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%I #51 Oct 12 2023 01:53:52
%S 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,
%T 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,
%U 7,3,2,1,3,2,5,10,1,10,6,38,3,3,4,1,41,2
%N The number of positive k < (2n-1)^2 such that (2^k - 1)/(2n - 1)^2 is an integer.
%C a(n) is the number of integers k in range 1 .. A016754(n-1)-1 such that A000225(k) is an integral multiple of A016754(n-1). - _Antti Karttunen_, Nov 15 2014
%C 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
%C From _Charlie Neder_, Oct 18 2018: (Start)
%C 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.
%C 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)
%H Kevin P. Thompson, <a href="/A246702/b246702.txt">Table of n, a(n) for n = 1..1560</a> (terms 1..128 from Antti Karttunen)
%F a(n) = floor( 4*n*(n-1) / A002326(2*n*(n-1)) ). - _Max Alekseyev_, Oct 11 2023
%e a(2) = 1 because (2^6 - 1)/(2*2 - 1)^2 = 7 is integer and 6 < 9.
%e a(3) = 1 because (2^20 - 1)/(2*3 - 1)^2 = 41943 is integer and 20 < 25.
%e 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.
%p A246702 := proc(n)
%p local a,klim,k ;
%p a := 0 ;
%p klim := (2*n-1)^2 ;
%p for k from 1 to klim-1 do
%p if modp(2^k-1,klim) = 0 then
%p a := a+1 ;
%p end if;
%p end do:
%p a ;
%p end proc:
%p seq(A246702(n),n=1..80) ; # _R. J. Mathar_, Nov 15 2014
%t 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];
%t Table[A246702[n], {n, 1, 84}] (* _Jean-François Alcover_, Oct 04 2017, translated from _R. J. Mathar_'s Maple code *)
%o (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
%o (PARI) a(n)=my(t=(2*n-1)^2,m=Mod(1,t)); sum(k=1,t-1,m*=2;m==1) \\ _Charles R Greathouse IV_, Nov 16 2014
%o (PARI) a246702(n) = my(m=(2*n-1)^2); (m-1)\znorder(Mod(2,m)); \\ _Max Alekseyev_, Oct 11 2023
%Y A246703 gives the positions of records.
%Y Cf. A000225, A002326, A016754, A049094, A237043.
%K nonn
%O 1,4
%A _Juri-Stepan Gerasimov_, Nov 15 2014
%E Corrected by _R. J. Mathar_, Nov 15 2014
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