A236847 -id:A236847 - cp's OEIS frontend

A234742 Product of the binary encodings of the irreducible factors (with multiplicity) of the polynomial over GF(2) whose encoding is n.

0, 1, 2, 3, 4, 9, 6, 7, 8, 21, 18, 11, 12, 13, 14, 27, 16, 81, 42, 19, 36, 49, 22, 39, 24, 25, 26, 63, 28, 33, 54, 31, 32, 93, 162, 91, 84, 37, 38, 99, 72, 41, 98, 75, 44, 189, 78, 47, 48, 77, 50, 243, 52, 57, 126, 55, 56, 117, 66, 59, 108, 61, 62, 147, 64, 441

Offset: 0

Antti Karttunen, Jan 22 2014

"Product" refers to the ordinary multiplication of integers.
Differs from A235042 and A236837 for the first time at n=25, where a(n)=25, while A235042(25)=5 and A236837(25)=0. Thus A234741(A234742(n)) = n up to n=24.
a(n) >= n. [All terms of the table A061858 are nonnegative as the product of multiplying two numbers with carries is never less than when multiplying them without carries.]
Specifically, for all n, a(A091209(n)) > A091209(n).
a(A091209(n)) is always composite and, by the above inequality, larger than A091209(n), which implies that none of the terms of A091209 occur in this sequence. Cf. also A236844.
Starting with various terms (primes) in A235033 and iterating the map A234742, we get 5 -> 9 -> 21 -> 49 -> 77 -> 177 -> 333 = a(333).
Another example: 17 -> 81 -> 169 -> 309 -> 721 = a(721).
Does every chain of such iterations eventually reach a fixed point? (One of the terms of A235035.) Or do some of them manage to avoid such "traps" indefinitely? (Note how the terms of A235035 seem to get rarer, but only rather slowly.)
Starting from 23, we get the sequence: 23, 39, 99, 279, 775, 1271, 3003, 26411, 45059, ... which reaches its fixed point, 3643749709604450870616156947649219, after 55 iterations. - M. F. Hasler, Feb 18 2014. [This is now sequence A244323. See also A260729, A260735 and A260441.] - Antti Karttunen, Aug 05 2015
Note also that when coming backwards from some term of such a chain by iterating A234741, we may not necessarily end at the same term we started from.

			3 has binary representation '11', which encodes the polynomial X + 1, which is irreducible in GF(2)[X], so the result is just a(3)=3.
5 has binary representation '101' which encodes the polynomial X^2 + 1, which is reducible in the polynomial ring GF(2)[X], factoring as (X+1)(X+1), i.e., 5 = A048720(3,3), as 3 ('11' in binary) encodes the polynomial (X+1), irreducible in GF(2)[X]. 3*3 = 9, thus a(5)=9.
9 has binary representation '1001', which encodes the polynomial X^3 + 1, which factors (in GF(2)[X]!) as (X+1)(X^2+X+1), i.e., 9 = A048720(3,7) (7, '111' in binary, encodes the other factor polynomial X^2+X+1). 3*7 = 21, thus a(9)=21.
25 has binary representation '11001', which encodes the polynomial X^4 + X^3 + 1, which is irreducible in GF(2)[X], so the result is just a(25)=25.

Antti Karttunen, Table of n, a(n) for n = 0..8192
Index entries for sequences related to binary expansion of n

A235035 gives the k for which a(k)=k.
A236853(n) gives the number of times n occurs in this sequence.
A236842 gives the same sequence sorted and with duplicates removed, A236844 gives the numbers that do not occur here, A236845 gives numbers that occur more than once, A236846 the least inverse and A236847 the greatest inverse. A236850 gives such k that a(k) = A236837(k).
Cf. A014580, A048720, A234741, A236379, A236837, A236851 & A236852, A115857, A115872.
Cf. also A260712, A260713, A260716 and A244323, A260729, A260735, A260441 (iterations starting from various terms of A236844).

A234742(n)=factorback(subst(lift(factor(Mod(1,2)*Pol(binary(n)))),x,2)) \\ M. F. Hasler, Feb 18 2014, corrected Andrew Howroyd, Aug 01 2018

To compute a(n): factor the polynomial over GF(2) encoded by n, into its irreducible factors; in other words, find a unique multiset of terms i, j, ..., k (not necessarily distinct) from A014580 for which i x j x ... x k = n, where x stands for the carryless multiplication A048720. Then a(n) = i*j*...*k is the product of those terms with ordinary multiplication. Because of the effect of the carry-bits in the latter, the result is always greater than or equal to n, so we have a(n) >= n for all n.
a(2n) = 2*a(n).
a(A235035(n)) = A235035(n).
A236379(n) = a(n) - n.
For all n, a(n) >= A236837(n).

A236844 Numbers that do not occur as results of "upward" remultiplication (GF(2)[X] -> N) of any number; numbers not present in A234742.

