cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 10 results.

A302037 A bigomega (A001222) analog based on the Ludic sieve (A255127): a(1) = 0; for n > 1, a(n) = 1 + a(A302032(n)).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Apr 01 2018

Keywords

Crossrefs

Cf. A003309 (gives the positions of terms <= 1), A302038 (gives the positions of 2's).
Cf. A302031 (an omega-analog), A253557.

Programs

Formula

a(1) = 0; for n > 1, a(n) = 1 + a(A302032(n)).
a(n) = A000120(A269388(n)).
a(n) = A001222(A302026(n)).

A032742 a(1) = 1; for n > 1, a(n) = largest proper divisor of n (that is, for n>1, maximum divisor d of n in range 1 <= d < n).

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 1, 4, 3, 5, 1, 6, 1, 7, 5, 8, 1, 9, 1, 10, 7, 11, 1, 12, 5, 13, 9, 14, 1, 15, 1, 16, 11, 17, 7, 18, 1, 19, 13, 20, 1, 21, 1, 22, 15, 23, 1, 24, 7, 25, 17, 26, 1, 27, 11, 28, 19, 29, 1, 30, 1, 31, 21, 32, 13, 33, 1, 34, 23, 35, 1, 36, 1, 37, 25, 38, 11, 39, 1, 40
Offset: 1

Views

Author

Patrick De Geest, May 15 1998

Keywords

Comments

It seems that a(n) = Max_{j=n+1..2n-1} gcd(n,j). - Labos Elemer, May 22 2002
This is correct: No integer in the range [n+1, 2n-1] has n as its divisor, but certainly at least one multiple of the largest proper divisor of n will occur there (e.g., if it is n/2, then at n + (n/2)). - Antti Karttunen, Dec 18 2014
The slopes of the visible lines made by the points in the scatter plot are 1/2, 1/3, 1/5, 1/7, ... (reciprocals of primes). - Moosa Nasir, Jun 19 2022

Crossrefs

Maximal GCD of k positive integers with sum n for k = 2..10: this sequence (k=2,n>=2), A355249 (k=3), A355319 (k=4), A355366 (k=5), A355368 (k=6), A355402 (k=7), A354598 (k=8), A354599 (k=9), A354601 (k=10).

Programs

  • Haskell
    a032742 n = n `div` a020639 n  -- Reinhard Zumkeller, Oct 03 2012
    
  • Maple
    A032742 :=proc(n) option remember; if n = 1 then 1; else numtheory[divisors](n) minus {n} ; max(op(%)) ; end if; end proc: # R. J. Mathar, Jun 13 2011
    1, seq(n/min(numtheory:-factorset(n)), n=2..1000); # Robert Israel, Dec 18 2014
  • Mathematica
    f[n_] := If[n == 1, 1, Divisors[n][[-2]]]; Table[f[n], {n, 100}] (* Vladimir Joseph Stephan Orlovsky, Mar 03 2010 *)
    Join[{1},Divisors[#][[-2]]&/@Range[2,80]] (* Harvey P. Dale, Nov 29 2011 *)
    a[n_] := n/FactorInteger[n][[1, 1]]; Array[a, 100] (* Amiram Eldar, Nov 26 2020 *)
    Table[Which[n==1,1,PrimeQ[n],1,True,Divisors[n][[-2]]],{n,80}] (* Harvey P. Dale, Feb 02 2022 *)
  • PARI
    a(n)=if(n==1,1,n/factor(n)[1,1]) \\ Charles R Greathouse IV, Jun 15 2011
    
  • Python
    from sympy import factorint
    def a(n): return 1 if n == 1 else n//min(factorint(n))
    print([a(n) for n in range(1, 81)]) # Michael S. Branicky, Jun 21 2022
  • Scheme
    (define (A032742 n) (/ n (A020639 n))) ;; Antti Karttunen, Dec 18 2014
    

