A124124 -id:A124124 - cp's OEIS frontend

A038761 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=9.

1, 9, 53, 309, 1801, 10497, 61181, 356589, 2078353, 12113529, 70602821, 411503397, 2398417561, 13979001969, 81475594253, 474874563549, 2767771787041, 16131756158697, 94022765165141, 548004834832149, 3194006243827753, 18616032628134369, 108502189524978461

Offset: 0

Barry E. Williams, May 02 2000

Bisection of A048654. - Lambert Klasen (lambert.klasen(AT)gmx.de), Nov 24 2004

This gives part of the (increasingly sorted) positive solutions y to the Pell equation x^2 - 2*y^2 = +7. For the x solutions see A038762. For the other part of solutions see A101386 and A253811. - Wolfdieter Lang, Feb 05 2015

			A038762(3)^2 - 2*a(4)^2 = 2547^2 - 2*1801^2 = +7. - _Wolfdieter Lang_, Feb 05 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
M. J. DeLeon, Pell's Equation and Pell Number Triples, Fib. Quart., 14(1976), pp. 456-460.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001109, A001541, A001542, A001652, A001653, A038762, A046090, A048654, A055997, A101386, A124124, A253811.

I:=[1, 9]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..40]]; // Vincenzo Librandi, Nov 16 2011

a[0]:=1: a[1]:=9: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..19); # Zerinvary Lajos, Jul 26 2006

LinearRecurrence[{6,-1},{1,9},40] (* Vincenzo Librandi, Nov 16 2011 *)

a(n)=([0,1; -1,6]^n*[1;9])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

a(n) = (9*((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) = sqrt(2*(A038762(n))^2-14)/2.
For n>1, a(n)-4a(n-1)=A001541(n)-A001542(n-2); e.g. 309-4*53=97=99-2. - Charlie Marion, Nov 12 2003
For n>0, a(n)=A046090(n)+A001653(n)+A001652(n-1)=A055997(n+1)+A001652(n-1); e.g., 309=120+169+20. - Charlie Marion, Oct 11 2006
G.f.: (1+3*x)/(1-6*x+x^2). - Philippe Deléham, Nov 03 2008
a(n) = third binomial transform of 1,6,8,48,64,384. - Al Hakanson (hawkuu(AT)gmail.com), Aug 15 2009
a(n)^2 + 2^2 = A124124(2*n+1)^2 + (A124124(2*n+1)+1)^2. - Hermann Stamm-Wilbrandt, Aug 31 2014
a(n) = irrational part of z(n) = (3 + sqrt(2))*(3 + 2*sqrt(2))^n, n >= 0. z(n) gives only part of the general positive solutions to the Pell equation x^2 - 2*y^2 = 7. See the Nagell reference in A038762 on how to find z(n), and a comment above. - Wolfdieter Lang, Feb 05 2015
a(n) = S(n, 6) + 3*S(n-1, 6), n >= 0, with the Chebyshev S-polynomials evaluated at x=6. See S(n-1, 6) = A001109(n). - Wolfdieter Lang, Mar 30 2015
E.g.f.: exp(3*x)*(2*cosh(2*sqrt(2)*x) + 3*sqrt(2)*sinh(2*sqrt(2)*x))/2. - Stefano Spezia, Mar 16 2024

Edited: Replaced the unspecific Pell comment. Moved a formula from the comment section to the formula section. - Wolfdieter Lang, Feb 05 2015

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

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

A012132 Numbers z such that x(x+1) + y(y+1) = z*(z+1) is solvable in positive integers x,y.

Original entry on oeis.org

3, 6, 8, 10, 11, 13, 15, 16, 18, 20, 21, 23, 26, 27, 28, 31, 33, 36, 37, 38, 40, 41, 43, 44, 45, 46, 48, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 66, 67, 68, 71, 73, 74, 75, 76, 77, 78, 80, 81, 83, 86, 88, 89, 91, 92, 93

