A242411 -id:A242411 - cp's OEIS frontend

A243055 Difference between the indices of the smallest and the largest prime dividing n: If n = p_i * ... * p_k, where p_i <= ... <= p_k, where p_h = A000040(h), then a(n) = (k-i), a(1) = 0 by convention.

0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 1, 0, 3, 1, 0, 0, 1, 0, 2, 2, 4, 0, 1, 0, 5, 0, 3, 0, 2, 0, 0, 3, 6, 1, 1, 0, 7, 4, 2, 0, 3, 0, 4, 1, 8, 0, 1, 0, 2, 5, 5, 0, 1, 2, 3, 6, 9, 0, 2, 0, 10, 2, 0, 3, 4, 0, 6, 7, 3, 0, 1, 0, 11, 1, 7, 1, 5, 0, 2, 0, 12, 0, 3, 4, 13, 8, 4, 0, 2

Offset: 1

Antti Karttunen, May 31 2014

nonn

For n>=1, A100484(n+1) gives the position where n occurs for the first time (setting also the records for the sequence).

a(n) = the difference between the largest and the smallest parts of the partition having Heinz number n. We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product(p_j-th prime, j=1...r) (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 2, 4, 10] we get 2*2*3*7*29 = 2436. Example: a(57) = 6; indeed, the partition having Heinz number 57 = 3*19 is [2, 8]. - Emeric Deutsch, Jun 04 2015

Antti Karttunen, Table of n, a(n) for n = 1..10000
Index entries for sequences computed from indices in prime factorization

Differs from A242411 for the first time at n=30.
A000961 gives the positions of zeros.
Cf. A243056, A241917, A241919, A049084, A027748, A055396, A061395, A100484, A215366.

with(numtheory):
a:= n-> `if`(n=1, 0, (f-> pi(max(f[]))-pi(min(f[])))(factorset(n))):
seq(a(n), n=1..100);  # Alois P. Heinz, Jun 04 2015

a[1]=0; a[n_] := Function[{f}, PrimePi[Max[f]] - PrimePi[Min[f]]][FactorInteger[n][[All, 1]]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Jul 29 2015, after Alois P. Heinz *)

from sympy import primepi, primefactors
def A243055(n): return primepi(max(p:=primefactors(n),default=0))-primepi(min(p,default=0)) # Chai Wah Wu, Oct 10 2023

(define (A243055 n) (- (A061395 n) (A055396 n)))

If n = p_i * ... * p_k, where p_i <= ... <= p_k are not necessarily distinct primes (sorted into nondescending order) in the prime factorization of n, where p_i = A000040(i), then a(n) = (k-i).

a(n) = A061395(n) - A055396(n).

A286470 a(n) = maximal gap between indices of successive primes in the prime factorization of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 1, 0, 3, 1, 0, 0, 1, 0, 2, 2, 4, 0, 1, 0, 5, 0, 3, 0, 1, 0, 0, 3, 6, 1, 1, 0, 7, 4, 2, 0, 2, 0, 4, 1, 8, 0, 1, 0, 2, 5, 5, 0, 1, 2, 3, 6, 9, 0, 1, 0, 10, 2, 0, 3, 3, 0, 6, 7, 2, 0, 1, 0, 11, 1, 7, 1, 4, 0, 2, 0, 12, 0, 2, 4, 13, 8, 4, 0, 1, 2, 8, 9, 14, 5, 1, 0, 3, 3, 2, 0, 5, 0, 5, 1, 15, 0, 1, 0, 2, 10, 3, 0, 6, 6, 9, 4, 16, 3, 1

Offset: 1

Antti Karttunen, May 13 2017

nonn

			For n = 70 = 2*5*7 = prime(1)*prime(3)*prime(4), the largest index difference occurs between prime(1) and prime(3), thus a(70) = 3-1 = 2.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A001221, A032742, A055396, A073490, A243055, A286455, A286471, A286472.
Cf. A286469 (version which considers the index of the smallest prime as the initial gap).
Cf. A000961 (positions of zeros).
Differs from A242411 for the first time at n=70, where a(70) = 2, while A242411(70) = 1.

