A076294 -id:A076294 - cp's OEIS frontend

A076296 Consider all Pythagorean triples (X,X+7,Z); sequence gives X values.

-3, 0, 5, 8, 21, 48, 65, 140, 297, 396, 833, 1748, 2325, 4872, 10205, 13568, 28413, 59496, 79097, 165620, 346785, 461028, 965321, 2021228, 2687085, 5626320, 11780597, 15661496, 32792613, 68662368, 91281905, 191129372, 400193625, 532029948, 1113983633

Offset: 0

Henry Bottomley, Oct 05 2002

First two terms included for consistency with A076293.
From Klaus Brockhaus, Feb 18 2009: (Start)
Lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
Lim_{n -> infinity} a(n)/a(n-1) = (9+4*sqrt(2))/7 for n mod 3 = {1, 2}.
Lim_{n -> infinity} a(n)/a(n-1) = (3+2*sqrt(2)) / ((9+4*sqrt(2))/7)^2 for n mod 3 = 0. (End)
For the generic case x^2 + (x+p)^2 = y^2 with p=2*m^2-1 a prime number in A066436, m >= 2, the x values are given by the sequence defined by: a(n) = 6*a(n-3) - a(n-6) + 2p with a(1)=0, a(2)=2m+1, a(3)=6m^2-10m+4, a(4)=3p, a(5)=6m^2+10m+4, a(6)=40m^2-58m+21. Y values are given by the sequence defined by: b(n)=6*b(n-3)-b(n-6) with b(1)=p, b(2)=2m^2+2m+1, b(3)=10m^2-14m+5, b(4)=5p, b(5)=10m^2+14m+5, b(6)=58m^2-82m+29. - Mohamed Bouhamida, Sep 09 2009
For the generic case x^2 + (x + p)^2 = y^2 with p = 2*m^2 - 1 a prime number, m>=2, the first three consecutive solutions are: (0;p), (2*m+1; 2*m^2+2*m+1), (6*m^2-10*m+4; 10*m^2-14*m+5) and the other solutions are defined by: (X(n); Y(n))= (3*X(n-3)+2*Y(n-3)+p; 4*X(n-3)+3*Y(n-3)+2*p). - Mohamed Bouhamida, Aug 20 2019

			8 is in the sequence as the shorter leg of the (8,15,17) triangle.

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

Cf. A001652, A076293, A076294, A076295.

Cf. A156035 (decimal expansion of 3+2*sqrt(2)), A156649 (decimal expansion of (9+4*sqrt(2))/7). - Klaus Brockhaus, Feb 18 2009

I:=[-3,0,5,8,21,48,65]; [n le 7 select I[n] else Self(n-1) +6*Self(n-3) -6*Self(n-4) -Self(n-6) +Self(n-7): n in [1..30]]; // G. C. Greubel, May 04 2018

CoefficientList[Series[(3-3x-5x^2-21x^3+5x^4+3x^5+4x^6)/(-1+x+6x^3-6x^4-x^6+x^7),{x,0,50}],x] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2012 *)
LinearRecurrence[{1,0,6,-6,0,-1,1}, {-3,0,5,8,21,48,65}, 50] (* T. D. Noe, Feb 07 2012 *)

x='x+O('x^30); Vec((-3+3*x+5*x^2+21*x^3-5*x^4-3*x^5-4*x^6)/((1-x)*(1-6*x^3 +x^6))) \\ G. C. Greubel, May 04 2018

a(n) = 6a(n-3) - a(n-6) + 14 = (A076293(n) - 7)/2.
a(n) = sqrt(A076294(n)^2 - A076295(n)^2) = A076295(n) - 7.
a(3*n+1) = 7*A001652(n).
From Mohamed Bouhamida, Jul 06 2007: (Start)
a(n) = 5*(a(n-3) + a(n-6)) - a(n-9) + 28.
a(n) = 7*(a(n-3) - a(n-6)) + a(n-9). (End)
G.f.: (-3 + 3*x + 5*x^2 + 21*x^3 - 5*x^4 - 3*x^5 - 4*x^6)/((1-x)*(1 - 6*x^3 + x^6)). - Klaus Brockhaus, Feb 18 2009

More terms from Klaus Brockhaus, Feb 18 2009

A076293 Numbers k where the root mean square (RMS) of k and 7 is an integer, i.e., sqrt((k^2 + 7^2)/2) is an integer.

