A201633 Numbers k such that Sum_{j=0..3} (k + j)^2 is a triangular number.
11, 28, 424, 1001, 14453, 34054, 491026, 1156883, 16680479, 39300016, 566645308, 1335043709, 19249260041, 45352186138, 653908196134, 1540639285031, 22213629408563, 52336383504964, 754609491695056, 1777896399883793, 25634509088223389, 60396141212544046
Offset: 1
Examples
For n=3: a(3)=424; 424^2+425^2+426^2+427^2=724206. u(3)=A201632(3)=1203; t(1203)=1203*1204/2=724206.
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).
Programs
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Mathematica
LinearRecurrence[{1,34,-34,-1,1},{11,28,424,1001,14453},30] (* Harvey P. Dale, Apr 16 2013 *) -
PARI
Vec(x*(x^4+x^3-22*x^2-17*x-11)/((x-1)*(x^2-6*x+1)*(x^2+6*x+1)) + O(x^30)) \\ Colin Barker, Dec 21 2014
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Python
from functools import cache @cache def a(n): if n < 6: return [11, 28, 424, 1001, 14453][n-1] return a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5) print([a(n) for n in range(1, 23)]) # Michael S. Branicky, Nov 28 2021
Formula
G.f.: (11*x+17*x^2+22*x^3-x^4-x^5)/((1-x)*(1-34*x^2+x^4)). [corrected by Georg Fischer, May 11 2019]
a(n+4) = 34*a(n-2) - a(n-4) + 48; r=sqrt(2).
a(n+5) = a(n+4) + 34*a(n+3) - 34*a(n+2) - a(n+1) + a(n).
Eigenvalues ej: {1,(3+2r),-(3+2r),(3-2*r),-(3-2*r)}.
a(n+1) = (k1*e1+k2*e2^n+k3*e3^n+k4*e4^n+k5*e5^n)/16 for k1=-24, k2=70+50r, k3=30+21r, k4=70-50r, k5=30-21r.
Extensions
More terms from Colin Barker, Dec 21 2014
Comments