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A053662
Numbers k such that 2k+1 divides k!+1.
3
3, 5, 9, 21, 23, 33, 39, 51, 63, 65, 81, 89, 95, 99, 113, 131, 173, 183, 191, 209, 215, 221, 239, 245, 251, 261, 281, 285, 299, 303, 309, 315, 341, 345, 363, 369, 371, 393, 411, 419, 431, 443, 473, 495, 509, 525, 543, 545, 561, 575, 593, 645, 659, 683, 711
OFFSET
1,1
COMMENTS
k+1 divides k!+1 gives primes-1 by Wilson's Theorem. For the present sequence, there are 309 terms below 5000, compared with 669 primes (309/669 = 0.461...). There are 553 terms below 10000, compared with 1229 primes (553/1229 = 0.449...). - Ed Pegg Jr, Dec 05 2001
LINKS
FORMULA
a(n) >> n log n. - Charles R Greathouse IV, Apr 16 2024
MAPLE
A053662:=n->`if`(n!+1 mod (2*n+1) = 0, n, NULL): seq(A053662(n), n=1..1000); # Wesley Ivan Hurt, Dec 01 2015
MATHEMATICA
Drop[Union[Table[If[IntegerQ[(n!+1)/(2n+1)], n], {n, 1, 1000}]], -1] (* Ed Pegg Jr, Dec 05 2001 *)
Select[Range[1000], Mod[#! +1, 2*# +1] == 0 &] (* G. C. Greubel, May 18 2019 *)
PROG
(PARI) for(n=1, 10^3, if((n!+1)%(2*n+1)==0, print1(n, ", ")) ) \\ G. C. Greubel, May 18 2019
(Magma) [n: n in [1..1000] | (Factorial(n)+1) mod (2*n+1) eq 0 ]; // G. C. Greubel, May 18 2019
(Sage) [n for n in (1..1000) if Mod(factorial(n)+1, 2*n+1)==0 ] # G. C. Greubel, May 18 2019
(GAP) Filtered([1..1000], n-> (Factorial(n)+1) mod (2*n+1)=0) # G. C. Greubel, May 18 2019
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Chris K. Caldwell, Feb 16 2000
STATUS
approved