A247216 -id:A247216 - cp's OEIS frontend

A247229 Minimum of the greatest prime factors of (i^prime(n)-1)/(i-1), when i runs through all integers in [2, prime(n)].

3, 7, 11, 127, 89, 8191, 30839, 363889, 178481, 2089, 4404047, 616318177, 164511353, 2099863, 13264529, 20394401

Offset: 1

Vladimir Shevelev, Nov 27 2014

If in the definition "greatest" is replaced with "smallest", then we obtain A035095 (see comment there).

a(n) = {mini = 0; p = prime(n); for (i=2, p, f = factor((i^p-1)/(i-1)); gpf = f[#f~, 1]; if (! mini, mini = gpf, mini = min(mini, gpf));); mini;} \\ Michel Marcus, Dec 09 2014

a(n) == 1 (mod prime(n)).

More terms from Peter J. C. Moses, Nov 27 2014
a(13)-a(14) from Michel Marcus, Dec 09 2014
a(15)-a(16) from Jinyuan Wang, Mar 03 2020

A247230 Maximum of the greatest prime factors of (i^prime(n)-1)/(i-1), when i runs through all integers in [2, prime(n)].

Original entry on oeis.org

3, 13, 71, 55987, 12207031, 16148168401, 50544702849929377, 109912203092239643840221, 26552618219228090162977481, 582595509837473004489665028935473297399, 568972471024107865287021434301977158534824481, 4673016293261684390446904937094157950067612234753

Offset: 1

Vladimir Shevelev, Nov 27 2014

nonn

If in the definition "maximum" is replaced with "minimum," then we obtain A247229.

Cf. A247216, A247229.

a(n) = {maxi = 0; p = prime(n); for (i=2, p, f = factor((i^p-1)/(i-1)); gpf = f[#f~, 1]; if (! maxi, maxi = gpf, maxi = max(maxi, gpf));); maxi;} \\ Michel Marcus, Dec 09 2014

a(n)==1 (mod prime(n)).

More terms from Peter J. C. Moses, Nov 27 2014

cp's OEIS Frontend

A247229 Minimum of the greatest prime factors of (i^prime(n)-1)/(i-1), when i runs through all integers in [2, prime(n)].

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