A088167 Number of earlier occurring divisors of n; a(1)=1.
1, 1, 2, 3, 2, 5, 2, 5, 3, 7, 2, 8, 2, 8, 6, 9, 2, 12, 2, 11, 5, 10, 2, 16, 5, 10, 5, 11, 2, 21, 2, 15, 6, 12, 8, 19, 2, 14, 4, 24, 2, 21, 2, 18, 11, 15, 2, 28, 3, 23, 5, 17, 2, 24, 11, 24, 6, 17, 2, 37, 2, 19, 9, 24, 8, 29, 2, 23, 7, 31, 2, 41, 2, 23, 13, 25, 8, 29, 2, 38, 7, 24, 2, 40
Offset: 1
Keywords
A354606 a(1) = 1; for n > 1, a(n) is number of terms in the first n-1 terms of the sequence that have the same number of divisors as a(n-1).
1, 1, 2, 1, 3, 2, 3, 4, 1, 4, 2, 5, 6, 1, 5, 7, 8, 2, 9, 3, 10, 3, 11, 12, 1, 6, 4, 4, 5, 13, 14, 5, 15, 6, 7, 16, 1, 7, 17, 18, 2, 19, 20, 3, 21, 8, 9, 6, 10, 11, 22, 12, 4, 7, 23, 24, 1, 8, 13, 25, 8, 14, 15, 16, 2, 26, 17, 27, 18, 5, 28, 6, 19, 29, 30, 2, 31, 32, 7, 33, 20, 8, 21, 22, 23, 34
Offset: 1
Comments
After 250000 terms the most common number of divisors of all terms are 4, 8, 2, 12, 16 divisors. These correspond to the uppermost five lines of the attached image. It is unknown if these stay the most common or are passed by numbers with more divisors as n gets arbitrarily large.
See A355606 for the indices where a(n) = 1.
Examples
a(6) = 2 as a(5) = 3 which has two divisors, and the total number of terms in the first five terms with two divisors is two, namely a(3) = 2 and a(5) = 3.
Links
- Scott R. Shannon, Table of n, a(n) for n = 1..10000
- Scott R. Shannon, Image of the first 250000 terms. The green line is y = n.
- Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20.
Programs
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Mathematica
nn = 120; c[] := 0; a[1] = j = 1; Do[k = ++c[DivisorSigma[0, j]]; Set[{a[n], j}, {k, k}], {n, 2, nn}]; Array[a, nn] (* _Michael De Vlieger, Dec 12 2024 *)
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Python
from sympy import divisor_count from collections import Counter def f(n): return divisor_count(n) def aupton(nn): an, fan, alst, inventory = 1, 1, [1], Counter([1]) for n in range(2, nn+1): an = inventory[fan] fan = f(an) alst.append(an) inventory.update([fan]) return alst print(aupton(86)) # Michael S. Branicky, Jul 09 2022
A362031 a(1) = 1; for n > 1, a(n) is number of terms in the first n-1 terms of the sequence that have the same number of prime factors, counted with multiplicity, as a(n-1).
1, 1, 2, 1, 3, 2, 3, 4, 1, 4, 2, 5, 6, 3, 7, 8, 1, 5, 9, 4, 5, 10, 6, 7, 11, 12, 2, 13, 14, 8, 3, 15, 9, 10, 11, 16, 1, 6, 12, 4, 13, 17, 18, 5, 19, 20, 6, 14, 15, 16, 2, 21, 17, 22, 18, 7, 23, 24, 3, 25, 19, 26, 20, 8, 9, 21, 22, 23, 27, 10, 24, 4, 25, 26, 27, 11, 28, 12, 13, 29, 30, 14, 28
Offset: 1
Keywords
Comments
After 1 million terms the most common numbers for the number of prime factors of the terms are 3, 2, 4, and 5. These correspond to the uppermost four lines of the attached image. It is unknown if these stay the most common or are passed by numbers with more prime factor as n gets arbitrarily large.
See A362033 for the indices where a(n) = 1.
