A005867 -id:A005867 - cp's OEIS frontend

A121406 a(1) = a(2) = 0; a(3) = 2; for n >= 4, a(n) = (prime(n-1)-2)*a(n-1), where prime(n) is the n-th prime.

0, 0, 2, 6, 30, 270, 2970, 44550, 757350, 15904350, 429417450, 12453106050, 435858711750, 16998489758250, 696938080088250, 31362213603971250, 1599472893802533750, 91169954946744423750, 5379027341857921001250

Offset: 1

Dennis R. Martin (dennis.martin(AT)dptechnology.com), Jul 28 2006, Dec 05 2006

Also number of distinct twin prime eliminations which can be attributed to a particular lowest prime factor prime(n) over primorial intervals of prime(n)#. That is, it is the number of composite numbers having prime(n) for their lowest prime factor within any interval of width prime(n)# starting after prime(n) which are adjacent to the center post of a twin prime candidate for which that twin prime candidate is not also eliminated by a prime factor less than prime(n). Or put simply, it is the number of twin prime eliminations by prime(n) within intervals of its primorial that are left after subtracting out the double eliminations that can be attributed to previous prime factors.

Sum_{ n >= 1 } a(n)/A002110(n) converges to 1/6. That is, (2 / 30) + (6 / 210) + (30 / 2310) + (270 / 30030) + (2970 / 510510) + ... = (1 / 6).

			The prime factors prime(1) = 2 and prime(2) = 3 cannot eliminate any twin prime candidates, therefore a(1) = a(2) = 0. The prime factor prime(3) = 5 will eliminate a(3) = 2 twin prime candidates in every interval of prime(3)# = 30 starting after prime(3). For example, the composites 25 and 35 eliminate the twin prime candidate pairs centered at 24 and 36, respectively, while the composites 55 and 65 eliminate the twin prime candidates centered at 54 and 66.
For the prime factor prime(4) = 7, there will be 8 composites having prime(4) for their lowest prime factor within every interval of prime(4)# = 210 starting after 7. For instance, the composites {49, 77, 91, 119, 133, 161, 203, 217} are adjacent to and eliminate the twin prime candidates centered at {48, 78, 90, 120, 132, 162, 204, 216}. However, 2 of those 8 are already eliminated by prime(3), those being the candidates centered at 204 and 216, since 205 and 215 obviously are composites having 5 for their lowest prime factor. And in the next interval of prime(4)# = 210 the pattern repeats. The composites {259, 287, 301, 329, 343, 371, 413, 427} all have 7 for their lowest prime factor and they eliminate the twin prime candidate pairs centered at {258, 288, 300, 330, 342, 372, 414, 426}. But the ones centered at 414 and 426 are also adjacent to 415 and 425, which have 5 for their lowest prime factor and thus can be considered to have already been eliminated. a(4) = 8 - 2 = 6.
For prime(5) = 11, there are 48 composites that have 11 for their lowest prime factor over any interval of prime(5)# = 2310 starting after 11. Those 48 composites are all adjacent to a twin prime candidate center post, but 12 of those candidates are eliminated by prime(3) (the ones corresponding to the centers 144, 186, 474, 516, 804, 1134, 1176, 1506, 1794, 1836, 2124 and 2166) and 6 are eliminated by prime(4) (those corresponding to the candidate centered at 120, 342, 582, 1728, 1968 and 2190). That is a total of 18 out of those 48 in every interval of 2310 that are eliminated by a prime factor less than prime(5), therefore a(5) = 48 - 18 = 30.
But then 30 = 6(7-2) and 6 = 2(5-2). By continuing to count the twin prime eliminations in this manner, it can be deduced that each subsequent term is found by multiplying the previous term by the previous prime minus 2.

Dennis Martin, Table of n, a(n) for n = 1..100
Dennis R. Martin, On the Infinite Series Characterizing the Elimination of Twin Prime Candidates [Broken link]
Dennis R. Martin, On the Infinite Series Characterizing the Elimination of Twin Prime Candidates [Cached copy, with permission of the author]

Cf. A002110, A005867, A111490, A121407.

nxt[{n_,a_}]:={n+1,a(Prime[n]-2)}; Join[{0,0},Transpose[NestList[nxt,{3,2},20]][[2]]] (* Harvey P. Dale, Sep 23 2015 *)

a(1) = a(2) = 0; a(3) = 2; for n >= 4, a(n) = (prime(n-1)-2)*a(n-1), where prime(n) is the n-th prime.

