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A201632 If the sum of the squares of 4 consecutive numbers is a triangular number t(u), then a(n) is its index u.

%I A201632 #22 Dec 10 2024 11:48:47
%S A201632 35,83,1203,2835,40883,96323,1388835,3272163,47179523,111157235,
%T A201632 1602714963,3776073843,54445129235,128275353443,1849531679043,
%U A201632 4357585943235,62829631958243,148029646716563,2134357954901235,5028650402419923
%N A201632 If the sum of the squares of 4 consecutive numbers is a triangular number t(u), then a(n) is its index u.
%C A201632 Sum_{(e(n)+j)^2,j=0..3} = a(n)*(a(n)+1)/2=t(a(n)) give the Pell equation c(n)^2 - 32*d(n)^2 = 41 with 2*a(n) + 1 = c(n) and e(n) + 1.5 = d(n). e(n) = A201633(n).
%C A201632 In general, for the sum of the squares of k consecutive numbers, one get an analog sequence with k in {4, 5, 6, 7, 11, 15, 17, 19, 23,...}. It gives the Pell equation c(n)^2 - 8k*d(n)^2 = 4*binomial((k+1),3) + 1 with 2*a(n) + 1 = c(n) and e(n) + (k-1)/2 = d(n).
%H A201632 Harvey P. Dale, <a href="/A201632/b201632.txt">Table of n, a(n) for n = 1..1000</a>
%H A201632 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,34,-34,-1,1).
%F A201632 G.f.: (35*x+48*x^2-70*x^3+3*x^5)/((1-x)*(1-34*x^2+x^4)).
%F A201632 a(n+4) = 34*a(n+2) - a(n) + 16.
%F A201632 a(n+5) = a(n+4) + 34*a(n+3) - 34*a(n+2) - a(n+1) + a(n).
%F A201632 eigenvalues ej: {1,(3+2r),-(3+2r),(3-2r),-(3-2r)}.
%F A201632 a(n+1) = (k1*e1 + k2*e2^n + k3*e3^n + k4*e4^n + k5*e5^n)/4 for k1=-2; k2=50+35r; k3=21+15r; k4=50-35r; k5=21-15r, where r = sqrt(2).
%e A201632 For n=2: a(2)=83; t(83)=83*84/2=3486.
%e A201632 A201633(2)=e(2)=28; 28^2+29^2+30^2+31^2=3486.
%t A201632 LinearRecurrence[{1,34,-34,-1,1},{35,83,1203,2835,40883},30] (* _Harvey P. Dale_, Dec 10 2024 *)
%Y A201632 Cf. A201633, A218395.
%K A201632 nonn,easy
%O A201632 1,1
%A A201632 _Paul Weisenhorn_, Jan 09 2013
%E A201632 Corrected by _R. J. Mathar_, Jun 14 2016