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 18 results. Next

A251719 a(n) = the least k such that A250474(k) > n.

Original entry on oeis.org

1, 1, 1, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10
Offset: 1

Views

Author

Antti Karttunen, Dec 15 2014

Keywords

Comments

Provided that A250474 is strictly increasing (implied for example if either Legendre's or Brocard's conjecture is true) then all natural numbers occur in this sequence, in order, and after three 1's, each n+1 appears for the first time at A250474(n). Thus from n=2 onward, each n occurs A251723(n-1) times.
With the same provision, we have for n>1: a(n) = smallest positive integer k such that A083221(k, n) is a semiprime and A083221(k+1, n) = A003961(A083221(k, n)), where A003961 shifts the prime factorization one step towards larger primes, thus the latter value is also a semiprime.

Crossrefs

Formula

Equally: a(1) = a(2) = a(3) = 1; and for n>=4: a(n) = the largest k such that A250474(k-1) <= n.
Other identities. For all n >= 1:
a(n) >= A251718(n) >= A251717(n).
a(n) = A055396(A251724(n)), or equally, A251724(n) = A083221(a(n), n). [This sequence gives the row-index of the first "settled semiprime" in column n of the sieve of Eratosthenes.]

A251723 First differences of A054272, A250473 and A250474: a(n) = A054272(n+1) - A054272(n).

Original entry on oeis.org

1, 4, 5, 14, 8, 21, 10, 26, 46, 15, 56, 43, 19, 45, 79, 77, 31, 89, 65, 29, 105, 74, 113, 162, 88, 41, 86, 41, 99, 353, 98, 164, 48, 298, 57, 181, 185, 127, 197, 194, 75, 355, 76, 143, 74, 462, 478, 167, 81, 165, 269, 89, 437, 274, 273, 291, 90, 291, 198, 98, 511, 734, 219, 106, 214, 783, 340, 578, 124, 240, 362, 488, 380, 379, 251, 393, 529, 261, 530, 669, 150, 708, 150
Offset: 1

Views

Author

Antti Karttunen, Dec 15 2014

Keywords

Crossrefs

One less than A050216, the first differences of A000879.

Formula

a(n) = A054272(n+1) - A054272(n).
a(n) = A256447(n) + A256448(n). [Cf. also A256449.]

A256447 Number of integers in range (prime(n)^2)+1 .. (prime(n)*prime(n+1)) whose smallest prime factor is at least prime(n): a(n) = A250477(n) - A250474(n).

Original entry on oeis.org

2, 3, 3, 7, 5, 9, 6, 13, 23, 9, 28, 22, 12, 24, 39, 37, 17, 44, 32, 16, 53, 37, 53, 76, 46, 23, 43, 20, 49, 161, 48, 82, 23, 142, 27, 91, 90, 66, 103, 97, 41, 181, 41, 74, 39, 228, 228, 86, 45, 86, 130, 44, 217, 134, 141, 138, 46, 148, 106, 47, 261, 355, 116, 53, 109, 387, 166, 284, 65, 119, 181, 243, 198, 195, 122, 190, 268, 125, 265, 330, 78
Offset: 1

Views

Author

Antti Karttunen, Mar 29 2015

Keywords

Comments

a(n) = number of integers in range [(prime(n)^2)+1, (prime(n) * prime(n+1))] whose smallest prime factor is at least prime(n).
All the terms are strictly positive, because at least for the last number in the range we have A020639(prime(n)*prime(n+1)) = prime(n).
See the conjectures in A256448.