A121407 Number of double eliminations of twin prime candidates within primorial intervals of p(n)#. That is, it is the number of twin prime candidates for which each half of the pair is composite, where one of those composites has p(n) for its lowest prime factor and the other composite has a prime less than p(n) for its lowest prime factor.

Original entry on oeis.org

0, 0, 0, 2, 18, 210, 2790, 47610, 901530, 20591010, 592452630, 18202996350, 667760974650, 27146297697750, 1157142993063750, 53925515761020750, 2835489033177050250, 166057836818071448250, 10054640164031031318750

Offset: 1

Dennis R. Martin (dennis.martin(AT)dptechnology.com), Jul 28 2006

			The prime factors p(1) = 2 and p(2) = 3 cannot eliminate any twin prime candidates, therefore a(1) = a(2) = a(3) = 0.
For the prime factor p(4) = 7, there will be 8 composites having p(4) for their lowest prime factor within every interval of p(4)# = 210 starting after 7. For instance, the composites {49, 77, 91, 119, 133, 161, 203, 217} are adjacent to and eliminate the twin prime candidates centered at {48, 78, 90, 120, 132, 162, 204, 216}. However, 2 of those 8 are already eliminated by p(3), those being the candidates centered at 204 and 216, since 205 and 215 obviously are composites having 5 for their lowest prime factor. Therefore a(4) = 2 because there are 2 double eliminations by 7 and by a prime less than 7 within each interval of p(4)# = 210.
For p(5) = 11, there are 48 composites that have 11 for their lowest prime factor over any interval of p(5)# = 2310 starting after 11. Those 48 composites are all adjacent to a twin prime candidate center post, but 12 of those candidates are eliminated by p(3) (the ones corresponding to the centers 144, 186, 474, 516, 804, 1134, 1176, 1506, 1794, 1836, 2124 and 2166) and 6 are eliminated by p(4) (those corresponding to the candidate centered at 120, 342, 582, 1728, 1968 and 2190). Therefore a(5) = 12 + 6 = 18.

Dennis Martin, Table of n, a(n) for n = 1..100

Cf. A002110, A005867, A121406.

a(1) = a(2) = 0; for n >= 3, a(n) = t(n) - e'(n) = (p(n-1)-1)*t(n-1) - (p(n-1)-2)*e'(n-1), where p(n) is n-th prime, t(n) is given by sequence A005867 and e'(n) is given by sequence A121406.

A174909 Triangle T(n,i) whose n-th row gives the number of numbers in any prime(n)# consecutive numbers whose smallest prime factor is prime(n-i+1).

Original entry on oeis.org

1, 1, 3, 2, 5, 15, 8, 14, 35, 105, 48, 88, 154, 385, 1155, 480, 624, 1144, 2002, 5005, 15015, 5760, 8160, 10608, 19448, 34034, 85085, 255255, 92160, 109440, 155040, 201552, 369512, 646646, 1616615, 4849845, 1658880, 2119680, 2517120, 3565920

Offset: 1

T. D. Noe, Apr 01 2010

Here prime(n)# denotes the product of the first n primes. Row n begins with A005867(n-1). The other n-1 terms in row n are prime(n) times the previous row. The sum of the terms in row n is cototient(prime(n)#), which is A053144(n), and which equals prime(n)#-A005867(n). This sequence is a generalization of a comment in A005867 by Dennis Martin.

			For n=3, we have prime(n)=5 and any range of 2*3*5=30 consecutive numbers has 2 numbers whose smallest prime factor is 5, 5 numbers whose smallest prime factor is 3, and 15 numbers whose smallest prime factor is 2.
From _Bob Selcoe_, Oct 12 2017: (Start)
Triangle starts:
n/i  1   2   3    4    5     6
1    1
2    1   3
3    2   5   15
4    8  14   35  105
5   48  88  154  385  1155
6  480 624 1144 2002  5005 15015
(End)

Cf. A002110, A005867 (first column), A020639, A053144, A070826 (main diagonal).

Cf. A293558 (transpose).

t={{1}}; q=2; Do[p=Prime[n]; t=AppendTo[t, Join[{(q-1)*t[[ -1,1]]}, p*t[[ -1]]]]; q=p, {n,2,9}]; Flatten[t]
(* Second program: *)
Block[{nn = 8, s}, s = Array[FactorInteger[#][[1, 1]] &, Product[Prime@i, {i, nn}]]; Table[With[{P = Product[Prime@ k, {k, n}]}, Count[Take[s, P], ?(# == Prime[n - i + 1] &)]], {n, nn}, {i, n}]] (* _Michael De Vlieger, Oct 14 2017 *)

A218245 Nicolas's sequence, whose nonnegativity is equivalent to the Riemann hypothesis.