Formula

a(n) = n / A020639(n).
Other identities and observations:
A054576(n) = a(a(n)); A117358(n) = a(a(a(n))) = a(A054576(n)); a(A008578(n)) = 1, a(A002808(n)) > 1. - Reinhard Zumkeller, Mar 10 2006
a(n) = A130064(n) / A006530(n). - Reinhard Zumkeller, May 05 2007
a(m)*a(n) < a(m*n) for m and n > 1. - Reinhard Zumkeller, Apr 11 2008
a(m*n) = max(m*a(n), n*a(m)). - Robert Israel, Dec 18 2014
From Antti Karttunen, Mar 31 2018: (Start)
a(n) = n - A060681(n).
For n > 1, a(n) = A003961^(r)(A246277(n)), where r = A055396(n)-1 and A003961^(r)(n) stands for shifting the prime factorization of n by r positions towards larger primes.
For all n >= 1, A276085(a(A276086(n))) = A276151(n).
(End)
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Sum_{k>=1} A005867(k-1)/(prime(k)*A002110(k)) = 0.165049... . - Amiram Eldar, Nov 19 2022

Extensions

Definition clarified by N. J. A. Sloane, Dec 26 2022

A003309 Ludic numbers: apply the same sieve as Eratosthenes, but cross off every k-th remaining number.

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107, 115, 119, 121, 127, 131, 143, 149, 157, 161, 173, 175, 179, 181, 193, 209, 211, 221, 223, 227, 233, 235, 239, 247, 257, 265, 277, 283, 287, 301, 307, 313
Offset: 1

Views

Author

Keywords

Comments

The definition can obviously only be applied from k = a(2) = 2 on: for k = 1, all remaining numbers would be deleted. - M. F. Hasler, Nov 02 2024

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Without the initial 1 occurs as the leftmost column in arrays A255127 and A260717.
Cf. A003310, A003311, A100464, A100585, A100586 (variants).
Cf. A192503 (primes in sequence), A192504 (nonprimes), A192512 (number of terms <= n).
Cf. A192490 (characteristic function).
Cf. A192607 (complement).
Cf. A260723 (first differences).
Cf. A255420 (iterates of f(n) = A003309(n+1) starting from n=1).
Subsequence of A302036.
Cf. A237056, A237126, A237427, A235491, A255407, A255408, A255421, A255422, A260435, A260436, A260741, A260742 (permutations constructed from Ludic numbers).
Cf. also A000959, A008578, A255324, A254100, A272565 (Ludic factor of n), A297158, A302032, A302038.
Cf. A376237 (ludic factorial: cumulative product), A376236 (ludic Fortunate numbers).

Programs

  • Haskell
    a003309 n = a003309_list !! (n - 1)
    a003309_list = 1 : f [2..] :: [Int]
       where f (x:xs) = x : f (map snd [(u, v) | (u, v) <- zip [1..] xs,
                                                 mod u x > 0])
    -- Reinhard Zumkeller, Feb 10 2014, Jul 03 2011
    
  • Maple
    ludic:= proc(N) local i, k,S,R;
      S:= {$2..N};
      R:= 1;
      while nops(S) > 0 do
        k:= S[1];
        R:= R,k;
        S:= subsop(seq(1+k*j=NULL, j=0..floor((nops(S)-1)/k)),S);
      od:
    [R];
    end proc:
    ludic(1000); # Robert Israel, Feb 23 2015
  • Mathematica
    t = Range[2, 400]; r = {1}; While[Length[t] > 0, k = First[t]; AppendTo[r, k]; t = Drop[t, {1, -1, k}];]; r (* Ray Chandler, Dec 02 2004 *)
  • PARI
    t=vector(399,x,x+1); r=[1]; while(length(t)>0, k=t[1];r=concat(r,[k]);t=vector((length(t)*(k-1))\k,x,t[(x*k+k-2)\(k-1)])); r \\ Phil Carmody, Feb 07 2007
    
