A077442 -id:A077442 - cp's OEIS frontend

A101386 Expansion of g.f.: (5 - 3x)/(1 - 6x + x^2).

5, 27, 157, 915, 5333, 31083, 181165, 1055907, 6154277, 35869755, 209064253, 1218515763, 7102030325, 41393666187, 241259966797, 1406166134595, 8195736840773, 47768254910043, 278413792619485, 1622714500806867, 9457873212221717, 55124524772523435, 321289275422918893

Offset: 0

Creighton Dement, Jan 23 2005

A floretion-generated sequence relating to NSW numbers and numbers n such that (n^2 - 8)/2 is a square. It is also possible to label this sequence as the "tesfor-transform of the zero-sequence" under the floretion given in the program code, below. This is because the sequence "vesseq" would normally have been A046184 (indices of octagonal numbers which are also a square) using the floretion given. This floretion, however, was purposely "altered" in such a way that the sequence "vesseq" would turn into A000004. As (a(n)) would not have occurred under "natural" circumstances, one could speak of it as the transform of A000004.
Floretion Algebra Multiplication Program FAMP code: - tesforseq[ + 3'i - 2'j + 'k + 3i' - 2j' + k' - 4'ii' - 3'jj' + 4'kk' - 'ij' - 'ji' + 3'jk' + 3'kj' + 4e], Note: vesforseq = A000004, lesforseq = A002315, jesforseq = A077445
From Wolfdieter Lang, Feb 05 2015: (Start)
All positive solutions x = a(n) of the (generalized) Pell equation x^2 - 2*y^2 = +7 based on the  fundamental solution (x2,y2) = (5,3) of the second class of (proper) solutions. The corresponding y solutions are given by y(n) = A253811(n).
All other positive solutions come from the first class of (proper) solutions based on the fundamental solution (x1,y1) = (3,1). These are given in A038762 and A038761.
All solutions of this Pell equation are found in A077443(n+1) and A077442(n), for n >= 0. See, e.g., the Nagell reference on how to find all solutions.
(End)

T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

Colin Barker, Table of n, a(n) for n = 0..1000
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 55, 56.
M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.
Tanya Khovanova, Recursive Sequences
Morris Newman, Daniel Shanks, and H. C. Williams, Simple groups of square order and an interesting sequence of primes, Acta Arith., 38 (1980/1981) 129-140. MR82b:20022.
Eric Weisstein's World of Mathematics, NSW Number.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000004, A000129, A001109, A002315, A038761, A038762, A046184, A054490, A164737.

Cf. A077442, A077443(n+1), A077445, A253811.

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!((5 - 3*x)/(1-6*x+x^2))); // G. C. Greubel, Jul 26 2018

A101386:= (n) -> simplify(5*ChebyshevU(n, 3) - 3*ChebyshevU(n-1, 3)); seq( A101386(n), n = 0..30); # G. C. Greubel, Mar 17 2020

CoefficientList[ Series[(5-3x)/(1-6x+x^2), {x,0,30}], x] (* Robert G. Wilson v, Jan 29 2005 *)
LinearRecurrence[{6,-1},{5,27},30] (* Harvey P. Dale, Apr 23 2016 *)

Vec((5-3*x)/(1-6*x+x^2) + O(x^30)) \\ Colin Barker, Feb 05 2015

[5*chebyshev_U(n,3) -3*chebyshev_U(n-1,3) for n in (0..30)] # G. C. Greubel, Mar 17 2020

a(n) = A002315(n) + A077445(n+1). Note: the offset of A077445 is 1.
a(n+1) - a(n) = 2*A054490(n+1).
a(n) = 6*a(n-1) - a(n-2), a(0)=5, a(1)=27. - Philippe Deléham, Nov 17 2008
From Al Hakanson (hawkuu(AT)gmail.com), Aug 17 2009: (Start)
a(n) = ((5+sqrt(18))*(3 + sqrt(8))^n + (5-sqrt(18))*(3 - sqrt(8))^n)/2.
Third binomial transform of A164737. (End)
a(n) = rational part of z(n), with z(n) = (5+3*sqrt(2))*(3+2*sqrt(2))^n, n >= 0, the general positive solutions of the second class of proper solutions. See the preceding formula. - Wolfdieter Lang, Feb 05 2015
a(n) = 5*A001109(n+1) - 3*A001109(n). - G. C. Greubel, Mar 17 2020
a(n) = Pell(2*n+2) + 3*Pell(2*n+1), where Pell(n) = A000129(n). - G. C. Greubel, Apr 17 2020
E.g.f.: exp(3*x)*(5*cosh(2*sqrt(2)*x) + 3*sqrt(2)*sinh(2*sqrt(2)*x)). - Stefano Spezia, Mar 16 2024

More terms from Robert G. Wilson v, Jan 29 2005

A124174 Sophie Germain triangular numbers tr: 2*tr+1 is also a triangular number.

