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A178147
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Sum of squares d^2 of distinct divisors of n, d in {2, 3, 5}.
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0
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0, 4, 9, 4, 25, 13, 0, 4, 9, 29, 0, 13, 0, 4, 34, 4, 0, 13, 0, 29, 9, 4, 0, 13, 25, 4, 9, 4, 0, 38, 0, 4, 9, 4, 25, 13, 0, 4, 9, 29, 0, 13, 0, 4, 34, 4, 0, 13, 0, 29, 9, 4, 0, 13, 25, 4, 9, 4, 0, 38, 0, 4, 9, 4, 25, 13, 0, 4, 9, 29, 0, 13, 0, 4, 34, 4, 0, 13, 0
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OFFSET
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1,2
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COMMENTS
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The sequence is periodic with period {0 4 9 4 25 13 0 4 9 29 0 13 0 4 34 4 0 13 0 29 9 4 0 13 25 4 9 4 0 38} of length 30.
A generalization: let B={b_1,...,b_t} be a set of t positive (not necessarily distinct) integers and m>=0 an integer.
For m>=0, let A(n)=Sum d^m over divisors d of n which are elements of B (with the multiplicities as in B). Calculating directly values of
A(b_i),A(b_i+b_j),A(b_i+b_j+b_k),...,
A(b_1+...+b_t), for the other values of A(n) we have the recursion:
A(n)=Sum{1<=i<=t}A(n-b_i)- Sum{1<=i<j<=t}A(n-b_i-b_j)+...+((-1)^(t-1))*A(n-b_1-...-b_t), where we put A(n)=0, if n<0. Such sequence is lcm(b_1,...,b_t)-periodic.
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LINKS
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FORMULA
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a(n)= a(n-2) +a(n-3) -a(n-7)- a(n-8) +a(n-10), n>10.
By the comment, up to 10 it is sufficient to
calculate directly only values a(2)=4, a(3)=9, a(5)=25, a(7)=0, a(8)=4, a(10)=29.
For other n's we can use the recursion, accepting formally a(n)=0 for n<0. So a(1)=0; a(4)=a(2)+a(1)=4;a(6)=a(4)+a(3)=4+9=13,
a(9)=a(7)+a(6)-a(2)-a(1)=0+13-4+0=9.
a(n) = -2*a(n-1) -2*a(n-2) -a(n-3) +a(n-5) +2*a(n-6) +2*a(n-7) +a(n-8). - R. J. Mathar, Jul 13 2010
G.f. -x^2*(4+17*x+30*x^2+55*x^3+80*x^4+38*x^6+76*x^5) / ( (x-1)*(1+x)*(1+x+x^2)*(x^4+x^3+x^2+x+1) ). - R. J. Mathar, Dec 17 2012
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CROSSREFS
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KEYWORD
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nonn,easy,less
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AUTHOR
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STATUS
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approved
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