  • PARI
    A3309=[1]; next_A003309(n)=nn && break); n+!if(n=setsearch(A3309,n+1,1),return(A3309[n])) \\ Should be made more efficient if n >> max(A3309). - M. F. Hasler, Nov 02 2024
    {A003309(n) = while(n>#A3309, next_A003309(A3309[#A3309])); A3309[n]} \\ Should be made more efficient in case n >> #A3309. - M. F. Hasler, Nov 03 2024
    
  • PARI
    upto(nn)= my(r=List([1..nn]), p=1); while(p++<#r, my(k=r[p], i=p); while((i+=k)<=#r, listpop(~r, i); i--)); Vec(r); \\ Ruud H.G. van Tol, Dec 13 2024
    
  • Python
    remainders = [0]
    ludics = [2]
    N_MAX = 313
    for i in range(3, N_MAX) :
        ludic_index = 0
        while ludic_index < len(ludics) :
            ludic = ludics[ludic_index]
            remainder = remainders[ludic_index]
            remainders[ludic_index] = (remainder + 1) % ludic
            if remainders[ludic_index] == 0 :
                break
            ludic_index += 1
        if ludic_index == len(ludics) :
            remainders.append(0)
            ludics.append(i)
    ludics = [1] + ludics
    print(ludics)
    # Alexandre Herrera, Aug 10 2023
    
  • Python
    def A003309(): # generator of the infinite list of ludic numbers
        L = [2, 3]; yield 1; yield 2; yield 3
        while k := len(L)//2: # could take min{k | k >= L[-1-k]-1}
            for j in L[-1-k::-1]: k += 1 + k//(j-1)
            L.append(k+2); yield k+2
    A003309_upto = lambda N=99: [t for t,_ in zip(A003309(),range(N))]
    # M. F. Hasler, Nov 02 2024
  • Scheme
    (define (A003309 n) (if (= 1 n) n (A255127bi (- n 1) 1))) ;; Code for A255127bi given in A255127.
    ;; Antti Karttunen, Feb 23 2015
    

Formula

Complement of A192607; A192490(a(n)) = 1. - Reinhard Zumkeller, Jul 05 2011
From Antti Karttunen, Feb 23 2015: (Start)
a(n) = A255407(A008578(n)).
a(n) = A008578(n) + A255324(n).
(End)

Extensions

More terms from David Applegate and N. J. A. Sloane, Nov 23 2004

A302042 A032742 analog for a nonstandard factorization process based on the sieve of Eratosthenes (A083221).

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 1, 4, 3, 5, 1, 6, 1, 7, 5, 8, 1, 9, 1, 10, 9, 11, 1, 12, 5, 13, 7, 14, 1, 15, 1, 16, 15, 17, 7, 18, 1, 19, 11, 20, 1, 21, 1, 22, 21, 23, 1, 24, 7, 25, 25, 26, 1, 27, 25, 28, 27, 29, 1, 30, 1, 31, 13, 32, 11, 33, 1, 34, 33, 35, 1, 36, 1, 37, 17, 38, 11, 39, 1, 40, 39, 41, 1, 42, 35, 43, 35, 44, 1, 45, 49, 46, 45, 47, 13, 48
Offset: 1

Views

Author

Antti Karttunen, Mar 31 2018

Keywords

Comments

Like [A020639(n), A032742(n)] also ordered pair [A020639(n), a(n)] is unique for each n. Iterating n, a(n), a(a(n)), a(a(a(n))), ..., until 1 is reached, and taking the smallest prime factor (A020639) of each term gives a multiset of primes in ascending order, unique for each natural number n >= 1. Permutation pair A250245/A250246 maps between this non-standard prime factorization of n and the ordinary prime factorization of n.