Examples

			For n=1, we have in range [(prime(1)^2)+1, (prime(1) * prime(2))], that is, in range [5,6], two numbers, 5 and 6, whose smallest prime factor (A020639) is at least 2, thus a(1) = 2.
For n=2, we have in range [10, 15] three numbers, {11, 13, 15}, whose smallest prime factor is at least 3, thus a(2) = 3.
For n=3, we have in range [26, 35] three numbers, {29, 31, 35}, whose smallest prime factor is at least prime(3) = 5, thus a(3) = 3.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Count[Range[Prime[n]^2 + 1, Prime[n] Prime[n + 1]],
      x_ /; Min[First /@ FactorInteger[x]] >=
    Prime@n]; Array[f, 81] (* Michael De Vlieger, Mar 30 2015 *)
  • Scheme
    (define (A256447 n) (- (A250477 n) (A250474 n)))

Formula

a(n) = A250477(n) - A250474(n).
a(n) = A251723(n) - A256448(n).
a(n) = A256448(n) + A256449(n).
a(n) = A256468(n) + 1.
Other identities. For all n >= 1:
a(n+1) = A256446(n) - A256448(n).

A256448 a(n) = A250474(n+1) - A250477(n).

Original entry on oeis.org

-1, 1, 2, 7, 3, 12, 4, 13, 23, 6, 28, 21, 7, 21, 40, 40, 14, 45, 33, 13, 52, 37, 60, 86, 42, 18, 43, 21, 50, 192, 50, 82, 25, 156, 30, 90, 95, 61, 94, 97, 34, 174, 35, 69, 35, 234, 250, 81, 36, 79, 139, 45, 220, 140, 132, 153, 44, 143, 92, 51, 250, 379, 103, 53, 105, 396, 174, 294, 59, 121, 181, 245, 182, 184, 129, 203, 261, 136, 265, 339, 72
Offset: 1

Views

Author

Antti Karttunen, Mar 29 2015

Keywords

Comments

a(n) tells how many more positive integers there are <= prime(n+1)^2 whose smallest prime factor is at least prime(n+1), as compared to how many positive integers there are <= (prime(n) * prime(n+1)) whose smallest prime factor is at least prime(n).
Conjecture 1: for n >= 2, a(n) > 0.
Conjecture 2: ratio a(n)/A256447 converges towards 1. See the associated plots in A256447 and A256449 and comments in A050216.
As what comes to the second conjecture, it's not necessarily true. See the plots linked into A256468. - Antti Karttunen, Mar 30 2015

Examples

			For n=1, the respective primes are prime(1) = 2 and prime(2) = 3, and the ranges in question are [1, 9] and [1, 6]. The former range contains 4 such numbers whose lpf (A020639) is at least 3, namely {3, 5, 7, 9}, while the latter range contains 5 such numbers whose lpf is at least 2, namely {2, 3, 4, 5, 6}, thus a(1) = 4 - 5 = -1.
For n=2, the respective primes are prime(2) = 3 and prime(3) = 5, and the ranges in question are [1, 25] and [1, 15]. The former range contains 8 such numbers whose lpf is at least 5, namely {5, 7, 11, 13, 17, 19, 23, 25}, while the latter range contains 7 such numbers whose lpf is at least 3, namely {3, 5, 7, 9, 11, 13, 15}, thus a(2) = 8 - 7 = 1.
For n=3, the respective primes are prime(3) = 5 and prime(4) = 7, and the ranges in question are [1, 49] and [1, 35]. The former range contains 13 such numbers whose lpf is at least 7, namely {7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49}, while the latter range contains 11 such numbers whose lpf is at least 5, namely {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35}, thus a(3) = 13 - 11 = 2.
		

Crossrefs

Programs

Formula

a(n) = A256469(n) - 2.
a(n) = A250474(n+1) - A250477(n).
a(n) = A251723(n) - A256447(n).
a(n) = A256446(n) - A256447(n+1).
a(n) = A256447(n) - A256449(n).