Table[If[Or[n == 1, PrimeNu@ n == 1], 0, Max@ Differences@ PrimePi[FactorInteger[n][[All, 1]]]], {n, 120}] (* Michael De Vlieger, May 16 2017 *)

from sympy import primepi, isprime, primefactors, divisors
def a049084(n): return primepi(n)*(1*isprime(n))
def a055396(n): return 0 if n==1 else a049084(min(primefactors(n)))
def x(n): return 1 if n==1 else divisors(n)[-2]
def a(n): return 0 if n==1 or len(primefactors(n))==1 else max(a055396(x(n)) - a055396(n), a(x(n))) # Indranil Ghosh, May 17 2017

(define (A286470 n) (cond ((or (= 1 n) (= 1 (A001221 n))) 0) (else (max (- (A055396 (A032742 n)) (A055396 n)) (A286470 (A032742 n))))))

a(1) = 0, for n > 1, if A001221(n) = 1 [when n is a prime power], a(n) = 0, otherwise a(n) = max((A055396(A032742(n))-A055396(n)), a(A032742(n))).

For all n >= 1, a(n) <= A243055(n).

Definition corrected by Zak Seidov, May 16 2017

A241917 If n is a prime with index i, p_i, a(n) = i, (with a(1)=0), otherwise difference (i-j) of the indices of the two largest primes p_i, p_j, i >= j in the prime factorization of n: a(n) = A061395(n) - A061395(A052126(n)).

Original entry on oeis.org

0, 1, 2, 0, 3, 1, 4, 0, 0, 2, 5, 1, 6, 3, 1, 0, 7, 0, 8, 2, 2, 4, 9, 1, 0, 5, 0, 3, 10, 1, 11, 0, 3, 6, 1, 0, 12, 7, 4, 2, 13, 2, 14, 4, 1, 8, 15, 1, 0, 0, 5, 5, 16, 0, 2, 3, 6, 9, 17, 1, 18, 10, 2, 0, 3, 3, 19, 6, 7, 1, 20, 0, 21, 11, 0, 7, 1, 4, 22, 2, 0, 12, 23

Offset: 1

Antti Karttunen, May 13 2014

nonn

Note: the two largest primes in the multiset of prime divisors of n are equal for all numbers that are in A070003, thus, after a(1)=0, A070003 gives the positions of the other zeros in this sequence.

Cf. A052126, A061395, A070003, A122111, A241909.
Cf. A049084, A027746.
Cf. A241919, A242411, A243055 for other variants.

a241917 n = i - j where
            (i:j:_) = map a049084 $ reverse (1 : a027746_row n)
-- Reinhard Zumkeller, May 15 2014

A241917(n) = if(isprime(n), primepi(n), if(1>=omega(n), 0, my(f=factor(n)); if(f[#f~,2]>1, 0, primepi(f[#f~,1])-primepi(f[(#f~)-1,1])))); \\ Antti Karttunen, Jul 10 2024

from sympy import primefactors, primepi
def a061395(n): return 0 if n==1 else primepi(primefactors(n)[-1])
def a052126(n): return 1 if n==1 else n/primefactors(n)[-1]
def a(n): return 0 if n==1 else a061395(n) - a061395(a052126(n)) # Indranil Ghosh, May 19 2017

(define (A241917 n) (- (A061395 n) (A061395 (A052126 n))))

a(n) = A061395(n) - A061395(A052126(n)).

