A166063 -id:A166063 - cp's OEIS frontend

A007310 Numbers congruent to 1 or 5 mod 6.

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

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

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

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

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

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Andreas Enge, William Hart, and Fredrik Johansson, Short addition sequences for theta functions, arXiv:1608.06810 [math.NT], 2016-2018.
B. W. J. Irwin, Constants Whose Engel Expansions are the k-rough Numbers.
L. B. W. Jolley, Summation of Series, Dover, 1961
Cedric A. B. Smith, Prime factors and recurring duodecimals, Math. Gaz. 59 (408) (1975) 106-109.
William A. Stein's The Modular Forms Database, PARI-readable dimension tables for Gamma_0(N).
Eric Weisstein's World of Mathematics, Rough Number.
Eric Weisstein's World of Mathematics, Pi Formulas. [_Jaume Oliver Lafont_, Oct 23 2009]
Index entries for sequences related to smooth numbers [_Michael B. Porter_, Oct 09 2009]
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

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).

Filtered([1..150],n->n mod 6=1 or n mod 6=5); # Muniru A Asiru, Dec 19 2018

a007310 n = a007310_list !! (n-1)
a007310_list = 1 : 5 : map (+ 6) a007310_list
-- Reinhard Zumkeller, Jan 07 2012

[n: n in [1..250] | n mod 6 in [1, 5]]; // Vincenzo Librandi, Feb 12 2016

seq(seq(6*i+j,j=[1,5]),i=0..100); # Robert Israel, Sep 08 2014

Select[Range[200], MemberQ[{1, 5}, Mod[#, 6]] &] (* Harvey P. Dale, Aug 27 2013 *)
a[n_] := (6 n + (-1)^n - 3)/2; a[rem156, 60] (* Robert G. Wilson v, May 26 2014 from a suggestion by N. J. A. Sloane *)
Flatten[Table[6n + {1, 5}, {n, 0, 24}]] (* Alonso del Arte, Feb 06 2016 *)
Table[2*Floor[3*n/2] - 1, {n, 1000}] (* Mikk Heidemaa, Feb 11 2016 *)

isA007310(n) = gcd(n,6)==1 \\ Michael B. Porter, Oct 09 2009

A007310(n)=n\2*6-(-1)^n \\ M. F. Hasler, Oct 31 2014

\\ given an element from the sequence, find the next term in the sequence.
nxt(n) = n + 9/2 - (n%6)/2 \\ David A. Corneth, Nov 01 2016

def A007310(n): return (n+(n>>1)<<1)-1 # Chai Wah Wu, Oct 10 2023

[i for i in range(150) if gcd(6,i) == 1] # Zerinvary Lajos, Apr 21 2009

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)

A005867 a(0) = 1; for n > 0, a(n) = (prime(n)-1)*a(n-1).

Original entry on oeis.org

1, 1, 2, 8, 48, 480, 5760, 92160, 1658880, 36495360, 1021870080, 30656102400, 1103619686400, 44144787456000, 1854081073152000, 85287729364992000, 4434961926979584000, 257227791764815872000, 15433667505888952320000

Offset: 0

N. J. A. Sloane

Local minima of Euler's phi function. - Walter Nissen
Number of potential primes in a modulus primorial(n+1) sieve. - Robert G. Wilson v, Nov 20 2000
Let p=prime(n) and let p# be the primorial (A002110), then it can be shown that any p# consecutive numbers have exactly a(n-1) numbers whose lowest prime factor is p. For a proof, see the "Proofs Regarding Primorial Patterns" link. For example, if we let p=7 and consider the interval [101,310] containing 210 numbers, we find the 8 numbers 119, 133, 161, 203, 217, 259, 287, 301. - Dennis Martin (dennis.martin(AT)dptechnology.com), Jul 16 2006
From Gary W. Adamson, Apr 21 2009: (Start)
Equals (-1)^n * (1, 1, 1, 2, 8, 48, ...) dot (-1, 2, -3, 5, -7, 11, ...).
a(6) = 480 = (1, 1, 1, 2, 8, 48) dot (-1, 2, -3, 5, -7, 11) = (-1, 2, -3, 10, -56, 528). (End)
It can be proved that there are at least T prime numbers less than N, where the recursive function T is: T = N - N*Sum_{i=0..T(sqrt(N))} A005867(i)/A002110(i). This can show for example that at least 0.16*N numbers are primes less than N for 29^2 > N > 23^2. - Ben Paul Thurston, Aug 23 2010
First column of A096294. - Eric Desbiaux, Jun 20 2013
Conjecture: The g.f. for the prime(n+1)-rough numbers (A000027, A005408, A007310, A007775, A008364, A008365, A008366, A166061, A166063) is x*P(x)/(1-x-x^a(n)+x^(a(n)+1)), where P(x) is an order a(n) polynomial with symmetric coefficients (i.e., c(0)=c(n), c(1)=c(n-1), ...). - Benedict W. J. Irwin, Mar 18 2016
a(n)/A002110(n+1) (primorial(n+1)) is the ratio of natural numbers whose smallest prime factor is prime(n+1); i.e., prime(n+1) coprime to A002110(n). So the ratio of even numbers to natural numbers = 1/2; odd multiples of 3 = 1/6; multiples of 5 coprime to 6 (A084967) = 2/30 = 1/15; multiples of 7 coprime to 30 (A084968) = 8/210 = 4/105; etc. - Bob Selcoe, Aug 11 2016
The 2-adic valuation of a(n) is A057773(n), being sum of the 2-adic valuations of the product terms here. - Kevin Ryde, Jan 03 2023
For n > 1, a(n) is the number of prime(n+1)-rough numbers in [1, primorial(prime(n))]. - Alexandre Herrera, Aug 29 2023

