A006452 -id:A006452 - cp's OEIS frontend

A216162 Sequences A006452 and A216134 interlaced.

1, 0, 1, 1, 2, 4, 4, 9, 11, 26, 23, 55, 64, 154, 134, 323, 373, 900, 781, 1885, 2174, 5248, 4552, 10989, 12671, 30590, 26531, 64051, 73852, 178294, 154634, 373319, 430441, 1039176, 901273, 2175865, 2508794, 6056764, 5253004, 12681873, 14622323, 35301410

Offset: 0

Raphie Frank, Sep 07 2012

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

Cf. A000129.
For some k in n:
a(2n) = A006452 (k^2 - 1 is triangular).
a(2n + 1) = A216134 (T_k and 2T_k + 1 are triangular).
a(2n + 1) - a(2n) = A006451 (T_k + 1 is square).
a(2n + 1) + a(2n) = A124124 (T_k and (T_k - 1)/2 are triangular).
a(4n + 1) + a(4n + 2) = A001108 (T_k is square).
a(4n + 3) + a(4n + 4) = A001652 (T_k and 2T_k are triangular).
Sum(a(n)) - 1 = A048776 for even n (the second partial summation of the Pell numbers).

Vec((-1-x^3+5*x^4-3*x^5-2*x^6+x^7-2*x^8+x^9)/((x-1)*(1+x)*(x^4-2*x^2-1)*(x^4+2*x^2-1))+O(x^99)) \\ Charles R Greathouse IV, Jun 12 2015

(a(2n) + a(2n - 1)) - (a(2n - 2) + a(2n - 3)) = A000129(n); n>1.
It follows that sqrt(2) = lim n --> infinity ((a(2n + 2) + a(2n + 1)) - (a(2n - 2) + a(2n - 3)))/((a(2n + 2) + a(2n + 1)) - (a(2n) + a(2n - 1))).
G.f. ( -1-x^3+5*x^4-3*x^5-2*x^6+x^7-2*x^8+x^9 ) / ( (x-1)*(1+x)*(x^4-2*x^2-1)*(x^4+2*x^2-1) ). - R. J. Mathar, Sep 08 2012

Edited by N. J. A. Sloane, May 24 2021

A006451 Numbers k such that k*(k+1)/2 + 1 is a square.

Original entry on oeis.org

0, 2, 5, 15, 32, 90, 189, 527, 1104, 3074, 6437, 17919, 37520, 104442, 218685, 608735, 1274592, 3547970, 7428869, 20679087, 43298624, 120526554, 252362877, 702480239, 1470878640, 4094354882, 8572908965, 23863649055, 49966575152

Offset: 0

N. J. A. Sloane and Jeffrey Shallit

A. J. Gottlieb, How four dogs meet in a field, etc., Technology Review, Problem J/A2, Jul/August 1973 pp. 73-74; solution Jan 1974 (see link).
Jeffrey Shallit, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
A. J. Gottlieb, How four dogs meet in a field, etc. (scanned copy)
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
J. Shallit, Letter to N. J. A. Sloane, Oct. 1975
Hermann Stamm-Wilbrandt, 4 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A000124, A006452, A006454, A098586, A098790.

Cf. numbers m such that k*A000217(m)+1 is a square: this sequence for k=1; m=0 for k=2; A233450 for k=3; A001652 for k=4; A129556 for k=5; A001921 for k=6. - Bruno Berselli, Dec 16 2013

a006451 n = a006451_list !! n
a006451_list = 0 : 2 : 5 : 15 : map (+ 2)
   (zipWith (-) (map (* 6) (drop 2 a006451_list)) a006451_list)
-- Reinhard Zumkeller, Jan 10 2012

N:= 100: # to get a(0) to a(N)
A[0]:= 0: A[1]:= 2: A[2]:= 5: A[3]:= 15:
for n from 4 to N do A[n]:= 6*A[n-2] - A[n-4] + 2 od:
seq(A[n],n=0..N); # Robert Israel, Aug 26 2014

LinearRecurrence[{1,6,-6,-1,1},{0,2,5,15,32},30] (* Harvey P. Dale, Jul 17 2013 *)
Select[Range[10^6], IntegerQ@ Sqrt[# (# + 1)/2 + 1] &] (* Michael De Vlieger, Apr 25 2017 *)

for(n=1,10000,t=n*(n+1)/2+1;if(issquare(t), print1(n,", "))) \\ Joerg Arndt, Oct 10 2009

