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

35, 83, 1203, 2835, 40883, 96323, 1388835, 3272163, 47179523, 111157235, 1602714963, 3776073843, 54445129235, 128275353443, 1849531679043, 4357585943235, 62829631958243, 148029646716563, 2134357954901235, 5028650402419923

Offset

1,1

Comments

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).

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).

Links

Table of n, a(n) for n=1..20.

Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Formula

G.f.: (35*x+48*x^2-70*x^3+3*x^5)/((1-x)*(1-34*x^2+x^4)).

a(n+4) = 34*a(n+2) - a(n) + 16.

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),(3-2r),-(3-2r)}.

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).

Example

For n=2: a(2)=83; t(83)=83*84/2=3486.
A201633(2)=e(2)=28; 28^2+29^2+30^2+31^2=3486.

Crossrefs

Cf. A201633, A218395.

Sequence in context: A201068 A111144 A267555 * A044173 A044554 A156126

Adjacent sequences: A201629 A201630 A201631 * A201633 A201634 A201635

Keyword

nonn,easy

Author

Paul Weisenhorn, Jan 09 2013

Extensions

Corrected by R. J. Mathar, Jun 14 2016

Status

approved