Original entry on oeis.org

2, 1, 1, 1, 1, 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, 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

Offset: 1

Jonathan Sondow, Oct 24 2012

nonn

a(n) = floor(p(n)#/phi(p(n)#) - log(log(p(n)#))*exp(gamma)), where p(n)# is the n-th primorial, phi is Euler's totient function, and gamma is Euler's constant.
J.-L. Nicolas proved that all terms are >= 0 if and only if the Riemann hypothesis (RH) is true. In fact, results in his 2012 paper imply that RH is equivalent to a(n) = 0 for n > 6. Nicolas's refinement of this result is in A233825.
He also proved that if RH is false, then infinitely many terms are >= 0 and infinitely many terms are < 0.
See Nicolas's sequence A216868 for references, links, and additional cross-refs.

			p(2)# = 2*3 = 6 and phi(6) = 2, so a(2) = [6/2 - log(log(6))*e^gamma] = [3-0.58319...*1.78107...] = [3-1.038...] = 1.

Cf. A216868, A233825.

primorial[n_] := Product[Prime[k], {k, n}]; Table[ With[{p = primorial[n]}, Floor[N[p/EulerPhi[p] - Log[Log[p]]*Exp[EulerGamma]]]], {n, 1, 100}]

a(n) = [p(n)#/phi(p(n)#) - log(log(p(n)#))*exp(gamma)].

a(n) = [A002110(n)/A005867(n) - log(log(A002110(n)))*e^gamma].

A249747 a(n) = floor((A002110(n) * A054272(n)) / A001248(n)).

Original entry on oeis.org

1, 2, 8, 51, 496, 6041, 97155, 1746481, 38377034, 1053921489, 31722366805, 1127475187757, 45429396874080, 1910408631449923, 87682336584597009, 4571067440374822934, 260160909199262899454, 15823372061924831882182, 1034588557961336117180784, 72606463908572608290939197, 5235472173106695729625747152, 407296805992490241506213234700

Offset: 1

Antti Karttunen, Dec 08 2014

nonn

Antti Karttunen, Table of n, a(n) for n = 1..100

Cf. A000010, A001248, A002110, A005867, A054272.

default(primelimit, 2^31 + 2^30);
A002110(n) = prod(i=1, n, prime(i));
A054272(n) = 1 + primepi(prime(n)^2) - n;
A249747(n) = (A054272(n)*A002110(n))\(prime(n)^2);
for(n=1, 100, write("b249747.txt", n, " ", A249747(n)));

a(n) = floor((A002110(n) * A054272(n)) / A001248(n)).

The ratio a(n) / A005867(n) seems to stay near 1. Note that A005867(n) = A000010(A002110(n)). See also formulas in A054272.

A286942 Irregular triangle read by rows: numbers 1 <= k <= (A002110(n) - 1) where gcd(k, A002110(n - 1)) = 1.

Original entry on oeis.org

1, 2, 1, 3, 5, 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 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

Offset: 1

Jamie Morken and Michael De Vlieger, May 16 2017

From Michael De Vlieger, May 18 2017: (Start)
Row n of a(n) is the list of numbers 1 <= k <= A002110(n) that are coprime to A002110(n-1).
A286941(n) and A279864(n) are subsets of a(n) such that the terms of the rows of each sequence combined and sorted comprise all the terms of a(n).
Row lengths = A005867(n) + A005867(n-1): {2, 3, 10, 56, 528, 6240, 97920, ...}.
1 is coprime to all n thus delimits the rows of a(n).
The smallest prime q in row n of a(n) is gpf(primorial(n)) = A006530(A002110(n)) = prime(n) by definition of primorial.
The smallest composite x in row n of a(n) is q^2 = A001248(n).
The Kummer number A057588(n) = A002110(n) - 1 is the largest term in row n of a(n). (End)

			The triangle starts:
1, 2;
1, 3, 5;
1, 5, 7, 11, 13, 17, 19, 23, 25, 29
Example1:
To find row n of the irregular triangle A286942, take a running sum for each value in the irregular triangle row n-1 of A286941 with A002110(n-1) b-1 times, where b is the largest prime factor in A002110(n).
For example to find row 3 of A286942: Take a running sum for both 1 and 5 in row n-1 of A286941 with A002110(3-1)=6, 5-1=4 times, where b is the largest prime factor 5 in A002110(3).
Result:
1 5
7 11
13 17
19 23
25 29
Equal to row 3 of A286942: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29.
Example2:
To find row n of the irregular triangle A279864, multiply each value in row n-1 of A286941 with the largest prime factor b in A002110(n).
Example for n=3: b=5.
1*5=5
5*5=25
Example3:
To find row n of the irregular triangle A286941, remove the values that are in row n of the irregular triangle A279864 from the values that are in row n of the irregular triangle A286942.
For n=3.
A286942 row n = 1, 5, 7, 11, 13, 17, 19, 23, 25, 29.
A279864 row n = 5, 25.
Removing values 5, 25 from the values in A286942 row n gives row n of A286941: 1, 7, 11, 13, 17, 19, 23, 29.