A261079 Sum of index differences between prime factors of n, summed over all unordered pairs of primes present (with multiplicity) in the prime factorization of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 2, 0, 3, 1, 0, 0, 2, 0, 4, 2, 4, 0, 3, 0, 5, 0, 6, 0, 4, 0, 0, 3, 6, 1, 4, 0, 7, 4, 6, 0, 6, 0, 8, 2, 8, 0, 4, 0, 4, 5, 10, 0, 3, 2, 9, 6, 9, 0, 7, 0, 10, 4, 0, 3, 8, 0, 12, 7, 6, 0, 6, 0, 11, 2, 14, 1, 10, 0, 8, 0, 12, 0, 10, 4, 13, 8, 12, 0, 6, 2, 16, 9, 14, 5, 5, 0, 6, 6, 8, 0, 12, 0, 15, 4, 15, 0, 6, 0, 8, 10, 12, 0, 14, 6, 18, 8, 16, 3, 10

Offset: 1

Antti Karttunen, Sep 23 2015

nonn

			For n = 1 the prime factorization is empty, thus there is nothing to sum, so a(1) = 0.
For n = 6 = 2*3 = prime(1) * prime(2), a(6) = 1 because the (absolute value of) difference between prime indices of 2 and 3 is 1.
For n = 10 = 2*5 = prime(1) * prime(3), a(10) = 2 because the difference between prime indices of 2 and 5 is 2.
For n = 12 = 2*2*3 = prime(1) * prime(1) * prime(2), a(12) = 2 because the difference between prime indices of 2 and 3 is 1, and the pair (2,3) occurs twice as one can pick either one of the two 2's present in the prime factorization to be a pair of a single 3. Note that the index difference between 2 and 2 is 0, thus the pair (2,2) of prime divisors does not contribute to the sum.
For n = 36 = 2*2*3*3, a(36) = 4 because the index difference between 2 and 3 is 1, and the prime factor pair (2,3) occurs 2^2 = four times in total. As the index difference is zero between 2 and 2 as well as between 3 and 3, the pairs (2,2) and (3,3) do not contribute to the sum.

Antti Karttunen, Table of n, a(n) for n = 1..10000
Index entries for sequences computed from indices in prime factorization

Cf. A000720.
Cf. A000961 (positions of zeros), A006094 (positions of ones).
Cf. A243055, A242411.
Cf. also A260737.
A055396 gives minimum prime index, maximum A061395.
A112798 list prime indices, length A001222, sum A056239.
A304818 adds up partial sums of reversed prime indices, row sums of A359361.
A318283 adds up partial sums of prime indices, row sums of A358136.
Cf. A029931, A316413, A325362, A326836, A326844, A355536, A358137, A359362.

Table[Function[p, Total@ Map[Function[b, Times @@ {First@ Differences@ PrimePi@ b, Count[Subsets[p, {2}], c_ /; SameQ[c, b]]}], Subsets[Union@ p, {2}]]][Flatten@ Replace[FactorInteger@ n, {p_, e_} :> ConstantArray[p, e], 2]], {n, 120}] (* Michael De Vlieger, Mar 08 2017 *)

a(n) = A304818(n) - A318283(n). - Gus Wiseman, Jan 09 2023

a(n) = 2*A304818(n) - A359362(n). - Gus Wiseman, Jan 09 2023

A241919 If n is a prime power, p_i^e, a(n) = i, (with a(1)=0), otherwise difference (i-j) of the indices of the two largest distinct primes p_i, p_j, i > j in the prime factorization of n: a(n) = A061395(n) - A061395(A051119(n)).

Original entry on oeis.org

0, 1, 2, 1, 3, 1, 4, 1, 2, 2, 5, 1, 6, 3, 1, 1, 7, 1, 8, 2, 2, 4, 9, 1, 3, 5, 2, 3, 10, 1, 11, 1, 3, 6, 1, 1, 12, 7, 4, 2, 13, 2, 14, 4, 1, 8, 15, 1, 4, 2, 5, 5, 16, 1, 2, 3, 6, 9, 17, 1, 18, 10, 2, 1, 3, 3, 19, 6, 7, 1, 20, 1, 21, 11, 1, 7, 1, 4, 22, 2, 2, 12, 23

Offset: 1

Antti Karttunen, May 13 2014

nonn

See A242411 and A241917 for other variants.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A241917, A242411, A051119, A061395, A122111.

