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 17 results. Next

A055396 Smallest prime dividing n is a(n)-th prime (a(1)=0).

Original entry on oeis.org

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

Views

Author

Henry Bottomley, May 15 2000

Keywords

Comments

Grundy numbers of the game in which you decrease n by a number prime to n, and the game ends when 1 is reached. - Eric M. Schmidt, Jul 21 2013
a(n) = the smallest part 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(21) = 2; indeed, the partition having Heinz number 21 = 3*7 is [2,4]. - Emeric Deutsch, Jun 04 2015
a(n) is the number of numbers whose largest proper divisor is n, i.e., for n>1, number of occurrences of n in A032742. - Stanislav Sykora, Nov 04 2016
For n > 1, a(n) gives the number of row where n occurs in arrays A083221 and A246278. - Antti Karttunen, Mar 07 2017

Examples

			a(15) = 2 because 15=3*5, 3<5 and 3 is the 2nd prime.
		

References

  • John H. Conway, On Numbers and Games, 2nd Edition, p. 129.

Crossrefs

Programs

  • Haskell
    a055396 = a049084 . a020639  -- Reinhard Zumkeller, Apr 05 2012
    
  • Maple
    with(numtheory):
    a:= n-> `if`(n=1, 0, pi(min(factorset(n)[]))):
    seq(a(n), n=1..100);  # Alois P. Heinz, Aug 03 2013
  • Mathematica
    a[1] = 0; a[n_] := PrimePi[ FactorInteger[n][[1, 1]] ]; Table[a[n], {n, 1, 96}](* Jean-François Alcover, Jun 11 2012 *)
  • PARI
    a(n)=if(n==1,0,primepi(factor(n)[1,1])) \\ Charles R Greathouse IV, Apr 23 2015
    
  • Python
    from sympy import primepi, isprime, primefactors
    def a049084(n): return primepi(n)*(1*isprime(n))
    def a(n): return 0 if n==1 else a049084(min(primefactors(n))) # Indranil Ghosh, May 05 2017

Formula

From Reinhard Zumkeller, May 22 2003: (Start)
a(n) = A049084(A020639(n)).
A000040(a(n)) = A020639(n); a(n) <= A061395(n).
(End)
From Antti Karttunen, Mar 07 2017: (Start)
A243055(n) = A061395(n) - a(n).
a(A276086(n)) = A257993(n).
(End)

A007310 Numbers congruent to 1 or 5 mod 6.

Original entry on oeis.org

1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 121, 125, 127, 131, 133, 137, 139, 143, 145, 149, 151, 155, 157, 161, 163, 167, 169, 173, 175
Offset: 1