Offset: 1

Sander van Rijnswou (sander(AT)win.tue.nl)

nonn

Theorem (Sierpinski, 1963): n is a term iff n^2+(n+1)^2 is a composite number. - N. J. A. Sloane, Feb 29 2020
For n > 1, A047219 is a subset of this sequence. This is because n^2 + (n+1)^2 is divisible by 5 if n is (1 or 3) mod 5 (also see A027861). - Dmitry Kamenetsky, Sep 02 2008
From Hermann Stamm-Wilbrandt, Sep 10 2014: (Start)
For n > 0, A212160 is a subset of this sequence (n^2 + (n+1)^2 is divisible by 13 if n == (2 or 10) (mod 13)).
For n >= 0, A212161 is a subset of this sequence (n^2 + (n+1)^2 is divisible by 17 if n == (6 or 10) (mod 17)).
The above are for divisibility by 5, 13, 17; notation (1,3,5), (2,10,13), (6,10,17). Divisibility by p for a and p-a-1; notation (a,p-a-1,p). These are the next tuples: (8,20,29), (15,21,37), (4,36,41), (11,41,53), ... . The corresponding sequences are a subset of this sequence (8,20,37,49,66,78,... for (8,20,29)). These sequences have no entries in the OEIS yet. For any prime of the form 4*k+1 there is exactly one of these tuples/sequences.
For n > 1, A000217 (triangular numbers) is a subset of this sequence (3,6,10,15,...); z=A000217(n), y=z-1, x=n.
For n > 0, A001652 is a subset of this sequence; z=A001652(n), x=y=A053141(n).
For n > 1, A001108(=A115598) is a subset of this sequence; z=A001108(n), x=A076708(n), y=x+1.
For n > 0, A124124(2*n+1)(=A098790(2*n)) is a subset of this sequence (6,37,218,...); z=A124124(2*n+1), x=a(n)-1, y=a(n)+1, a(m) = 6*a(m-1) - a(m-2) + 2, a(0)=0, a(1)=4.
(End)

Aviezri S. Fraenkel, Diophantine equations involving generalized triangular and tetrahedral numbers, pp. 99-114 of A. O. L. Atkin and B. J. Birch, editors, Computers in Number Theory. Academic Press, NY, 1971.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
H. Finner and K. Strassburger, Increasing sample sizes do not necessarily increase the power of UMPU-tests for 2 X 2-tables, Metrika, 54, 77-91, (2001).
Heiko Harborth, Fermat-like binomial equations, Applications of Fibonacci numbers, Proc. 2nd Int. Conf., San Jose/Ca., August 1986, 1-5 (1988).
W. Sierpinski, On triangular numbers which are sums of two smaller triangular numbers, (Polish), Wiadom. Mat. (2) 7 (1963), 27-28. See MR0182602.

Complement of A027861. - Michael Somos, Jun 08 2000
Cf. A000217, A047219, A027861, A166080.
Cf. also A212160, A212161, A001652, A001108, A115598, A124124, A098790.

Select[Range[100], !PrimeQ[#^2 + (#+1)^2]& ] (* Jean-François Alcover, Jan 17 2013, after Michael Somos *)

More terms and references from Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Feb 09 2000

A098790 a(n) = 2*a(n-1) + a(n-2) + 1, a(0) = 1, a(1) = 2.

Original entry on oeis.org

1, 2, 6, 15, 37, 90, 218, 527, 1273, 3074, 7422, 17919, 43261, 104442, 252146, 608735, 1469617, 3547970, 8565558, 20679087, 49923733, 120526554, 290976842, 702480239, 1695937321, 4094354882, 9884647086, 23863649055, 57611945197

Offset: 0

Creighton Dement, Oct 30 2004

nonn

Previous name was: a(n) = A048739(n) - A000129(n).

Partial sums of Pell numbers A000129 except omit next-to-last Pell number. E.g., 37 = 0+1+2+5+12+29 - 12.

M. Bicknell-Johnson and G. E. Bergum, The Generalized Fibonacci Numbers {C(n)}, C(n)=C(n-1)+C(n-2)+K, Applications of Fibonacci Numbers, 1986, pp. 193-205.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Bicknell, A Primer on the Pell Sequence and related sequences, Fibonacci Quarterly, Vol. 13, No. 4, 1975, pp. 345-349.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434.
Hermann Stamm-Wilbrandt, 4 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (3, -1, -1).