G.f.: x*(-2-3*x+2*x^2+x^3)/(x-1)/(x^2+2*x-1)/(x^2-2*x-1). Conjectured (correctly) by Simon Plouffe in his 1992 dissertation.
a(n) = 6*a(n-2) - a(n-4) + 2 with a(0)=0, a(1)=2, a(2)=5, a(3)=15. - Zak Seidov, Apr 15 2008
a(n) = 3*a(n-2) + 4*sqrt((a(n-2)^2 + a(n-2))/2 + 1) + 1 with a(0) = 0, a(1) = 2. - Raphie Frank, Feb 02 2013
a(n) = a(n-1) + 6*a(n-2) - 6*a(n-3) - a(n-4) + a(n-5); a(0)=0, a(1)=2, a(2)=5, a(3)=15, a(4)=32. - Harvey P. Dale, Jul 17 2013
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>5. - Hermann Stamm-Wilbrandt, Aug 26 2014
a(2*n+1) = A098790(2*n+1). - Hermann Stamm-Wilbrandt, Aug 26 2014
a(2*n) = A098586(2*n-1), for n>0. - Hermann Stamm-Wilbrandt, Aug 27 2014
a(n) = 8*sqrt(T(a(n-2)) + 1) + a(n-4) where T(a(n)) = A000217(a(n)), and a(-1) = -1, a(0)=0, a(1)=2, a(2)=5. - Vladimir Pletser, Apr 29 2017

More terms from Larry Reeves (larryr(AT)acm.org), Feb 07 2001

Edited by N. J. A. Sloane, Oct 24 2009, following discussions by several correspondents in the Sequence Fans Mailing List, Oct 10 2009

A006454 Solution to a Diophantine equation: each term is a triangular number and each term + 1 is a square.

Original entry on oeis.org

0, 3, 15, 120, 528, 4095, 17955, 139128, 609960, 4726275, 20720703, 160554240, 703893960, 5454117903, 23911673955, 185279454480, 812293020528, 6294047334435, 27594051024015, 213812329916328, 937385441796000, 7263325169820735, 31843510970040003, 246739243443988680

Offset: 0

N. J. A. Sloane, Jeffrey Shallit

Alternative definition: a(n) is triangular and a(n)/2 is the harmonic average of consecutive triangular numbers. See comments and formula section of A005563, of which this sequence is a subsequence. - Raphie Frank, Sep 28 2012
As with the Sophie Germain triangular numbers (A124174), 35 = (a(n) - a(n-6))/(a(n-2) - a(n-4)). - Raphie Frank, Sep 28 2012
Sophie Germain triangular numbers of the second kind as defined in A217278. - Raphie Frank, Feb 02 2013
Triangular numbers m such that m+1 is a square. - Bruno Berselli, Jul 15 2014
From Vladimir Pletser, Apr 30 2017: (Start)
Numbers a(n) which are the triangular number T(b(n)), where b(n) is the sequence A006451(n) of numbers n such that T(n)+1 is a square.
a(n) also gives the x solutions of the 3rd-degree Diophantine Bachet-Mordell equation y^2 = x^3 + K, with y = T(b(n))*sqrt(T(b(n))+1) = A285955(n) and K = T(b(n))^2 = A285985(n), the square of the triangular number of b(n) = A006451(n).
Also: This sequence is a subsequence of A000217(n), namely A000217(A006451(n)). (End)

			From _Raphie Frank_, Sep 28 2012: (Start)
35*(528 - 15) + 0 = 17955 = a(6),
35*(4095 - 120) + 3 = 139128 = a(7),
35*(17955 - 528) + 15 = 609960 = a(8),
35*(139128 - 4095) + 120 = 4726275 = a(9). (End)
From _Raphie Frank_, Feb 02 2013: (Start)
a(7) = 139128 and a(9) = 4726275.
a(9) = (2*(sqrt(8*a(7) + 1) - 1)/2 + 3*sqrt(a(7) + 1) + 1)^2 - 1 = (2*(sqrt(8*139128 + 1) - 1)/2 + 3*sqrt(139128 + 1) + 1)^2 - 1 = 4726275.
a(9) = 1/2*((3*(sqrt(8*a(7) + 1) - 1)/2 + 4*sqrt(a(7) + 1) + 1)^2 + (3*(sqrt(8*a(7) + 1) - 1)/2 + 4*sqrt(a(7) + 1) + 1)) = 1/2*((3*(sqrt(8*139128 + 1) - 1)/2 + 4*sqrt(139128 + 1) + 1)^2 + (3*(sqrt(8*139128 + 1) - 1)/2 + 4*sqrt(139128 + 1) + 1)) = 4726275. (End)
From _Vladimir Pletser_, Apr 30 2017: (Start)
For n=2, b(n)=5, a(n)=15
For n=5, b(n)=90, a(n)= 4095
For n = 3, A006451(n) = 15. Therefore, A000217(A006451(n)) = A000217(15) = 120. (End)

