: (Start)
a(n) = (2*(-1)^n + (1+sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1))/4 for n > 0.
a(n) = 5*a(n-1) + 5*a(n-2) - a(n-3) for n > 3. (End)
From G. C. Greubel, Feb 11 2023: (Start)
a(n) = (1/2)*(A001109(n+1) + A001109(n) + (-1)^n) - [n=0].
a(n) = (A001333(2*n+1) + (-1)^n)/2 - [n=0]. (End)
E.g.f.: exp(-x)*(1 + exp(4*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)))/2 - 1. - Stefano Spezia, Aug 03 2024

More terms from Sascha Kurz, Jan 23 2003

A065113 Sum of the squares of the a(n)-th and the (a(n)+1)st triangular numbers (A000217) is a perfect square.

Original entry on oeis.org

6, 40, 238, 1392, 8118, 47320, 275806, 1607520, 9369318, 54608392, 318281038, 1855077840, 10812186006, 63018038200, 367296043198, 2140758220992, 12477253282758, 72722761475560, 423859315570606, 2470433131948080, 14398739476117878, 83922003724759192

Offset: 1

Robert G. Wilson v, Nov 12 2001

The sequence of square roots of the sum of the squares of the n-th and the (n+1)st triangular numbers is A046176.

			T6 = 21 and T7 = 28, 21^2 + 28^2 = 441 + 784 = 1225 = 35^2.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
M. Ulas, On certain diophantine equations related to triangular and tetrahedral numbers, arXiv:0811.2477 [math.NT] (2008)
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A001652, A002315, A003499 (first differences), A065651.

CoefficientList[ Series[2*(x - 3)/(-1 + 7x - 7x^2 + x^3), {x, 0, 24} ], x]
LinearRecurrence[{7,-7,1},{6,40,238},41] (* Harvey P. Dale, Dec 27 2011 *)

a(n)=-1+subst(poltchebi(abs(n+1))-poltchebi(abs(n)),x,3)/2

Vec(2*x*(3-x)/((1-6*x+x^2)*(1-x)) + O(x^40)) \\ Colin Barker, Mar 05 2016

a(n) = 2*A001652(n) = -1 + A002315(n).
a(n) - a(n-1) = A003499(n).
From Michael Somos, Apr 07 2003: (Start)
G.f.: 2*x*(3-x)/((1-6*x+x^2)*(1-x)).
a(n) = 6*a(n-1) - a(n-2) + 4.
a(-1-n) = -a(n) - 2. (End)
a(1)=6, a(2)=40, a(3)=238, a(n) = 7*a(n-1)-7*a(n-2)+a(n-3). - Harvey P. Dale, Dec 27 2011
a(n)^2 + (a(n)+2)^2 = A075870(n+1)^2 = A165518(n+1). - Joerg Arndt, Feb 15 2012
a(n) = (-2-(3-2*sqrt(2))^n*(-1+sqrt(2))+(1+sqrt(2))*(3+2*sqrt(2))^n)/2. - Colin Barker, Mar 05 2016
From Klaus Purath, Sep 05 2021: (Start)
(a(n+1) - a(n) - a(n-1) + a(n-2))/8 = A005319(n), for n >= 3.
((a(n) - a(n-1))^2)/2 - 2 = A005319(n)^2 = 2*A132592(n), for n>= 2.
a(n) = A265278(2*n+1).
a(n) = A293004(2*n+1).
a(n) = A213667(2*n).
a(n) = Sum_{k=1..n} A003499(k). (End)

A082639 Numbers k such that 2k(k+2) is a square.

Original entry on oeis.org

0, 2, 16, 98, 576, 3362, 19600, 114242, 665856, 3880898, 22619536, 131836322, 768398400, 4478554082, 26102926096, 152139002498, 886731088896, 5168247530882, 30122754096400, 175568277047522, 1023286908188736

Offset: 1

James R. Buddenhagen, May 15 2003

Even-indexed terms are squares. Their square roots form sequence A005319. Odd-indexed terms divided by 2 are squares. Their square roots form the sequence A002315. (Index starts at 0.)