			a(3): the mod 30 prime remainder set sieve representation yields the remainder set: {1, 7, 11, 13, 17, 19, 23, 29}, 8 elements.

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

T. D. Noe, Table of n, a(n) for n = 0..99
Larry Deering, The Black Key Sieve, Box 275, Bellport NY 11713-0275, 1998.
Alphonse de Polignac, Six propositions arithmologiques déduites du crible d'Ératosthène, Nouvelles annales de mathématiques : journal des candidats aux écoles polytechnique et normale, Série 1, Tome 8 (1849), pp. 423-429. See p. 425.
Frank Ellermann, Illustration for A002110, A005867, A038110, A060753.
Ken Hicks and Kevin Ward, Series and Product Relations Made from Primes, arXiv:2108.03268 [math.NT], 2021.
Dennis Martin, Proofs Regarding Primorial Patterns [via Internet Archive Wayback-machine]
Dennis Martin, Proofs Regarding Primorial Patterns [Cached copy, with permission of the author]
Francis E. Masat, Letter to N. J. A. Sloane with attachment: "A note on prime number sequences" (unpublished manuscript), Apr. 1991.
Travis Near, Improving MATLAB's isprime performance without arbitrary-precision arithmetic, arXiv:2108.04791 [cs.MS], 2021.
John K. Sellers, Distribution of twin primes in repeating sequences of prime factors, arXiv:2108.00288 [math.GM], 2021. See Table 1 p. 11.
Andrew V. Sutherland, Order Computations in Generic Groups, Ph. D. Dissertation, Math. Dept., M.I.T., 2007.

Cf. A000010, A002110, A006093, A054640, A058254, A055768, A070918, A101301, A058251, A060753.
Cf. A057773 (2-adic valuation).
Column 1 of A281890.
Cf. A016035, A345974.

a005867 n = a005867_list !! n
a005867_list = scanl (*) 1 a006093_list
-- Reinhard Zumkeller, May 01 2013

A005867 := proc(n)
    mul(ithprime(j)-1,j=1..n) ;
end proc: # Zerinvary Lajos, Aug 24 2008, R. J. Mathar, May 03 2017

Table[ Product[ EulerPhi[ Prime[ j ] ], {j, 1, n} ], {n, 1, 20} ]
RecurrenceTable[{a[0]==1,a[n]==(Prime[n]-1)a[n-1]},a,{n,20}] (* Harvey P. Dale, Dec 09 2013 *)
EulerPhi@ FoldList[Times, 1, Prime@ Range@ 18] (* Michael De Vlieger, Mar 18 2016 *)

for(n=0, 22, print1(prod(k=1,n, prime(k)-1), ", "))

a(n) = phi(product of first n primes) = A000010(A002110(n)).
a(n) = Product_{k=1..n} (prime(k)-1) = Product_{k=1..n} A006093(n).
Sum_{n>=0} a(n)/A002110(n+1) = 1. - Bob Selcoe, Jan 09 2015
a(n) = A002110(n)-((1/A000040(n+1) - A038110(n+1)/A038111(n+1))*A002110(n+1)). - Jamie Morken, Mar 27 2019
a(n) = |Sum_{k=0..n} A070918(n,k)|. - Alois P. Heinz, Aug 18 2019
a(n) = A058251(n)/A060753(n+1). - Jamie Morken, Apr 25 2022
a(n) = A002110(n) - A016035(A002110(n)) - 1 for n >= 1. - David James Sycamore, Sep 07 2024
Sum_{n>=0} 1/a(n) = A345974. - Amiram Eldar, Jun 26 2025

Offset changed to 0, Name changed, and Comments and Examples sections edited by T. D. Noe, Apr 04 2010

A007775 Numbers not divisible by 2, 3 or 5.