Edward J. Barbeau, Pell's Equation, New York: Springer-Verlag, 2003, p. 17, Exercise 1.2.
Allan J. Gottlieb, How four dogs meet in a field, and other problems, Technology Review, Jul/August 1973, pp. 73-74.
Vladimir Pletser, On some solutions of the Bachet-Mordell equation for large parameter values, to be submitted, April 2017.
Jeffrey Shallit, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vladimir Pletser, Table of n, a(n) for n = 0..1000 (first 60 terms from Vincenzo Librandi)
M.A. Bennett and A. Ghadermarzi, Data on Mordell's curve.
Michael A. Bennett and Amir Ghadermarzi, Mordell's equation : a classical approach, arXiv:1311.7077 [math.NT], 2013.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Jeffrey Shallit, Letter to N. J. A. Sloane, Oct. 1975.
K. B. Subramaniam, Almost Square Triangular Numbers, The Fibonacci Quarterly, Vol. 37, No. 3 (1999), pp. 194-197.
Eric Weisstein's World of Mathematics, Mordell Curve.
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. sqrt(a(n) + 1) = A006452(n + 1) = A216162(2n + 2) and (sqrt(8a(n) + 1) - 1)/2 = A006451.
Cf. A217278, A124174, A216134. - Raphie Frank, Feb 02 2013
Subsequence of A182334.
Cf. A001109, A245031.
Cf. A285955, A285985, A000217, A006451, A081119, A054504. - Vladimir Pletser, Apr 30 2017

I:=[0,3,15,120,528,4095]; [n le 6 select I[n] else 35*(Self(n-2) - Self(n-4)) + Self(n-6): n in [1..30]]; // Vincenzo Librandi, Dec 21 2015

A006454:=-3*z*(1+4*z+z**2)/(z-1)/(z**2-6*z+1)/(z**2+6*z+1); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation
restart: bm2:=-1: bm1:=0: bp1:=2: bp2:=5: print ('0,0','1,3','2,15'); for n from 3 to 1000 do b:= 8*sqrt((bp1^2+bp1)/2+1)+bm2; a:=b*(b+1)/2; print(n,a); bm2:=bm1; bm1:=bp1; bp1:=bp2; bp2:=b; end do: # Vladimir Pletser, Apr 30 2017

Clear[a]; a[0] = a[1] = 1; a[2] = 2; a[3] = 4; a[n_] := 6a[n - 2] - a[n - 4]; Array[a, 40]^2 - 1 (* Vladimir Joseph Stephan Orlovsky, Mar 03 2011 *)
LinearRecurrence[{1,34,-34,-1,1},{0,3,15,120,528},30] (* Harvey P. Dale, Feb 18 2023 *)

concat(0, Vec(3*x*(1 + 4*x + x^2) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)) + O(x^30))) \\ Colin Barker, Apr 30 2017

a(n) = A006451(n)*(A006451(n)+1)/2.
a(n) = A006452(n)^2 - 1. - Joerg Arndt, Mar 04 2011
a(n) = 35*(a(n-2) - a(n-4)) + a(n-6). - Raphie Frank, Sep 28 2012
From Raphie Frank, Feb 01 2013: (Start)
a(0) = 0, a(1) = 3, and a(n+2) = (2x + 3y + 1)^2 - 1  = 1/2*((3x + 4y + 1)^2 + (3x + 4y + 1)) where x = (sqrt(8*a(n) + 1) - 1)/2 = A006451(n) =  1/2*(A216134(n + 1) + A216134(n - 1)) and y = sqrt(a(n) + 1) = A006452(n + 1) = 1/2*(A216134(n + 1) - A216134(n - 1)).
Note that A216134(n + 1) = x + y, and A216134(n + 3) = (2x + 3y + 1) + (3x + 4y + 1) = (5x + 7y + 2), where A216134 gives the indices of the Sophie Germain triangular numbers. (End)
a(n) = (1/64)*(((4 + sqrt(2))*(1 -(-1)^(n+1)*sqrt(2))^(2* floor((n+1)/2)) + (4 - sqrt(2))*(1 + (-1)^(n+1)*sqrt(2))^(2*floor((n+1)/2))))^2 - 1. - Raphie Frank, Dec 20 2015
From Vladimir Pletser, Apr 30 2017: (Start)
Since b(n) = 8*sqrt(T(b(n-2))+1)+ b(n-4) = 8*sqrt((b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4), with b(-1)=-1, b(0)=0, b(1)=2, b(2)=5 (see A006451) and a(n) = T(b(n)) (this sequence), we have:
a(n) = ((8*sqrt((b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4))*(8*sqrt((b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4)+1)/2). (End)
From Colin Barker, Apr 30 2017: (Start)
G.f.: 3*x*(1 + 4*x + x^2) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)).
a(n) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5) for n > 4.
(End)
a(n)  = (A001109(n/2+1) - 2*A001109(n/2))^2 - 1 if n is even, and (A001109((n+3)/2) - 4*A001109((n+1)/2))^2 - 1 if n is odd (Subramaniam, 1999). - Amiram Eldar, Jan 13 2022

Better description from Harvey P. Dale, Jan 28 2001
More terms from Larry Reeves (larryr(AT)acm.org), Feb 07 2001
Minor edits by N. J. A. Sloane, Oct 24 2009

A077443 Numbers k such that (k^2 - 7)/2 is a square.