Examples
a(6) = 2 as the number of prime factors of a(5) = A001222(a(5)) = A001222(3) = 1, and there are two previous terms, a(3) and a(5), that have one prime factor. a(9) = 1 as the number of prime factors of a(8) = A001222(a(8)) = A001222(4) = 2, and there is only one term, a(8), that has two prime factors.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
- Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^16, with a color function representing Omega(a(n-1)), where black = 0, red = 1, orange = 2, ..., magenta = 14.
- Scott R. Shannon, Image of the first 1 million terms.
Programs
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Mathematica
nn = 120; c[] = 0; j = a[1] = c[0] = 1; m = 0; Do[Set[k, c[m]]; (Set[{a[n], j, m}, {k, k, #}]; c[#]++) &[PrimeOmega[k]], {n, 2, nn}]; Array[a, nn] (* _Michael De Vlieger, Apr 06 2023 *)
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Python
from sympy import factorint from itertools import islice from collections import Counter def A362031gen(): # generator of terms an, c, d = 1, Counter(), dict() while True: yield an pf = d[an] if an in d else sum(factorint(an).values()) c[pf] += 1 an = c[pf] print(list(islice(A362031gen(), 83))) # Michael S. Branicky, Apr 06 2023
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Python
from itertools import islice from sympy import primeomega def A362031_gen(): # generator of terms a, b, c = {}, {}, 1 while True: yield c d = b[c] = b.get(c,primeomega(c)) c = a[d] = a.get(d,0)+1 A362031_list = list(islice(A362031_gen(),20)) # Chai Wah Wu, Apr 08 2023
A359034 a(n+1) is the sum of the number of terms in all groups of contiguous terms that add up to a(n); a(1)=1.
1, 1, 2, 3, 3, 4, 4, 5, 3, 5, 4, 6, 6, 7, 7, 8, 10, 11, 4, 7, 9, 9, 10, 12, 13, 14, 13, 15, 8, 11, 7, 10, 13, 16, 19, 18, 18, 19, 19, 20, 7, 11, 8, 12, 14, 14, 15, 9, 11, 9, 12, 15, 10, 14, 16, 20, 14, 17, 17, 18, 22, 22, 23, 22, 24, 23, 23, 24, 24, 25, 28, 27, 22
Offset: 1
Keywords
Comments
If strongly smoothened, this sequence displays growth. This growth appears to be caused by the number of groups which is increasing by growing length of the sequence roughly proportional to n^(1/2). But the length of the groups appears to be nearly uninfluenced by this. - Thomas Scheuerle, Dec 14 2022
Examples
a(17) is 10 because in the sequence so far (1, 1, 2, 3, 3, 4, 4, 5, 3, 5, 4, 6, 6, 7, 7, 8), these are the ways of adding contiguous terms to get a(16)=8: (2, 3, 3); (4, 4); (5, 3); (3, 5); (8). This is 10 terms in total, so a(17) is 10. Notice groups (5,3) and (3,5) overlap.
Links
- Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
- Thomas Scheuerle, Scatter plot of a(n) per number of groups found. For n = 1...20000. This is the mean number of contiguous terms for each n.
- Thomas Scheuerle, Scatter plot, number of groups of contiguous terms for n = 1...20000
Programs
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MATLAB
function a = A359034( max_n ) a = [1 1]; for n = 3:max_n s = 1; e = 1; sm = 1; c = 0; while e < n-1 while sm < a(n - 1) && e < (n - 1) e = e + 1; sm = sm + a(e); end if sm == a(n - 1) c = c + (e - s) + 1; end s = s + 1; e = s; sm = a(s); end a(n) = c + 1; end end % Thomas Scheuerle, Dec 14 2022
A362061 a(1) = 1; for n > 1, a(n) is number of terms in the first n-1 terms of the sequence that have the same number of distinct prime factors as a(n-1).