A078898 Number of times the smallest prime factor of n is the smallest prime factor for numbers <= n; a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 3, 1, 4, 2, 5, 1, 6, 1, 7, 3, 8, 1, 9, 1, 10, 4, 11, 1, 12, 2, 13, 5, 14, 1, 15, 1, 16, 6, 17, 3, 18, 1, 19, 7, 20, 1, 21, 1, 22, 8, 23, 1, 24, 2, 25, 9, 26, 1, 27, 4, 28, 10, 29, 1, 30, 1, 31, 11, 32, 5, 33, 1, 34, 12, 35, 1, 36, 1, 37, 13, 38, 3, 39, 1, 40, 14, 41, 1, 42, 6, 43
Offset: 0

Views

Author

Reinhard Zumkeller, Dec 12 2002

Keywords

Comments

From Antti Karttunen, Dec 06 2014: (Start)
For n >= 2, a(n) tells in which column of the sieve of Eratosthenes (see A083140, A083221) n occurs in. A055396 gives the corresponding row index.
(End)

Crossrefs

Programs

  • Haskell
    import Data.IntMap (empty, findWithDefault, insert)
    a078898 n = a078898_list !! n
    a078898_list = 0 : 1 : f empty 2 where
       f m x = y : f (insert p y m) (x + 1) where
               y = findWithDefault 0 p m + 1
               p = a020639 x
    -- Reinhard Zumkeller, Apr 06 2015
  • Maple
    N:= 1000: # to get a(0) to a(N)
    Primes:= select(isprime, [2,seq(2*i+1,i=1..floor((N-1)/2))]):
    A:= Vector(N):
    for p in Primes do
      t:= 1:
      A[p]:= 1:
      for n from p^2 to N by p do
        if A[n] = 0 then
           t:= t+1:
           A[n]:= t
        fi
      od
    od:
    0,1,seq(A[i],i=2..N); # Robert Israel, Jan 04 2015
  • Mathematica
    Module[{nn=90,spfs},spfs=Table[FactorInteger[n][[1,1]],{n,nn}];Table[ Count[ Take[spfs,i],spfs[[i]]],{i,nn}]] (* Harvey P. Dale, Sep 01 2014 *)
  • PARI
    \\ Not practical for computing, but demonstrates the sum moebius formula:
    A020639(n) = { if(1==n,n,vecmin(factor(n)[, 1])); };
    A055396(n) = { if(1==n,0,primepi(A020639(n))); };
    A002110(n) = prod(i=1, n, prime(i));
    A078898(n) = { my(k,p); if(1==n, n, k = A002110(A055396(n)-1); p = A020639(n); sumdiv(k, d, moebius(d)*(n\(p*d)))); };
    \\ Antti Karttunen, Dec 05 2014
    
  • Scheme
    ;; With memoizing definec-macro.
    (definec (A078898 n) (if (< n 2) n (+ 1 (A078898 (A249744 n)))))
    ;; Much better for computing. Needs also code from A249738 and A249744. - Antti Karttunen, Dec 06 2014
    

Formula

Ordinal transform of A020639 (Lpf). - Franklin T. Adams-Watters, Aug 28 2006
From Antti Karttunen, Dec 05-08 2014: (Start)
a(0) = 0, a(1) = 1, a(n) = 1 + a(A249744(n)).
a(0) = 0, a(1) = 1, a(n) = sum_{d | A002110(A055396(n)-1)} moebius(d) * floor(n / (A020639(n)*d)).
a(0) = 0, a(1) = 1, a(n) = sum_{d | A002110(A055396(n)-1)} moebius(d) * floor(A032742(n) / d).
[Instead of Moebius mu (A008683) one could use Liouville's lambda (A008836) in the above formulas, because all primorials (A002110) are squarefree. A020639(n) gives the smallest prime dividing n, and A055396 gives its index].
a(0) = 0, a(1) = 1, a(2n) = n, a(2n+1) = a(A250470(2n+1)). [After a similar recursive formula for A246277. However, this cannot be used for computing the sequence, unless a definition for A250470(n) is found which doesn't require computing the value of A078898(n).]
For n > 1: a(n) = A249810(n) - A249820(n).
(End)
Other identities:
a(2*n) = n.
For n > 1: a(n)=1 if and only if n is prime.
For n > 1: a(n) = A249808(n, A055396(n)) = A249809(n, A055396(n)).
For n > 1: a(n) = A246277(A249818(n)).
From Antti Karttunen, Jan 04 2015: (Start)
a(n) = 2 if and only if n is a square of a prime.
For all n >= 1: a(A251728(n)) = A243055(A251728(n)) + 2. That is, if n is a semiprime of the form prime(i)*prime(j), prime(i) <= prime(j) < prime(i)^2, then a(n) = (j-i)+2.
(End)
a(A000040(n)^2) = 2; a(A000040(n)*A000040(n+1)) = 3. - Reinhard Zumkeller, Apr 06 2015
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = Sum_{k>=1} (A038110(k)/A038111(k))^2 = 0.2847976823663... . - Amiram Eldar, Oct 26 2024