Original entry on oeis.org

3, 5, 13, 27, 75, 157, 437, 915, 2547, 5333, 14845, 31083, 86523, 181165, 504293, 1055907, 2939235, 6154277, 17131117, 35869755, 99847467, 209064253, 581953685, 1218515763, 3391874643, 7102030325, 19769294173, 41393666187, 115223890395, 241259966797, 671574048197

Offset: 1

Gregory V. Richardson, Nov 06 2002

Lim_{n -> inf} a(n)/a(n-2) = 3 + 2*sqrt(2) = R1*R2. Lim_{k -> inf} a(2*k-1)/a(2*k) = (9 + 4*sqrt(2))/7 = R1 = A156649 (ratio #1). Lim_{k -> inf} a(2*k)/a(2*k-1) = (11 + 6*sqrt(2))/7 = R2 (ratio #2).
Also gives solutions > 3 to the equation x^2-4 = floor(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 14 2004
From Paul Curtz, Dec 15 2012: (Start)
a(n-1) and A006452(n) are companions. Like A000129 and A001333.
Reduced mod 10 this is a sequence of period 12: 3, 5, 3, 7, 5, 7, 7, 5, 7, 3, 5, 3.
(End)
The Pisano periods (periods of the sequence reducing a(n) modulo m) for m>=1 are 1,  1,  8,  4, 12,  8,  6,  4, 24, 12, 24,  8, 28, ... R. J. Mathar, Dec 15 2012
Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 56 = 0. - Colin Barker, Feb 08 2014
From Wolfdieter Lang, Feb 05 2015: (Start)
a(n+1) gives for n >= 0 all positive x solutions of the (generalized) Pell equation x^2 - 2*y^2 = +7.
The corresponding y solutions are given in A077442(n), n >= 0. The, e.g., the Nagell reference for finding all solutions.
Because the primitive Pythagorean triangle (3,4,5) is the only one with the sum of legs equal to 7 all positive solutions (x(n),y(n)) = (a(n+1),A077442(n)) of the Pell equation x^2 - 2*y^2 = +7 satisfy x(n) - y(n) < y(n) if n >= 1; only the first solution (x(0),y(0)) = (3,2) satisfies 3-1 > 1. Proof: Primitive Pythagorean triangles are characterized by the positive integer pairs [u,v] with  u+v odd, gcd(u,v) = 1 and u > v. See the Niven et al. reference, Theorem 5.5, p. 232. The leg sum is L = (u+v)^2 - 2*v^2. With L = 7, x = u+v and y = v, every solution (x(n),y(n)) with x(n)-y(n) = u(n) > v(n) = y(n) will correspond to a primitive Pythagorean triangle. Note that because of gcd(x,y) = 1 also gcd(u,v) = 1. But there is only one such triangle with L=7, namely the one with  [u(0),v(0)] = [2,1]. All other solutions with n >= 1 must therefore satisfy x(n)-y(n) < y(n). (End)
For n > 0, a(n+1) is the n-th almost Lucas-cobalancing number of first type (see Tekcan and Erdem). - Stefano Spezia, Nov 25 2022

			a(3)^2 - 2*A077442(2)^2 = 13^2 - 2*9^2 = +7. - _Wolfdieter Lang_, Feb 05 2015

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.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.
Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991.

Vincenzo Librandi, Table of n, a(n) for n = 1..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.
J. J. O'Connor and E. F. Robertson, History of Pell's Equation
J. P. Robertson, Solving the Generalized Pell Equation
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A000129, A001333, A006452, A038761, A038762, A077442, A101386, A124124, A156649, A176981, A216134, A253811.

LinearRecurrence[{0,6,0,-1},{3,5,13,27},50] (* Sture Sjöstedt, Oct 09 2012 *)

a(2n+1) = A038762(n). a(2n) = A101386(n-1).
The same recurrences hold for the odd and the even indices: a(n+2) = 6*a(n) - a(n-2), a(n+1) = 3*a(n) + 2*(2*a(n)^2-14)^0.5 - Richard Choulet, Oct 11 2007
O.g.f.: -x*(x-1)*(3*x^2+8*x+3) / ( (x^2+2*x-1)*(x^2-2*x-1) ). - R. J. Mathar, Nov 23 2007
If n is even a(n) = (1/2)*(3+sqrt(2))*(3+2*sqrt(2))^(-n/2) + (1/2)*(3-sqrt(2))*(3-2*sqrt(2))^(-n/2); if n is odd a(n) = (1/2)*(3+sqrt(2))*(3+2*sqrt(2))^((n-1)/2) + (1/2)*(3-sqrt(2))*(3-2*sqrt(2))^((n-1)/2). - Antonio Alberto Olivares, Apr 20 2008
a(n) = A000129(n+1) + (-1)^n*A176981(n-1), n>1. - R. J. Mathar, Jul 03 2011
a(n) = A000129(n+1) -(-1)^n*A000129(n-2), rephrasing the formula above. - Paul Curtz, Dec 07 2012
a(n) = sqrt(8*A216134(n)^2 + 8*A216134(n) + 9) = 2*A124124(n) + 1. - Raphie Frank, May 24 2013
E.g.f.: cosh(sqrt(2)*x)*(3*cosh(x) - sinh(x)) + sqrt(2)*(2*cosh(x) - sinh(x))*sinh(sqrt(2)*x) - 3. - Stefano Spezia, Nov 25 2022

