A212552 -id:A212552 - cp's OEIS frontend

A125135 Triangle read by rows in which row n lists prime factors of p^p - 1 where p = prime(n).

3, 2, 13, 2, 2, 11, 71, 2, 3, 29, 4733, 2, 5, 15797, 1806113, 2, 2, 3, 53, 264031, 1803647, 2, 2, 2, 2, 10949, 1749233, 2699538733, 2, 3, 3, 109912203092239643840221, 2, 11, 461, 1289, 831603031789, 1920647391913

Offset: 1

N. J. A. Sloane, Jan 21 2007

			Triangle begins:
3;
2, 13;
2, 2, 11, 71;
2, 3, 29, 4733;
2, 5, 15797, 1806113;
2, 2, 3, 53, 264031, 1803647;
2, 2, 2, 2, 10949, 1749233, 2699538733;
2, 3, 3, 109912203092239643840221;
2, 11, 461, 1289, 831603031789, 1920647391913;
2, 2, 7, 59, 16763, 84449, 2428577, 14111459, 58320973, 549334763;
...
n=4: p=7, 7^7-1 = 823542 = 2*3*29*4733 gives row 4.

Sam Wagstaff, Factorizations of p^p - 1 for most p < 180

Cf. A006486, A088730, A125136, A212552, A214812.

for p in [ n : n in [1..100] | IsPrime(n) ] do "\nDoing p =", p; n := p^p -1; Factorisation(n); end for; // John Cannon

T:= n-> (p-> sort(map(i-> i[1]$i[2], ifactors(p^p-1)[2]))[])(ithprime(n)):
seq(T(n), n=1..10);  # Alois P. Heinz, May 20 2022

A214812 Largest prime factor of (p^p-1)/(p-1) where p = prime(n).

Original entry on oeis.org

3, 13, 71, 4733, 1806113, 1803647, 2699538733, 109912203092239643840221, 1920647391913, 549334763, 568972471024107865287021434301977158534824481, 41903425553544839998158239, 5926187589691497537793497756719, 19825223972382274003506149120708429799166030881820329892377241, 194707033016099228267068299180244011637

Offset: 1

N. J. A. Sloane, Jul 31 2012

nonn

Daniel Suteu, Table of n, a(n) for n = 1..33
T. S. Motzkin, Sorting numbers for cylinders and other classification numbers, in Combinatorics, Proc. Symp. Pure Math. 19, AMS, 1971, pp. 167-176. [Annotated, scanned copy]

Cf. A001039, A088807, A125135, A212552, A214811.

FactorInteger[#][[-1,1]]&/@Table[(p^p-1)/(p-1),{p,Prime[Range[15]]}] (* Harvey P. Dale, Aug 27 2016 *)

a(n) = my(p=prime(n)); vecmax(factor((p^p-1)/(p-1))[,1]); \\ Daniel Suteu, May 26 2022

a(n) = A006530(A001039(n)). - Daniel Suteu, May 26 2022

A329065 Smallest m_0 such that A118106(m_0) = n; smallest m_0 such that if we write m_0 = Product_{i=1..t} p_i^e_i, then lcm_{1<=i,j<=t, i!=j} ord(p_i,p_j^e_j) = n, where ord(a,r) is the multiplicative order of a modulo r.

Original entry on oeis.org

1, 6, 14, 10, 55, 18, 203, 34, 146, 22, 46, 26, 689, 86, 302, 51, 5759, 38, 955, 50, 98, 69, 94, 288, 505, 5462, 327, 58, 466, 77, 9305, 384, 5447, 309, 142, 74, 446, 2933, 158, 246, 3403, 129, 862, 115, 543, 141, 4702, 119, 5713, 453, 206, 106, 5671, 162, 605, 928, 687, 118

Offset: 1

Jianing Song, Nov 03 2019

nonn

For n != 1, 6, a(n) <= 2*A112927(n): suppose n != 1, 6, by Zsigmondy's theorem, 2^n - 1 has at least one primitive factor p. Here a primitive factor p means that ord(2,p) = n, where ord(a,r) is the multiplicative order of a modulo r. So we have A118106(2p) = lcm(ord(p,2),ord(2,p)) = lcm(1,n) = n. Specially, we have A118106(2*A112927(n)) = n for n != 1, 6.

There is another way to construct m such that A118106(m) = n > 1 (and usually this way generates smaller m's than the way above): let q be any prime factor of n, again, by Zsigmondy's theorem, q^n - 1 has at least one primitive factor p unless (n,q) = (6,2). Note that q^(p-1) == 1 (mod p), so q^gcd(p-1,n) == 1 (mod p). But n is the smallest positive number such that q^n == 1 (mod p), so gcd(p-1,n) = n. So we have A118106(pq) = lcm(ord(p,q),ord(q,p)) = lcm(1,n) = n. For example, if n = 5, then q = 5, p = 11, m = 55 (the way above gives A118106(62) = 5); if n = 7, then q = 7, p = 29, m = 203 (the way above gives A118106(254) = 7); if n = 13, then q = 13, p = 53, m = 689 (the way above gives A118106(16382) = 13). This gives a(q) <= q*A212552(q) for primes q.

			A118106(203) = 7; for any m < 203, A118106(m) is not equal to 7, so a(7) = 203.

Cf. A118106, A112927, A212552.

a(n) = for(k=1, oo, if(A118106(k)==n, return(k))) \\ See A118106 for its program

A354226 a(n) is the number of distinct prime factors of (p^p - 1)/(p - 1) where p = prime(n).

Original entry on oeis.org

1, 1, 2, 2, 2, 3, 3, 1, 4, 7, 1, 7, 5, 3, 3, 5, 3, 4, 6, 4, 10, 5, 4, 6, 6, 9, 5, 4, 5, 8, 6, 4, 11

Offset: 1

Luis H. Gallardo, May 20 2022

a(34) > 3, and depends on the full factorization of the 296-digit composite number (139^139 - 1)/138. - Tyler Busby, Jan 22 2023

Sequence continues as ?, 8, ?, 5, 8, 4, 5, ?, 8, ?, 8, 7, 6, 3, 3, ..., where ? represents uncertain terms. - Tyler Busby, Jan 22 2023

			a(3)=2, since (5^5 - 1)/(5 - 1) = 11*71.

factordb, Status of (139^139 - 1)/138.

Cf. A000040, A001039, A001221, A088790, A125135, A212552, A214812.

a(n) = my(p=prime(n)); omega((p^p-1)/(p-1)); \\ Michel Marcus, May 22 2022

from sympy import factorint, prime
def a(n): p = prime(n); return len(factorint((p**p-1)//(p-1)))
print([a(n) for n in range(1, 12)]) # Michael S. Branicky, May 23 2022

a(n) = A001221(A001039(n)).

a(18)-a(33) from Amiram Eldar, May 20 2022

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A125135 Triangle read by rows in which row n lists prime factors of p^p - 1 where p = prime(n).

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A214812 Largest prime factor of (p^p-1)/(p-1) where p = prime(n).

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A329065 Smallest m_0 such that A118106(m_0) = n; smallest m_0 such that if we write m_0 = Product_{i=1..t} p_i^e_i, then lcm_{1<=i,j<=t, i!=j} ord(p_i,p_j^e_j) = n, where ord(a,r) is the multiplicative order of a modulo r.

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PARI

A354226 a(n) is the number of distinct prime factors of (p^p - 1)/(p - 1) where p = prime(n).

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