Michael De Vlieger, Table of n, a(n) for n = 1..6839 (rows 1 <= n <= 6).
Eric Weisstein's World of Mathematics, Relatively Prime - _Michael De Vlieger_, May 18 2017
Seqfan, Formula for the sequence.

Cf. A002110, A005867, A279864, A286941, A286942.

Table[Select[Range@ #2, Function[k, CoprimeQ[k, #1]]] & @@ Map[Times @@ # &, {Most@ #, #}] &@ Prime@ Range@ n, {n, 4}] // Flatten (* Michael De Vlieger, May 18 2017 *)

a(n) = union(A286941(n), A279864(n)) where n consists of all terms in row n of each sequence. - Michael De Vlieger, May 18 2017

More terms from Michael De Vlieger, May 18 2017

A309804 a(n) is the coefficient of x^n in the polynomial Product_{i=1..n+4} (prime(i)*x-1).

Original entry on oeis.org

1, 28, 652, 16186, 414849, 11970750, 411154568, 14802996860, 617651235401, 28112591190218, 1330940558814492, 68134228016658366, 3888046744502816953, 244783216404832868510, 15878401438954693327808, 1123935467586630569656024, 83970858613393528568199649

Offset: 0

Alexey V. Bazhin, Aug 17 2019

Cf. A000040, A002110, A024451, A070918, A309802, A309803, A033999, A007504, A024447, A024448, A024449, A054640, A005867, A238146, A260613.

a:= n-> coeff(mul(ithprime(i)*x-1, i=1..n+4), x, n):
seq(a(n), n=0..20);  # Alois P. Heinz, Aug 19 2019

a[n_] := CoefficientList[Series[Product[Prime[i]*x - 1, {i, 1, n+4}], {x, 0, 25}], x] [[n+1]]; Array[a, 17, 0] (* Amiram Eldar, Aug 24 2019 *)

a(n) = polcoef(prod(i=1, n+4, prime(i)*x-1), n); \\ Michel Marcus, Aug 25 2019

a(n) = [x^n] Product_{i=1..n+4} (prime(i)*x-1).
a(n) = abs(A070918(n+4,4)).
a(n) = abs(A238146(n+4,n)) for n>0.
a(n) = A260613(n+4,n).

A335284 Numbers k > 1 such that, if p is the least prime dividing k, k is less than or equal to the product of all prime numbers up to (and including) p.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 25, 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

Offset: 1

Javier Múgica, May 30 2020

nonn

The sequence A279864 contains the same terms as this one in different order, namely, sorted according to their least prime factor.
A number k > 1 belongs to this sequence if k <= A002110(A055396(k)) = A034386(A020639(k)). This condition approaches log(k) <= p as k -> infinity, p being the least prime factor of k.
All prime numbers belong to this sequence. Squares of prime numbers are included starting at 5^2; cubes are included starting at 11^3, and so on. That is, for all m there exists a p(m) such that all m-th powers of prime numbers from p(m)^m onwards belong to the sequence.
For large N the number of integers 1 < k <= N which belong to this sequence is ~ e^(-gamma)*N/log(log(N)), where gamma is Euler's constant: A001620.
Let p = p_r denote the r-th prime number and P_r = A034386(p) (the product of primes <= p). This sequence contains 1*2*4*...*(p_(r-1)-1) = A005867(r-1) elements whose least prime factor is p. These are distributed symmetrically about P_r/2, the first ones being p and, for p >= 5, p^2, and the last one being P_r-p.

			The least prime factor of 77 is 7, and 77 < 2*3*5*7 = 210, therefore 77 belongs to the sequence.

Javier Múgica, Table of n, a(n) for n = 1..19720 (terms <= 100000)

Cf. A034386, A020639.
A279864 contains the same terms as this sequence in a different order.
Contains A308966. Both sequences agree in their first 38 terms.

isok(k) = if (k>1, my(p=vecmin(factor(k)[,1])); k <= prod(j=1, primepi(p), prime(j))); \\ Michel Marcus, May 31 2020

Asymptotic expression for a(n): e^(gamma)*n*(log(log(n))+O(1)), where gamma is Euler's constant: A001620.