Cf. A049084, A027748.

a241919 1 = 0
a241919 n = i - j where
            (i:j:_) = map a049084 $ reverse (1 : a027748_row n)
-- Reinhard Zumkeller, May 15 2014

from sympy import factorint, primefactors, primepi
def a061395(n): return 0 if n==1 else primepi(primefactors(n)[-1])
def a053585(n):
    if n==1: return 1
    p = primefactors(n)[-1]
    return p**factorint(n)[p]
def a051119(n): return n/a053585(n)
def a(n): return a061395(n) - a061395(a051119(n)) # Indranil Ghosh, May 19 2017

(define (A241919 n) (- (A061395 n) (A061395 (A051119 n))))

a(n) = A061395(n) - A061395(A051119(n)).

A243056 If n is the i-th prime, p_i = A000040(i), then a(n) = i, otherwise the difference between the indices of the smallest and the largest prime dividing n: for n = p_i * ... * p_k, where p_i <= ... <= p_k, a(n) = (k-i); a(1) = 0 by convention.

Original entry on oeis.org

0, 1, 2, 0, 3, 1, 4, 0, 0, 2, 5, 1, 6, 3, 1, 0, 7, 1, 8, 2, 2, 4, 9, 1, 0, 5, 0, 3, 10, 2, 11, 0, 3, 6, 1, 1, 12, 7, 4, 2, 13, 3, 14, 4, 1, 8, 15, 1, 0, 2, 5, 5, 16, 1, 2, 3, 6, 9, 17, 2, 18, 10, 2, 0, 3, 4, 19, 6, 7, 3, 20, 1, 21, 11, 1, 7, 1, 5, 22, 2, 0, 12, 23, 3, 4, 13, 8, 4, 24, 2

Offset: 1

Antti Karttunen, May 31 2014

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000

Useful when computing A243057 or A243059.
A025475 (prime powers that are not primes) gives the positions of zeros.
Differs from A241917 for the first time at n=18.
Cf. A243055, A241919, A242411, A049084, A027748, A055396, A061395.

(define (A243056 n) (if (prime? n) (A049084 n) (A243055 n)))

a(1) = 0, for n>1, if n = A000040(i), a(n) = i, otherwise a(n) = A061395(n) - A055396(n) = A243055(n).

A286471 If n is noncomposite, then a(n) = 0, otherwise 1 + difference between indices of the two smallest (not necessarily distinct) prime factors of n.

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 0, 1, 1, 3, 0, 1, 0, 4, 2, 1, 0, 2, 0, 1, 3, 5, 0, 1, 1, 6, 1, 1, 0, 2, 0, 1, 4, 7, 2, 1, 0, 8, 5, 1, 0, 2, 0, 1, 1, 9, 0, 1, 1, 3, 6, 1, 0, 2, 3, 1, 7, 10, 0, 1, 0, 11, 1, 1, 4, 2, 0, 1, 8, 3, 0, 1, 0, 12, 2, 1, 2, 2, 0, 1, 1, 13, 0, 1, 5, 14, 9, 1, 0, 2, 3, 1, 10, 15, 6, 1, 0, 4, 1, 1, 0, 2, 0, 1, 2, 16, 0, 1, 0, 3, 11, 1, 0, 2, 7, 1, 1, 17

Offset: 1

Antti Karttunen, May 11 2017

nonn

			For n = 1, 2 and 3, which are all noncomposite numbers, a(n) = 0.
For n = 4 = 2*2 = prime(1)*prime(1), the difference 1-1 = 0, plus one is 1, thus a(4) = 1.
For n = 6 = 2*3 = prime(1)*prime(2), the difference 2-1 = 1, plus one is 2, thus a(6) = 2.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A001222, A032742, A055396.