Extensions

a(0) = 0 prepended for recurrence's sake by Antti Karttunen, Dec 06 2014

A083221 Sieve of Eratosthenes arranged as an array and read by antidiagonals as A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ...

Original entry on oeis.org

2, 4, 3, 6, 9, 5, 8, 15, 25, 7, 10, 21, 35, 49, 11, 12, 27, 55, 77, 121, 13, 14, 33, 65, 91, 143, 169, 17, 16, 39, 85, 119, 187, 221, 289, 19, 18, 45, 95, 133, 209, 247, 323, 361, 23, 20, 51, 115, 161, 253, 299, 391, 437, 529, 29, 22, 57, 125, 203, 319, 377, 493, 551, 667
Offset: 2

Views

Author

Yasutoshi Kohmoto, Jun 05 2003

Keywords

Comments

This is permutation of natural numbers larger than 1.
From Antti Karttunen, Dec 19 2014: (Start)
If we assume here that a(1) = 1 (but which is not explicitly included because outside of the array), then A252460 gives an inverse permutation. See also A249741.
For navigating in this array:
A055396(n) gives the row number of row where n occurs, and A078898(n) gives its column number, both starting their indexing from 1.
A250469(n) gives the number immediately below n, and when n is an odd number >= 3, A250470(n) gives the number immediately above n. If n is a composite, A249744(n) gives the number immediately left of n.
First cube of each row, which is {the initial prime of the row}^3 and also the first number neither a prime or semiprime, occurs on row n at position A250474(n).
(End)
The n-th row contains the numbers whose least prime factor is the n-th prime: A020639(T(n,k)) = A000040(n). - Franklin T. Adams-Watters, Aug 07 2015

Examples

			The top left corner of the array:
   2,   4,   6,    8,   10,   12,   14,   16,   18,   20,   22,   24,   26
   3,   9,  15,   21,   27,   33,   39,   45,   51,   57,   63,   69,   75
   5,  25,  35,   55,   65,   85,   95,  115,  125,  145,  155,  175,  185
   7,  49,  77,   91,  119,  133,  161,  203,  217,  259,  287,  301,  329
  11, 121, 143,  187,  209,  253,  319,  341,  407,  451,  473,  517,  583
  13, 169, 221,  247,  299,  377,  403,  481,  533,  559,  611,  689,  767
  17, 289, 323,  391,  493,  527,  629,  697,  731,  799,  901, 1003, 1037
  19, 361, 437,  551,  589,  703,  779,  817,  893, 1007, 1121, 1159, 1273
  23, 529, 667,  713,  851,  943,  989, 1081, 1219, 1357, 1403, 1541, 1633
  29, 841, 899, 1073, 1189, 1247, 1363, 1537, 1711, 1769, 1943, 2059, 2117
  ...
		

Crossrefs

Transpose of A083140.
One more than A249741.
Inverse permutation: A252460.
Column 1: A000040, Column 2: A001248.
Row 1: A005843, Row 2: A016945, Row 3: A084967, Row 4: A084968, Row 5: A084969, Row 6: A084970.
Main diagonal: A083141.
First semiprime in each column occurs at A251717; A251718 & A251719 with additional criteria. A251724 gives the corresponding semiprimes for the latter. See also A251728.
Permutations based on mapping numbers between this array and A246278: A249817, A249818, A250244, A250245, A250247, A250249. See also: A249811, A249814, A249815.
Also used in the definition of the following arrays of permutations: A249821, A251721, A251722.