Original entry on oeis.org

1, 7, 17, 23, 49, 103, 137, 287, 601, 799, 1673, 3503, 4657, 9751, 20417, 27143, 56833, 118999, 158201, 331247, 693577, 922063, 1930649, 4042463, 5374177, 11252647, 23561201, 31322999, 65585233, 137324743, 182563817, 382258751, 800387257, 1064059903

Offset: 0

Henry Bottomley, Oct 05 2002

			17 is in the sequence since sqrt((17^2 + 7^2)/2) = 13 is an integer.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
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.
Index entries for linear recurrences with constant coefficients, signature (0,0,6,0,0,-1).

Cf. A002315, A076294, A076295, A076296.

Column[LinearRecurrence[{0, 0, 6, 0, 0, -1}, {1, 7, 17, 23, 49, 103}, 35] ] (* Vincenzo Librandi, Jul 30 2017 *)

Vec((x+1)*(x^2+3*x+1)^2/(x^6-6*x^3+1) + O(x^100)) \\ Colin Barker, Sep 14 2014

a(n) = 6a(n-3) - a(n-6) = sqrt(2*A076294(n)^2 - 49) = A076295(n) + A076296(n).
a(3n+1) = 7*A002315(n).
G.f.: (x+1)*(x^2+3*x+1)^2 / (x^6-6*x^3+1). - Colin Barker, Sep 14 2014

More terms from Colin Barker, Sep 14 2014

A076295 Consider all Pythagorean triples (Y-7,Y,Z); sequence gives Y values.

Original entry on oeis.org

4, 7, 12, 15, 28, 55, 72, 147, 304, 403, 840, 1755, 2332, 4879, 10212, 13575, 28420, 59503, 79104, 165627, 346792, 461035, 965328, 2021235, 2687092, 5626327, 11780604, 15661503, 32792620, 68662375, 91281912, 191129379, 400193632, 532029955, 1113983640

Offset: 0

Henry Bottomley, Oct 05 2002

nonn

First two terms included for consistency with A076293.

			15 is in the sequence as the longer leg of the (8,15,17) triangle.

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

Cf. A046090, A076293, A076294, A076296.

LinearRecurrence[{1,0,6,-6,0,-1,1},{4,7,12,15,28,55,72},40] (* Harvey P. Dale, Feb 02 2012 *)

a(n) =6a(n-3)-a(n-6)-14 =(A076293(n)+7)/2 =sqrt(A076294(n)^2-A076296(n)^2) =A076296(n)+7.
a(3n+1) = 7*A046090(n).
a(0)=4, a(1)=7, a(2)=12, a(3)=15, a(4)=28, a(5)=55, a(6)=72, a(n)= a(n-1)+ 6*a(n-3)-6*a(n-4)-a(n-6)+a (n-7). - Harvey P. Dale, Feb 02 2012
G.f.: -(3*x^6-3*x^5-5*x^4-21*x^3+5*x^2+3*x+4) / ((x-1)*(x^6-6*x^3+1)). - Colin Barker, Sep 14 2014

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A076296 Consider all Pythagorean triples (X,X+7,Z); sequence gives X values.

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A076293 Numbers k where the root mean square (RMS) of k and 7 is an integer, i.e., sqrt((k^2 + 7^2)/2) is an integer.

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A076295 Consider all Pythagorean triples (Y-7,Y,Z); sequence gives Y values.

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Programs

Mathematica

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