Examples

			For n = 66, A020639(66) [its smallest prime factor] is 2. Because A055396(66) = A000720(2) = 1, a(66) is just A078898(66) = 66/2 = 33.
For n = 33, A020639(33) = 3 and A055396(33) = 2, so a(33) = A250469(A078898(33)) = A250469(6) = 15. [15 is under 6 in array A083221].
For n = 15, A020639(15) = 3 and A055396(15) = 2, so a(15) = A250469(A078898(15)) = A250469(3) = 5. [5 is under 3 is array A083221].
For n = 5, A020639(5) = 5 and A055396(5) = 3, so a(5) = A250469(A250469(A078898(5))) = A250469(A250469(1)) = 1.
Collecting the primes given by A020639 we get a multiset of factors: [2, 3, 3, 5]. Note that 2*3*3*5 = 90 = A250246(66).
If we start from n = 66, iterating the map n -> A302044(n) [instead of n -> A302042(n)] and apply A020639 to each term obtained we get just a single instance of each prime: [2, 3, 5]. Then by applying A302045 to the same terms we get the corresponding exponents (multiplicities) of those primes: [1, 2, 1].
		

Crossrefs

Cf. also following analogs: A302041 (omega), A253557 (bigomega), A302043, A302044, A302045 (exponent of the least prime present), A302046 (prime signature filter), A302050 (Moebius mu), A302051 (tau), A302052 (char.fun of squares), A302039, A302055 (Arith. derivative).

Programs

  • PARI
    \\ Assuming A250469 and its inverse A268674 have been precomputed, then the following is fast enough:
    A302042(n) = if(1==n,n,my(k=0); while((n%2), n = A268674(n); k++); n = n/2; while(k>0, n = A250469(n); k--); (n));
    
  • PARI
    A020639(n) = if(n>1, if(n>n=factor(n, 0)[1, 1], n, factor(n)[1, 1]), 1); \\ From A020639
    A078898(n) = if(n<=1,n, my(spf=A020639(n),k=1,m=n/spf); while(m>1,if(A020639(m)>=spf,k++); m--); (k));
    \\ Faster if we precompute A078898 as an ordinal transform of A020639:
    up_to = 65537;
    ordinal_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), pt); for(i=1, length(invec), if(mapisdefined(om,invec[i]), pt = mapget(om, invec[i]), pt = 0); outvec[i] = (1+pt); mapput(om,invec[i],(1+pt))); outvec; };
    v078898 = ordinal_transform(vector(up_to,n,A020639(n)));
    A078898(n) = v078898[n];
    A302042(n) = if((1==n)||isprime(n),1,my(c = A078898(n), p = prime(-1+primepi(A020639(n))+primepi(A020639(c))), d = A078898(c), k=0); while(d, k++; if((1==k)||(A020639(k)>=p),d -= 1)); (k*p));

Formula

For n > 1, a(n) = A250469^(r)(A078898(n)), where r = A055396(n)-1 and A250469^(r)(n) stands for applying r times the map x -> A250469(x), starting from x = n.
a(n) = n - A302043(n).

A272565 Smallest ludic factor of n.

Original entry on oeis.org

1, 2, 3, 2, 5, 2, 7, 2, 3, 2, 11, 2, 13, 2, 3, 2, 17, 2, 5, 2, 3, 2, 23, 2, 25, 2, 3, 2, 29, 2, 7, 2, 3, 2, 5, 2, 37, 2, 3, 2, 41, 2, 43, 2, 3, 2, 47, 2, 5, 2, 3, 2, 53, 2, 11, 2, 3, 2, 7, 2, 61, 2, 3, 2, 5, 2, 67, 2, 3, 2, 71, 2, 13, 2, 3, 2, 77, 2, 5, 2, 3
Offset: 1

Views

Author

Max Barrentine, May 09 2016

Keywords

Comments

This sequence is somewhat analogous to the smallest prime factor of n (A020639). However, each natural number has only one ludic factor, because once it is crossed off in the k-th step of the sieve process, it is not a member of the terms considered in the (k+1)-th step.
On the other hand, by iteratively invoking A302032 it is possible to factor n to its constituent "Ludic factors", with each natural number having a unique such decomposition, analogous to prime factorization of n. See comments and examples given in A302032. - Antti Karttunen, Apr 08 2018
The "ludic factor" here is the k which either yields one of the ludic numbers A003309, or is used to cross out a non ludic number. In that case, this "ludic factor" often does not divide n, see A276569. But in the usual sieve of Eratosthenes, the fact that numbers are crossed out from the list does not mean they don't have other factors, so exactly the same could be considered here, which makes disputable the assertion that numbers have only one ludic factor. - M. F. Hasler, Nov 03 2024