Programs

  • Mathematica
    lim = 11; a = Table[Take[Prime[n] Select[Range[lim^2], GCD[# Prime@ n, Product[Prime@ i, {i, 1, n - 1}]] == 1 &], lim], {n, lim}]; Flatten[Table[a[[i, n - i + 1]], {n, lim}, {i, n}]] (* Michael De Vlieger, Jan 04 2016, after Yasutoshi Kohmoto at A083140 *)

Extensions

More terms from Hugo Pfoertner, Jun 13 2003

A050216 Number of primes between (prime(n))^2 and (prime(n+1))^2, with a(0) = 2 by convention.

Original entry on oeis.org

2, 2, 5, 6, 15, 9, 22, 11, 27, 47, 16, 57, 44, 20, 46, 80, 78, 32, 90, 66, 30, 106, 75, 114, 163, 89, 42, 87, 42, 100, 354, 99, 165, 49, 299, 58, 182, 186, 128, 198, 195, 76, 356, 77, 144, 75, 463, 479, 168, 82, 166, 270, 90, 438, 275, 274, 292, 91, 292, 199, 99
Offset: 0

Views

Author

Keywords

Comments

The function in Brocard's Conjecture, which states that for n >= 2, a(n) >= 4.
The lines in the graph correspond to prime gaps of 2, 4, 6, ... . - T. D. Noe, Feb 04 2008
Lengths of blocks of consecutive primes in A000430 (union of primes and squares of primes). - Reinhard Zumkeller, Sep 23 2011
In the n-th step of the sieve of Eratosthenes, all multiples of prime(n) are removed. Then a(n) gives the number of new primes obtained after the n-th step. - Jean-Christophe Hervé, Oct 27 2013
More precisely, after the n-th step, one is sure to have eliminated all composites less than prime(n+1)^2, since any composite N has a prime factor <= sqrt(N). It is in exactly this (restricted) sense that a(n) yields the number of "new primes" (additional numbers known to be prime) after the n-th step. But one knows after the n-th step also that all remaining numbers between prime(n+1)^2 and prime(n+1)*(prime(n+1)+2) are prime: By construction they don't have a factor less than prime(n+1) and they don't have a factor prime(n+1) so the least prime factor could be prime(n+2) >= prime(n+1)+2. For example, after eliminating multiples of 3 in the 2nd step, one has (2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 31, 35, ...) and one knows that all remaining numbers strictly in between 5^2=25 and 5*(5+2)=35 are prime, too. - M. F. Hasler, Dec 31 2014
Numerically, the slope of the lowest "ray" m(n) = min {a(k); k>n}, seems to converge to a value somewhere in the range 1.75 < m(n)/n < 1.8; with m(n)/n > 1.7 for n > 900, m(n)/n > 1.75 for n > 2700. - M. F. Hasler, Dec 31 2014
Legendre's conjecture (see A014085) would imply that a(n) >= 2 for all n and that sequences A054272, A250473 and A250474 were thus strictly increasing (see the Wikipedia article about Brocard's conjecture). - Antti Karttunen, Jan 01 2015
a(n) >= 4 up to at least n = 4*10^5. - Eric W. Weisstein, Jan 13 2025

Examples

			There are 2 primes less than 2^2, there are 2 primes between 2^2 and 3^2, 5 primes between 3^2 and 5^2, etc. [corrected by Jonathan Sperry, Aug 30 2013]
		

References

  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 183.

Crossrefs

First differences of A000879.
One more than A251723.
Cf. A380135 (High water marks for number of primes between prime(n)^2 and prime(n+1)^2).
Cf. A380136 (Positions of the high water marks for number of primes between prime(n)^2 and prime(n+1)^2).