Views

Author

C. Christofferson (Magpie56(AT)aol.com)

Keywords

Comments

Numbers n such that phi(4n) = phi(3n). - Benoit Cloitre, Aug 06 2003
Or, numbers relatively prime to 2 and 3, or coprime to 6, or having only prime factors >= 5; also known as 5-rough numbers. (Edited by M. F. Hasler, Nov 01 2014: merged with comments from Zak Seidov, Apr 26 2007 and Michael B. Porter, Oct 09 2009)
Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 38 ).
Numbers k such that k mod 2 = 1 and (k+1) mod 3 <> 1. - Klaus Brockhaus, Jun 15 2004
Also numbers n such that the sum of the squares of the first n integers is divisible by n, or A000330(n) = n*(n+1)*(2*n+1)/6 is divisible by n. - Alexander Adamchuk, Jan 04 2007
Numbers n such that the sum of squares of n consecutive integers is divisible by n, because A000330(m+n) - A000330(m) = n*(n+1)*(2*n+1)/6 + n*(m^2+n*m+m) is divisible by n independent of m. - Kaupo Palo, Dec 10 2016
A126759(a(n)) = n + 1. - Reinhard Zumkeller, Jun 16 2008
Terms of this sequence (starting from the second term) are equal to the result of the expression sqrt(4!*(k+1) + 1) - but only when this expression yields integral values (that is when the parameter k takes values, which are terms of A144065). - Alexander R. Povolotsky, Sep 09 2008
For n > 1: a(n) is prime if and only if A075743(n-2) = 1; a(2*n-1) = A016969(n-1), a(2*n) = A016921(n-1). - Reinhard Zumkeller, Oct 02 2008
A156543 is a subsequence. - Reinhard Zumkeller, Feb 10 2009
Numbers n such that ChebyshevT(x, x/2) is not an integer (is integer/2). - Artur Jasinski, Feb 13 2010
If 12*k + 1 is a perfect square (k = 0, 2, 4, 10, 14, 24, 30, 44, ... = A152749) then the square root of 12*k + 1 = a(n). - Gary Detlefs, Feb 22 2010
A089128(a(n)) = 1. Complement of A047229(n+1) for n >= 1. See A164576 for corresponding values A175485(a(n)). - Jaroslav Krizek, May 28 2010
Cf. property described by Gary Detlefs in A113801 and in Comment: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (with h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 6). Also a(n)^2 - 1 == 0 (mod 12). - Bruno Berselli, Nov 05 2010 - Nov 17 2010
Numbers n such that ( Sum_{k = 1..n} k^14 ) mod n = 0. (Conjectured) - Gary Detlefs, Dec 27 2011
From Peter Bala, May 02 2018: (Start)
The above conjecture is true. Apply Ireland and Rosen, Proposition 15.2.2. with m = 14 to obtain the congruence 6*( Sum_{k = 1..n} k^14 )/n = 7 (mod n), true for all n >= 1. Suppose n is coprime to 6, then 6 is a unit in Z/nZ, and it follows from the congruence that ( Sum_{k = 1..n} k^14 )/n is an integer. On the other hand, if either 2 divides n or 3 divides n then the congruence shows that ( Sum_{k = 1..n} k^14 )/n cannot be integral. (End)
A126759(a(n)) = n and A126759(m) < n for m < a(n). - Reinhard Zumkeller, May 23 2013
(a(n-1)^2 - 1)/24 = A001318(n), the generalized pentagonal numbers. - Richard R. Forberg, May 30 2013
Numbers k for which A001580(k) is divisible by 3. - Bruno Berselli, Jun 18 2014
Numbers n such that sigma(n) + sigma(2n) = sigma(3n). - Jahangeer Kholdi and Farideh Firoozbakht, Aug 15 2014
a(n) are values of k such that Sum_{m = 1..k-1} m*(k-m)/k is an integer. Sums for those k are given by A062717. Also see Detlefs formula below based on A062717. - Richard R. Forberg, Feb 16 2015
a(n) are exactly those positive integers m such that the sequence b(n) = n*(n + m)*(n + 2*m)/6 is integral, and also such that the sequence c(n) = n*(n + m)*(n + 2*m)*(n + 3*m)/24 is integral. Cf. A007775. - Peter Bala, Nov 13 2015
Along with 2, these are the numbers k such that the k-th Fibonacci number is coprime to every Lucas number. - Clark Kimberling, Jun 21 2016
This sequence is the Engel expansion of 1F2(1; 5/6, 7/6; 1/36) + 1F2(1; 7/6, 11/6; 1/36)/5. - Benedict W. J. Irwin, Dec 16 2016
The sequence a(n), n >= 4 is generated by the successor of the pair of polygonal numbers {P_s(4) + 1, P_(2*s - 1)(3) + 1}, s >= 3. - Ralf Steiner, May 25 2018
The asymptotic density of this sequence is 1/3. - Amiram Eldar, Oct 18 2020
Also, the only vertices in the odd Collatz tree A088975 that are branch values to other odd nodes t == 1 (mod 2) of A005408. - Heinz Ebert, Apr 14 2021
From Flávio V. Fernandes, Aug 01 2021: (Start)
For any two terms j and k, the product j*k is also a term (the same property as p^n and smooth numbers).
From a(2) to a(phi(A033845(n))), or a((A033845(n))/3), the terms are the totatives of the A033845(n) itself. (End)
Also orders n for which cyclic and semicyclic diagonal Latin squares exist (see A123565 and A342990). - Eduard I. Vatutin, Jul 11 2023
If k is in the sequence, then k*2^m + 3 is also in the sequence, for all m > 0. - Jules Beauchamp, Aug 29 2024

Examples

			G.f. = x + 5*x^2 + 7*x^3 + 11*x^4 + 13*x^5 + 17*x^6 + 19*x^7 + 23*x^8 + ...
		

References

  • K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, Springer-Verlag, 1980.