More terms from Richard Choulet, Oct 11 2007

Edited: replaced n by a(n) in the name. Moved Pell remarks to the comment section. Added cross references. - Wolfdieter Lang, Feb 05 2015

A216134 Numbers k such that 2 * A000217(k) + 1 is triangular.

Original entry on oeis.org

0, 1, 4, 9, 26, 55, 154, 323, 900, 1885, 5248, 10989, 30590, 64051, 178294, 373319, 1039176, 2175865, 6056764, 12681873, 35301410, 73915375, 205751698, 430810379, 1199208780, 2510946901, 6989500984, 14634871029, 40737797126, 85298279275, 237437281774

Offset: 0

Raphie Frank, Sep 01 2012

Numbers n such that 2*triangular(n) + 1 is a triangular number. Equivalently, numbers n such that n^2 + n + 1 is a triangular number. - Alex Ratushnyak, Apr 18 2013

For n > 0, a(n) is the n-th almost cobalancing number of first type (see Tekcan and Erdem). - Stefano Spezia, Nov 25 2022

Colin Barker, Table of n, a(n) for n = 0..1000
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Wikipedia, Pell numbers
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A124174, A000129, A001333, A006451, A006452, A124124, A079496.

Cf. A000217, A069017 (triangular numbers of the form k^2 + k + 1).

LinearRecurrence[{1, 6, -6, -1, 1}, {0, 1, 4, 9, 26}, 40] (* T. D. Noe, Sep 03 2012 *)

Vec( x*(1+3*x-x^2-x^3)/((1-x)*(1+2*x-x^2)*(1-2*x-x^2)) + O(x^66) ) \\ Joerg Arndt, Aug 13 2014

isok(n) = ispolygonal(n*(n+1) + 1, 3); \\ Michel Marcus, Aug 13 2014

G.f.: x*(1+3*x-x^2-x^3)/((1-x)*(1+2*x-x^2)*(1-2*x-x^2)). - R. J. Mathar, Sep 08 2012
sqrt(2) = lim_{k->infinity} ((a(2k+1) + a(2k) + 1)/2)/(a(2k+1) - a(2k)) = lim_{k->infinity} A001333(2k + 1)/A000129(2k + 1).
1 + (sqrt 2) = lim_{k->infinity} (a(2k + 1) - a(2k))/(a(2k + 1) - 2*a(2k) +  a(2k - 1)) = lim_{k->infinity} A000129(2k + 1)/A000129(2k).
1 + 1/(sqrt 2) = lim_{k->infinity} (a(2k+1) - a(2k))/(a(2k) - a(2k - 1)) = lim_{k->infinity} A000129(2k + 1)/A001333(2k).
a(n) = (2*A000129(n) + (-1)^n*(A000129(2*floor(n/2) - 1) - (-1)^n)/2). - Raphie Frank, Jan 04 2013
From Raphie Frank, Jan 04 2013: (Start)
A124174(n) = a(n)*(a(n) + 1)/2.
A079496(n) = a(n + 1) - a(n).
A000129(2n) = a(2n) - 2*a(2n - 1) +  a(2n - 2).
A000129(2n) = a(2n + 1) - 2*a(2n) +  a(2n - 1).
A000129(2n + 1) = a(2n + 1) - a(2n).
A001333(2n) = a(2n) - a(2n - 1).
A001333(2n + 1) = (a(2n + 1) + a(2n) + 1)/2.
A006451(n + 1) = (a(n + 2) + a(n))/2.
A006452(n + 2) = (a(n + 2) - a(n))/2.
A124124(n + 2) = (a(n + 2) + a(n))/2 + (a(n + 2) - a(n)).
(End)
a(n + 2) = sqrt(8*a(n)^2 + 8*a(n) + 9) + 3*a(n) + 1; a(0) = 0, a(1) = 1. - Raphie Frank, Feb 02 2013
a(n) = (3/8 + sqrt(2)/4)*(1 + sqrt(2))^n + (-1/8 - sqrt(2)/8)*(-1 + sqrt(2))^n + (3/8 - sqrt(2)/4)*(1 - sqrt(2))^n + (-1/8 + sqrt(2)/8)*(-1 - sqrt(2))^n - 1/2. - Robert Israel, Aug 13 2014
E.g.f.: (1/4)*(-2*cosh(x) - 2*sinh(x) + 2*cosh(sqrt(2)*x)*(cosh(x) + 2*sinh(x)) + sqrt(2)*(cosh(x) + 3*sinh(x))*sinh(sqrt(2)*x)). - Stefano Spezia, Dec 10 2019

A124124 Nonnegative integers n such that 2n^2 + 2n - 3 is square.