Programs

  • Haskell
    import Data.List (group)
    a050216 n = a050216_list !! (n-1)
    a050216_list =
       map length $ filter (/= [0]) $ group $ map a010051 a000430_list
    -- Reinhard Zumkeller, Sep 23 2011
    
  • Maple
    A050216 := proc(n)
        local p,pn ;
        if n = 0 then
            2;
        else
            p := ithprime(n) ;
            pn := nextprime(p) ;
            numtheory[pi](pn^2)-numtheory[pi](p^2) ;
        end if;
    end proc:
    seq(A050216(n),n=0..40) ; # R. J. Mathar, Jan 27 2025
  • Mathematica
    -Subtract @@@ Partition[PrimePi[Prime[Range[20]]^2], 2, 1] (* Eric W. Weisstein, Jan 10 2025 *)
  • PARI
    a(n)={n||return(2);primepi(prime(n+1)^2)-primepi(prime(n)^2)} \\ M. F. Hasler, Dec 31 2014

Formula

For all n >= 1, a(n) = A256468(n) + A256469(n). - Antti Karttunen, Mar 30 2015
Limit_{N->oo} (Sum_{n=1..N} a(n)) / (Sum_{n=1..N} prime(n)) = 1. - Alain Rocchelli, Sep 30 2023

Extensions

Edited by N. J. A. Sloane, Nov 15 2009

A249821 Square array of permutations: A(row,col) = A246277(A083221(row,col)), read by antidiagonals A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ... .

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 3, 2, 1, 5, 5, 3, 2, 1, 6, 4, 5, 3, 2, 1, 7, 7, 7, 5, 3, 2, 1, 8, 11, 11, 7, 5, 3, 2, 1, 9, 6, 13, 11, 7, 5, 3, 2, 1, 10, 13, 17, 13, 11, 7, 5, 3, 2, 1, 11, 17, 4, 17, 13, 11, 7, 5, 3, 2, 1, 12, 10, 19, 19, 17, 13, 11, 7, 5, 3, 2, 1, 13, 19, 23, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1, 14, 9, 6, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1, 15, 8, 29, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1
Offset: 1

Views

Author

Antti Karttunen, Nov 06 2014

Keywords

Comments

Permutation A249817 preserves the smallest prime factor of n, i.e., A055396(A249817(n)) = A055396(n), in other words, keeps all the terms that appear on any row of A246278 on the same row of A083221. Permutations in this table are induced by changes that A249817 does onto each row of the latter table, thus permutation on row r of this table can be used to sort row r of A246278 into ascending order. I.e., A246278(r, A(r,c)) = A083221(r,c) [the corresponding row in the Sieve of Eratosthenes, where each row appears in monotone order].
The multi-set of cycle-sizes of permutation A249817 is a disjoint union of cycle-sizes of all permutations in this array. For example, A249817 has a 7-cycle (33 39 63 57 99 81 45) which originates from the 7-cycle (6 7 11 10 17 14 8) of A064216, which occurs as the second row in this table.
On each row, 4 is the first composite number (and the first term less than previous, apart from row 1), and on row n it occurs in position A250474(n). This follows because A001222(A246277(n)) = A001222(n)-1 and because on each row of A083221 (see A083140) all terms between the square of prime (second term on each row) and the first cube (of the same prime, this cube mapping in this array to 4) are nonsquare semiprimes (A006881), this implies that the corresponding terms in this array must be primes.
Also, as the smaller prime factor of the terms on row n of A083221 is constant, A020639(n), and for all i < j: A246277(p_{i} * p_{j}) < A246277(p_i * p_{j+1}), the primes on any row appear in monotone order.

Examples

			The top left corner of the array:
1, 2, 3, 4, 5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, ...
1, 2, 3, 5, 4,  7, 11,  6, 13, 17, 10, 19,  9,  8, 23, 29, 14, 15, 31, ...
1, 2, 3, 5, 7, 11, 13, 17,  4, 19, 23,  6, 29, 31, 37, 41,  9, 43, 10, ...
1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,  4, 41, 43, 47, 53, 59, ...
...
		

