A100564 -id:A100564 - cp's OEIS frontend

A264907 a(n) is the smallest "cyclic" integer, k, that has exactly n prime factors.

1, 2, 15, 255, 5865, 146965, 3380195, 125067215, 7378965685, 494390700895, 36090521165335, 2571956263189313, 187752807212819849, 18212022299643525353, 1839414252263996060653, 196817324992247578489871, 21453088424154986055395939, 2981979290957543061700035521

Offset: 0

Geoffrey Critzer, Nov 28 2015

nonn

The cyclic numbers are given in A003277.
a(n) = k if k = p_1*p_2*...*p_n  where the p_i are distinct primes and no p_j-1 is divisible by any p_i and k is the smallest such integer.
a(n) < a(n+1) for all n. For some n, also a(n) | a(n+1). Note that the divisors of any cyclic number are cyclic. - Jeppe Stig Nielsen, May 22 2021
Are the prime factors of a(n) always a "normal sequence of primes" in the sense of A100564? Equivalently, can you always find a(n) by starting from a suitable smallest prime p_1, and then pick the subsequent prime factors greedily (under the condition that p_j - 1 is not divisible by any p_i) until you have n primes? If yes, then it is easy to calculate a(n), as all one needs to do is find the optimal starting prime. - Jeppe Stig Nielsen, May 23 2021

			The prime factorizations for terms a(1)..a(12) are:
2
3, 5
3, 5, 17
3, 5, 17, 23
5, 7, 13, 17, 19
5, 7, 13, 17, 19, 23
5, 7, 13, 17, 19, 23, 37
5, 7, 13, 17, 19, 23, 37, 59
5, 7, 13, 17, 19, 23, 37, 59, 67
5, 7, 13, 17, 19, 23, 37, 59, 67, 73
7, 11, 13, 17, 19, 31, 37, 41, 47, 59, 61
7, 11, 13, 17, 19, 31, 37, 41, 47, 59, 61, 73
Corrected by _Jeppe Stig Nielsen_, May 22 2021.
146965 = 5*7*13*17*19 is cyclic. Since it is the smallest example with 5 primes, 146965 = a(5). It is not a multiple of a(4) = 3*5*17*23. - _Jeppe Stig Nielsen_, May 22 2021

Cf. A003277, A100564.

n=0;for(m=1,+oo,if(gcd(m,eulerphi(m))==1&&omega(m)==n,print1(m,", ");n++)) \\ slow, from Jeppe Stig Nielsen, May 22 2021

N=0;for(n=0,+oo,a=+oo;forsubset([N,n],x,m=prod(j=1,n,prime(x[j]));mJeppe Stig Nielsen, May 22 2021

Wrong terms a(5), a(6), a(7), a(8), a(10), a(12) corrected, and more terms added, and a(0)=1 preprended by Jeppe Stig Nielsen, May 22 2021

a(17) from Jeppe Stig Nielsen, May 22 2021

A287060 Primes a(n) such that a(1) = 2, a(2) = 3, and a(n) is the smallest prime greater than a(n-1) such that (a(n) - 1)/2 is not divisible by a(m) for all m < n.

Original entry on oeis.org

2, 3, 11, 47, 59, 71, 83, 107, 131, 179, 191, 227, 239, 251, 311, 347, 431, 443, 467, 491, 563, 587, 599, 647, 719, 743, 839, 911, 971, 1019, 1031, 1091, 1103, 1151, 1187, 1259, 1283, 1307, 1319, 1367, 1427, 1451, 1511, 1523, 1559, 1571, 1583, 1619, 1667

Offset: 1

Amiram Eldar, May 19 2017

nonn

As in A100564, the number of terms in this sequence which do not exceed x is ~ (1 + o(1)) x/(logx loglogx), thus the sum of the their reciprocals diverges.

			5 and 7 are not in the sequence since a(1) | (5 - 1)/2 and a(2) | (7 - 1)/2.
a(3) = 11 is in the sequence since (11 - 1)/2 = 5 is not divisible by 2 or 3.

Jean-Marie De Koninck and Florian Luca, Analytic Number Theory: Exploring the Anatomy of Integers, American Mathematical Society, 2012, Problem 15.1, p. 263.

Amiram Eldar, Table of n, a(n) for n = 1..1000

Cf. A100564.

a[1] = 2; a[2] = 3; a[n_] := a[n] = Block[{k=PrimePi[a[n - 1]] + 1, t=Table[a[i], {i, n-1}]}, While[Union[ Mod[(Prime[k] - 1)/2, t]][[1]] == 0, k++]; Prime[k]]; Table[a[n], {n, 49}]

isok(p, va) = {q = (p-1)/2; for (k=1, #va, if (!(q % va[k]), return (0));); return (1);}
lista(nn) = {va = [2, 3]; print1(va[1], ", " va[2], ", "); for (n=3, nn, forprime(p=nextprime(vecmax(va)+1),, if (isok(p, va), va = concat(va, p); print1(p, ", "); break);););} \\ Michel Marcus, May 21 2017

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A264907 a(n) is the smallest "cyclic" integer, k, that has exactly n prime factors.

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A287060 Primes a(n) such that a(1) = 2, a(2) = 3, and a(n) is the smallest prime greater than a(n-1) such that (a(n) - 1)/2 is not divisible by a(m) for all m < n.

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