Crossrefs

Cf. A003309 (ludic numbers), A020639 (least prime factor), A027748 (prime factors of n), A192607, A255127, A260738, A276440, A276568, A276569, A302032.
Cf. A276347, A276447, A276448 (ludic factor is equal, less than or greater than the smallest prime factor).
Cf. A260739 (ordinal transform), A302036 (numbers with all Ludic factors equal).
Cf. A264940 (analogous version for lucky numbers).

Programs

Formula

From Antti Karttunen, Sep 11 2016: (Start)
a(n) = A003309(1 + A260738(n)).
For all n >= 1, a(A276347(n)) = A020639(A276347(n)). (End)
From M. F. Hasler, Nov 04 2024: (Start)
To rephrase the above: By definition, k is in A276347 iff a(k) = A020639(k).
Particular cases: a(2n) = 2 and a(6n-3) = 3 for all n. (End)

Extensions

Added "smallest" in the definition because the explanation of "only one..." in the first comment might be disputable. - M. F. Hasler, Nov 03 2024

A302026 Permutation of natural numbers mapping "Ludic factorization" to ordinary factorization: a(1) = 1, a(2n) = 2*a(n), a(A269379(n)) = A003961(a(n)).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 25, 20, 27, 22, 19, 24, 23, 26, 21, 28, 29, 30, 49, 32, 45, 34, 35, 36, 31, 50, 33, 40, 37, 54, 41, 44, 81, 38, 43, 48, 125, 46, 75, 52, 47, 42, 121, 56, 63, 58, 77, 60, 53, 98, 39, 64, 55, 90, 59, 68, 135, 70, 61, 72, 169, 62, 51, 100, 67, 66, 175, 80, 99, 74, 71, 108, 343, 82, 105, 88
Offset: 1

Views

Author

Antti Karttunen, Apr 03 2018

Keywords

Comments

See comments and examples in A302032 to see how Ludic factorization proceeds.

Crossrefs

Cf. A302025 (inverse permutation).
Cf. A005940, A250246, A269172, A269388 (similar or related permutations).

Programs

Formula

a(1) = 1, a(2n) = 2*a(n), a(2n+1) = A003961(a(A269380(2n+1))).
a(n) = A250246(A269172(n)).
a(n) = A005940(1+A269388(n)).
Other identities. For all n >= 1:
A001221(a(n)) = A302031(n).
A001222(a(n)) = A302037(n).

A302025 Permutation of natural numbers mapping ordinary factorization to "Ludic factorization": a(1) = 1, a(2n) = 2*a(n), a(A003961(n)) = A269379(a(n)).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 23, 20, 27, 22, 25, 24, 19, 26, 21, 28, 29, 30, 37, 32, 39, 34, 35, 36, 41, 46, 63, 40, 43, 54, 47, 44, 33, 50, 53, 48, 31, 38, 75, 52, 61, 42, 65, 56, 99, 58, 67, 60, 71, 74, 57, 64, 95, 78, 77, 68, 135, 70, 83, 72, 89, 82, 51, 92, 59, 126, 91, 80, 45, 86, 97, 108, 155, 94, 147, 88
Offset: 1

Views

Author

Antti Karttunen, Apr 03 2018

Keywords

Comments

See comments and examples in A302032 to see how Ludic factorization proceeds.

Crossrefs

Cf. A302026 (inverse permutation).
Cf. A156552, A250245, A269171, A269387 (similar or related permutations).

Programs

Formula

a(1) = 1, a(2n) = 2*a(n), a(2n+1) = A269379(a(A064989(2n+1))).
a(n) = A269171(A250245(n)).
a(n) = A269387(A156552(n)).

