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.

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

Author

Henry Bottomley, Oct 05 2002

Keywords

• nonn
• easy

Examples

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

Links

• 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).

Crossrefs

Cf. A002315, A076294, A076295, A076296.

Programs

Mathematica

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

PARI

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

Formula

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

Extensions

More terms from Colin Barker, Sep 14 2014