Original entry on oeis.org

1, 2, 6, 13, 37, 78, 218, 457, 1273, 2666, 7422, 15541, 43261, 90582, 252146, 527953, 1469617, 3077138, 8565558, 17934877, 49923733, 104532126, 290976842, 609257881, 1695937321, 3551015162, 9884647086, 20696833093, 57611945197, 120629983398, 335787024098

Offset: 1

John W. Layman, Nov 29 2006

First differences are apparently in A143608. [R. J. Mathar, Jul 17 2009]

Alternative definition: T_n and (T_n - 1)/2 are triangular numbers. - Raphie Frank, Sep 06 2012

Michael De Vlieger, Table of n, a(n) for n = 1..2613
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.
Hermann Stamm-Wilbrandt, 4 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A001108, A001652, A001653, A006452, A008844, A046090, A046172, A077442, A098790, A216134.

A124124 := proc(n)
coeftayl(x*(1+x-2*x^2+x^3+x^4)/((1-x)*(x^2-2*x-1)*(x^2+2*x-1)), x=0, n);
end proc:
seq(A124124(n), n=1..20); # Wesley Ivan Hurt, Aug 04 2014
# Alternative:
a[1]:= 1: a[2]:= 2: a[3]:= 6:
for n from 4 to 1000 do
a[n]:= (3 + 2*(n mod 2))*(a[n-1]-a[n-2])+a[n-3]
od:
seq(a[n],n=1..100); # Robert Israel, Aug 13 2014

LinearRecurrence[{1,6,-6,-1,1},{1,2,6,13,37},40] (* Harvey P. Dale, Nov 05 2011 *)
CoefficientList[Series[(1 + x - 2*x^2 + x^3 + x^4)/((1 - x)*(x^2 - 2*x - 1)*(x^2 + 2*x - 1)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Aug 04 2014 *)

for(n=1,10^10,if(issquare(2*n^2+2*n-3),print1(n,", "))) \\ Derek Orr, Aug 13 2014

It appears that a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) if n is even, a(n) = 5*a(n-1)-5*a(n-2)+a(n-3) if n is odd. Can anyone confirm this?
Corrected and confirmed (using the g.f.) by Robert Israel, Aug 27 2014
2*a(n) = sqrt(7+2*A077442(n-1)^2)-1. - R. J. Mathar, Dec 03 2006
a(n) = a(n-1)+6*a(n-2)-6*a(n-3)-a(n-4)+a(n-5). G.f.: -x*(1+x-2*x^2+x^3+x^4)/((x-1)*(x^2-2*x-1)*(x^2+2*x-1)). [R. J. Mathar, Jul 17 2009]
For n>0, a(2n-1) = 2*A001653(n) - A046090(n-1) and a(2n) = 2*A001653(n) + A001652(n-1). - Charlie Marion, Jan 03 2012
From Raphie Frank, Sep 06 2012: (Start)
If y = A006452(n), then a(n) = 2y + ((sqrt(8y^2 - 7) - 1)/2 - (1 - sgn(n))).
Also see A216134 [a(n) = y + ((sqrt(8y^2 - 7) - 1)/2 - (1 - sgn(n)))].
(End)
From Hermann Stamm-Wilbrandt, Aug 27 2014: (Start)
a(2*n+2) = A098586(2*n).
a(2*n+1) = A098790(2*n).
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>6. (End)
a(2*n+1)^2 + (a(2*n+1)+1)^2 = A038761(n)^2 + 2^2. - Hermann Stamm-Wilbrandt, Aug 31 2014

More terms from Harvey P. Dale, Feb 07 2011

More terms from Wesley Ivan Hurt, Aug 04 2014

A237599 Positive integers k such that x^2 - 6xy + y^2 + k = 0 has integer solutions.

Original entry on oeis.org

4, 7, 8, 16, 23, 28, 31, 32, 36, 47, 56, 63, 64, 68, 71, 72, 79, 92, 100, 103, 112, 119, 124, 127, 128, 136, 144, 151, 164, 167, 175, 184, 188, 191, 196, 199, 200, 207, 223, 224, 239, 248, 252, 256, 263, 271, 272, 279, 284, 287, 288, 292, 311, 316, 324, 328

Offset: 1

Colin Barker, Feb 10 2014

nonn

Nonnegative numbers of the form 8x^2 - y^2. - Jon E. Schoenfield, Jun 03 2022

			4 is in the sequence because x^2 - 6xy + y^2 + 4 = 0 has integer solutions, for example (x, y) = (1, 5).