Original entry on oeis.org

0, 1, 10, 45, 351, 1540, 11935, 52326, 405450, 1777555, 13773376, 60384555, 467889345, 2051297326, 15894464365, 69683724540, 539943899076, 2367195337045, 18342198104230, 80414957735001, 623094791644755, 2731741367653000, 21166880717817451, 92798791542467010

Offset: 1

Zak Seidov, Dec 04 2006

Sophie Germain triangular numbers are one of an infinite number of triangular number sets tr where 2*tn^2*tr + tn is a triangular number: tr and tn both also being triangular numbers with tn being held constant. For the present numbers, a(n) = tr, 8*(2*tr + 1) + 1 = 16*tr + 9 is also a square, the square root of which is 2*y+1 with y being the argument of the triangular number 2*tr + 1. Now (1/2)*(y^2+y) = a^2 + a + 1 from the definition of Sophie Germain triangular numbers. Multiply both sides by 4 and subtract 3 to get 2*y^2 + 2*y - 3 = 4*a^2 + 4*a + 1 (a square). Cf. A124124: Numbers y such that 2*y^2 + 2*y - 3 is a square. The values y are the same y such that 2*y+1 = sqrt(16*tr + 9). - Kenneth J Ramsey, Jun 25 2011

Values of k such that 2*k+1 and 9*k+1 are both triangular numbers. - Colin Barker, Jun 29 2016

Vincenzo Librandi, Table of n, a(n) for n = 1..100
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A005384, A077442, A124124.

Cf. A216134, A000129.

I:=[0,1,10,45]; [n le 4 select I[n] else 34*Self(n-2)-Self(n-4)+11: n in [1..30]]; // Vincenzo Librandi, Sep 29 2011

a:= n-> (Matrix([[10, 1, 0, 0, 1]]). Matrix(5, (i, j)-> if i=j-1 then 1 elif j=1 then [1, 34, -34, -1, 1][i] else 0 fi)^n)[1, 4]: seq(a(n), n=1..30); # Alois P. Heinz, Apr 27 2009

LinearRecurrence[{1,34,-34,-1,1}, {0,1,10,45,351},30] (* Harvey P. Dale, Sep 28 2011 *)

a=[0, 1, 10, 45, 351];for(n=5,20,a=concat(a,a[#a]+34*a[#a-1]- 34*a[#a-2]-a[#a-3]+a[#a-4]));a \\ Charles R Greathouse IV, Sep 29 2011

a(n) = (A124124(n)^2 + A124124(n)-2)/4.
a(n) = 35*(a(n-2) - a(n-4)) + a(n-6).
From Peter Pein, Dec 04 2006: (Start)
a(n) = -11/32 + (-3 - 2*sqrt(2))^n/64 + (5*(3 - 2*sqrt(2))^n)/32 + (-3 - 2*sqrt(2))^n/(32*sqrt(2)) - (5*(3 - 2*sqrt(2))^n)/(32*sqrt(2)) + (-3 + 2*sqrt(2))^n/64 - (-3 + 2*sqrt(2))^n/(32*sqrt(2)) + (5*(3 + 2*sqrt(2))^n)/32 + (5*(3 + 2*sqrt(2))^n)/(32*sqrt(2));
O.g.f.: (x*(1 + 9*x + x^2))/((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2));
E.g.f.: (-22*exp(x) + exp(-3*x+2*x*sqrt(2))*(1-sqrt(2)) - 5*exp(3*x-2*x*sqrt(2))*(-2 + sqrt(2)) + exp(-3*x-2*x*sqrt(2))*(1+sqrt(2)) + 5*exp(3*x+2*x*sqrt(2))*(2+sqrt(2)))/64.  (End)
a(n) = 34*a(n-2) - a(n-4) + 11. - Kieren MacMillan, Nov 08 2008
a(n) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5) with a(0)=0, a(1)=1, a(2)=10, a(3)=45, a(4)=351. - Harvey P. Dale, Sep 28 2011
a(n) = x*(x + 1)/2 where x = A216134(n) = (2*A000129(n) + (-1)^n*(A000129(2*floor(n/2) - 1) - (-1)^n)/2). - Raphie Frank, Jan 04 2013
a(n+2) = 1/2*((3/2*sqrt(8*a(n) + 1) + sqrt(16*a(n) + 9) - 1/2)*(3/2*sqrt(8*a(n) + 1) + sqrt(16*a(n) + 9) + 1/2)); a(0) = 0, a(1) = 1. - Raphie Frank, Jan 29 2013