Original entry on oeis.org

1, 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

Offset: 1

N. J. A. Sloane

Also numbers n such that the sum of the 4th powers of the first n positive integers is divisible by n, or A000538(n) = n*(n+1)(2*n+1)(3*n^2+3*n-1)/30 is divisible by n. - Alexander Adamchuk, Jan 04 2007
Also the 7-rough numbers: positive integers that have no prime factors less than 7. - Michael B. Porter, Oct 09 2009
a(n) mod 3 has period 8, repeating [1,1,2,1,2,1,2,2] = (n mod 2) + floor(((n-1) mod 8)/7) - floor(((n-2) mod 8)/7) + 1. floor(a(n)/3) is the set of numbers k such that k is congruent to {0,2,3,4,5,6,7,9} mod 10 = floor((5*n-2)/4)-floor((n mod 8)/6). - Gary Detlefs, Jan 08 2012
Numbers k such that C(k+3,3)==1 (mod k) and C(k+5,5)==1 (mod k). - Gary Detlefs, Sep 15 2013
a(n) mod 30 has period 8 repeating [1, 7, 11, 13, 17, 19, 23, 29]. The mean of these 8 numbers is 120/8 = 15. (a(n)-15) mod 30 has period 8 repeating [-14, -8, -4, -2, 2, 4, 8, 14]. One half of the absolute value produces the symmetric sequence [7, 4, 2, 1, 1, 2, 4, 7] = A061501(((n-1) + 16) mod 8). - Gary Detlefs, Sep 24 2013
a(n) are exactly those positive integers m such that the sequence b(n) = n*(n + m)*(n + 2*m)*(n + 3*m)(n + 4*m)/120 is integral. Cf. A007310. - Peter Bala, Nov 13 2015
The asymptotic density of this sequence is 4/15. - Amiram Eldar, Sep 30 2020
If a(n) + a(n+1) = 0 (mod 30), then a(n-j) + a(n+j+1) = a(n) + a(n+1) for each j in [1, n-1]. - Alexandre Herrera, Jun 27 2023

N. J. A. Sloane, Table of n, a(n) for n = 1..8000
Peter Bala, A note on A007775.
Abel Jansma, E_8 Symmetry Structures in the Ising model, Master's thesis, University of Amsterdam, 2018.
Eric Weisstein's World of Mathematics, Rough Number.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,1,-1).
Index entries for sequences related to smooth numbers

Cf. A000538, A054403, A145011 (first differences).
For k-rough numbers with other values of k, see A000027, A005408, A007310, A007775, A008364, A008365, A008366, A166061, A166063.
Complement is A080671.
For digital root of Fibonacci numbers indexed by this sequence, see A227896.

a007775 n = a007775_list !! (n-1)
a007775_list = 1 : filter ((> 5) . a020639) [1..]
-- Reinhard Zumkeller, Jan 06 2013

I:=[1, 7, 11, 13, 17, 19, 23, 29, 31]; [n le 9 select I[n] else Self(n-1) +Self(n-8) - Self(n-9): n in [1..80]]; // G. C. Greubel, Oct 22 2018

for i from 1 to 500 do if gcd(i,30) = 1 then print(i); fi; od;
for k from 1 to 300 do if ((k^2 mod 48=1) or (k^2 mod 48=25)) and ((k^2 mod 120=1) or (k^2 mod 120=49)) then print(k) fi od. # Gary Detlefs, Dec 30 2011

Select[ Range[ 300 ], GCD[ #1, 30 ]==1& ]
Select[Range[250], Mod[#, 2]>0&&Mod[#, 3]>0&&Mod[#, 5]>0&] (* Vincenzo Librandi, Feb 08 2014 *)
a[ n_] := Quotient[ n, 8, 1] 30 + {1, 7, 11, 13, 17, 19, 23, 29}[[Mod[n, 8, 1]]]; (* Michael Somos, Jun 02 2014 *)
LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 1, -1}, {1, 7, 11, 13, 17, 19, 23, 29, 31}, 100] (* Mikk Heidemaa, Dec 07 2017 *)
Cases[Range@1000, x_ /; NoneTrue[Array[Prime, 3], Divisible[x, #] &]] (* Mikk Heidemaa, Dec 07 2017 *)
CoefficientList[ Series[(x^8 + 6x^7 + 4x^6 + 2x^5 + 4x^4 + 2x^3 + 4x^2 + 6x + 1)/((x - 1)^2 (x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)), {x, 0, 55}], x] (* Robert G. Wilson v, Dec 07 2017 *)

isA007775(n) = gcd(n,30)==1 \\ Michael B. Porter, Oct 09 2009

{a(n) = n\8 * 30 + [ -1, 1, 7, 11, 13, 17, 19, 23][n%8 + 1]} /* Michael Somos, Feb 05 2011 */

{a(n) = n\8 * 6 + 9 + 3 * (n+1)\2 * 2 - max(5, (n-2)%8) * 2} /* Michael Somos, Jun 02 2014 */