N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)

Cf. A001653 (k = 4), A006452 (k = 7), A001541 (k = 8), A075870 (k = 16), A156066 (k = 23), A217975 (k = 28), A003499 (k = 32), A075841 (k = 36), A077443 (k = 56).
Cf. A031363, A084917, A237351, A237606, A237609, A237610.
For primes see A007522 and A141175.
For a list of sequences giving numbers and/or primes represented by binary quadratic forms, see the "Binary Quadratic Forms and OEIS" link.

A077446 Numbers k such that 2*k^2 + 14 is a square.

Original entry on oeis.org

1, 5, 11, 31, 65, 181, 379, 1055, 2209, 6149, 12875, 35839, 75041, 208885, 437371, 1217471, 2549185, 7095941, 14857739, 41358175, 86597249, 241053109, 504725755, 1404960479, 2941757281, 8188709765, 17145817931, 47727298111

Offset: 1

Gregory V. Richardson, Nov 09 2002

The equation "2*n^2 + 14 is a square" is a version of the generalized Pell Equation x^2 - D*y^2 = C where x^2 - 2*y^2 = 14.
Numbers n such that (ceiling(sqrt(n*n/2)))^2 = (7+n^2)/2. - Ctibor O. Zizka, Nov 09 2009
From Wolfdieter Lang, Feb 26 2015: (Start)
This sequence gives all positive solutions x = a(n+1), n >= 0, of the Pell equation x^2 - 2*y^2 = -7. For the corresponding y-solutions see y(n) = 2*A006452(n+2) = A077447(n+1)/2. This implies that X^2 - 2*Y^2 = 14 has the general solutions (X(n),Y(n)) = (2*y(n), x(n)). See the first comment above.
For the positive first class solutions see (A054490(n), 2*A038723(n)) and for the second class solutions (A255236(n), 2*A038725(n+1)). (End)
For n > 0, a(n) is the n-th almost Lucas-balancing number of second type (see Tekcan and Erdem). - Stefano Spezia, Nov 26 2022

			n = 3: (A077447(3))^2 - 2*a(3)^2 = 16^2 - 2*11^2  = 14;
a(3)^2 - 2*(2*A006452(3+1))^2 = 11^2 - 2*(2*4)^2 = -7. - _Wolfdieter Lang_, Feb 26 2015

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.

Michael De Vlieger, Table of n, a(n) for n = 1..2612
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.
J. J. O'Connor and E. F. Robertson, Pell's Equation
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Eric Weisstein's World of Mathematics, Pell Equation
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A001109, A006452, A038723, A038725, A049310, A054490, A077447, A155765, A156035, A156649, A255236.

LinearRecurrence[{0,6,0,-1},{1,5,11,31},50] (* Sture Sjöstedt, Oct 08 2012 *)

2*(a(n))^2 + 14 = (A077447(n))^2.
Lim. n-> Inf. a(n)/a(n-2) = 5.8284271247461... = 3 + 2*sqrt(2) = A156035 = RG (Great Ratio).
Lim. k-> Inf. a(2*k+1)/a(2*k) = 2.09383632135605... = (9 + 4*sqrt(2))/7 = A156649 = R1 (Ratio 1).
Lim. k -> Inf. a(2*k)/a(2*k-1) = 2.78361162489122432754 = (11 + 6*sqrt(2))/7 = R2 (Ratio 2); RG = R1*R2.
a(2*k-1) = [ 2*[(3+2*Sqrt(2))^n - (3-2*Sqrt(2))^n] - [(3+2*Sqrt(2))^(n-1) - (3-2*Sqrt(2))^(n-1)] + [(3+2*Sqrt(2))^(n-2) - (3-2*Sqrt(2))^(n-2)] ] / (4*Sqrt(2)) a(2*k) = [ 5*[(3+2*Sqrt(2))^n - (3-2*Sqrt(2))^n] + [(3+2*Sqrt(2))^(n-1) - (3-2*Sqrt(2))^(n-1)] ] / (4*Sqrt(2)).
a(n) = 6*a(n-2) - a(n-4).
G.f.: x*(1+x)*(x^2+4*x+1) / ( (x^2+2*x-1)*(x^2-2*x-1) ). - R. J. Mathar, Jul 03 2011
a(n) = 6*a(n-2) - a(n-4) with a(1)=1, a(2)=5, a(3)=11, a(4)=31. - Sture Sjöstedt, Oct 08 2012
Bisection: a(2*k+1) = S(k, 6) + 5*S(k-1, 6), a(2*k) = 5*S(k-1, 6) + S(k-2, 6), with the Chebyshev polynomials S(n, x) (A049310) with S(-2, x) = -1, S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n+1). See A054490 and A255236, and the given g.f.s. - Wolfdieter Lang, Feb 26 2015
E.g.f.: 1 - cosh(sqrt(2)*x)*(cosh(x) - 3*sinh(x)) - sqrt(2)*(cosh(x) - 2*sinh(x))*sinh(sqrt(2)*x). - Stefano Spezia, Nov 26 2022
a(n) = a(n-1) + 2*A217975(n-1) + A123335(n-2) - A123335(n-3) for n > 1 and with A123335(-1) = 1. - Vladimir Pletser, Aug 30 2025

A226069 Primes p such that p-1 is a square and p-2 is a triangular number.

