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A201633
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Numbers k such that Sum_{j=0..3} (k + j)^2 is a triangular number.
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3
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11, 28, 424, 1001, 14453, 34054, 491026, 1156883, 16680479, 39300016, 566645308, 1335043709, 19249260041, 45352186138, 653908196134, 1540639285031, 22213629408563, 52336383504964, 754609491695056, 1777896399883793, 25634509088223389, 60396141212544046
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OFFSET
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1,1
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COMMENTS
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Sum_{j=0..3} (a(n)+j)^2 = u(n)*(u(n)+1)/2 = t(u(n)) with A201632(n) = u(n) give the Pell equation c(n)^2 - 32*d(n)^2 = 41. 2*u(n)+1 = c(n) and a(n) + 1.5 = d(n).
Also integers k such that k^2 + (k+1)^2 + (k+2)^2 + (k+3)^2 is equal to a hexagonal number. - Colin Barker, Dec 21 2014
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LINKS
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FORMULA
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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.
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EXAMPLE
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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.
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MATHEMATICA
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LinearRecurrence[{1, 34, -34, -1, 1}, {11, 28, 424, 1001, 14453}, 30] (* Harvey P. Dale, Apr 16 2013 *)
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PROG
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(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
(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)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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