Crossrefs

A005408 \ A016945. Union of A016921 and A016969; union of A038509 and A140475. Essentially the same as A038179. Complement of A047229. Subsequence of A186422.
Cf. A000330, A001580, A002194, A019670, A032528 (partial sums), A038509 (subsequence of composites), A047209, A047336, A047522, A056020, A084967, A090771, A091998, A144065, A175885-A175887.
For k-rough numbers with other values of k, see A000027, A005408, A007775, A008364-A008366, A166061, A166063.
Cf. A126760 (a left inverse).
Row 3 of A260717 (without the initial 1).
Cf. A105397 (first differences).

Programs

Formula

a(n) = (6*n + (-1)^n - 3)/2. - Antonio Esposito, Jan 18 2002
a(n) = a(n-1) + a(n-2) - a(n-3), n >= 4. - Roger L. Bagula
a(n) = 3*n - 1 - (n mod 2). - Zak Seidov, Jan 18 2006
a(1) = 1 then alternatively add 4 and 2. a(1) = 1, a(n) = a(n-1) + 3 + (-1)^n. - Zak Seidov, Mar 25 2006
1 + 1/5^2 + 1/7^2 + 1/11^2 + ... = Pi^2/9 [Jolley]. - Gary W. Adamson, Dec 20 2006
For n >= 3 a(n) = a(n-2) + 6. - Zak Seidov, Apr 18 2007
From R. J. Mathar, May 23 2008: (Start)
Expand (x+x^5)/(1-x^6) = x + x^5 + x^7 + x^11 + x^13 + ...
O.g.f.: x*(1+4*x+x^2)/((1+x)*(1-x)^2). (End)
a(n) = 6*floor(n/2) - 1 + 2*(n mod 2). - Reinhard Zumkeller, Oct 02 2008
1 + 1/5 - 1/7 - 1/11 + + - - ... = Pi/3 = A019670 [Jolley eq (315)]. - Jaume Oliver Lafont, Oct 23 2009
a(n) = ( 6*A062717(n)+1 )^(1/2). - Gary Detlefs, Feb 22 2010
a(n) = 6*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i), with n > 1. - Bruno Berselli, Nov 05 2010
a(n) = 6*n - a(n-1) - 6 for n>1, a(1) = 1. - Vincenzo Librandi, Nov 18 2010
Sum_{n >= 1} (-1)^(n+1)/a(n) = A093766 [Jolley eq (84)]. - R. J. Mathar, Mar 24 2011
a(n) = 6*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011
a(n) = 3*n + ((n+1) mod 2) - 2. - Gary Detlefs, Jan 08 2012
a(n) = 2*n + 1 + 2*floor((n-2)/2) = 2*n - 1 + 2*floor(n/2), leading to the o.g.f. given by R. J. Mathar above. - Wolfdieter Lang, Jan 20 2012
1 - 1/5 + 1/7 - 1/11 + - ... = Pi*sqrt(3)/6 = A093766 (L. Euler). - Philippe Deléham, Mar 09 2013
1 - 1/5^3 + 1/7^3 - 1/11^3 + - ... = Pi^3*sqrt(3)/54 (L. Euler). - Philippe Deléham, Mar 09 2013
gcd(a(n), 6) = 1. - Reinhard Zumkeller, Nov 14 2013
a(n) = sqrt(6*n*(3*n + (-1)^n - 3)-3*(-1)^n + 5)/sqrt(2). - Alexander R. Povolotsky, May 16 2014
a(n) = 3*n + 6/(9*n mod 6 - 6). - Mikk Heidemaa, Feb 05 2016
From Mikk Heidemaa, Feb 11 2016: (Start)
a(n) = 2*floor(3*n/2) - 1.
a(n) = A047238(n+1) - 1. (suggested by Michel Marcus) (End)
E.g.f.: (2 + (6*x - 3)*exp(x) + exp(-x))/2. - Ilya Gutkovskiy, Jun 18 2016
From Bruno Berselli, Apr 27 2017: (Start)
a(k*n) = k*a(n) + (4*k + (-1)^k - 3)/2 for k>0 and odd n, a(k*n) = k*a(n) + k - 1 for even n. Some special cases:
k=2: a(2*n) = 2*a(n) + 3 for odd n, a(2*n) = 2*a(n) + 1 for even n;
k=3: a(3*n) = 3*a(n) + 4 for odd n, a(3*n) = 3*a(n) + 2 for even n;
k=4: a(4*n) = 4*a(n) + 7 for odd n, a(4*n) = 4*a(n) + 3 for even n;
k=5: a(5*n) = 5*a(n) + 8 for odd n, a(5*n) = 5*a(n) + 4 for even n, etc. (End)
From Antti Karttunen, May 20 2017: (Start)
a(A273669(n)) = 5*a(n) = A084967(n).
a((5*n)-3) = A255413(n).
A126760(a(n)) = n. (End)
a(2*m) = 6*m - 1, m >= 1; a(2*m + 1) = 6*m + 1, m >= 0. - Ralf Steiner, May 17 2018
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(3) (A002194).
Product_{n>=2} (1 + (-1)^n/a(n)) = Pi/3 (A019670). (End)