Vec(x*(1+6*x+4*x^2+2*x^3+4*x^4+2*x^5+4*x^6+6*x^7+x^8)/((1+x)*(x^2+1)*(x^4+1)*( x-1)^2) + O(x^100)) \\ Altug Alkan, Nov 16 2015

def A007775(n): return ((m:=n-1)<<2|1)-(m>>2&-2)+(2,0,-2,0)[m-1>>1&3] # Chai Wah Wu, Feb 02 2025

a = lambda n: ((((n-1)<< 2)-((n-1)>>2))|1) + ((((n-1)<<1)-((n-1)>> 1)) & 2)
print([a(n) for n in (1..56)]) # after Andrew Lelechenko, Peter Luschny, Jul 08 2017

A141256(a(n)) = n+1. - Reinhard Zumkeller, Jun 17 2008
From R. J. Mathar, Feb 27 2009: (Start)
a(n+8) = a(n) + 30.
a(n) = a(n-1) + a(n-8) - a(n-9).
G.f.: x*(1 + 6*x + 4*x^2 + 2*x^3 + 4*x^4 + 2*x^5 + 4*x^6 + 6*x^7 + x^8)/((1 + x)*(x^2 + 1)*(x^4 + 1)*(x-1)^2). (End)
a(n) = 4*n - 3 - 2*floor((n-1)/8) + (1 + (-1)^floor((n-2)/2))*(-1)^floor((n-2)/4), n >= 1. - Timothy Hopper, Mar 14 2010
a(1 - n) = -a(n). - Michael Somos, Feb 05 2011
Numbers k such that ((k^2 mod 48=1) or (k^2 mod 48=25)) and ((k^2 mod 120=1) or (k^2 mod 120=49)). - Gary Detlefs, Dec 30 2011
Numbers k such that k^2 mod 30 is 1 or 19. - Gary Detlefs, Dec 31 2011
a(n) = 3*(floor((5*n-2)/4) - floor((n mod 8)/6)) + (n mod 2) + floor(((n-1) mod 8)/7) - floor(((n-2) mod 8)/7) + 1. - Gary Detlefs, Jan 08 2012
a(n) = 4*n - 3 + 2*(floor((n+6)/8) - floor((n+4)/8) - floor((n+2)/8) + floor(n/8) - floor((n-1)/8)), n >= 1. From the o.g.f. given above by R. J. Mathar (with the denominator written as (1-x^8)*(1-x)), and a two-step reduction of the floor functions. Compare with Hopper's and Detlefs's formulas above. - Wolfdieter Lang, Jan 26 2012
a(n) = (6*f(n) - 3 + (-1)^f(n))/2, where f(n)= n + floor(n/4)+ floor(((n+4) mod 8)/6). - Gary Detlefs, Sep 15 2013
a(n) = 30*floor((n-1)/8) + 15 + 2*f((n-1) mod 8 + 16)*(-1)^floor(((n+3) mod 8)/4), where f(n) = (n*(n+1)/2+1) mod 10. - Gary Detlefs, Sep 24 2013
a(n) = 3*n + 6*floor(n/8) + (n mod 2) - 2*floor(((n-2) mod 8)/6) - 2*floor(((n-2) mod 8)/7) + 1. - Gary Detlefs, Jun 01 2014
a(n+1) = ((n << 2 - n >> 2) || 1) + ((n << 1 - n >> 1) && 2), where << and >> are bitwise left and right shifts, || and && are bitwise "or" and "and". - Andrew Lelechenko, Jul 08 2017
a(n) = 2*n + 2*floor(1/2 + (7*n)/8) + 2*(91 mod (2 - ((3*n)/4 + n^2/4) mod 2)) - 3 (n > 0). - Mikk Heidemaa, Dec 06 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(23 + sqrt(5) - sqrt(6*(5 + sqrt(5))))*Pi/15. - Amiram Eldar, Dec 13 2021

A008364 11-rough numbers: not divisible by 2, 3, 5 or 7.

