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A246702
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The number of positive k < (2n-1)^2 such that (2^k - 1)/(2n - 1)^2 is an integer.
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6
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0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 10, 2, 1, 1, 1, 6, 3, 2, 1, 9, 2, 3, 3, 2, 2, 6, 1, 13, 9, 1, 1, 10, 5, 1, 3, 2, 8, 3, 2, 2, 1, 1, 10, 3, 8, 7, 9, 2, 2, 3, 1, 2, 26, 1, 3, 9, 4, 2, 9, 4, 1, 6, 1, 18, 9, 1, 7, 3, 2, 1, 3, 2, 5, 10, 1, 10, 6, 38, 3, 3, 4, 1, 41, 2
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OFFSET
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1,4
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COMMENTS
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Conjecture: the positions of 1's, a(k)=1, are exactly given by the 2k-1 which are elements of A167791. - Antti Karttunen, Nov 15 2014
It would appear that, if 2k-1 is in A167791, then so is (2k-1)^2, and so a(k) = 1 would follow by definition.
Conjecture: Let B be the first value such that (2k-1)^2 divides 2^B - 1. Then either 2k-1 divides B, or 2k-1 is a Wieferich prime (A001220). (End)
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LINKS
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FORMULA
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EXAMPLE
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a(2) = 1 because (2^6 - 1)/(2*2 - 1)^2 = 7 is integer and 6 < 9.
a(3) = 1 because (2^20 - 1)/(2*3 - 1)^2 = 41943 is integer and 20 < 25.
a(3) = 2 because (2^21 - 1)/(2*4 - 1)^2 = 42799 is integer and 21 < 49; and also (2^42 - 1)/(2*4 - 1)^2 = 89756051247 is integer and 42 < 49.
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MAPLE
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local a, klim, k ;
a := 0 ;
klim := (2*n-1)^2 ;
for k from 1 to klim-1 do
if modp(2^k-1, klim) = 0 then
a := a+1 ;
end if;
end do:
a ;
end proc:
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MATHEMATICA
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A246702[n_] := Module[{a, klim, k}, a = 0; klim = (2*n-1)^2; For[k = 1, k <= klim-1, k++, If[Mod[2^k-1, klim] == 0, a = a+1]]; a];
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PROG
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(Scheme) (define (A246702 n) (let ((u (A016754 (- n 1)))) (let loop ((k (- u 1)) (s 0)) (cond ((zero? k) s) ((zero? (modulo (A000225 k) u)) (loop (- k 1) (+ s 1))) (else (loop (- k 1) s)))))) ;; Antti Karttunen, Nov 15 2014
(PARI) a246702(n) = my(m=(2*n-1)^2); (m-1)\znorder(Mod(2, m)); \\ Max Alekseyev, Oct 11 2023
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CROSSREFS
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A246703 gives the positions of records.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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