A156572 Squares of the form k^2+(k+23)^2 with integer k.
289, 529, 1369, 4225, 13225, 42025, 139129, 444889, 1423249, 4721929, 15108769, 48344209, 160402225, 513249025, 1642275625, 5448949489, 17435353849, 55789022809, 185103876169, 592288777609, 1895184495649, 6288082836025
Offset: 1
Examples
4225 = 65^2 is of the form k^2+(k+23)^2 with k = 33: 33^2+56^2 = 4225. Hence 4225 is in the sequence.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (1,0,34,-34,0,-1,1).
Crossrefs
Programs
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Mathematica
LinearRecurrence[{1,0,34,-34,0,-1,1}, {289,529,1369,4225,13225,42025,139129}, 30] (* Harvey P. Dale, Mar 21 2020 *) -
PARI
{forstep(n=-8, 1800000, [1, 3], if(issquare(a=2*n*(n+23)+529), print1(a, ",")))} -
Sage
def f(n,p,q): return p*chebyshev_U(n,17) - q*chebyshev_U(n-1,17) def a(n): if (n%3==0): return -289*bool(n==0) + (1/4)*(529 + 3*f(n/3, 209, 5457)) elif (n%3==1): return (1/4)*(529 + 3*f((n-1)/3, 209, 1649)) else: return (1/4)*(529 + 3*f((n-2)/3, 529, 529)) [a(n) for n in (1..30)] # G. C. Greubel, Jan 04 2022
Formula
a(n) = 34*a(n-3) - a(n-6) - 4232 for n > 6; a(1)=289, a(2)=529, a(3)=1369, a(4)=4225, a(5)=13225, a(6)=42025.
a(n) = A156567(n)^2.
G.f.: x*(289 +240*x +840*x^2 -6970*x^3 +840*x^4 +240*x^5 +289*x^6)/((1-x)*(1 -34*x^3 +x^6)).
Limit_{n -> infinity} a(n)/a(n-3) = 17 + 12*sqrt(2).
Limit_{n -> infinity} a(n)/a(n-1) = ((627 + 238*sqrt(2))/23^2)^2 for n mod 3 = 1.
Limit_{n -> infinity} a(n)/a(n-1) = ((27 + 10*sqrt(2))/23)^2 for n mod 3 = {0, 2}.
a(n) = -289*[n=0] + (529/4) + (3/4)*( f(n/3, 209, 5457)*(n mod 3 = 1) + f((n-1)/3, 209, 1649)*(n mod 3 = 1) + f((n-2)/2, 529, 529)*(n mod 3 = 2) ), where f(n, p, q) = p*ChebyshevU(n, 17) - q*ChebyshevU(n-1, 17). - G. C. Greubel, Jan 04 2022
Extensions
Revised by Klaus Brockhaus, Feb 16 2009
G.f. corrected, third comment and cross-references edited by Klaus Brockhaus, Sep 22 2009
Comments