More terms from Alois P. Heinz, Apr 27 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

A253811 Part of the y solutions of the Pell equation x^2 - 2*y^2 = +7.

Original entry on oeis.org

3, 19, 111, 647, 3771, 21979, 128103, 746639, 4351731, 25363747, 147830751, 861620759, 5021893803, 29269742059, 170596558551, 994309609247, 5795261096931, 33777256972339, 196868280737103, 1147432427450279, 6687726283964571, 38978925276337147, 227185825374058311

Offset: 0

Wolfdieter Lang, Feb 05 2015

All positive solutions y = a(n) of the (generalized) Pell equation x^2 - 2*y^2 = +7 based on the fundamental solution (x2,y2) = (5,3) of the second class of (proper) solutions. The corresponding x solutions are given by x(n) = A101386(n).
All other positive solutions come from the first class of (proper) solutions based on the fundamental solution (x1,y1) = (3,1). These are given in A038762 and A038761.
All solutions of this Pell equation are found in A077443(n+1) and A077442(n), for n >= 0. See the Nagell reference on how to find all solutions.

			A101386(2)^2 - 2*a(2) = 157^2 - 2*111^2 = +7.

T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

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

Cf. A001109, A038761, A038762, A077442, A077443, A101386.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((x+3)/(x^2-6*x+1))); // G. C. Greubel, Jul 26 2018

LinearRecurrence[{6,-1}, {3,19}, 30] (* or *) CoefficientList[Series[ (x+3)/(x^2-6*x+1), {z, 0, 50}], x]  (* G. C. Greubel, Jul 26 2018 *)

Vec((x+3)/(x^2-6*x+1) + O(x^100)) \\ Colin Barker, Feb 05 2015

a(n) = irrational part of z(n), where z(n) = (5+3*sqrt(2))*(3+2*sqrt(2))^n, n >= 0, the general positive solutions of the second class of proper solutions.
From Colin Barker, Feb 05 2015: (Start)
a(n) = 6*a(n-1) - a(n-2).
G.f.: (x+3) / (x^2-6*x+1). (End)
a(n) = 3*A001109(n+1) + A001109(n). - R. J. Mathar, Feb 05 2015
E.g.f.: exp(3*x)*(6*cosh(2*sqrt(2)*x) + 5*sqrt(2)*sinh(2*sqrt(2)*x))/2. - Stefano Spezia, Mar 16 2024

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

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

A052454 Positive integer values of k such that 10*k^2 - 9 is a square.

Original entry on oeis.org

1, 3, 13, 25, 111, 493, 949, 4215, 18721, 36037, 160059, 710905, 1368457, 6078027, 26995669, 51965329, 230804967, 1025124517, 1973314045, 8764510719, 38927735977, 74933968381, 332820602355, 1478228842609, 2845517484433, 12638418378771

Offset: 1

John W. Layman, May 20 2003

			25 is a term of the sequence since 10*25^2-9 = 6241 = 79^2.

Paolo Xausa, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,38,0,0,-1).

Cf. A077442, A082651, A221874.

LinearRecurrence[{0, 0, 38, 0, 0, -1}, {1, 3, 13, 25, 111, 493}, 50] (* Paolo Xausa, Mar 18 2024 *)

a(n) = 38*a(n-3)-a(n-6).

G.f.: x*(1-x)*(1+4*x+17*x^2+4*x^3+x^4)/(1-38*x^3+x^6). [Colin Barker, Jun 13 2012]

More terms from Bruno Berselli, Jan 29 2013

A258149 Triangle of the absolute difference of the two legs (catheti) of primitive Pythagorean triangles.