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

A083140 Sieve of Eratosthenes arranged as an array and read by antidiagonals in the up direction; n-th row has property that smallest prime factor is prime(n).

Original entry on oeis.org

2, 3, 4, 5, 9, 6, 7, 25, 15, 8, 11, 49, 35, 21, 10, 13, 121, 77, 55, 27, 12, 17, 169, 143, 91, 65, 33, 14, 19, 289, 221, 187, 119, 85, 39, 16, 23, 361, 323, 247, 209, 133, 95, 45, 18, 29, 529, 437, 391, 299, 253, 161, 115, 51, 20, 31, 841, 667, 551, 493, 377, 319, 203, 125, 57, 22
Offset: 2

Views

Author

Yasutoshi Kohmoto, Jun 05 2003

Keywords

Comments

A permutation of natural numbers >= 2.
The proportion of integers in the n-th row of the array is given by A005867(n-1)/A002110(n) = A038110(n)/A038111(n). - Peter Kagey, Jun 03 2019, based on comments by Jamie Morken and discussion with Tom Hanlon.
The proportion of the integers after the n-th row of the array is given by A005867(n)/A002110(n). - Tom Hanlon, Jun 08 2019

Examples

			Array begins:
   2   4   6   8  10  12  14  16  18  20  22  24 .... (A005843 \ {0})
   3   9  15  21  27  33  39  45  51  57  63  69 .... (A016945)
   5  25  35  55  65  85  95 115 125 145 155 175 .... (A084967)
   7  49  77  91 119 133 161 203 217 259 287 301 .... (A084968)
  11 121 143 187 209 253 319 341 407 451 473 517 .... (A084969)
  13 169 221 247 299 377 403 481 533 559 611 689 .... (A084970)
		

Crossrefs

Cf. A083141 (main diagonal), A083221 (transpose), A004280, A038179, A084967, A084968, A084969, A084970, A084971.
Arrays of integers grouped into rows by various criteria:
by greatest prime factor: A125624,
by lowest prime factor: this sequence (upward antidiagonals), A083221 (downward antidiagonals),
by number of distinct prime factors: A125666,
by number of prime factors counted with multiplicity: A078840,
by prime signature: A095904,
by ordered prime signature: A096153,
by number of divisors: A119586,
by number of 1's in binary expansion: A066884 (upward), A067576 (downward),
by distance to next prime: A192179.

Programs

  • Mathematica
    a = Join[ {Table[2n, {n, 1, 12}]}, Table[ Take[ Prime[n]*Select[ Range[100], GCD[ Prime[n] #, Product[ Prime[i], {i, 1, n - 1}]] == 1 &], 12], {n, 2, 12}]]; Flatten[ Table[ a[[i, n - i]], {n, 2, 12}, {i, n - 1, 1, -1}]]
    (* second program: *)
    rows = 12; Clear[T]; Do[For[m = p = Prime[n]; k = 1, k <= rows, m += p, If[ FactorInteger[m][[1, 1]] == p, T[n, k++] = m]], {n, rows}]; Table[T[n - k + 1, k], {n, rows}, {k, n}] // Flatten (* Jean-François Alcover, Mar 08 2016 *)

Extensions

More terms from Hugo Pfoertner and Robert G. Wilson v, Jun 13 2003

A010694 Period 2: repeat (2,4).

