A201633 -id:A201633 - cp's OEIS frontend

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

Paul Weisenhorn, Jan 09 2013

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).

			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.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A201633, A218395.

LinearRecurrence[{1,34,-34,-1,1},{35,83,1203,2835,40883},30] (* Harvey P. Dale, Dec 10 2024 *)

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).

Corrected by R. J. Mathar, Jun 14 2016

A252747 Numbers n such that the hexagonal number H(n) is equal to the sum of four consecutive squares.

Original entry on oeis.org

18, 42, 602, 1418, 20442, 48162, 694418, 1636082, 23589762, 55578618, 801357482, 1888036922, 27222564618, 64137676722, 924765839522, 2178792971618, 31414815979122, 74014823358282, 1067178977450618, 2514325201209962, 36252670417341882, 85413042017780418

Offset: 1

Colin Barker, Dec 21 2014

Also positive integers y in the solutions to 8*x^2-4*y^2+24*x+2*y+28 = 0, the corresponding values of x being A201633.

			18 is in the sequence because H(18) = 630 = 121+144+169+196 = 11^2+12^2+13^2+14^2.

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).

Cf. A000290, A000384, A201633,

Vec(-2*x*(x^4-26*x^2+12*x+9)/((x-1)*(x^2-6*x+1)*(x^2+6*x+1)) + O(x^100))

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

G.f.: -2*x*(x^4-26*x^2+12*x+9) / ((x-1)*(x^2-6*x+1)*(x^2+6*x+1)).

cp's OEIS Frontend

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

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