Original entry on oeis.org

1, 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, 241, 247

Offset: 1

N. J. A. Sloane

The first A005867(4) = 48 terms give the reduced residue system for the 4th primorial number 210 = A002110(4).
This sequence is closed under multiplication: any product of terms is also a term. - Labos Elemer, Feb 26 2003
Conjecture: these are numbers n such that (Sum_{k=1..n} k^4) mod n = 0 and (Sum_{k=1..n} k^6) mod n = 0. - Gary Detlefs, Dec 20 2011
From Peter Bala, May 03 2018: (Start)
The above conjecture is true. Let m be even and let the m-th Bernoulli number be written in reduced form as Bernoulli(m) = N(m)/D(m). Apply Ireland and Rosen, Proposition 15.2.2, to show the congruence D(m)*( Sum_{k = 1..n} k^m )/n = N(m) (mod n) holds for all n >= 1. It follows easily from this congruence that ( Sum_{k = 1..n} k^m )/n is integral iff n is coprime to D(m). Now Bernoulli(4) = -1/(2*3*5) and Bernoulli(6) = 1/(2*3*7) so the numbers n such that both (Sum_{k=1..n} k^4) mod n = 0 and (Sum_{k=1..n} k^6) mod n = 0 are exactly those numbers coprime to the primes 2, 3, 5 and 7, that is, the 11-rough numbers. (End)
Conjecture: these are numbers n such that (n^6 mod 210 = 1) or (n^6 mod 210 = 169). - Gary Detlefs, Dec 30 2011
The second Detlefs conjecture above is true and extremely easy to verify with some basic properties of congruences: take the terms of this sequence up to 209 and compute their sixth powers modulo 210: there should only be 1's and 169's there. Then take the complement of this sequence up to 210, where you will see no instances of 1 or 169. - Alonso del Arte, Jan 12 2014
It is well-known that the product of 7 consecutive integers is divisible by 7!. Conjecture: This sequence is exactly the set of positive values of r such that ( Product_{k = 0..6} n + k*r )/7! is an integer for all n. - Peter Bala, Nov 14 2015
From Ruediger Jehn, Nov 05 2020: (Start)
This conjecture is true. The first part of the proof deals with numbers not in A008364, i.e., numbers which are divisible by p (p either 2, 3, 5, 7). Let r = p*s and n = 1, then (Product_{k = 0..6} n + k*r) is not divisible by p, because none of the factors 1 + k*p*s are divisible by p. Hence dividing the product by 7! does not return an integer.
The second part deals with numbers in A008364. If r and q are coprime, then for any i < q there exists k < q with (k*r mod q) = i. From this, it also follows that for any n there exists k < q with ((n + k*r) mod q) = 0. But this means that Product_{k = 0..6} n + k*r is divisible by all numbers from 2 to 7 because there is always a factor that is divisible. We still have to show that the product is also divisible by 2 times 3 times 4 times 6. If the k_1 with ((n + k_1*r) mod 4) = 0 is even, then (n mod 2) = ((n + 2*r) mod 2) = ((n + 4*r) mod 2) = ((n + 6*r) mod 2) = 0. If this k_1 is odd, then ((n + r) mod 2) = ((n + 3*r) mod 2) = ((n + 5*r) mod 2) = 0. In both cases there are at least 2 other factors divisible by 2. If the k_2 with ((n + k_2*r) mod 6) = 0 is smaller than 4, then ((n + (k_2 + 3)*r) mod 3) = 0. Otherwise, ((n + (k_2 - 3)*r) mod 3) = 0. In both cases there is at least 1 other factor divisible by 3. And therefore Product_{k = 0..6} n + k*r is divisible by 7! for any n.
(End)

Diatomic sequence of 4th prime: A. de Polignac (1849), J. Dechamps (1907).
Dickson L. E., History of the Theory of Numbers, Vol. 1, p. 439, Chelsea, 1952.
K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, Springer-Verlag, 1980.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Alphonse de Polignac, Recherches Nouvelles sur les Nombres Premiers, in Comptes Rendus de l'Académie des Sciences, October 15 1849. See also Rectification.
Alphonse de Polignac, Six propositions arithmologiques déduites du crible d'Ératosthène, Nouvelles annales de mathématiques : journal des candidats aux écoles polytechnique et normale, Série 1, Tome 8 (1849), pp. 423-429.
Han-Lin Li, Shu-Cherng Fang, and Way Kuo, The Periodic Table of Primes, Advances in Pure Mathematics, Volume 14, Issue 5, May 2024.
Eric Weisstein's World of Mathematics, Rough Number
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-1).
Index entries for sequences related to smooth numbers

First differences give A049296. Cf. A002110, A048597.
For k-rough numbers with other values of k, see A000027, A005408, A007310, A007775, A008364, A008365, A008366, A166061, A166063. - Michael B. Porter, Oct 10 2009
Cf. A005867, A092695, A210679, A080672 (complement).

a008364 n = a008364_list !! (n-1)
a008364_list = 1 : filter ((> 7) . a020639) [1..]
-- Reinhard Zumkeller, Mar 26 2012

for i from 1 to 500 do if gcd(i,210) = 1 then print(i); fi; od;
t1:=[]; for i from 1 to 1000 do if gcd(i,210) = 1 then t1:=[op(t1),i]; fi; od: t1;
S:= (j,n)-> sum(k^j,k=1..n): for n from 1 to 247 do if (S(4,n) mod n = 0) and (S(6,n) mod n = 0) then print(n) fi od; # Gary Detlefs, Dec 20 2011