Original entry on oeis.org

2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2
Offset: 0

Views

Author

Keywords

Comments

Simple continued fraction expansion of 1 + sqrt(3/2) = A176051. - R. J. Mathar, Mar 08 2012
Number of linear characters of dihedral group of order 2(n+1). - Eric M. Schmidt, Feb 12 2013
a(n) is the n-th increment between two consecutive elements of the wheel in the wheel factorization with the basis {2, 3}. See A038179. - Wojciech Raszka, May 10 2019
In base 3, make a sequence such that after the initial term 2, each term is the sum of the squares of the digits of the previous term. That's this sequence (see A000216 for the base 10 version). - Alonso del Arte, Mar 19 2020

Crossrefs

Programs

Formula

From R. J. Mathar, Aug 28 2008: (Start)
a(n) = 2 * A000034(n).
G.f.: 2(1 + 2x)/((1 - x)(1 + x)). (End)
a(n) = a(n-2) for n >= 2. - Jaume Oliver Lafont, Mar 20 2009
a(n) = 2^(n+1) mod 6. - Roderick MacPhee, Mar 31 2011

A055399 Number of stages of sieve of Eratosthenes needed to identify n as prime or composite.

Original entry on oeis.org

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

Views

Author

Henry Bottomley, May 15 2000

Keywords

Comments

Primes are known as primes actually one step before a(n): at step k of the sieve, multiples of prime(k) are removed, the smallest integer removed being prime(k)^2; every remaining integer less than prime(k+1)^2 will then never be removed, and it is newly known at step k for those between prime(k)^2 and prime(k+1)^2. For example, at step 3, multiples of prime(3) = 5 are removed and remaining integers after this step are prime up to prime(4)^2 = 49; then, 29, 31, 37, 41, 43, 47 are known as prime at step 3. - Jean-Christophe Hervé, Nov 01 2013

Examples

			a(7)=2 because 7 is not removed by the first two stages of the sieve, but is less than the square of the second prime (though not the square of the first); a(35)=3 because 35 is removed in the third stage as a multiple of 5.
		

Crossrefs

Programs

  • Mathematica
    a[n_ /; !PrimeQ[n]] := PrimePi[ FactorInteger[n][[1, 1]]]; a[n_ /; PrimeQ[n]] := PrimePi[ NextPrime[ Sqrt[n]]]; Table[a[n], {n, 3, 107}](* Jean-François Alcover, Jun 11 2012, after formula *)

Formula

If n is composite, a(n) = A055396(n); if n is prime, a(n) = A056811(n)+1. [Corrected by Charles R Greathouse IV, Sep 03 2013]
a(n) = A010051(n)*(A056811(n)+1)+(1-A010051(n))*A055396(n). - Jean-Christophe Hervé, Nov 01 2013

A249735 Odd bisection of A003961: Replace in 2n-1 each prime factor p(k) with prime p(k+1).

Original entry on oeis.org

1, 5, 7, 11, 25, 13, 17, 35, 19, 23, 55, 29, 49, 125, 31, 37, 65, 77, 41, 85, 43, 47, 175, 53, 121, 95, 59, 91, 115, 61, 67, 275, 119, 71, 145, 73, 79, 245, 143, 83, 625, 89, 133, 155, 97, 187, 185, 161, 101, 325, 103, 107, 385, 109, 113, 205, 127, 203, 425, 209, 169, 215, 343, 131, 235, 137, 253, 875, 139, 149, 265, 221, 217, 605, 151
Offset: 1

Views

Author

Antti Karttunen, Nov 23 2014

Keywords

Comments

This has the same terms as A007310 (Numbers congruent to 1 or 5 mod 6), but in different order. Apart from 1, they are the numbers that occur below the first two rows of arrays like A246278 and A083221 (A083140).

Crossrefs

Cf. A249734 (the other bisection of A003961).
Cf. also A007310 (A038179), A249746.

Programs

Formula

a(n) = A003961(2n - 1).
a(n) = A007310(A249746(n)). [Permutation of A007310, Numbers congruent to 1 or 5 mod 6.]
Other identities. For all n >= 1:
A007310(n) = a(A249745(n)).
A246277(5*a(A048673(n))) = n.
A246277(5*a(n)) = A064216(n).

A144065 Values of k such that the expression sqrt(4!*(k+1) + 1) yields an integer.