Cf. also A241917, A241919, A242411, A243055, A243056, A286472.

Table[If[Length@ # < 2, 0, First@ Differences@ PrimePi@ Take[#, 2] + 1] &@ Flatten[FactorInteger[n] /. {p_, e_} /; p > 0 :> ConstantArray[p, e]], {n, 118}] (* Michael De Vlieger, May 12 2017 *)

from sympy import primepi, isprime, primefactors, divisors
def a049084(n): return primepi(n)*(1*isprime(n))
def a055396(n): return 0 if n==1 else a049084(min(primefactors(n)))
def a(n): return 0 if n==1 or isprime(n) else 1 + a055396(divisors(n)[-2]) - a055396(n) # Indranil Ghosh, May 12 2017

(define (A286471 n) (if (or (= 1 n) (= 1 (A001222 n))) 0 (+ 1 (- (A055396 (A032742 n)) (A055396 n)))))

If n is noncomposite, then a(n) = 0, otherwise a(n) = 1 + A055396(A032742(n)) - A055396(n).

A297173 Smallest difference between indices of prime divisors of n, or 0 if n is a prime power.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 1, 0, 3, 1, 0, 0, 1, 0, 2, 2, 4, 0, 1, 0, 5, 0, 3, 0, 1, 0, 0, 3, 6, 1, 1, 0, 7, 4, 2, 0, 1, 0, 4, 1, 8, 0, 1, 0, 2, 5, 5, 0, 1, 2, 3, 6, 9, 0, 1, 0, 10, 2, 0, 3, 1, 0, 6, 7, 1, 0, 1, 0, 11, 1, 7, 1, 1, 0, 2, 0, 12, 0, 1, 4, 13, 8, 4, 0, 1, 2, 8, 9, 14, 5, 1, 0, 3, 3, 2, 0, 1, 0, 5, 1

Offset: 1

Antti Karttunen, Mar 03 2018

nonn

			For n = 130 = 2*5*13 = prime(1)*prime(3)*prime(6), the smallest difference between indices is 3-1 = 2, thus a(130) = 2.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences computed from indices in prime factorization

Cf. A000720, A073490, A073491.

Cf. also A100573, A243055, A242411, A261079, A286470, A286471.

A297173(n) = if(omega(n)<=1,0,my(ps=factor(n)[,1]); vecmin(vector((#ps)-1,i,primepi(ps[i+1])-primepi(ps[i]))));

a(A073491(n)) <= 1.

cp's OEIS Frontend

A243055 Difference between the indices of the smallest and the largest prime dividing n: If n = p_i * ... * p_k, where p_i <= ... <= p_k, where p_h = A000040(h), then a(n) = (k-i), a(1) = 0 by convention.

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A286470 a(n) = maximal gap between indices of successive primes in the prime factorization of n.

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A241917 If n is a prime with index i, p_i, a(n) = i, (with a(1)=0), otherwise difference (i-j) of the indices of the two largest primes p_i, p_j, i >= j in the prime factorization of n: a(n) = A061395(n) - A061395(A052126(n)).

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A261079 Sum of index differences between prime factors of n, summed over all unordered pairs of primes present (with multiplicity) in the prime factorization of n.

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A241919 If n is a prime power, p_i^e, a(n) = i, (with a(1)=0), otherwise difference (i-j) of the indices of the two largest distinct primes p_i, p_j, i > j in the prime factorization of n: a(n) = A061395(n) - A061395(A051119(n)).

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A243056 If n is the i-th prime, p_i = A000040(i), then a(n) = i, otherwise the difference between the indices of the smallest and the largest prime dividing n: for n = p_i * ... * p_k, where p_i <= ... <= p_k, a(n) = (k-i); a(1) = 0 by convention.

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A286471 If n is noncomposite, then a(n) = 0, otherwise 1 + difference between indices of the two smallest (not necessarily distinct) prime factors of n.

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