Lower of two terms with difference 2 in A088827. - Ophir Spector, Nov 06 2024

Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A001541, A002315, A005319, A088827.

a[0] = 0; a[1] = 2; a[n_] := a[n] = 6a[n - 1] - a[n - 2] + 4; Table[ a[n], {n, 0, 20}]
LinearRecurrence[{7,-7,1},{0,2,16},30] (* Harvey P. Dale, Nov 21 2015 *)

a(n) = A001541(n) - 1.
a(n) = (1/2)*(s^n + t^n) - 1, where s = 3 + 2*sqrt(2), t = 3 - 2*sqrt(2). Note: s=1/t. a(n) = 6*a(n-1) - a(n-2) + 4, a(0)=0, a(1)=2.
a(n) = 1/kappa(sqrt(2)/A001542(n)); a(n) = 1/kappa(sqrt(8)/A005319(n)) where kappa(x) is the sum of successive remainders by computing the Euclidean algorithm for (1, x). - Thomas Baruchel, Nov 29 2003
G.f.: -2*x^2*(x+1)/((x-1)*(x^2-6*x+1)). - Colin Barker, Nov 22 2012

More terms from Robert G. Wilson v, May 15 2003

A082981 Start with the sequence S(0)={1,1} and for k>0 define S(k) to be I(S(k-1)) where I denotes the operation of inserting, for i=1,2,3..., the term a(i)+a(i+1) between any two terms for which 4a(i+1)<=5a(i). The listed terms are the initial terms of the limit of this process as k goes to infinity.

Original entry on oeis.org

1, 2, 3, 4, 9, 14, 19, 24, 53, 82, 111, 140, 309, 478, 647, 816, 1801, 2786, 3771, 4756, 10497, 16238, 21979, 27720, 61181, 94642, 128103, 161564, 356589, 551614, 746639, 941664, 2078353, 3215042, 4351731, 5488420, 12113529, 18738638, 25363747

Offset: 1

John W. Layman, May 28 2003

nonn

Conjectures:
(1) the section {a(2n+1)}={1,3,9,19,53,111,...} is A077442, the terms of which are solutions of ax^2+7 = a square,
(2) the section {a(4n+1)}={1,9,53,309,1801,...} is A038761,
(3) the section {a(4n+2)}={2,14,82,478,2786,...} is A077444, the terms of which are solutions of 2x^2+8 = a square,
(4) the sequence {a(4n+2)/2}={1,7,41,239,1393,...} is A002315, the terms of which are solutions of 2x^2+2 = a square,
(5) the section {a(4n+4)}={4,24,140,816,4756,...} is A005319, the terms of which are solutions of 2x^2+4=a square,
(6) the sequence {a(4n+4)/4}={1,6,35,204,1189,...} is A001109, the terms of which are solutions of 8x^2+1=a square.

Ivan Neretin, Table of n, a(n) for n = 1..1000
John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types, June 2003 [Broken link]
John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types, June 2003 [local copy, corrected]
John W. Layman, Sequences Generated by Age-Determined Insertion Trees, Jan 2006
John W. Layman, Sequences Generated by Age-Determined Insertion Trees, Jan 2006 [Local copy]

Cf. A001109, A002315, A005319, A038761, A077442, A077444.

Most@Nest[If[#[[-2]] >= 4 #[[-1]], Append[Most@#, #[[-1]] + #[[-2]]], Insert[#, #[[-1]] + #[[-2]], -2]] &, {1, 1}, 47] (* Ivan Neretin, Apr 27 2017 *)

It appears that a(n)=6a(n-4)-a(n-8).

Empirical g.f.: x*(x+1)^2*(x^2+1)^2/((x^4-2*x^2-1)*(x^4+2*x^2-1)). - Colin Barker, Nov 06 2014

A089499 a(0)=0; a(1)=1; a(2n) = 4*Sum_{k=0..n} a(2k-1); a(2n+1) = a(2n) + a(2n-1).

Original entry on oeis.org

0, 1, 4, 5, 24, 29, 140, 169, 816, 985, 4756, 5741, 27720, 33461, 161564, 195025, 941664, 1136689, 5488420, 6625109, 31988856, 38613965, 186444716, 225058681, 1086679440, 1311738121, 6333631924, 7645370045, 36915112104, 44560482149

Offset: 0

Charlie Marion, Nov 11 2003

1, 4, 5, 24, 29, 140, ...= numerators in convergents to (sqrt(8) - 2) = continued fraction [0; 1, 4, 1, 4, 1, 4, ...]; where sqrt(8) - 2 = 0.828427124... = the inradius of a right triangle with hypotenuse 6, legs sqrt(32) and 2. Denominators of convergents to [0; 1, 4, 1, 4, 1, 4, ...] = A041011 starting (1, 5, 6, 29, 35, ...). - Gary W. Adamson, Dec 22 2007

This is a strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n,m)) for all natural numbers n and m. - Peter Bala, May 12 2014

J. L. Ramirez, F. Sirvent, A q-Analogue of the Bi-Periodic Fibonacci Sequence, J. Int. Seq. 19 (2016) # 16.4.6, t_n at a=4, b=1.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A041011.

