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@ -204,3 +204,61 @@ Proof.
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apply LEMMA. rewrite H4 in H7. contradiction H7.
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reflexivity.
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Qed.
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Lemma tm_step_palindromic_even_center :
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forall (n : nat) (hd a tl : list bool),
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tm_step n = hd ++ a ++ (rev a) ++ tl
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-> 0 < length a
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-> even (length (hd ++ a)) = true.
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Proof.
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intros n hd a tl. intros H I.
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assert (J: a = removelast a ++ [ last a false ]).
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apply app_removelast_last.
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assert (K: {a=nil} + {~ a=nil}). apply list_eq_dec. apply bool_dec.
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destruct K. rewrite e in I. inversion I. assumption.
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rewrite J in H. rewrite rev_app_distr in H.
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rewrite <- app_assoc in H. rewrite <- app_assoc in H.
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rewrite app_assoc in H.
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replace
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([last a false] ++ rev [last a false] ++ rev (removelast a) ++ tl)
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with ([last a false; last a false] ++ (rev (removelast a) ++ tl)) in H.
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apply tm_step_consecutive_identical in H. rewrite J.
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rewrite app_assoc. rewrite app_length. rewrite Nat.even_add.
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rewrite <- Nat.negb_odd. rewrite H. reflexivity.
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reflexivity.
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Qed.
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(*
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TODO: les palindromes de longueur 16 sont centrés autour
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des valeurs de la suite ci-dessous A056196
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TODO: erratum 216 est un centre possible ! (216 = 2^3 3^3)
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A056196 Numbers n such that A055229(n) = 2. +30
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2
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8, 24, 32, 40, 56, 72, 88, 96, 104, 120, 128, 136, 152, 160, 168, 184, 200, 224, 232,
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248, 264, 280, 288, 296, 312, 328, 344, 352, 360, 376, 384, 392, 408, 416, 424, 440,
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456, 472, 480, 488, 504, 512, 520, 536, 544, 552, 568, 584, 600, 608, 616, 632, 640
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(list; graph; refs; listen; history; text; internal format)
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OFFSET 1,1
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COMMENTS By definition, the largest square divisor and squarefree part of these
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numbers have GCD = 2.
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Different from A036966. E.g., 81 is not here because A055229(81) = 1.
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Numbers of the form 2^(2*k+1) * m, where k >= 1 and m is an odd number
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whose prime factorization contains only exponents that are either 1 or
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even. The asymptotic density of this sequence is (1/12) * Product_{p odd
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prime} (1-1/(p^2*(p+1))) = A065465 / 11 = 0.08013762179319734335... -
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Amiram Eldar, Dec 04 2020, Nov 25 2022
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LINKS Robert Israel, Table of n, a(n) for n = 1..10000
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EXAMPLE 88 is here because 88 has squarefree part 22, largest square divisor 4,
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and A055229(88) = gcd(22, 4) = 2.
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*)
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