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Number of distinct prime factors p of n such that p^p is a divisor of n.
44

%I #37 Mar 29 2023 17:50:15

%S 0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,1,1,0,0,0,1,0,0,

%T 0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,

%U 0,0,0,1,0,0,0,1,0,0,0,1,1,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,2,0,0,0,1,0,0,0,1,0,0,0,1

%N Number of distinct prime factors p of n such that p^p is a divisor of n.

%C Average value is A094289 = 0.28735...; attains record values on A076265, in particular a(A076265(n)) = n.

%H Reinhard Zumkeller, <a href="/A129251/b129251.txt">Table of n, a(n) for n = 1..10000</a>

%H Horst Alzer and Man Kam Kwong, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL26/Alzer/alzer15.html">On Sándor's Inequality for the Riemann Zeta Function</a>, J. Int. Seq. (2023) Vol. 26, Article 23.3.6.

%F a(A048103(n)) = 0, a(A100716(n)) > 0.

%F a(n) << sqrt(log n)/log log n. - _Charles R Greathouse IV_, Sep 14 2015

%F From _Antti Karttunen_, Aug 18 2016: (Start)

%F These formulas use Iverson bracket, which gives 1 as its value if the condition given inside [ ] is true and 0 otherwise:

%F a(1) = 0, for n > 1, a(n) = a(A028234(n)) + [A067029(n) >= A020639(n)].

%F Or, for n > 1, a(n) = a(A028234(n)) + [0 = n mod (A020639(n)^A020639(n))].

%F (End)

%F a(n) = Sum_{d|n} [rad(d) = Omega(d)*[omega(d) = 1]], where [ ] is the Iverson bracket. - _Wesley Ivan Hurt_, Feb 09 2022

%F Additive with a(p^e) = 1 if e >= p, and 0 otherwise. - _Amiram Eldar_, Nov 07 2022

%e Since 15 = 3^1 * 5^1, a(15) = 0. But 16 = 2^4 is divisible by 2^2, so a(16) = 1. - _Michael B. Porter_, Aug 18 2016

%t {0}~Join~Table[Count[FactorInteger[n][[All, 1]], _?(Mod[n, #^#] == 0 &)], {n, 2, 120}] (* _Michael De Vlieger_, Oct 30 2019 *)

%o (PARI) a(n)=my(s,t,v); forprime(p=2,, v=valuation(n,p); if(v, n/=p^v; if(v>=p, s++), if(p^p>n, return(s)))) \\ _Charles R Greathouse IV_, Sep 14 2015

%o (Scheme, two variants)

%o (define (A129251 n) (if (= 1 n) 0 (+ (A129251 (A028234 n)) (if (zero? (modulo n (expt (A020639 n) (A020639 n)))) 1 0))))

%o (define (A129251 n) (if (= 1 n) 0 (+ (A129251 (A028234 n)) (if (>= (A067029 n) (A020639 n)) 1 0))))

%o ;; _Antti Karttunen_, Aug 18 2016

%Y Cf. A129252, A020639, A028234, A001221, A051674, A067029, A094289.

%Y Cf. A048103 (indices of zeros), A100716 (nonzeros).

%Y Differs from A276077 for the first time at n=625, where a(625) = 0, while A276077(625) = 1.

%K nonn,easy

%O 1,108

%A _Reinhard Zumkeller_, Apr 07 2007

%E Data section filled up to 120 terms by _Antti Karttunen_, Aug 18 2016