Cf. A006451, A000129, A048739, A124124, A024537, A074323, A052542.

a[0] = 1; a[1] = 2; a[n_] := a[n] = 2a[n - 1] + a[n - 2] + 1; Table[ a[n], {n, 0, 28}] (* Robert G. Wilson v, Nov 04 2004 *)
LinearRecurrence[{3,-1,-1},{1,2,6},31] (* Harvey P. Dale, Oct 15 2011 *)
CoefficientList[Series[(x^2 - x + 1)/((1 - x) (1 - 2 x - x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 14 2014 *)

a(n) = 2*a(n-1) + a(n-2) + 1, a(0) = 1, a(1) = 2.
G.f.: (x^2-x+1)/((1-x)(1-2x-x^2)).
a(n+1) = - A024537(n+1) + 2*A048739(n+1) - 2*A048739(n).
a(n) = - A024537(n) + A052542(n+1).
Partial sums of A074323. - Paul Barry, Mar 11 2007
a(n) = (sqrt(2)+1)^n*(3/4+sqrt(2)/4)+(sqrt(2)-1)^n*(3/4-sqrt(2)/4)*(-1)^n-1/2; - Paul Barry, Mar 11 2007
a(0)=1, a(1)=2, a(2)=6, a(n)=3*a(n-1)-a(n-2)-a(n-3). [Harvey P. Dale, Oct 15 2011]
a(2*n) = A124124(2*n+1). - Hermann Stamm-Wilbrandt, Aug 03 2014
a(2*n+1) = A006451(2*n+1). - Hermann Stamm-Wilbrandt, Aug 26 2014
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>5. - Hermann Stamm-Wilbrandt, Aug 26 2014
2*a(n) = A135532(n+1)-1. - R. J. Mathar, Jan 13 2023

More terms from Robert G. Wilson v, Nov 04 2004
Definition edited by N. J. A. Sloane, Aug 03 2014
New name from existing formula by Joerg Arndt, Aug 13 2014

A216162 Sequences A006452 and A216134 interlaced.

Original entry on oeis.org

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

A098586 a(n) = (1/2) * (5*P(n+1) + P(n) - 1), where P(k) are the Pell numbers A000129.

Original entry on oeis.org

2, 5, 13, 32, 78, 189, 457, 1104, 2666, 6437, 15541, 37520, 90582, 218685, 527953, 1274592, 3077138, 7428869, 17934877, 43298624, 104532126, 252362877, 609257881, 1470878640, 3551015162, 8572908965, 20696833093, 49966575152, 120629983398, 291226541949

Offset: 0

Creighton Dement, Oct 03 2004

Colin Barker, Table of n, a(n) for n = 0..1000
Hermann Stamm-Wilbrandt, 4 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (3,-1,-1).

Cf. A006451, A124124.

I:=[2,5,13]; [n le 3 select I[n] else 3*Self(n-1) - Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Feb 03 2018

A:= LREtools[REtoproc](a(n) = 3*a(n-1) - a(n-2) - a(n-3), a(n), {a(0)=2, a(1)=5, a(2)=13}):
seq(A(n),n=0..100); # Robert Israel, Aug 26 2014

LinearRecurrence[{3, -1, -1}, {2, 5, 13}, 28] (* Hermann Stamm-Wilbrandt, Aug 26 2014 *)
CoefficientList[Series[(2-x)/((1-x)*(1-2*x-x^2)), {x,0,50}], x] (* G. C. Greubel, Feb 03 2018 *)

Vec((2-x)/((1-x)*(1-2*x-x^2)) + O(x^50)) \\ Colin Barker, Mar 16 2016

a(n) = 3*a(n-1) - a(n-2) - a(n-3) with a(0)=2, a(1)=5, a(2)=13. - Hermann Stamm-Wilbrandt, Aug 26 2014
G.f.: (2-x)/((1-x)*(1-2*x-x^2)). - Robert Israel, Aug 26 2014
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>5. - Hermann Stamm-Wilbrandt, Aug 27 2014
a(2*n-1) = A006451(2*n), for n>0. - Hermann Stamm-Wilbrandt, Aug 27 2014
a(2*n) = A124124(2*n+2). - Hermann Stamm-Wilbrandt, Aug 27 2014
a(n) = (-2+(5-3*sqrt(2))*(1-sqrt(2))^n + (1+sqrt(2))^n*(5+3*sqrt(2)))/4. - Colin Barker, Mar 16 2016

Formula supplied by Thomas Baruchel, Oct 03 2004

More terms from Emeric Deutsch, Nov 17 2004

cp's OEIS Frontend

A038761 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=9.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

Extensions

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A012132 Numbers z such that x*(x+1) + y*(y+1) = z*(z+1) is solvable in positive integers x,y.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

Extensions

A098790 a(n) = 2*a(n-1) + a(n-2) + 1, a(0) = 1, a(1) = 2.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A216162 Sequences A006452 and A216134 interlaced.

A012132 Numbers z such that x(x+1) + y(y+1) = z*(z+1) is solvable in positive integers x,y.