Numerator[NestList[(4/(4+#))&,0,60]] (* Vladimir Joseph Stephan Orlovsky, Apr 13 2010 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^n*[0;1;4;5])[1,1] \\ Charles R Greathouse IV, Nov 13 2015

For n > 0, a(n) = A001333(n) + A084068(n-1)*(-1)^n.
a(n)*a(n+1) = A046729(n).
a(2n+1) = A001653(n); a(2n) = A005319(n).
a(1) = 1, a(2n) = 4*a(2n-1) + a(2n-2); a(2n-1) = a(2n-2) + a(2n-3). Given the 2 X 2 matrix X = [1, 4; 1, 5], [a(2n-1), a(2n)] = top row of X^n. The sequence starting (1, 4, 5, 24, 29, ...) = numerators in continued fraction [0; 1, 4, 1, 4, 1, 4, ...] = (sqrt(8) - 2) = 0.828427124... E.g., X^3 = [29, 140; 35, 169], where 29/35, 140/169 are convergents to (sqrt(8)-2). - Gary W. Adamson, Dec 22 2007
From R. J. Mathar, Jul 08 2009: (Start)
a(n) = A000129(n)*A000034(n+1).
a(n) = 6*a(n-2) - a(n-4).
G.f.: -x*(-1-4*x+x^2)/((x^2-2*x-1)*(x^2+2*x-1)). (End)
From Peter Bala, May 12 2014: (Start)
a(2*n + 1) = A041011(2*n + 1); a(2*n) = 4*A041011(2*n).
For n odd, a(n) = (alpha^n - beta^n)/(alpha - beta), and for n even, a(n) = 4*(alpha^n - beta^n)/(alpha^2 - beta^2), where alpha = 1 + sqrt(2) and beta = 1 - sqrt(2).
a(n) = Product_{j = 1..floor(n/2)} ( 4 + 4*cos^2(j*Pi/n) ) for n >= 1. (End)

Corrected by T. D. Noe, Nov 08 2006

Definition corrected by Jonathan Sondow, Jun 06 2014

A227972 Two column recursive array A(n,k), relating expressions based on half-squares (A007590) to each other and several other sequences, read by rows.

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 3, 4, 5, 7, 7, 10, 17, 24, 29, 41, 41, 58, 99, 140, 169, 239, 239, 338, 577, 816, 985, 1393, 1393, 1970, 3363, 4756, 5741, 8119, 8119, 11482, 19601, 27720, 33461, 47321, 47321, 66922, 114243, 161564, 195025, 275807, 275807, 390050, 665857, 941664, 1136689, 1607521