A302034 A028234 analog for a factorization process based on the Ludic sieve (A255127); Discard all instances of the (smallest) Ludic factor A272565(n) from n.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 1, 3, 1, 7, 5, 1, 1, 9, 1, 5, 1, 11, 1, 3, 1, 13, 7, 7, 1, 15, 1, 1, 5, 17, 7, 9, 1, 19, 11, 5, 1, 21, 1, 11, 1, 23, 1, 3, 1, 25, 19, 13, 1, 27, 1, 7, 7, 29, 11, 15, 1, 31, 13, 1, 11, 33, 1, 17, 5, 35, 1, 9, 1, 37, 17, 19, 1, 39, 7, 5, 11, 41, 1, 21, 1, 43, 35, 11, 1, 45, 1, 23, 1, 47, 13, 3, 1, 49, 23, 25, 1, 51, 13, 13, 19
Offset: 1

Views

Author

Antti Karttunen, Apr 01 2018

Keywords

Comments

Iterating n, a(n), a(a(n)), a(a(a(n))), ..., until 1 is reached, and taking the Ludic factor (A272565) of each term gives a sequence of distinct Ludic numbers (A003309) in ascending order, while applying A302035 to the same terms gives the corresponding "exponents" of these Ludic factors in this nonstandard "Ludic factorization of n", unique for each natural number n >= 1. Permutation pair A302025/A302026 maps between this Ludic factorization and the ordinary prime factorization of n. See also comments and examples in A302032.

Crossrefs

Cf. A302036 (gives the positions of 1's).
Cf. also A028234, A302044.

Programs

  • PARI
    \\ Assuming A269379 and its inverse A269380 have been precomputed, then the following is reasonably fast:
    A302034(n) = if(1==n,n,my(k=0); while((n%2), n = A269380(n); k++); n = (n/2^valuation(n, 2)); while(k>0, n = A269379(n); k--); (n));

Formula

For n > 1, a(n) = A269379^(r)(A000265(A260739(n))), where r = A260738(n)-1 and A269379^(r)(n) stands for applying r times the map x -> A269379(x), starting from x = n.
a(n) = A302025(A028234(A302026(n))).

A302035 a(1) = 0, for n > 1, a(n) = A001511(A260739(n)); Number of instances of (the smallest) Ludic factor A272565(n) in n.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Apr 01 2018

Keywords

Comments

An A067029 analog for "Ludic factorization": iterating the map n -> A302034(n) until 1 is reached, and taking the Ludic factor (A272565) of each term gives a sequence of distinct Ludic numbers (A003309) in ascending order, while applying this function (A302035) to those terms gives the corresponding "exponents" of those Ludic factors, that is, the count of consecutive occurrences of each when iterating the map n -> A302032(n), which gives the same factors with repetitions. Permutation pair A302025/A302026 maps between the Ludic factorization and the ordinary prime factorization of n. See also comments and examples in A302032.

Crossrefs

Formula

a(1) = 0; for n > 1, a(n) = A001511(A260739(n)).
For n > 1, a(n) = A302025(A067029(A302026(n))).

A302038 Semiludic numbers or "Ludic semiprimes": numbers n for which A302037(n) = 2.

Original entry on oeis.org

4, 6, 9, 10, 14, 15, 19, 22, 26, 27, 31, 34, 35, 39, 46, 50, 55, 58, 59, 63, 65, 73, 74, 75, 82, 86, 94, 95, 99, 101, 103, 106, 113, 122, 133, 134, 135, 142, 145, 147, 154, 155, 163, 166, 167, 171, 178, 182, 185, 187, 194, 203, 205, 214, 219, 230, 238, 242, 243, 245, 253, 254, 255, 262, 263, 269, 271, 279, 286, 293, 298, 299, 311
Offset: 1

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Author

Antti Karttunen, Apr 01 2018

Keywords

Comments

An analog for A001358 based on the Ludic factorization (see A302032).

Crossrefs

Cf. A254100 (a subsequence).
Cf. also A001358, A302036.

Programs

  • PARI
    k=0; for(n=1,512,if(2==A302037(n),k++;print1(n,","))); \\ Other code as in A302037.
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