A129251 Number of distinct prime factors p of n such that p^p is a divisor of n.

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, 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, 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

Offset: 1

Author

Reinhard Zumkeller, Apr 07 2007

Keywords

• nonn
• easy

Comments

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

Examples

			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
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Horst Alzer and Man Kam Kwong, On Sándor's Inequality for the Riemann Zeta Function, J. Int. Seq. (2023) Vol. 26, Article 23.3.6.

Crossrefs

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

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

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

Programs

Mathematica

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

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

Scheme

(define (A129251 n) (if (= 1 n) 0 (+ (A129251 (A028234 n)) (if (zero? (modulo n (expt (A020639 n) (A020639 n)))) 1 0))))
;; Antti Karttunen, Aug 18 2016

Scheme

(define (A129251 n) (if (= 1 n) 0 (+ (A129251 (A028234 n)) (if (>= (A067029 n) (A020639 n)) 1 0))))
;; Antti Karttunen, Aug 18 2016

Formula

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

a(n) << sqrt(log n)/log log n. - Charles R Greathouse IV, Sep 14 2015

From Antti Karttunen, Aug 18 2016: (Start)

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

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

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

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

Additive with a(p^e) = 1 if e >= p, and 0 otherwise. - Amiram Eldar, Nov 07 2022

Extensions

Data section filled up to 120 terms by Antti Karttunen, Aug 18 2016