1, 1, 2, 1, 3, 2, 3, 4, 5, 6, 1, 4, 7, 8, 9, 10, 2, 11, 12, 3, 13, 14, 4, 15, 5, 16, 17, 18, 6, 7, 19, 20, 8, 21, 9, 22, 10, 11, 23, 24, 12, 13, 25, 26, 14, 15, 16, 27, 28, 17, 29, 30, 1, 5, 31, 32, 33, 18, 19, 34, 20, 21, 22, 23, 35, 24, 25, 36, 26, 27, 37, 38, 28, 29, 39, 30, 2, 40, 31, 41, 42
Offset: 1
Keywords
Comments
After 5 million terms the most common numbers for the number of distinct prime factors of the terms are 3, 2, 4, 1, and 5, although it is likely these change as n increases.
See A362062 for the indices where a term with k distinct prime factors first appears.
Examples
a(9) = 5 as the number of distinct prime factors of a(8) = A001221(a(8)) = A001221(4) = 1, and there are five previous terms, a(3), a(5) a(6), a(7) and a(8), that have one prime factor. a(11) = 1 as the number of distinct prime factors of a(10) = A001221(a(10)) = A001221(6) = 2, and there is only one term, a(10), that has two prime factors.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
- Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20, with a color function representing omega(a(n-1)), where black = 0, red = 1, yellow = 2, ..., magenta = 6.
- Scott R. Shannon, Image of the first 250000 terms.
Programs
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Mathematica
nn = 120; c[] = 0; j = a[1] = c[0] = 1; m = 0; Do[Set[k, c[m]]; (Set[{a[n], j, m}, {k, k, #}]; c[#]++) &[PrimeNu[k]], {n, 2, nn}]; Array[a, nn] (* _Michael De Vlieger, Apr 06 2023 *)
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Python
from itertools import islice from sympy import primefactors from collections import Counter def A362061gen(): # generator of terms an, c, d = 1, Counter(), dict() while True: yield an dpf = d[an] if an in d else len(primefactors(an)) c[dpf] += 1 an = c[dpf] print(list(islice(A362061gen(), 81))) # Michael S. Branicky, Apr 06 2023
A362077 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that has not yet appeared that is a multiple of Omega(a(n-1)).
1, 2, 3, 4, 6, 8, 9, 10, 12, 15, 14, 16, 20, 18, 21, 22, 24, 28, 27, 30, 33, 26, 32, 5, 7, 11, 13, 17, 19, 23, 25, 34, 36, 40, 44, 39, 38, 42, 45, 48, 35, 46, 50, 51, 52, 54, 56, 60, 64, 66, 57, 58, 62, 68, 63, 69, 70, 72, 55, 74, 76, 75, 78, 81, 80, 65, 82, 84, 88, 92, 87, 86, 90, 96, 102, 93
Offset: 1
Keywords
Comments
Other than the first three terms the only other primes in the first 500000 terms are the consecutive terms a(24)..a(30) = 5, 7, 11, 13, 17, 19, 23. It is unknown if more exist.
In the same range the fixed points are 1, 2, 3, 4, and 48559, although it is possible more exist.
Examples
a(4) = 4 as Omega(a(3)) = A001222(3) = 1, and 4 is the smallest unused number that is a multiple of 1. a(10) = 15 as Omega(a(9)) = A001222(12) = 3, and 15 is the smallest unused number that is a multiple of 3.
Links
- Scott R. Shannon, Table of n, a(n) for n = 1..10000
- Scott R. Shannon, Image of the first 200000 terms. The green line is a(n) = n.
Programs
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Python
from sympy import primeomega from itertools import count, islice def A362077_gen(): # generator of terms a, b = {1,2}, 2 yield from (1,2) while True: for b in count(p:=primeomega(b),p): if b not in a: yield b a.add(b) break A362077_list = list(islice(A362077_gen(),20)) # Chai Wah Wu, Apr 11 2023
A362178 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that has not yet appeared that is a multiple of omega(a(n-1)).