Original entry on oeis.org

1, 0, 7, 7, 0, 17, 0, 1, 0, 31, 23, 0, 0, 0, 49, 0, 17, 0, 23, 0, 71, 47, 0, 7, 0, 41, 0, 97, 0, 41, 0, 7, 0, 0, 0, 127, 79, 0, 31, 0, 0, 0, 89, 0, 161, 0, 73, 0, 17, 0, 47, 0, 119, 0, 199, 119, 0, 0, 0, 1, 0, 73, 0, 0, 0, 241

Offset: 2

Wolfdieter Lang, Jun 10 2015

For primitive Pythagorean triangles characterized by certain (n,m) pairs and references see A225949.
Here a(n,m) = 0 for non-primitive Pythagorean triangles, and for primitive Pythagorean triangles a(n,m) = abs(n^2 - m^2 - 2*n*m) = abs((n-m)^2 - 2*m^2).
The number of non-vanishing entries in row n is A055034(n).
D(n,m):= n^2 - m^2 - 2*n*m >= 0 if 1 <= m <= floor(n/(sqrt(2)+1)), and D(n,m) < 0 if n/(sqrt(2)+1)+1 <= m <= n-1, for n >= 2.
The Pell equation (n-m)^2 - 2*m^2 = +/- N is important here to find the representations of +N or -N in the triangle D(n,m). For instance, odd primes N have to be of the +1 (mod 8) (A007519) or -1 (mod 8) (A007522) form, that is, from A001132. See the Nagell reference, Theorem 110, p. 208 with Theorem 111, pp. 210-211. E.g., N = +7 appears for m = 1, 3, 9, 19, 53, ... (A077442) for n = 4, 8, 22, 46, 128, ... (2*A006452).
  N = -7 appears for n = 3, 9, 19, 53, 111, ... (A077442) and m = 2, 4, 8, 22, 46, ... (2*A006452).
For the  signed version 2*n*m - (n^2 - m^2) see A278717. - Wolfdieter Lang, Nov 30 2016

			The triangle a(n,m) begins:
n\m   1  2  3  4  5  6  7   8   9  10  11 ...
2:    1
3:    0  7
4:    7  0 17
5:    0  1  0 31
6:   23  0  0  0 49
7:    0 17  0 23  0 71
8:   47  0  7  0 41  0 97
9:    0 41  0  7  0  0  0 127
10:  79  0 31  0  0  0 89   0 161
11:   0 73  0 17  0 47  0 119   0 199
12: 119  0  0  0  1  0 73   0   0   0 241
...
a(2,1) = |1^2 - 2*1^2| = 1 for the primitive Pythagorean triangle (pPt) [3,4,5] with |3-4| = 1.
a(3,2) = |1^2 - 2*2^2| = 7 for the pPt [5,12,13] with |5 - 12| = 7.
a(4,1) = |3^2 - 2*1^2| = 7 for the pPt [15, 8, 17] with |15 - 8| = 7.

See also A225949.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, pp. 208, 210-211.

Cf. A249866, A222946, A225949, A222951, A258150, A278717 (signed).

a[n_, m_] /; n > m >= 1 && CoprimeQ[n, m] && (-1)^(n+m) == -1 := Abs[n^2 - m^2 - 2*n*m]; a[, ] = 0; Table[a[n, m], {n, 2, 12}, {m, 1, n-1}] // Flatten (* Jean-François Alcover, Jun 16 2015, after given formula *)

a(n,m) = abs(n^2 - m^2 -2*n*m) = abs((n-m)^2 - 2*m^2) if n > m >= 1, gcd(n,m) = 1, and n and m are integers of opposite parity (i.e., (-1)^(n+m) = -1); otherwise a(n,m) = 0.

A133216 Integers that are simultaneously triangular (A000217) and decagonal (A001107).

Original entry on oeis.org

0, 1, 10, 1540, 11935, 1777555, 13773376, 2051297326, 15894464365, 2367195337045, 18342198104230, 2731741367653000, 21166880717817451, 3152427171076225351, 24426562006163234620, 3637898223680596402450, 28188231388231654934425, 4198131397700237172202345

Offset: 1

Richard Choulet, Oct 11 2007; Ant King, Nov 04 2011

nonn

Positive terms are of the form (m^2-9)/16 where m runs over the elements of A077443 that are congruent to 5 modulo 8. Correspondingly, for n>1, sqrt(16*a(n)+9) form a subsequence of A077443, while sqrt(8*a(n)+1) form a subsequence of A077442 with indices congruent to 2,3 modulo 4. [Max Alekseyev]

			The initial terms of the sequences of triangular (A000217) and decagonal (A001107) numbers are 0, 1, 3, 6, 10, 15, ... and 0, 1, 10, 27, ... respectively. As the third number which is common to both sequences is 10, we have a(3) = 10.

