The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A344380 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
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A344380

Complement of A344378 in A172186.
(history; published version)

#26 by N. J. A. Sloane at Fri Jun 18 12:37:13 EDT 2021

STATUS

editing

approved

#25 by N. J. A. Sloane at Fri Jun 18 12:36:54 EDT 2021

NAME

Complement to of A344378 in A172186.

STATUS

proposed

editing

Discussion

Fri Jun 18

12:37

N. J. A. Sloane: "of" sounds better in definition here

#24 by René Gy at Tue May 25 12:55:47 EDT 2021

STATUS

editing

proposed

#23 by René Gy at Tue May 25 12:50:46 EDT 2021

MATHEMATICA

list = Select[Range[2 L + m-1], PrimeQ[#] && Mod[L, (# - 1)/2] == 0 &];

STATUS

proposed

editing

#22 by René Gy at Mon May 24 13:24:44 EDT 2021

STATUS

editing

proposed

#21 by René Gy at Mon May 24 13:24:17 EDT 2021

EXAMPLE

14 belongs to the sequence, because it is squarefree, and 1+2^(2k)+3^(2k)+4^(2k)+5^(2k)+6^(2k)+7^(2k)+8^(2k)+9^(2k)+10^(2k)+11^(2k)+12^(2k)+13^(2k)+14^(2k)is always divisible by 29 when 14 does not divide k, and when 14 divides k, it is divisible by 13 or by 7.

STATUS

proposed

editing

#20 by René Gy at Mon May 17 12:10:32 EDT 2021

STATUS

editing

proposed

#19 by René Gy at Mon May 17 12:10:10 EDT 2021

DATA

6, 14, 38, 42, 57, 65, 70, 93, 106, 114, 118, 138, 154, 158, 182, 186, 190, 205, 210, 217, 218, 222, 266, 266, 277, 281, 285, 309, 326

STATUS

proposed

editing

#18 by Jon E. Schoenfield at Sun May 16 23:36:40 EDT 2021

STATUS

editing

proposed

#17 by Jon E. Schoenfield at Sun May 16 23:36:37 EDT 2021

COMMENTS

Belong Terms belong to A172186 but not to A344378. Even though a(n)*(a(n)+1)*(2*a(n)+1) is squarefree, Sum_{1<=j<=1..a(n)} j^{(2k} ) always has a prime divisor which is smaller than 2*a(n)+3, whatever k. For the integers m such that m*(m+1)*(2*m+1) is non squarefree, nonsquarefree, Sum_{1<=j<=1..m} j^{(2k} ) always has a prime divisor which is smaller than 2*m+3, whatever k, because it is divisible by any prime p such that p^2 divides m*(m+1)*(2*m+1).

STATUS

proposed

editing