1, 2, 3, 4, 5, 6, 8, 7, 9, 10, 12, 14, 16, 11, 13, 15, 18, 20, 22, 24, 26, 28, 30, 21, 32, 17, 19, 23, 25, 27, 29, 31, 33, 34, 36, 38, 40, 42, 39, 44, 46, 48, 50, 52, 54, 56, 58, 60, 45, 62, 64, 35, 66, 51, 68, 70, 57, 72, 74, 76, 78, 63, 80, 82, 84, 69, 86, 88, 90, 75, 92, 94, 96, 98, 100
Offset: 1
Keywords
Comments
Unlike A362077 numerous primes appear in the sequence; in the first 500000 terms there are seventy-four in total. In the same range there are twelve fixed points, the last being 57. It is unknown whether more exist.
Examples
a(5) = 5 as omega(a(4)) = A001221(4) = 1, and 5 is the smallest unused number that is a multiple of 1. a(7) = 8 as omega(a(6)) = A001221(6) = 2, and 8 is the smallest unused number that is a multiple of 2.
Links
- Scott R. Shannon, Image of the first 500000 terms. The green line is a(n) = n.
Programs
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Python
from itertools import count, islice from sympy import primenu def A362178_gen(): # generator of terms a, b = {1,2}, 2 yield from (1,2) while True: for b in count(p:=primenu(b),p): if b not in a: yield b a.add(b) break A362178_list = list(islice(A362178_gen(),20)) # Chai Wah Wu, Apr 12 2023
A364934 a(n+1) = 1 + number of previous terms that share a factor > 1 with a(n); a(1) = 2.
2, 2, 3, 2, 4, 5, 2, 6, 8, 8, 9, 4, 10, 12, 14, 13, 2, 14, 15, 8, 16, 17, 2, 18, 22, 20, 23, 2, 22, 23, 3, 8, 24, 29, 2, 26, 28, 28, 29, 3, 10, 32, 31, 2, 32, 33, 13, 4, 34, 36, 42, 43, 2, 38, 39, 17, 4, 40, 43, 3, 15, 21, 21, 22, 43, 4, 43, 5, 9, 19, 3, 20, 48, 58, 48, 60, 63, 28, 52, 53, 2, 51, 28
Offset: 1
Keywords
Comments
There are prominent lines that have more terms, their coefficients are approximately: 0.519, 0.329, 0.689, 0.188, 0.615, ... (see the frequency link). They seem to be distorted prime harmonic lines: 1/2, 1/3, 2/3, 1/5, 3/5, ... from A016035.
It appears limsup a(n)/n is approximately 0.83.
Examples
[2,*] 1 term shares a factor with 2, so a(2) = 1+1 = 2. [2,2,*] 2 terms share a factor with 2, so a(3) = 1+2 = 3. [2,2,3,*] 1 term shares a factor with 3, so a(4) = 1+1 = 2. [2,2,3,2,*] 3 terms share a factor with 2, so a(5) = 1+3 = 4. [2,2,3,2,4,*] 4 terms share a factor with 4, so a(6) = 1+4 = 5.
Links
- Rok Cestnik, Table of n, a(n) for n = 1..1000
- Rok Cestnik, a(n)/n frequency for 3*10^5 terms.
- Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^16.
Programs
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Mathematica
nn = 120; c[] := 0; s = {}; a[1] = j = 2; Do[c[j]++; If[c[j] == 1, AppendTo[s, j]]; k = 1 + Total@ Map[c[#] Boole[GCD[#, j] > 1] &, s]; Set[{a[n], j}, {k, k}], {n, 2, nn}]; Array[a, nn] (* _Michael De Vlieger, Aug 16 2023 *)
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Python
from sympy.ntheory import primefactors a=[2]; p = [{2}]; for n in range(1,1000): a.append(sum(1 if p[-1].intersection(p[i]) else 0 for i in range(n))+1) p.append(set(primefactors(a[-1])))
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Python
from math import gcd, prod from sympy import factorint from itertools import islice from collections import Counter def agen(): # generator of terms a=[2]; f=2; c=Counter([f]) while True: yield a[-1] a.append(1 + sum(c[fi] for fi in c if gcd(f,fi)>1)) f=prod(factorint(a[-1]).keys()) c[f] += 1 a=list(islice(agen(), 1000)) # Michael S. Branicky, Aug 16 2023
Comments
Links
Crossrefs
Programs
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Mathematica
Formula