A178143 Sum of squares d^2 over the divisors d=2 and/or d=3 of n.
0, 4, 9, 4, 0, 13, 0, 4, 9, 4, 0, 13, 0, 4, 9, 4, 0, 13, 0, 4, 9, 4, 0, 13, 0, 4, 9, 4, 0, 13, 0, 4, 9, 4, 0, 13, 0, 4, 9, 4, 0, 13, 0, 4, 9, 4, 0, 13, 0, 4, 9, 4, 0, 13, 0, 4, 9, 4, 0, 13, 0, 4, 9, 4, 0, 13, 0, 4, 9, 4, 0, 13, 0, 4, 9, 4, 0, 13, 0, 4, 9, 4
Offset: 1
Examples
a(1)=0, a(2)=2^2=4 since 2|2, a(3)=3^2=9 since 3|3, a(4)=2^2=4 since 2|4.
Links
- Vladimir Shevelev,A recursion for divisor function over divisors belonging to a prescribed finite sequence of positive integers and a solution of the Lahiri problem for divisor function sigma_x(n), arXiv:0903.1743 [math.NT], 2009.
- Index entries for linear recurrences with constant coefficients, signature (-1,0,1,1).
Programs
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Magma
&cat [[0, 4, 9, 4, 0, 13]^^20]; // Wesley Ivan Hurt, Jul 05 2016
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Maple
seq(op([0, 4, 9, 4, 0, 13]), n=1..30); # Wesley Ivan Hurt, Jul 05 2016
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Mathematica
PadRight[{}, 100, {0, 4, 9, 4, 0, 13}] (* Wesley Ivan Hurt, Jul 05 2016 *) -
PARI
a(n)=[13,0,4,9,4,0][n%6+1] \\ Charles R Greathouse IV, May 21 2013
Formula
a(n) = Sum_{d|n, d=2 and/or d=3} d^2.
a(n) = -a(n-1) + a(n-3) + a(n-4) for n>4.
G.f.: x*(4+13*x+13*x^2) / ( (1-x)*(1+x)*(1+x+x^2) ).
a(n+6) = a(n).
a(n) = 4 * (1 + floor(n/2) - ceiling(n/2)) + 9 * (1 + floor(n/3) - ceiling(n/3)). - Wesley Ivan Hurt, May 20 2013
a(n) = 5 + 2*cos(n*Pi) + 6*cos(2*n*Pi/3). - Wesley Ivan Hurt, Jul 05 2016
Extensions
Replaced recurrence by a shorter one; added keyword:less - R. J. Mathar, May 28 2010
Comments