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 A268669 a(n) = polynomial quotient (computed over GF(2), result is its binary encoding) that is left after all instances of polynomial (X+1) have been factored out of the polynomial that is encoded by the binary expansion of n. (See comments for details). 9
 1, 2, 1, 4, 1, 2, 7, 8, 7, 2, 11, 4, 13, 14, 1, 16, 1, 14, 19, 4, 21, 22, 13, 8, 25, 26, 7, 28, 11, 2, 31, 32, 31, 2, 35, 28, 37, 38, 11, 8, 41, 42, 25, 44, 7, 26, 47, 16, 49, 50, 1, 52, 19, 14, 55, 56, 13, 22, 59, 4, 61, 62, 21, 64, 21, 62, 67, 4, 69, 70, 61, 56, 73, 74, 13, 76, 59, 22, 79, 16, 81, 82, 49, 84, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS When polynomials over GF(2) are encoded in the binary representation of n in a natural way where each polynomial b(n)*X^n+...+b(0)*X^0 over GF(2) is represented by the binary number b(n)*2^n+...+b(0)*2^0 in N (each coefficient b(k) is either 0 or 1), then a(n) = representation of the polynomial that is left as a quotient when all X+1 polynomials (encoded by 3, "11" in binary) have been divided out. Each a(n) is one of the "Garden of Eden" patterns of Rule-60 one-dimensional cellular automaton, a seed pattern which after A268389(n) generations yields the configuration encoded in binary expansion of n. No terms of A001969 occur so all terms are odious (in A000069). Each odious number occurs an infinitely many times. LINKS Antti Karttunen, Table of n, a(n) for n = 1..65537 Eric Weisstein's World of Mathematics, Rule 60 FORMULA If A006068(n) is odd, then a(n) = n, otherwise a(n) = a(A006068(n)/2). Other identities and observations. For all n >= 1: a(n) = A003188(A268670(n)). A010060(a(n)) = 1. [All terms are odious.] a(n) <= n. More precisely, a(A000069(n)) = A000069(n) and a(A001969(n)) < A001969(n). The equivalence of the following two formulas stems from the additive nature of Rule-60 cellular automaton. Or more plainly, because carryless binary multiplication A048720 distributes over carryless binary sum, XOR A003987: A048724^A268389(n) (a(n)) = n. [Starting from k = a(n), and iterating map k -> A048724(k) exactly A268389(n) times yields n back.] A048720(a(n),A048723(3,A268389(n))) = A048720(a(n),A001317(A268389(n))) = n. EXAMPLE For n = 5 ("101" in binary) which encodes polynomial x^2 + 1, we observe that it can be factored in ring GF(2)[X] as (X+1)(X+1), and thus a(5) = 1, because after dividing both instances of (X+1) off, we are left with the quotient polynomial 1 which is encoded by 1. For n = 8 ("1000" in binary) which encodes polynomial x^3, we observe that it is not divisible in ring GF(2)[X] by polynomial X+1, thus a(8) = 8. For n = 9 ("1001" in binary) which encodes polynomial x^3 + 1, we observe that it can be factored over GF(2) as (X+1)(X^2 + X + 1), and thus a(9) = 7, because the quotient polynomial X^2 + X + 1 is encoded by 7 ("111" in binary). MATHEMATICA f[n_] := If[OddQ@ #, n, f[#/2]] &@ Fold[BitXor, n, Quotient[n, 2^Range[BitLength@ n - 1]]]; Array[f, {85}] (* Michael De Vlieger, Feb 12 2016, after Jan Mangaldan at A006068 *) PROG (PARI) a(n) = {p = Pol(binary(n))*Mod(1, 2); q = (x+1)*Mod(1, 2); while (type(r = p/q) == "t_POL", p = r); np = Polrev(vector(poldegree(p)+1, k, k--; lift(polcoeff(p, k)))); subst(np, x, 2); } \\ Michel Marcus, Feb 12 2016 (Scheme) ;; This employs the given recurrence and uses memoization-macro definec: (definec (A268669 n) (if (odd? (A006068 n)) n (A268669 (/ (A006068 n) 2)))) (define (A268669 n) (let loop ((n n)) (let ((k (A006068 n))) (if (odd? k) n (loop (/ k 2)))))) ;; Computed in a loop, no memoization. CROSSREFS Cf. A001317 (positions of ones). Cf. A000069, A001969, A003188, A003987, A010060, A048720, A048723, A048724, A268384, A268670. Cf. A268389 (the highest exponent for (X+1)). Cf. also A136386. Sequence in context: A163618 A275392 A106616 * A030652 A224065 A077904 Adjacent sequences:  A268666 A268667 A268668 * A268670 A268671 A268672 KEYWORD nonn,base AUTHOR Antti Karttunen, Feb 10 2016 STATUS approved

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Last modified October 22 07:48 EDT 2019. Contains 328315 sequences. (Running on oeis4.)