Original entry on oeis.org

0, 1, 4, 6, 11, 14, 21, 25, 34, 39, 50, 56, 69, 76, 91, 99, 116, 125, 144, 154, 175, 186, 209, 221, 246, 259, 286, 300, 329, 344, 375, 391, 424, 441, 476, 494, 531, 550, 589, 609, 650, 671, 714, 736, 781, 804, 851, 875, 924, 949, 1000, 1026, 1079, 1106, 1161, 1189, 1246, 1275, 1334, 1364, 1425
Offset: 0

Views

Author

Keywords

Comments

Integers of form m*(m+5)/6 (nonnegative values of m are listed in A032766). - Bruno Berselli, Jul 18 2016

Crossrefs

Cf. sequences of the form m*(m+k)/(k+1) listed in A274978. [Bruno Berselli, Jul 25 2016]

Programs

  • Magma
    [(-3+2*(-1)^n*n+3*(-1)^n+6*n^2+18*n)/16: n in [0..60]]; // Vincenzo Librandi, Jul 16 2016
  • Maple
    seq(seq(((24*a+b)^2-25)/24, b=[5,7,11,13,17,19,23,25]),a=0..10); # Robert Israel, Jul 15 2016
  • Mathematica
    LinearRecurrence[{0,3,0,-3,0,1}, {0, 1, 4, 6, 11, 14}, 50] (* G. C. Greubel, Jul 15 2016 *)
    Select[Range[0,1500],IntegerQ[Sqrt[4!(#+1)+1]]&] (* Harvey P. Dale, Sep 20 2019 *)
  • PARI
    j=[];for(n=0, 300,if((floor(sqrt(4!*(n+1) + 1))) == ceil(sqrt(4!*(n+1) + 1)), j=concat(j, n))); j
    

Formula

a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6). - Jaume Oliver Lafont, Jan 21 2009
a(n) = (-3 + 2*(-1)^n*n + 3*(-1)^n + 6*n^2 + 18*n)/16. - Alexander R. Povolotsky, Jan 27 2009
a(n) = A001318(n+1) - 1. - Peter Bala, Mar 22 2009
G.f.: x*(1 + 3*x - x^3)/((1 + x)^2*(1 - x)^3). - Jaume Oliver Lafont, Aug 31 2009
a(n) = Sum_{i=1..n+3} numerator(i/2) - denominator(i/2). - Wesley Ivan Hurt, Feb 26 2017
Sum_{n>=1} 1/a(n) = (93+10*sqrt(3)*Pi)/75. - Amiram Eldar, Sep 22 2022

A055398 Result of fourth stage of sieve of Eratosthenes (after eliminating multiples of 2, 3, 5, 7).

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 121, 127, 131, 137, 139, 143, 149, 151, 157, 163, 167, 169, 173, 179, 181, 187, 191, 193, 197, 199, 209, 211, 221, 223, 227, 229, 233, 239
Offset: 1

Views

Author

Henry Bottomley, May 15 2000

Keywords

Comments

Essentially the same as A052424. - R. J. Mathar, Oct 13 2008

Crossrefs

Programs

  • Mathematica
    Join[{2,3,5,7},Select[Table[n,{n,2,500}],Mod[#,2]!=0&&Mod[#,3]!=0&&Mod[#,5]!=0&&Mod[#,7]!=0&]] (* Vladimir Joseph Stephan Orlovsky, Feb 18 2011 *)

Formula

a(n+4) = 2 * floor((3 * floor((10 * floor((9 * floor((56 * floor((55 * floor((54 * floor((53 * floor((52 * floor((51 * floor((50 * floor((49 * n + 1) / 48) + 13) / 49) + 20) / 50) + 24) / 51) + 31) / 52) + 35) / 53) + 42) / 54) + 54) / 55) + 1) / 8) + 8) / 9) + 1) / 2) + 1 for n>=1; see (22) in Diab link. - Michel Marcus, Dec 14 2020

A054403 Result of third stage of sieve of Eratosthenes (after eliminating multiples of 2, 3 and 5).

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 77, 79, 83, 89, 91, 97, 101, 103, 107, 109, 113, 119, 121, 127, 131, 133, 137, 139, 143, 149, 151, 157, 161, 163, 167, 169, 173, 179, 181, 187, 191, 193, 197, 199, 203, 209, 211
Offset: 1

Views

Author

Henry Bottomley, May 15 2000

Keywords

Crossrefs

Programs

Formula

2, 3, 5 and numbers 30k +/- 1, 7, 11, or 13.
a(n+3) = 2*floor((3*floor((10*floor((9*n+1)/8)+8)/9)+1)/2)+1 for n>=1; see (19) in Diab link. - Michel Marcus, Dec 14 2020
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