Crossrefs

Inverse permutations can be found from table A249822.
Row k+1 is a left-to-right composition of the first k rows of A251721.
Row 1: A000027 (an identity permutation), Row 2: A064216, Row 3: A249823, Row 4: A249825.
The initial growing part of each row converges towards A008578.

Programs

Formula

A(row,col) = A246277(A083221(row,col)).
A001222(A(row,col)) = A001222(A083221(row,col)) - 1. [This follows directly from the properties of A246277.]

A251728 Semiprimes p*q for which p <= q < p^2.

Original entry on oeis.org

4, 6, 9, 15, 21, 25, 35, 49, 55, 65, 77, 85, 91, 95, 115, 119, 121, 133, 143, 161, 169, 187, 203, 209, 217, 221, 247, 253, 259, 287, 289, 299, 301, 319, 323, 329, 341, 361, 377, 391, 403, 407, 437, 451, 473, 481, 493, 517, 527, 529, 533, 551, 559, 583, 589, 611, 629, 649, 667, 671, 689, 697, 703
Offset: 1

Views

Author

Antti Karttunen, Dec 16 2014

Keywords

Comments

Semiprimes p*q for which there exists r <= q such that r^k <= p <= q < r^(k+1), for some k >= 1, i.e., semiprimes whose both prime factors fit inside a semiopen range of two consecutive powers of some natural number r which itself is not greater than the larger prime factor. If such r exists, then it must be <= p (the smaller prime factor of n), which forces q to be less than p^2. On the other hand, when p <= q < p^2, then setting r = p and k = 1 satisfies the equation r^k <= p <= q < r^(k+1).
Assuming that A054272(n), the number of primes in interval [p(n), p(n)^2], is nondecreasing (implied for example if Legendre's or Brocard's conjecture is true), it follows that for any a(n), A003961(a(n)) is also in sequence. In other words, whenever prime(i)*prime(j) is in the sequence, then so is also prime(i+1)*prime(j+1).
From above would follow also that these are all the "settled semiprimes" that occur in a square array A083221 constructed from the sieve of Eratosthenes, from the level A251719 downward. Furthermore, this sequence would then be an infinite disjoint union of sequences of A003961-iterates starting from the initial values given in A251724.
See also the comments in the complementary sequence of semiprimes, A138511.
Composite numbers n with all prime factors greater than the cube root of n. - Doug Bell, Oct 27 2015
If "p <= q" in the definition were changed to "p < q" then the squares of primes (A001248) would be removed, yielding A138109. - Jon E. Schoenfield, Dec 27 2022

Crossrefs

An intersection of A251726 and A001358 (semiprimes).
Complement of A138511 in A001358.
A251724 after the initial 2 is a subsequence.

Programs

  • Haskell
    a251728 n = a251728_list !! (n-1)
    a251728_list = filter f [1..] where
                          f x = q < p ^ 2 && a010051' q == 1
                                where q = div x p; p = a020639 x
    -- Reinhard Zumkeller, Jan 06 2015
    
  • Mathematica
    fQ[n_] := Block[{pf = FactorInteger@ n, p, q}, p = pf[[1, 1]]; q = pf[[-1, 1]]; And[p <= q < p^2, PrimeOmega@ n == 2]]; Select[Range@ 720, fQ] (* Michael De Vlieger, Oct 27 2015 *)
  • PARI
    lista(nn) = forcomposite(n=1, nn, my(f = factor(n));if (#select(x->(x > n^(1/3)), f[,1]) == #f~, print1(n, ", "))); \\ Michel Marcus, Oct 27 2015
    
  • PARI
    list(lim)=my(v=List()); forprime(q=2,sqrtnint((lim\1)^2,3), forprime(p=sqrtint(q)+1,min(q,lim\q), listput(v,p*q))); Set(v) \\ Charles R Greathouse IV, Oct 27 2015
    
