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A145388
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Sum of (k,n)_* for k=1,2,...,n, where (k,n)_* is the greatest divisor of k which is a unitary divisor of n.
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5
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1, 3, 5, 7, 9, 15, 13, 15, 17, 27, 21, 35, 25, 39, 45, 31, 33, 51, 37, 63, 65, 63, 45, 75, 49, 75, 53, 91, 57, 135, 61, 63, 105, 99, 117, 119, 73, 111, 125, 135, 81, 195, 85, 147, 153, 135, 93, 155, 97, 147, 165, 175, 105, 159
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OFFSET
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1,2
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COMMENTS
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The sequence is the row sums of the following triangle of (k,n)_* with rows n and columns 1 <= k <= n (R. J. Mathar, Jun 01 2011):
1;
1, 2;
1, 1, 3;
1, 1, 1, 4;
1, 1, 1, 1, 5;
1, 2, 3, 2, 1, 6;
1, 1, 1, 1, 1, 1, 7;
1, 1, 1, 1, 1, 1, 1, 8;
1, 1, 1, 1, 1, 1, 1, 1, 9;
1, 2, 1, 2, 5, 2, 1, 2, 1, 10;
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11;
1, 1, 3, 4, 1, 3, 1, 4, 3, 1, 1, 12;
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13;
1, 2, 1, 2, 1, 2, 7, 2, 1, 2, 1, 2, 1, 14;
Sum_{k<=x} a(n) = Ax^2 log x + O(x^2) with A = Product(1 - 1/(p+1)^2) * 3/Pi^2 = 0.23584030... where the product is over the primes. That is, the average value of a(n) is A n log n. - Charles R Greathouse IV, Mar 21 2012
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LINKS
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FORMULA
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Multiplicative: a(p^e) = 2*p^e - 1 for every prime power p^e.
a(n) = Sum_{d|n, gcd(d, n/d) = 1} d * uphi(n/d), where uphi is A047994. - Amiram Eldar, May 29 2020
a(n) = Sum_{d|n} abs(A023900(d))*n/d. Verified for the first 10000 terms. - Mats Granvik, Feb 13 2021
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MAPLE
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A145388 := proc(n) option remember; local pf, p ; if n = 1 then 1; else pf := ifactors(n)[2] ; if nops(pf) = 1 then 2*n-1 ; else mul(procname(op(1, p)^op(2, p)), p=pf) ; end if; end if; end proc:
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MATHEMATICA
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f[p_, e_] := 2*p^e - 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, May 29 2020 *)
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PROG
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CROSSREFS
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KEYWORD
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mult,nonn
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AUTHOR
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STATUS
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approved
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