A201633 - OEIS

%I #37 Nov 28 2021 12:44:16

%S 11,28,424,1001,14453,34054,491026,1156883,16680479,39300016,

%T 566645308,1335043709,19249260041,45352186138,653908196134,

%U 1540639285031,22213629408563,52336383504964,754609491695056,1777896399883793,25634509088223389,60396141212544046

%N Numbers k such that Sum_{j=0..3} (k + j)^2 is a triangular number.

%C 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).

%C 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

%H Colin Barker, <a href="/A201633/b201633.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,34,-34,-1,1).

%F 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]

%F a(n+4) = 34*a(n-2) - a(n-4) + 48; r=sqrt(2).

%F a(n+5) = a(n+4) + 34*a(n+3) - 34*a(n+2) - a(n+1) + a(n).

%F Eigenvalues ej: {1,(3+2r),-(3+2r),(3-2*r),-(3-2*r)}.

%F 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.

%e For n=3: a(3)=424; 424^2+425^2+426^2+427^2=724206.

%e u(3)=A201632(3)=1203; t(1203)=1203*1204/2=724206.

%t LinearRecurrence[{1,34,-34,-1,1},{11,28,424,1001,14453},30] (* _Harvey P. Dale_, Apr 16 2013 *)

%o (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

%o (Python)

%o from functools import cache

%o @cache

%o def a(n):

%o     if n < 6: return [11, 28, 424, 1001, 14453][n-1]

%o     return a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5)

%o print([a(n) for n in range(1, 23)]) # _Michael S. Branicky_, Nov 28 2021

%Y Cf. A201632, A218395, A252747.

%K nonn,easy

%O 1,1

%A _Paul Weisenhorn_, Jan 09 2013

%E More terms from _Colin Barker_, Dec 21 2014