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If n = Product_{k >= 1} (p_k)^(c_k) where p_k is k-th prime and c_k >= 0 then a(n) = Sum_{k >= 1} c_k*((-1)^(k-1)).
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%I #55 Aug 21 2024 21:44:42

%S 0,1,-1,2,1,0,-1,3,-2,2,1,1,-1,0,0,4,1,-1,-1,3,-2,2,1,2,2,0,-3,1,-1,1,

%T 1,5,0,2,0,0,-1,0,-2,4,1,-1,-1,3,-1,2,1,3,-2,3,0,1,-1,-2,2,2,-2,0,1,2,

%U -1,2,-3,6,0,1,1,3,0,1,-1,1,1,0,1,1,0,-1,-1,5,-4,2,1,0,2,0,-2,4,-1,0,-2,3,0,2,0,4,1,-1,-1,4,-1,1,1,2,-1

%N If n = Product_{k >= 1} (p_k)^(c_k) where p_k is k-th prime and c_k >= 0 then a(n) = Sum_{k >= 1} c_k*((-1)^(k-1)).

%C Let p(n,x) be the completely additive polynomial-valued function such that p(1,x) = 0 and p(prime(n),x) = x^(n-1), like is defined in A206284 (although here we are not limited to just irreducible polynomials). Then a(n) is the value of the polynomial encoded in such a manner by n, when it is evaluated at x=-1. - The original definition rewritten and clarified by _Antti Karttunen_, Oct 03 2018

%C Positions of 0 give the values of n for which the polynomial p(n,x) is divisible by x+1. For related sequences, see the Mathematica section.

%C Also the number of odd prime indices of n minus the number of even prime indices of n (both counted with multiplicity), where a prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798. - _Gus Wiseman_, Oct 24 2023

%H Antti Karttunen, <a href="/A195017/b195017.txt">Table of n, a(n) for n = 1..65537</a>

%F Totally additive with a(p^e) = e * (-1)^(1+PrimePi(p)), where PrimePi(n) = A000720(n). - _Antti Karttunen_, Oct 03 2018

%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} = (-1)^(primepi(p)+1)/(p-1) = Sum_{k>=1} (-1)^(k+1)/A006093(k) = A078437 + Sum_{k>=1} (-1)^(k+1)/A036689(k) = 0.6339266524059... . - _Amiram Eldar_, Sep 29 2023

%F a(n) = A257991(n) - A257992(n). - _Gus Wiseman_, Oct 24 2023

%F a(n) = -Sum_{k=1..pi(n)} (-1)^k * valuation(n, prime(k)). - _Friedjof Tellkamp_, Aug 05 2024

%e The sequence can be read from a list of the polynomials:

%e p(n,x) with x = -1, gives a(n)

%e ------------------------------------------

%e p(1,x) = 0 0

%e p(2,x) = 1x^0 1

%e p(3,x) = x -1

%e p(4,x) = 2x^0 2

%e p(5,x) = x^2 1

%e p(6,x) = 1+x 0

%e p(7,x) = x^3 -1

%e p(8,x) = 3x^0 3

%e p(9,x) = 2x -2

%e p(10,x) = x^2 + 1 2.

%e (The list runs through all the polynomials whose coefficients are nonnegative integers.)

%t b[n_] := Table[x^k, {k, 0, n}];

%t f[n_] := f[n] = FactorInteger[n]; z = 200;

%t t[n_, m_, k_] := If[PrimeQ[f[n][[m, 1]]] && f[n][[m, 1]]

%t == Prime[k], f[n][[m, 2]], 0];

%t u = Table[Apply[Plus,

%t Table[Table[t[n, m, k], {k, 1, PrimePi[n]}], {m, 1,

%t Length[f[n]]}]], {n, 1, z}];

%t p[n_, x_] := u[[n]].b[-1 + Length[u[[n]]]]

%t Table[p[n, x] /. x -> 0, {n, 1, z/2}] (* A007814 *)

%t Table[p[2 n, x] /. x -> 0, {n, 1, z/2}] (* A001511 *)

%t Table[p[n, x] /. x -> 1, {n, 1, z}] (* A001222 *)

%t Table[p[n, x] /. x -> 2, {n, 1, z}] (* A048675 *)

%t Table[p[n, x] /. x -> 3, {n, 1, z}] (* A090880 *)

%t Table[p[n, x] /. x -> -1, {n, 1, z}] (* A195017 *)

%t z = 100; Sum[-(-1)^k IntegerExponent[Range[z], Prime[k]], {k, 1, PrimePi[z]}] (* _Friedjof Tellkamp_, Aug 05 2024 *)

%o (PARI) A195017(n) = { my(f); if(1==n, 0, f=factor(n); sum(i=1, #f~, f[i,2] * (-1)^(1+primepi(f[i,1])))); } \\ _Antti Karttunen_, Oct 03 2018

%Y Cf. A206284, A277322, A284010.

%Y For other evaluation functions of such encoded polynomials, see A001222, A048675, A056239, A090880, A248663.

%Y Cf. A006093, A036689, A078437.

%Y Zeros are A325698, distinct A325700.

%Y For sum instead of count we have A366749 = A366531 - A366528.

%Y A000009 counts partitions into odd parts, ranked by A066208.

%Y A035363 counts partitions into even parts, ranked by A066207.

%Y A112798 lists prime indices, reverse A296150, sum A056239.

%Y A257991 counts odd prime indices, even A257992.

%Y A300061 lists numbers with even sum of prime indices, odd A300063.

%Y Cf. A019507, A106529, A239241, A239261, A369180.

%K sign,easy

%O 1,4

%A _Clark Kimberling_, Feb 06 2012

%E More terms, name changed and example-section edited by _Antti Karttunen_, Oct 03 2018