  • Python
    from math import isqrt
    from sympy import primepi, primerange
    def A251728(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return int(n+x+((t:=primepi(s:=isqrt(x)))*(t-1)>>1)-sum(primepi(min(x//p,p**2)) for p in primerange(s+1)))
        return bisection(f,n,n) # Chai Wah Wu, Mar 05 2025

Formula

For all n >= 1, A078898(a(n)) = A243055(a(n)) + 2.
Limit_{n->oo} n*log(a(n))/a(n) = log(2). - Alain Rocchelli, Nov 10 2022

A054272 Number of primes in the interval [prime(n), prime(n)^2].

Original entry on oeis.org

2, 3, 7, 12, 26, 34, 55, 65, 91, 137, 152, 208, 251, 270, 315, 394, 471, 502, 591, 656, 685, 790, 864, 977, 1139, 1227, 1268, 1354, 1395, 1494, 1847, 1945, 2109, 2157, 2455, 2512, 2693, 2878, 3005, 3202, 3396, 3471, 3826, 3902, 4045, 4119, 4581, 5059
Offset: 1

Views

Author

Labos Elemer, May 05 2000

Keywords

Comments

These primes are candidates for fortunate numbers (A005235).
These are precisely the primes available for the solution of Aguilar's conjecture or Haga's conjecture in Carlos Rivera's The Prime Puzzles and Problems Connection, (conjecture 26). Aguilar's conjecture states that at least one prime will be available for placement on each row and column of a p X p square array. Haga's conjecture states that just p primes are required for such placement in any p X p array. - Enoch Haga, Jan 23 2002
Also number of times p_n (the n-th prime) occurs as the least prime factor (A020639) among numbers in range [(p_n)+1, ((p_n)^3)-1]. For n=1, p_1 = 2 and there are two even numbers in range [3, 7], namely 4 and 6, so a(1) = 2. See also A250474. - Antti Karttunen, Dec 05 2014
The number of consecutive primes after the leading 1 in the prime(n)-rough numbers. - Benedict W. J. Irwin, Mar 24 2016

Examples

			n=4, the zone in question is [7,49] and encloses a(4)=12 primes, as follows: {7,11,13,17,19,23,29,31,37,41,43,47}.
		

Crossrefs

One less than A250473, two less than A250474.
First differences: A251723.

Programs

  • Mathematica
    a[n_] := PrimePi[Prime[n]^2] - n + 1; Array[a, 50] (* Jean-François Alcover, Dec 07 2015 *)
  • PARI
    \\ A fast version:
    default(primelimit, 2^31 + 2^30);
    A054272(n) = 1 + primepi(prime(n)^2) - n;
    for(n=1, 5000, write("b054272.txt", n, " ", A054272(n)));
    \\ The following mirrors the given new formula. It is far from an optimal way to compute this sequence:
    allocatemem(234567890);
    A002110(n) = prod(i=1, n, prime(i));
    A054272(n) = { my(p2); p2 = prime(n)^2; sumdiv(A002110(n), d, moebius(d)*floor(p2/d)); };
    for(n=1, 22, print1(A054272(n),", ")); \\ Antti Karttunen, Dec 05 2014

Formula

a(n) = A000879(n) - n + 1.
From Antti Karttunen, Dec 05-08 2014: (Start)
a(n) = A250473(n) - 1 = A250474(n) - 2.
a(n) = sum_{d | A002110(n)} moebius(d) * floor((p_n)^2 / d). [Where p_n is the n-th prime (A000040(n)) and A002110(n) gives the product of the first n primes. Because the latter is always squarefree, one could also use Liouville's lambda (A008836) instead of Moebius mu (A008683).]
The ratio (a(n) * A002110(n)) / (A001248(n) * A005867(n)) stays near 1, which follows from the above summation formula. See also A249747.
(End)
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