Select[ Range[ 300 ], GCD[ #1, 210 ] == 1 & ]
Select[Range[250], Mod[#, 2]>0 && Mod[#, 3]>0 && Mod[#, 5]>0 && Mod[#, 7]>0 &] (* Vincenzo Librandi, Nov 16 2015 *)
Cases[Range@1000, x_ /; NoneTrue[Array[Prime, 4], Divisible[x, #] &]] (* Mikk Heidemaa, Dec 07 2017 *)
Select[Range[250],Union[Divisible[#,{2,3,5,7}]]=={False}&] (* Harvey P. Dale, Sep 24 2021 *)

isA008364(n) = gcd(n,210)==1 \\ Michael B. Porter, Oct 10 2009

Starting with a(49) = 211, a(n) = a(n-48) + 210. - Zak Seidov, Apr 11 2011
a(n) = a(n-1) + a(n-48) - a(n-49). - Charles R Greathouse IV, Dec 21 2011
A020639(a(n)) > 7. - Reinhard Zumkeller, Mar 26 2012
G.f.: x*(x^48 + 10*x^47 + 2*x^46 + 4*x^45 + 2*x^44 + 4*x^43 + 6*x^42 + 2*x^41 + 6*x^40 + 4*x^39 + 2*x^38 + 4*x^37 + 6*x^36 + 6*x^35 + 2*x^34 + 6*x^33 + 4*x^32 + 2*x^31 + 6*x^30 + 4*x^29 + 6*x^28 + 8*x^27 + 4*x^26 + 2*x^25 + 4*x^24 + 2*x^23 + 4*x^22 + 8*x^21 + 6*x^20 + 4*x^19 + 6*x^18 + 2*x^17 + 4*x^16 + 6*x^15 + 2*x^14 + 6*x^13 + 6*x^12 + 4*x^11 + 2*x^10 + 4*x^9 + 6*x^8 + 2*x^7 + 6*x^6 + 4*x^5 + 2*x^4 + 4*x^3 + 2*x^2 + 10*x + 1) / (x^49 - x^48 - x + 1). - Colin Barker, Sep 27 2013
a(n) = 35*n/8 + O(1). - Charles R Greathouse IV, Sep 14 2015
A007775 INTERSECT A206547. - R. J. Mathar, Apr 10 2024

New name from Charles R Greathouse IV, Dec 21 2011 based on comment from Michael B. Porter, Oct 10 2009

A008365 13-rough numbers: positive integers that have no prime factors less than 13.

Original entry on oeis.org

1, 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, 127, 131, 137, 139, 149, 151, 157, 163, 167, 169, 173, 179, 181, 191, 193, 197, 199, 211, 221, 223, 227, 229, 233, 239, 241, 247, 251, 257, 263, 269

Offset: 1

N. J. A. Sloane

For n > 1, the smallest prime factor of a(n) is >= 13.
Conjecture: Numbers n such that n^24 is congruent to {1,421,631,841} mod 2310. - Gary Detlefs, Dec 30 2011
This sequence is exactly the set of positive values of r such that ( Product_{k = 0..10} n + k*r )/11! is an integer for all n. - Peter Bala, Nov 14 2015
The asymptotic density of this sequence is 16/77. - Amiram Eldar, Sep 30 2020

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Peter Bala, A property of p-rough numbers.
Benedict W. J. Irwin, Generating Function.
Eric Weisstein's World of Mathematics, Rough Number.
Index entries for linear recurrences with constant coefficients, order 481.
Index entries for sequences related to smooth numbers

For k-rough numbers with other values of k, see A000027, A005408, A007310, A007775, A008364, A008365, A008366, A166061, A166063.

a008365 n = a008365_list !! (n-1)
a008365_list = 1 : filter ((> 11) . a020639) [1..]
-- Reinhard Zumkeller, Jan 06 2013

for i from 1 to 500 do if gcd(i,2310) = 1 then print(i); fi; od;

Select[ Range[ 300 ], GCD[ #1, 2310 ]==1& ]

isA008365(n) = gcd(n,2310)==1 \\ Michael B. Porter, Oct 10 2009

G.f: x*P(x)/(1 - x - x^480 + x^481) where P(x) is a polynomial of degree 480. - Benedict W. J. Irwin, Mar 18 2016
77*n/16 - 13 < a(n) < 77*n/16 + 8. - Charles R Greathouse IV, Mar 21 2023
a(n) = a(n-1) + a(n-480) - a(n-481). - Charles R Greathouse IV, Mar 21 2023

New name following a comment of Michael B. Porter, Mar 21 2023

A008366 Smallest prime factor is >= 17.

Original entry on oeis.org

1, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283

Offset: 1

N. J. A. Sloane

Also the 17-rough numbers: positive integers that have no prime factors less than 17. - Michael B. Porter, Oct 10 2009
a(n) - (1001/192) n is periodic with period 5760. - Robert Israel, Mar 18 2016
From Peter Bala, May 12 2018: (Start)
The product of two 17-rough numbers is a 17-rough number and the prime factors of a 17-rough number are 17-rough numbers.
Let k equal either 13, 14, 15 or 16. Then the product of k numbers n*(n + a)*(n + 2*a)*...*(n + (k-1)*a) in arithmetical progression is divisible by k! for all integer n if and only if a is a 17-rough number.
The sequence terms satisfy the congruence x^60 = 1 (mod 30030), where 30030 = 2*3*5*7*11*13. (End)
The asymptotic density of this sequence is 192/1001. - Amiram Eldar, Sep 30 2020

Robert Israel, Table of n, a(n) for n = 1..10000
Peter Bala, A property of p-rough numbers.
Benedict Irwin, Generating Function.
Eric Weisstein's World of Mathematics, Rough Number.
Index entries for sequences related to smooth numbers

For k-rough numbers with other values of k, see A000027 A005408 A007310 A007775 A008364 A008365 A008366 A166061 A166063.

