A103477 -id:A103477 - cp's OEIS frontend

A103478 Positive integers k for which 1 + 5*2^(k+2) divides the Fermat number 1 + 2^2^k.

5, 23, 73, 125, 1945, 23471

Offset: 1

Serhat Sevki Dincer (mesti_mudam(AT)yahoo.com), Feb 07 2005

On Keller's linked page, to find the terms, you run through the tables and find all rows with k = 5 and with n exactly 2 greater than m, then that m belongs to this sequence. - Jeppe Stig Nielsen, Dec 04 2018

			a(1)=5 because 5 is the smallest positive integer k for which 1 + 5*2^(k+2) divides the Fermat number 1 + 2^2^k.

Wilfrid Keller, Prime factors k*2^n + 1 of Fermat numbers F_m

Cf. A000215, A083575, A103477, A103479.

Select[Range[1, 2000], Mod[1 + PowerMod[2, 2^#, 1 + 5*2^(# + 2)], 1 + 5*2^(# + 2)] == 0 &] (* Julien Kluge, Jul 08 2016 *)

isok(n) = Mod(2, 1+5*2^(n+2))^(2^n) + 1 == 0; \\ Michel Marcus, Apr 29 2016

A103479 Positive integers k for which 1 + 6*2^(k+2) divides the Fermat number 1 + 2^2^k.

Original entry on oeis.org

38, 2478782

Offset: 1

Serhat Sevki Dincer (mesti_mudam(AT)yahoo.com), Feb 07 2005

On Keller's linked page, to find the terms, you run through the tables and find all rows with k = 3 and with n exactly 3 greater than m, then that m belongs to this sequence. - Jeppe Stig Nielsen, Dec 04 2018

			a(1)=38 because 38 is the smallest positive integer k for which 1 + 6*2^(k+2) divides the Fermat number 1 + 2^2^k.

Wilfrid Keller, Prime factors k*2^n + 1 of Fermat numbers F_m

Cf. A103477, A103478.

aQ[n_] := PowerMod[2, 2^n, 1 + 6*2^(n+2)] == 6*2^(n+2); Select[Range[3000000], aQ] (* Amiram Eldar, Dec 04 2018 *)

isOK(n) = Mod(2, 1+3*2^(n+3))^(2^n) + 1 == 0 \\ Jeppe Stig Nielsen, Dec 03 2018

Sequence name trimmed by Jeppe Stig Nielsen, Dec 03 2018

cp's OEIS Frontend

A103478 Positive integers k for which 1 + 5*2^(k+2) divides the Fermat number 1 + 2^2^k.

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