Offset: 1

Richard R. Forberg, Aug 01 2013

nonn

The first column (k=1) holds the interleaved integer square roots of these two "Half-Square" expressions in ascending order:  floor(m^2/2 + 1) for m=>0 and floor(m^2/2 - 1) for m=>1.  The second column (k=2) holds the value of m that yields the corresponding integer square root.
The value of m for row n (at n mod 3 = 2) is the value of the square root for the next row (at n mod 3 = 0) which uses the other expression.
There are twice as many results for the expression floor(m^2/2 + 1) as for floor(m^2/2 - 1), interleaved consistently as two of every three results (as shown in the example below).
The first column, for n mod 3 = 1, produces A001541.
The first column, for n mod 3 = 2, produces A001653.
  NOTE: Interleaving of the two sequences above is A079496.
The first column, for n mod 3 = 0, produces A002315 (NSW Numbers).
  NOTE: Interleaving of A001541 and A002315 is A001333.
The second column, for n mod 3 = 1, produces A005319.
The second column, for n mod 3 = 2, produces A002315 (again).
  NOTE: Interleaving of the two sequences above is A143608.
The second column, for n mod 3 = 0, produces A075870.
  NOTE: Interleaving of A005319 and A075870 is A052542 = 2*A000129 (Pell)
The row sums at n mod 3 = 1 and n mod 3 = 0 are used in the recursion to produce values in subsequent rows of the array for both columns.
For rows at n mod 3 = 2,  the ascending interleaved combination of A(n,1) and the row sum (of the same row) produces A000129 (Pell Numbers).
Row sums also hold all the integer square roots (as given in A001542) of the Half-Squares, (A007590), at n mod 3 = 2, and the corresponding values of m in the next row at n mod 3 = 0, corresponding to A001541.
The value of the floor of half the row sum, for n mod 3 =0 and n mod 3 = 1, produces A048739, giving the partial sums of A000129 (Pell Numbers), for the Pell Numbers produced through the prior row at  n mod 3 = 2.
The value of half the row sum, for n mod 3 = 2, produces A001109 (without its initial 0).  This subsequence is also produced from finding the integer square roots of A083374. The value of the indices of that sequence where these roots occur is given by A002315 (NSW Numbers).
The differences of two entries in row n equals the row sum for row n-3, consistently for all rows n > 3.
The ratio of the two entries in the same row converges to sqrt(2).
The ratio of two entries in the same column (either k=1 or k=2) converge as follows:
      A(k,n)/A(k,n-1)--> sqrt(2)       for n mod 3 = 0,
                     --> sqrt(2) + 1   for n mod 3 = 1,
                     --> sqrt(2)/2 + 1 for n mod 3 = 2.
      A(k,n)/A(k,n-3)--> sqrt(8) + 3   for n mod 3 = 0, 1, or 2,
  That last line means: A001541, A001653, A002315, A005319 and A075870 all have the convergence ratio of sqrt(8) + 3 for adjacent terms. In addition alternating Pell Numbers also converge to that ratio.

			The two column array with row number n and the row sum. An extra column on the right shows which expression is applicable to get that row's values: either floor(m^2/2 + 1) indicated as "+1",  or floor(m^2/2 - 1) indicated as "-1". (NOTE: The value of n is immaterial, except as a row number).
The array begins:
Row         k=1         k=2                   Applicable "Half-Square"
n          (sqrt)       (m)         Row Sum        Expression
1            1           0               1             +1
2            1           1               2             +1
3            1           2               3             -1
4            3           4               7             +1
5            5           7              12             +1
6            7          10              17             -1
7           17          24              41             +1
8           29          41              70             +1
9           41          58              99             -1
10          99         140             239             +1
11         169         239             408             +1
12         239         338             577             -1
13         577         816            1393             +1
14         985        1393            2378             +1
15        1393        1970            3363             -1
16        3363        4756            8119             +1
17        5741        8119           13860             +1
18        8119       11482           19601             -1
19       19601       27720           47321             +1
20       33461       47321           80782             +1

Cf. A007590, A001541, A001653, A079496, A002315, A005319, A143608, A075870, A000129, A001109, A083374.

Initialize row 1 as A(1,1) = 1 and A(1,2) = 0, then:
  For rows at n mod 3 = 0:  A(n,1) = A(n-1, 2)
                            A(n,2) = A(n, 1) + A(n-2, 1)
  For rows at n mod 3 = 1:  A(n,1) = A(n-1, 1) + A(n-1, 2)
                            A(n,2) = A(n, 1) + A(n-1, 1)
  For rows at n mod 3 = 2:  A(n,1) = A(n-1,1) + A(n-3, 1)
                            A(n,2) = A(n-1,1) + A(n-1, 2)
Empirical g.f.: -x*(2*x^11-x^10-x^9+x^8-4*x^7+3*x^6-2*x^5-x^4-x^3-x^2-1) / ((x^6-2*x^3-1)*(x^6+2*x^3-1)). - Colin Barker, Aug 08 2013

Some additional comments by Richard R. Forberg, Aug 12 2013

A100434 Expansion of g.f. (1+x)(3+x)/(1+6x^2+x^4).

Original entry on oeis.org

3, 4, -17, -24, 99, 140, -577, -816, 3363, 4756, -19601, -27720, 114243, 161564, -665857, -941664, 3880899, 5488420, -22619537, -31988856, 131836323, 186444716, -768398401, -1086679440, 4478554083, 6333631924, -26102926097, -36915112104, 152139002499, 215157040700