Index entries for linear recurrences with constant coefficients, signature (1, 1154, -1154, -1, 1).

Cf. A000217, A001107, A133217, A133218, A077443, A077442.

LinearRecurrence[{1, 1154, -1154, -1, 1} , {0, 1, 10, 1540, 11935, 1777555}, 17] (* first term 0 corrected by Georg Fischer, Apr 02 2019 *)

a(n) = A000217(A133218(n)) = A001107(A133217(n)).
For n>5, a(n) = 1154*a(n-2) - a(n-4) + 396.
For n>6, a(n) = a(n-1) + 1154*a(n-2) - 1154*a(n-3) - a(n-4) + a(n-5).
For n>1, a(n) = 1/64 * ( (9 + 4* sqrt(2)*(-1)^n)*(1+sqrt(2))^(4*n-6) + (9 - 4* sqrt(2)*(-1)^n)*(1-sqrt(2))^(4*n-6) - 22).
a(n) = floor ( 1/64 * (9 + 4*sqrt(2)*(-1)^n) * (1+sqrt(2))^(4*n-6) ).
G.f.: (x^5 + 9*x^4 + 376*x^3 + 9*x^2 + x)/((1 - x)*(x^2 - 34*x + 1)*(x^2 + 34*x + 1)). [corrected by Peter Luschny, Apr 04 2019]
Lim (n -> Infinity, a(2n+1)/a(2n)) = (1/49)*(3649+2580*sqrt(2)).
Lim (n -> Infinity, a(2n)/a(2n-1)) = (1/49)*(193+132*sqrt(2)).

Entry revised by N. J. A. Sloane, Nov 06 2011

Term 0 prepended and entry revised accordingly by Max Alekseyev, Nov 06 2011

A133217 Indices of decagonal numbers (A001107) that are also triangular (A000217).

Original entry on oeis.org

0, 1, 2, 20, 55, 667, 1856, 22646, 63037, 769285, 2141390, 26133032, 72744211, 887753791, 2471161772, 30157495850, 83946756025, 1024467105097, 2851718543066, 34801724077436, 96874483708207, 1182234151527715, 3290880727535960, 40161159427864862

Offset: 1

Richard Choulet, Oct 11 2007; Ant King, Nov 04 2011

nonn

For n>0, a(n) = (A055979(n) - A056161(n))/2, with those two sequences related through the Diophantine equation 2x^2 + 3x + 2 = r^2. - Richard R. Forberg, Nov 24 2013

			The third number which is both decagonal (A001107) and triangular (A000217) is A133216(3)=10. As this is the second decagonal number, we have a(3) = 2.

Index entries for linear recurrences with constant coefficients, signature (1, 34, -34, -1, 1).

Cf. A000217, A001107, A077443, A077442, A133216, A133218.

LinearRecurrence[{1, 34, -34, -1, 1} , {0, 1, 2, 20, 55, 667}, 24] (* first term 0 corrected by Georg Fischer, Apr 02 2019 *)

For n>5, a(n) = 34*a(n-2) - a(n-4) - 12.
For n>6, a(n) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5).
For n>1, a(n) = 1/16 * ((2*sqrt(2) + (-1)^n)*(1 + sqrt(2))^(2*n - 3) - (2*sqrt(2) - (-1)^n)*(1 - sqrt(2))^(2*n - 3) + 6).
For n>1, a(n) = ceiling (1/16*(2*sqrt(2) + (-1)^n)*(1 + sqrt(2))^(2*n - 3)).
G.f.: ( 1 - 33*x^2 + 18*x^3 + 2*x^4 ) / ((1 - x ) * (1 - 6*x + x^2 ) * (1 + 6*x + x^2)).
lim (n -> Infinity, a(2n+1)/a(2n)) = 1/7*(43 + 30*sqrt(2)).
lim (n -> Infinity, a(2n)/a(2n-1)) = 1/7*(11 + 6*sqrt(2)).

Entry revised by Max Alekseyev, Nov 06 2011

cp's OEIS Frontend

A101386 Expansion of g.f.: (5 - 3*x)/(1 - 6*x + x^2).

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A124174 Sophie Germain triangular numbers tr: 2*tr+1 is also a triangular number.

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

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A253811 Part of the y solutions of the Pell equation x^2 - 2*y^2 = +7.

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

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A101386 Expansion of g.f.: (5 - 3x)/(1 - 6x + x^2).