Cf. A005867.

for i from 1 to 500 do if gcd(i,30030) = 1 then print(i); fi; od;

Select[ Range[ 300 ], GCD[ #1, 30030 ]==1& ]
Join[{1},Select[Range[300],FactorInteger[#][[1,1]]>=17&]] (* Harvey P. Dale, Mar 28 2020 *)

isA008366(n) = gcd(n,30030)==1 \\ Michael B. Porter, Oct 10 2009

Numbers n > 1 such that ((Sum_{k=1..n} k^10) mod n = 0) and ((Sum_{k=1..n} k^12) mod n = 0) (conjecture). - Gary Detlefs, Dec 27 2011
a(n) = a(n-1) + a(n-5760) - a(n-5761). - Vaclav Kotesovec, Mar 18 2016
G.f: x*P(x)/(1 - x - x^5760 + x^5761) where P(x) is a polynomial of degree 5760. - Benedict W. J. Irwin, Mar 23 2016
a(n) = (1001/192)*n + O(1), where the O(1) term is bounded by +/- 19. - Charles R Greathouse IV, Oct 13 2022
A008365 SETMINUS A084970 . - R. J. Mathar, Nov 05 2024

A166061 19-rough numbers: positive integers that have no prime factors less than 19.

Original entry on oeis.org

1, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271

Offset: 1

Michael B. Porter, Oct 05 2009

Or, positive integers relatively prime to 510510 = 2*3*5*7*11*13*17.

			437 = 19 * 23 is in the sequence since the two prime factors, 19 and 23, are not less than 19.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Rough Number
Index entries for sequences related to smooth numbers [From _Michael B. Porter_, Oct 10 2009]

k-rough numbers: A000027, A005408, A007310, A007775, A008364, A008365, A008366, A166061, A166063.

Join[{1},Select[Range[300],Min[Transpose[FactorInteger[#]][[1]]]>18&]] (* Harvey P. Dale, Feb 18 2015 *)

isA166061(n) = gcd(n,510510)==1 \\ Michael B. Porter, Oct 10 2009

a(n) = k*n + O(1) where k = 17017/3072 = 5.539388.... In particular, k*n - 31 < a(n) < k*n + 25. - Charles R Greathouse IV, Sep 24 2018

A332797 Numbers whose smallest prime factor is 23.

Original entry on oeis.org

23, 529, 667, 713, 851, 943, 989, 1081, 1219, 1357, 1403, 1541, 1633, 1679, 1817, 1909, 2047, 2231, 2323, 2369, 2461, 2507, 2599, 2921, 3013, 3151, 3197, 3427, 3473, 3611, 3749, 3841, 3979, 4117, 4163, 4393, 4439, 4531, 4577, 4853, 5129, 5221, 5267, 5359, 5497

Offset: 1

Frank Ellermann, Feb 24 2020

The asymptotic density of this sequence is 55296/7436429. - Amiram Eldar, Dec 06 2020

			a(2) = 23*23, a(3) = 23*29.

Emmanuel Desurvire, Classical and Quantum Information Theory: An Introduction for the Telecom Scientist, Cambridge University Press, 2009, table 20.5 p. 421.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A084967 (5), A084968 (7), A084969 (11), A084970 (13), A332799 (17), A332798 (19), A166063 (23-rough numbers).

23 * Select[Range[240], CoprimeQ[#, 9699690] &] (* Amiram Eldar, Feb 24 2020 *)
Select[Range[6000],FactorInteger[#][[1,1]]==23&] (* Harvey P. Dale, Aug 29 2025 *)

P = 23         ;  S = P
do N = P by 2 while length( S ) < 255
   do I = 1 until P = X
      X = PRIME( I )
      if P = X       then  leave I
      if N // X = 0  then  iterate N
   end I
   S = S || ',' P*N
end N
say S          ;  return S

a(n) = 23*A166063(n).

A224534 Prime numbers that are the sum of three distinct prime numbers.

Original entry on oeis.org

19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307

Offset: 1

T. D. Noe, Apr 15 2013

nonn

Similar to Goldbach's weak conjecture.
Primes in A124867, and by the comment in A124867 also the set of all primes >=19. - R. J. Mathar, Apr 19 2013
"Goldbach's original conjecture (sometimes called the 'ternary' Goldbach conjecture), written in a June 7, 1742 letter to Euler, states 'at least it seems that every number that is greater than 2 is the sum of three primes' (Goldbach 1742; Dickson 2005, p. 421). Note that here Goldbach considered the number 1 to be a prime, a convention that is no longer followed." [Weisstein] - Jonathan Vos Post, May 15 2013

			19 = 3 + 5 + 11.