Offset: 0

N. J. A. Sloane, Nov 21 2004, suggested by correspondence from Creighton Dement

From Creighton Dement, Dec 18 2004: (Start)
Define the following sequences:
b(2n) = c(2n+1), b(2n+1) = c(2n); (c(n)) = (1, -3, -7, 17, 41, -99, -239, 577, 1393, -3363, -8119, 19601, 47321). This is the sequence A001333, apart from signs. Then c(2n) = ((-1)^n)*A002315(n) and c(2n+1) = ((-1)^(n+1))*A001541(n+1).
(d(n)) = (2, 4, -10, -24, 58, 140, -338, -816, 1970, 4756, -11482, -27720). This is A052542, apart from signs. Also, d(2n) = ((-1)^n)*A075870(n), d(2n+1) = ((-1)^n)*A005319(n+1).
(e(n)) = (1, -1, -5, 5, 29, -29, -169, 169, 985, -985, -5741, 5741, 33461, -33461), e(2n) = d(2n)/2, e(2n+1) = - d(2n)/2.
(f(n)) = (2, 2, -12, -12, 70, 70, -408, -408, 2378, 2378, -13860, -13860, ) f(2n) = f(2n+1) = d(2n+1)/2.
(g(n)) = (0, -3, 0, 17, 0, -99, 0, 577, 0, -3363, 0, 19601, 0, -114243, 0, 665857), g(2n) = 0, g(2n+1) = c(2n+1).
Then a(2n) = - c(2n+1), a(2n+1) = d(2n+1) and we have the following conjectures: c(n) + d(n) = e(n) + f(n) = g(n) + a(n); c(n) + d(n) = b(n). In other words, the sequences (c(n) + d(n)) = (e(n) + f(n)) = (g(n) + h(n)) all represent the sequence c with even- and odd-indexed terms reversed. (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,-6,0,-1).

Bisections give A001541, A005319.

Cf. A000034, A001333, A002315, A052542, A075870, A126354.

I:=[3,4,-17,-24]; [n le 4 select I[n] else -6*Self(n-2)-Self(n-4): n in [1..40]]; // G. C. Greubel, Apr 09 2023

LinearRecurrence[{0,-6,0,-1}, {3,4,-17,-24}, 41] (* G. C. Greubel, Apr 09 2023 *)

@CachedFunction
def a(n): # a = A100434
    if (n<4): return (3,4,-17,-24)[n]
    else: return -6*a(n-2) - a(n-4)
[a(n) for n in range(41)] # G. C. Greubel, Apr 09 2023

a(n) = (-1)^floor(n/2)*A000034(n)*A126354(n+3). - R. J. Mathar, Mar 08 2009

a(n) = -2*a(n-1) - 3*a(n-2) if n is even; a(n) = (4*a(n-1) - a(n-2))/3 if n is odd. - R. J. Mathar, Jun 18 2014

A175033 Numbers n such that (ceiling(sqrt(nn/2)))^2 - nn/2 = 17/2.

Original entry on oeis.org

9, 15, 55, 89, 321, 519, 1871, 3025, 10905, 17631, 63559, 102761, 370449, 598935, 2159135, 3490849, 12584361, 20346159, 73347031, 118586105, 427497825, 691170471

Offset: 1

Ctibor O. Zizka, Nov 09 2009

Let (ceiling(sqrt(n*n/2)))^2 - n*n/2 = i. Then for i=1/2 we have A002315, for i=1 we have A005319, for i=2 we have A077444, for i=7/2 we have A077446, for i=4 we have A081554.

Conjecture: a(n) = 6*a(n-2) - a(n-4). - Charles R Greathouse IV, Apr 30 2016

Cf. A005319, A077444, A081554, A002315, A077446,

lista(nn)=for (n=1, nn, if ((ceil(sqrt(n*n/2)))^2 - n*n/2 == 17/2, print1(n, ", "));); \\ Michel Marcus, Jun 02 2013

forstep(n=9,1e9,2, if((sqrtint(n^2\2)+1)^2==(n^2+17)/2, print1(n", "))) \\ Charles R Greathouse IV, Apr 30 2016

More terms from Michel Marcus, Jun 02 2013

a(17)-a(22) from Charles R Greathouse IV, Apr 30 2016

cp's OEIS Frontend

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A065113 Sum of the squares of the a(n)-th and the (a(n)+1)st triangular numbers (A000217) is a perfect square.

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A082639 Numbers k such that 2*k*(k+2) is a square.

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A082639 Numbers k such that 2k(k+2) is a square.

A100434 Expansion of g.f. (1+x)(3+x)/(1+6x^2+x^4).

A175033 Numbers n such that (ceiling(sqrt(nn/2)))^2 - nn/2 = 17/2.