A157472 -id:A157472 - cp's OEIS frontend

A156567 Positive numbers y such that y^2 is of the form x^2+(x+23)^2 with integer x.

17, 23, 37, 65, 115, 205, 373, 667, 1193, 2173, 3887, 6953, 12665, 22655, 40525, 73817, 132043, 236197, 430237, 769603, 1376657, 2507605, 4485575, 8023745, 14615393, 26143847, 46765813, 85184753, 152377507, 272571133, 496493125

Offset: 1

Klaus Brockhaus, Feb 11 2009 , Feb 16 2009

nonn

(-8, a(1)) and(A118337(n), a(n+1)) are solutions (x, y) to the Diophantine equation x^2+(x+23)^2 = y^2.
lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (27+10*sqrt(2))/23 for n mod 3 = {0, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (627+238*sqrt(2))/23^2 for n mod 3 = 1.
For the generic case x^2+(x+p)^2=y^2 with p=m^2-2 a prime number in A028871, 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+2, a(3)=3m^2-10m+8, a(4)=3p, a(5)=3m^2+10m+8, a(6)=20m^2-58m+42.Y values are given by the sequence defined by: b(n)=6*b(n-3)-b(n-6) with b(1)=p, b(2)=m^2+2m+2, b(3)=5m^2-14m+10, b(4)=5p, b(5)=5m^2+14m+10, b(6)=29m^2-82m+58. [From Mohamed Bouhamida, Sep 09 2009]

			(-8, a(1)) = (-8, 17) is a solution: (-8)^2+(-8+23)^2 = 64+225 = 289 = 17^2.
(A118337(1), a(2)) = (0, 23) is a solution: 0^2+(0+23)^2 = 529 = 23^2.
(A118337(3), a(4)) = (33, 65) is a solution: 33^2+(33+23)^2 = 1089+3136 = 4225 = 65^2.

Cf. A118337, A156035 (decimal expansion of 3+2*sqrt(2)), A156571 (decimal expansion of (27+10*sqrt(2))/23), A157472 (decimal expansion of (627+238*sqrt(2))/23^2).

A156570 (first trisection), A156568 (second trisection), A156569 (third trisection).

{forstep(n=-8, 360000000, [1,3], if(issquare(2*n*(n+23)+529, &k), print1(k, ",")))}

a(n) = 6*a(n-3)-a(n-6) for n > 6; a(1)=17, a(2)=23, a(3)=37, a(4)=65, a(5)=115, a(6)=205.

G.f.: x*(1-x)*(17+40*x+77*x^2+40*x^3+17*x^4)/(1-6*x^3+x^6).

G.f. corrected, third and fourth comment edited, cross-reference added by Klaus Brockhaus, Sep 18 2009

A156572 Squares of the form k^2+(k+23)^2 with integer k.

Original entry on oeis.org

289, 529, 1369, 4225, 13225, 42025, 139129, 444889, 1423249, 4721929, 15108769, 48344209, 160402225, 513249025, 1642275625, 5448949489, 17435353849, 55789022809, 185103876169, 592288777609, 1895184495649, 6288082836025

Offset: 1

Klaus Brockhaus, Feb 11 2009

Square roots of k^2+(k+17)^2 are in A156567, values k are in A118337.

			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.

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).

Cf. A156567, A156575 (first trisection), A156573 (second trisection), A156574 (third trisection).

Cf. A118337, A156035 (decimal expansion of 3+2*sqrt(2)), A156164 (decimal expansion of 17+12*sqrt(2)), A156571 (decimal expansion of (27+10*sqrt(2))/23), A157472 (decimal expansion of (627+238*sqrt(2))/23^2).

LinearRecurrence[{1,0,34,-34,0,-1,1}, {289,529,1369,4225,13225,42025,139129}, 30] (* Harvey P. Dale, Mar 21 2020 *)

{forstep(n=-8, 1800000, [1, 3], if(issquare(a=2*n*(n+23)+529), print1(a, ",")))}

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

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

Revised by Klaus Brockhaus, Feb 16 2009

G.f. corrected, third comment and cross-references edited by Klaus Brockhaus, Sep 22 2009

cp's OEIS Frontend

A156567 Positive numbers y such that y^2 is of the form x^2+(x+23)^2 with integer x.

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