Original entry on oeis.org

2, 5, 17, 17957, 4726277, 23911673957

Offset: 1

Alex Ratushnyak, May 25 2013

nonn

Roots of squares: a(n)-1 = b(n)*b(n), b(n) = A226070(n).
Roots of triangular numbers: a(n)-2 = c(n)*(c(n)+1)/2, c(n) = A226071(n).
Primes of the form A006452(k)^2+1. a(7) is too large to include here (see b-file). - Max Alekseyev, Jan 30 2014

Max Alekseyev, Table of n, a(n) for n = 1..12

Cf. A226070-A226074.

#include 
#include 
#include 
#define TOP (1ULL<<34)
int main() {
  unsigned long long i, j, k, r, n=1;
  unsigned char *c = (unsigned char *)malloc(TOP/8);
  memset(c, 0, TOP/8);
  for (i=3; i < TOP*2; i+=2)
    if ((c[i>>4] & (1<<((i>>1) & 7)))==0) {
      ++n;
      if (i<(1ULL<<32))
        for (j=i*i>>1; j>3] |= 1 << (j&7);
    }
  //printf("%llu\n", n);
  for (i=1, j=i*i+1; j < TOP*2; i++,j=i*i+1)
    if(j==2 || ((j&1) && (c[j>>4] & (1<<((j>>1) & 7)))==0)) {
      k = j-2;
      r = sqrt(k*2);
      if (r*r+r==k*2) printf("%9llu %9llu %9llu\n", r, i, j);
    }
  free(c);
  return 0;
}

Terms a(7)-a(12) from Max Alekseyev, Jan 30 2014

A114697 Expansion of (1+x+x^2)/((1-x^2)(1-2x-x^2)); a Pellian-related sequence.

Original entry on oeis.org

1, 3, 9, 22, 55, 133, 323, 780, 1885, 4551, 10989, 26530, 64051, 154633, 373319, 901272, 2175865, 5253003, 12681873, 30616750, 73915375, 178447501, 430810379, 1040068260, 2510946901, 6061962063, 14634871029, 35331704122, 85298279275, 205928262673

Offset: 0

Creighton Dement, Feb 18 2006

Generating floretion: (- .5'j + .5'k - .5j' + .5k' + 'ii' - .5'ij' - .5'ik' - .5'ji' - .5'ki')*('i + 'j + i').

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-2,-1).

Cf. A000129, A002203, A005409, A100828, A111954, A113224.

Cf. A114647, A114688, A114689, A114695, A116699.

Table[(3*LucasL[n, 2] +10*Fibonacci[n, 2] -3 +(-1)^n)/4, {n,0,30}] (* G. C. Greubel, May 24 2021 *)

Vec((1+x+x^2)/((1-x^2)*(1-2*x-x^2)) + O(x^40)) \\ Colin Barker, Jun 24 2015

[(4*lucas_number1(n+2,2,-1) -2*lucas_number1(n+1,2,-1) -3 +(-1)^n)/4 for n in (0..30)] # G. C. Greubel, May 24 2021

a(n+2) - 2*a(n+1) + a(n) = A111955(n+2).
G.f.: (1+x+x^2)/((1-x)*(1+x)*(1-2*x-x^2)).
From Raphie Frank, Oct 01 2012: (Start)
a(2*n) = A216134(2*n+1).
a(2*n+1) = A006452(2*n+3)-1.
Lim_{n->infinity} a(n+1)/a(n) = A014176. (End)
a(n) = (2*A078343(n+2) - A010694(n))/4. - R. J. Mathar, Oct 02 2012
From Colin Barker, May 26 2016: (Start)
a(n) = ( 2*(-3 +(-1)^n) + (6-5*sqrt(2))*(1-sqrt(2))^n + (1+sqrt(2))^n*(6+5*sqrt(2)) )/8.
a(n) = 2*a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) for n>3. (End)
a(n) = (3*A002203(n) + 10*A000129(n) - 3 + (-1)^n)/4. - G. C. Greubel, May 24 2021

cp's OEIS Frontend

A216162 Sequences A006452 and A216134 interlaced.

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A006451 Numbers k such that k*(k+1)/2 + 1 is a square.

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A006454 Solution to a Diophantine equation: each term is a triangular number and each term + 1 is a square.

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A077443 Numbers k such that (k^2 - 7)/2 is a square.

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A216134 Numbers k such that 2 * A000217(k) + 1 is triangular.

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A124124 Nonnegative integers n such that 2n^2 + 2n - 3 is square.

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A237599 Positive integers k such that x^2 - 6xy + y^2 + k = 0 has integer solutions.

A114697 Expansion of (1+x+x^2)/((1-x^2)(1-2x-x^2)); a Pellian-related sequence.