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A334205
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Under the isomorphism defined in A329329, of polynomials in GF(2)[x,y] to positive integers, a(n) is the image of the polynomial that results when x+1 is substituted for x in the polynomial with image n.
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3
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1, 2, 6, 4, 10, 3, 210, 8, 36, 5, 22, 24, 858, 105, 15, 16, 1870, 72, 9699690, 40, 35, 11, 46, 12, 100, 429, 216, 840, 4002, 30, 7130, 32, 33, 935, 21, 9, 160660290, 4849845, 143, 20, 20746, 70, 1008940218, 88, 360, 23, 2569288370, 96, 44100, 200, 2805, 3432, 32589158477190044730, 108, 55, 420, 1616615, 2001, 118, 60, 21594, 3565
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OFFSET
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1,2
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COMMENTS
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Under the isomorphism (defined in A329329), A059897(.,.), A329329(.,.) and A003961(.) represent polynomial addition, multiplication and multiplication by x respectively; prime(i+1) represents the polynomial x^i.
The equivalent sequence with y+1 substituted for y is A268385.
Self-inverse permutation of natural numbers. Squarefree numbers are mapped to squarefree numbers, squares are mapped to squares, and in general the sequence permutes {m : A267116(m) = k} for any k.
The odd numbers represent the polynomials that have x as a factor. So the odd bisection's terms represent polynomials with (x+1) as a factor. They are a permutation of A268390.
A193231 is an equivalent sequence with respect to GF(2)[x]. See the formula showing A019565 as the related injective homomorphism, mapping the usual encoding of GF(2) polynomials in x to their equivalent A329329-defined representation.
(End)
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LINKS
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FORMULA
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a(n^2) = a(n)^2.
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EXAMPLE
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Calculation for n = 5. 5 = prime(3) = prime(2+1) is the image of the polynomial x^2. Substituting x+1 for x, this becomes (x+1)^2 = x^2 + (1+1)x + 1 = x^2 + 1, as 1 + 1 = 0 in GF(2). The image of x^2 + 1 is A059897(prime(3), prime(1)) = A059897(5, 2) = 10. So a(5) = 10. (Note that A059897 gives the same result as multiplication when its operands are different terms of A050376, such as prime numbers.)
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PROG
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(PARI)
A048675(n) = { my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; };
A225546(n) = if(1==n, 1, my(f=factor(n), u=#binary(vecmax(f[, 2])), prods=vector(u, x, 1), m=1, e); for(i=1, u, for(k=1, #f~, if(bitand(f[k, 2], m), prods[i] *= f[k, 1])); m<<=1); prod(i=1, u, prime(i)^A048675(prods[i])));
A193231(n) = { my(x='x); subst(lift(Mod(1, 2)*subst(Pol(binary(n), x), x, 1+x)), x, 2) }; \\ From A193231
A268385(n) = if(1==n, n, my(f=factor(n)); prod(i=1, #f~, f[i, 1]^A193231(f[i, 2])));
(PARI)
\\ This program is better for larger values. A048675 and A193231 as in above:
A019565(n) = {my(j, v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ From A019565
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CROSSREFS
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Cf. A268390 (ordered odd bisection).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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