H. A. Helfgott and David J. Platt, Numerical Verification of the Ternary Goldbach Conjecture up to 8.875e30, arXiv:1305.3062 [math.NT], 2013-2014.
H. A. Helfgott and David J. Platt, Numerical verification of the Ternary Goldbach Conjecture up to 8.875*10^30, Exp. Math. 22 (4) (2013) 406-409.
Eric W. Weisstein, Goldbach conjecture
Wikipedia, Goldbach's conjecture
Wikipedia, Goldbach's weak conjecture

Cf. A002372, A002375, A024684 (number of sums), A224535, A166063, A166061, A071621.

Union[Select[Total /@ Subsets[Prime[Range[2, 30]], {3}], PrimeQ]]

A246541 Take the squares of all P_(n+2)-rough numbers less than the (n+1)-th primorial and mod each by the (n+1)-th primorial. There will be a(n) different results.

Original entry on oeis.org

1, 2, 6, 30, 180, 1440, 12960, 142560, 1995840, 29937600, 538876800, 10777536000, 226328256000, 5205549888000, 135344297088000, 3924984615552000, 117749538466560000, 3885734769396480000, 136000716928876800000, 4896025809439564800000, 190945006568143027200000

Offset: 1

John B. Yin, Aug 29 2014

nonn

The P_(n+2)-rough numbers less than the (n+1)-th primorial also comprise the reduced residue system of the (n+1)-th primorial.

The conjectured formula from Jon E. Schoenfield is true. This can be seen by considering that each odd prime p has exactly (p+1)/2 quadratic residues (mod p), of which (p-1)/2 are nonzero. The P_(n+2)-rough numbers less than the (n+1)-th primorial comprise all combinations of nonzero residues modulo the first n+1 primes. So for each odd prime p, the p-1 nonzero residues map to (p-1)/2 (nonzero) residues after squaring. - Bert Dobbelaere, Aug 09 2023

			For n=2, P_(n+2) = 7.
The 7-rough numbers less than 2*3*5 are 1,7,11,13,17,19,23,29.
The squares of those numbers mod 2*3*5 are 1,19,1,19,19,1,19,1.
There are 2 different results: 1 and 19; so a(2) = 2.

Eric Weisstein's World of Mathematics, Primorial
Eric Weisstein's World of Mathematics, Rough Number

Cf. A002110 (primorial).
Cf. k-rough numbers A007310 (k=5), A007775 (k=7), A008364 (k=11), A008365 (k=13), A008366 (k=17), A166061 (k=19), A166063 (k=23).
Cf. A323739.

import java.util.TreeSet;
for(int z = 1; z < 10 ; z++) {
int n = z;
int numNumPerLine = 210;
int[] primes = {2,3,5,7,11,13,17,19,23,29,31,37,41,43};
int numRepeats = 1;
int numSpaces = 1;
for(int i = 0; i < n + 1; i++) {
numSpaces *= (primes[i] - 1);
}
int counter = 0;
long integerLength = 1;
for(int i = 0; i < n + 1; i++) {
integerLength *= primes[i];
}
TreeSet numResults = new TreeSet();
numSpaces/=2;
for(int i = 1; i < integerLength / 2; i+=2) {
boolean isInList = true;
for(int j = 1; j < n + 1; j++) {
if(i % primes[j] == 0) {
isInList = false;
}
}
if(isInList) {
long k = i % integerLength;
if(k != 0) {
long l = (k * k) % integerLength;
if(!numResults.contains(l)) {
numResults.add(l);
}
}
}
}
System.out.println(numResults.size());
}

a(n) = {hp = prod(k=1, n+1, prime(k)); rp = prod(k=1, n+2, prime(k)); v = []; for (i=1, hp, if (gcd(i, rp) == 1, nv = i^2 % hp; if (! vecsearch(v, nv), v = vecsort(concat(v, nv))););); #v;} \\ Michel Marcus, Sep 06 2014

Conjecture: a(n) = (1/2^n)*Product_{j=1..n} (prime(j+1)-1) = A005867(n+1)/2^n. - Jon E. Schoenfield, Feb 20 2019

a(n) = A323739(n+1). - Bert Dobbelaere, Aug 09 2023

cp's OEIS Frontend

A007310 Numbers congruent to 1 or 5 mod 6.

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A005867 a(0) = 1; for n > 0, a(n) = (prime(n)-1)*a(n-1).

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A007775 Numbers not divisible by 2, 3 or 5.

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PARI

PARI

PARI

PARI

Python

Sage

Formula

A008364 11-rough numbers: not divisible by 2, 3, 5 or 7.

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Haskell

Maple

Mathematica

PARI

Formula

Extensions

A008365 13-rough numbers: positive integers that have no prime factors less than 13.

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