A140716 Blocky integers, i.e., integers m > 1 such that there is a run of m consecutive integer squares the average of which is a square.
7, 25, 31, 49, 55, 73, 79, 97, 103, 121, 127, 145, 151, 169, 175, 193, 199, 217, 223, 241, 247, 265, 271, 289, 295, 313, 319, 337, 343, 361, 367, 385, 391, 409, 415, 433, 439, 457, 463, 481, 487, 505, 511, 529, 535, 553, 559, 577, 583
Offset: 1
Examples
7 is blocky because ((-3)^2 + (-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 + 3^2)/7 = 28/7 = 4 = 2^2.
Links
- David Lovler, Table of n, a(n) for n = 1..10000
- S. Marivani and others, Problem 11227: Consecutive Squares with a Square Average, Amer. Math. Monthly, 115, No. 6, 2008, 568-569.
- Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
Crossrefs
Programs
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Maple
a:=proc(n) if `mod`(n,24)=1 or `mod`(n,24)=-1 or `mod`(n,24)=7 or `mod`(n,24) =-7 then n else end if end proc: seq(a(n),n=2..600);
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Mathematica
Table[12*n - 2 + 3*(-1)^n, {n, 1, 50}] (* Vaclav Kotesovec, Nov 14 2017 *) -
PARI
a(n) = 12*n - 2 + 3*(-1)^n \\ David Lovler, Aug 09 2022
Formula
n is blocky if and only if n>1 and n (mod 24) = 1 or -1 or 7 or -7.
a(n) = 8*(s-1)+1 for n odd, a(n) = 8*(s-1)+7 for n even. - Artur Jasinski, Oct 30 2008
From R. J. Mathar, Nov 25 2008: (Start)
G.f.: x*(7+18*x-x^2)/((1+x)*(1-x)^2).
a(n) = a(n-2) + 24 = 12n - 2 + 3*(-1)^n. (End)
a(n) = a(n-1) + a(n-2) - a(n-3). - Colin Barker, May 12 2012
E.g.f.: (12*x - 2)*exp(x) + 3*exp(-x) - 1. - David Lovler, Aug 09 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = 1 - (1+sqrt(2)/2)*Pi/12 - arccoth(sqrt(3))/(2*sqrt(3)) - arcsinh(sqrt(2))/(2*sqrt(6)). - Amiram Eldar, Aug 23 2022
Comments