A371151 -id:A371151 - cp's OEIS frontend

A080085 Number of factors of 2 in the factorial of the n-th prime, counted with multiplicity.

1, 1, 3, 4, 8, 10, 15, 16, 19, 25, 26, 34, 38, 39, 42, 49, 54, 56, 64, 67, 70, 74, 79, 85, 94, 97, 98, 102, 104, 109, 120, 128, 134, 135, 145, 146, 152, 159, 162, 168, 174, 176, 184, 190, 193, 194, 206, 216, 222, 224, 228, 232, 236, 244, 255, 259, 265, 266, 273, 277

Offset: 1

Paul D. Hanna, Jan 26 2003

nonn

n-th prime minus number of 1's in binary representation of n-th prime. [Juri-Stepan Gerasimov, May 17 2010]

Antti Karttunen, Table of n, a(n) for n = 1..10000 (first 1000 terms from Vincenzo Librandi)

Cf. A276133 (first differences).
Cf. A000040, A080084, A080086, A080087, A294898, A371151.
Column 1 of array A379008, incremented by one.

lst={};Do[p=Prime[n];s=0;While[p>1,p=IntegerPart[p/2];s+=p;];AppendTo[lst,s],{n,5!}];lst (* Vladimir Joseph Stephan Orlovsky, Jul 28 2009 *)

vector(58, n, valuation(prime(n)!, 2)) \\ Arkadiusz Wesolowski, Feb 22 2014

a(n) = prime(n) - hammingweight(prime(n)); \\ Joerg Arndt, Feb 22 2014

a(n) = Sum_{k=1..L} floor( p_n /2^k ), where L = log(p_n)/log(2), where p_n is the n-th prime.
a(n) = A000040(n) - A014499(n). - Juri-Stepan Gerasimov, May 17 2010
a(n) = 1+A294898(A000040(n)). - Antti Karttunen, Dec 14 2024

A362333 Least nonnegative integer k such that (gpf(n)!)^k is divisible by n, where gpf(n) is the greatest prime factor of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 1, 1, 1, 1, 3, 2, 1, 3, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 4, 2, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 2, 4, 1, 1, 1, 1, 1, 1

Offset: 1

Pontus von Brömssen, Apr 16 2023

nonn

First differs from A088388 at n = 40.

			For n = 12, gpf(n)! = 3! = 6 is not divisible by 12, but (3!)^2 = 36 is divisible by 12, so a(12) = 2.

Cf. A006530, A051903, A057109, A088388, A371148, A371149, A371151, A371152.

from sympy import factorint
def A362333(n):
    f = factorint(n)
    gpf = max(f,default=None)
    a = 0
    for p in f:
        m = gpf
        v = 0
        while m >= p:
            m //= p
            v += m
        a = max(a,-(-f[p]//v))
    return a

a(n) > 1 if and only if n is in A057109.
a(n) <= A051903(n).
a(n) = ceiling(A371148(n)/A371149(n)). - Pontus von Brömssen, Mar 16 2024

A371148 Let n = Product_{j=1..k} p_j^e_j and gpf(n)! = Product_{j=1..k} p_j^f_j, where p_j = A000040(j) is the j-th prime and p_k = gpf(n) = A006530(n) is the greatest prime factor of n. a(n) is the numerator of the maximum of e_j/f_j.

Original entry on oeis.org

1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 1, 1, 1, 1, 3, 2, 1, 3, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 4, 2, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 4, 4, 1, 1, 1, 1, 1, 1, 1

Offset: 2

Pontus von Brömssen, Mar 13 2024

			For n = 80 = 2^4 * 3^0 * 5^1, gpf(80)! = 5! = 2^3 * 3^1 * 5^1. The ratios of the prime exponents are 4/3, 0/1, and 1/1, the greatest of which is 4/3, so a(80) = 4.

Pontus von Brömssen, Table of n, a(n) for n = 2..10000

Cf. A000040, A006530, A362333, A371149 (denominators), A371150, A371151.

from sympy import factorint,Rational
def A371148(n):
    f = factorint(n)
    gpf = max(f,default=None)
    a = 0
    for p in f:
        m = gpf
        v = 0
        while m >= p:
            m //= p
            v += m
        a = max(a,Rational(f[p],v))
    return a.p

A362333(n) = ceiling(a(n)/A371149(n)).

A371149 Let n = Product_{j=1..k} p_j^e_j and gpf(n)! = Product_{j=1..k} p_j^f_j, where p_j = A000040(j) is the j-th prime and p_k = gpf(n) = A006530(n) is the greatest prime factor of n. a(n) is the denominator of the maximum of e_j/f_j.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 2

Pontus von Brömssen, Mar 13 2024

			For n = 80 = 2^4 * 3^0 * 5^1, gpf(80)! = 5! = 2^3 * 3^1 * 5^1. The ratios of the prime exponents are 4/3, 0/1, and 1/1, the greatest of which is 4/3, so a(80) = 3.

Pontus von Brömssen, Table of n, a(n) for n = 2..10000

Cf. A000040, A006530, A362333, A371148 (numerators), A371150, A371151.

from sympy import factorint, Rational
def A371149(n):
    f = factorint(n)
    gpf = max(f, default=None)
    a = 0
    for p in f:
        m = gpf
        v = 0
        while m >= p:
            m //= p
            v += m
        a = max(a, Rational(f[p], v))
    return a.q

A362333(n) = ceiling(A371148(n)/a(n)).

cp's OEIS Frontend

A080085 Number of factors of 2 in the factorial of the n-th prime, counted with multiplicity.

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A362333 Least nonnegative integer k such that (gpf(n)!)^k is divisible by n, where gpf(n) is the greatest prime factor of n.

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A371148 Let n = Product_{j=1..k} p_j^e_j and gpf(n)! = Product_{j=1..k} p_j^f_j, where p_j = A000040(j) is the j-th prime and p_k = gpf(n) = A006530(n) is the greatest prime factor of n. a(n) is the numerator of the maximum of e_j/f_j.

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A371149 Let n = Product_{j=1..k} p_j^e_j and gpf(n)! = Product_{j=1..k} p_j^f_j, where p_j = A000040(j) is the j-th prime and p_k = gpf(n) = A006530(n) is the greatest prime factor of n. a(n) is the denominator of the maximum of e_j/f_j.

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