A335334 Sum of the integers in the reduced residue system of A002110(n).

Original entry on oeis.org

1, 6, 120, 5040, 554400, 86486400, 23524300800, 8045310873600, 4070927302041600, 3305592969257779200, 3074201461409734656000, 4094836346597766561792000, 6715531608420337161338880000, 12128250084807128913378017280000

Offset: 1

Jamie Morken, Jun 02 2020

nonn

Sum of the integers up to A002110(n) and coprime to A002110(n).

The sequence gives the sum of row n of A286941(n).

			For n = 3: A002110(3) = 30, the reduced residue system of 30 is {1, 7, 11, 13, 17, 19, 23, 29}. The sum is a(3) = 120.

Michael De Vlieger, Table of n, a(n) for n = 1..196
Eric Weisstein's World of Mathematics, Totative.

Cf. A023896, A002110, A005867, A070826, A038110, A058250, A286941.

n = 15;
A002110 = Drop[FoldList[Times, 1, Prime[Range[n]]], 1];
A005867 = Drop[EulerPhi@FoldList[Times, 1, Prime@Range@n], 1];
A002110*A005867/2
(* Second program: *)
Map[# EulerPhi[#]/2 &, FoldList[Times, Prime@ Range@ 14]] (* Michael De Vlieger, Apr 07 2021 *)

a(n) = my(P=factorback(primes(n))); P*eulerphi(P)/2; \\ Michel Marcus, Jun 02 2020

a(n) = A023896(A002110(n)).
a(n) = A002110(n)*A005867(n)/2 = A070826(n)*A005867(n).
a(n) = (A002110(n)*A038110(n+1)/2)*A058250(n).

A336016 a(n) is the number of primes q less than primorial(n) having k = 2 as the least exponent such that q^k == 1 (mod primorial(n)).

Original entry on oeis.org

0, 1, 3, 5, 8, 19, 22, 51, 89, 145, 263, 453, 851, 1575, 2880, 5469, 10338, 19115, 35782, 67569, 128601, 243600, 463840, 883589

Offset: 1

Michael De Vlieger, David James Sycamore, David A. Corneth, Jul 10 2020

a(n) = length of row n of A336015 for n > 1.

			a(4) = 5 as there are 5 primes q coprime to primorial(4) = 210 such that 2 is the least positive integer exponent k where q^k == 1 (mod 210). Those primes are 29, 41, 71, 139, 181 and indeed we have 29^2 == 1 (mod 210), 41^2 == 1 (mod 210), 71^2 == 1 (mod 210), 139^2 == 1 (mod 210) and 181^2 == 1 (mod 210) and no more below 210. So as these are five such primes in row 4, a(4) = 5. - _David A. Corneth_, Aug 15 2020

Cf. A000010, A002110, A005867, A131577, A336015.

Table[Block[{P = #, k = 0}, Do[If[MultiplicativeOrder[Prime@ i, P] == 2, k++], {i, PrimePi[n + 1], PrimePi[P - 1]}]; k] &@ Product[Prime@ j, {j, n}], {n, 8}]

a(n) = {if(n <= 2, return(n-1)); my(pp = vecprod(primes(n))/2, d = divisors(pp), res = 0); for(i = 1, #d, c = lift(chinese(Mod(-1, d[i]), Mod(1, pp/d[i]))); forstep(i = c, pp*2, pp, if(isprime(i), res++ ) ) ); res } \\ David A. Corneth, Aug 16 2020

New name from David A. Corneth, Aug 15 2020

cp's OEIS Frontend

A121406 a(1) = a(2) = 0; a(3) = 2; for n >= 4, a(n) = (prime(n-1)-2)*a(n-1), where prime(n) is the n-th prime.

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A174909 Triangle T(n,i) whose n-th row gives the number of numbers in any prime(n)# consecutive numbers whose smallest prime factor is prime(n-i+1).

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A218245 Nicolas's sequence, whose nonnegativity is equivalent to the Riemann hypothesis.

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A249747 a(n) = floor((A002110(n) * A054272(n)) / A001248(n)).

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A286942 Irregular triangle read by rows: numbers 1 <= k <= (A002110(n) - 1) where gcd(k, A002110(n - 1)) = 1.

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A309804 a(n) is the coefficient of x^n in the polynomial Product_{i=1..n+4} (prime(i)*x-1).

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A335284 Numbers k > 1 such that, if p is the least prime dividing k, k is less than or equal to the product of all prime numbers up to (and including) p.

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