Original entry on oeis.org

5, 10, 15, 17, 20, 23, 29, 30, 34, 35, 40, 43, 45, 46, 51, 53, 58, 60, 65, 68, 69, 70, 71, 79, 80, 83, 85, 86, 89, 90, 92, 95, 101, 102, 105, 106, 107, 113, 116, 119, 120, 125, 127, 129, 130, 135, 136, 138, 139, 140, 142, 149, 151, 153, 155, 158, 159, 160, 161

Offset: 1

Antti Karttunen, Jan 31 2014

nonn

Numbers that do not occur in A234742 (A236842).
This is a subsequence of A236848, thus all terms are divisible by at least one such prime which is reducible as polynomial over GF(2) (i.e. one of the primes in A091209).
A236835(7)=27 is the first member of A236835 which does not occur here. a(12)=43 is the first term here which does not occur in A236835.

Antti Karttunen, Table of n, a(n) for n = 1..19282

Complement: A236842.
A setwise difference of A236848 and A236849.
A091209 is a subsequence.
Positions of zeros in A236853, A236846, A236847 and A236862.
Cf. A236845.
Cf. also A236834.

For all n, A236379(a(n)) > 0.

A236846 Least inverse of A234742: a(n) = minimal k such that when it is remultiplied "upwards", from GF(2)[X] to N, the result is n, and 0 if no such k exists.

Original entry on oeis.org

0, 1, 2, 3, 4, 0, 6, 7, 8, 5, 0, 11, 12, 13, 14, 0, 16, 0, 10, 19, 0, 9, 22, 0, 24, 25, 26, 15, 28, 0, 0, 31, 32, 29, 0, 0, 20, 37, 38, 23, 0, 41, 18, 0, 44, 0, 0, 47, 48, 21, 50, 0, 52, 0, 30, 55, 56, 53, 0, 59, 0, 61, 62, 27, 64, 0, 58, 67, 0, 0, 0, 0, 40, 73, 74, 43, 76, 49, 46, 0, 0, 17, 82, 0, 36, 0, 0, 87, 88, 0, 0, 35

Offset: 0

Antti Karttunen, Jan 31 2014

nonn

Apart from zero, each term occurs at most once. 91 is the smallest positive integer not present in this sequence.

Note that in contrast to the reciprocal case, where A234742(n) >= A236837(n) for all n [the former sequence gives the absolute upper bound for the latter], here it is not guaranteed that A234741(n) <= a(n) whenever a(n) > 0. For example, a(25)=25 and A234741(25)=17, and 25-17 = 8. On the other hand, a(75)=43, but A234741(75)=51, and 43-51 = -8.

Antti Karttunen, Table of n, a(n) for n = 0..8192

Differs from A236847 for the first time at n=91, where a(91)=35, while A236847(91)=91.
A236844 gives the positions of zeros.
Cf. A234742.
Cf. also A236836, A236837.

(define (A236846 n) (let loop ((k n) (minv 0)) (cond ((zero? k) minv) ((= (A234742 k) n) (loop (- k 1) k)) (else (loop (- k 1) minv)))))

a(n) = minimal k such that A234742(k) = n, and 0 if no such k exists.

For all n, a(n) <= n.

A236851 Remultiply n first "upward", from GF(2)[X] to N, and then remultiply that result back "downward", from N to GF(2)[X]: a(n) = A234741(A234742(n)).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 17, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 51, 44, 45, 46, 47, 48, 49, 34, 51, 52, 53, 54, 39, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67

Offset: 0

Antti Karttunen, Feb 02 2014

nonn

			5 ('101' in binary) = 3 x 3 (3 = '11' in binary). 3 is in A091206, so it stays intact, and 3 x 3 = 5, thus a(5)=5.
25 ('11001' in binary) = 25 (25 is irreducible in GF(2)[X]). However, it divides as 5*5 in Z, so the result is 5 x 5 = 17, thus a(25)=17, 25 being the least n which is not fixed by this function.
43 ('101011' in binary) = 3 x 25, of which the latter factor divides to 5*5, thus the result is 3 x 5 x 5 = 3 x 17 = 15 x 5 = 51.

Antti Karttunen, Table of n, a(n) for n = 0..8192
Index entries for sequences operating on polynomials over GF(2)

A236850 gives the fixed points.

Cf. A234741, A234742, A236380, A236852, A236836-A236837, A236846-A236847.

(define (A236851 n) (A234741 (A234742 n)))

a(n) = A234741(A234742(n)).
To compute a(n): factor n as a polynomial over GF(2) (where n is mapped to such polynomials via the binary representation of n), that is, find first a unique multiset of terms i, j, ..., k from A014580 for which i x j x ... x k = n, where x stands for the carryless multiplication (A048720). Then divide from those i, j, ..., k the ones that are in A091214 (composite integers in N) to their constituent prime factors (in N), and multiply all back together (including the factors that are in A091206 and thus not changed) with the carryless multiplication (A048720).
Compare this to how primes are "broken" in a similar way in A235027 (cf. also A235145).

A236852 Remultiply n first "downward", from N to GF(2)[X], and then remultiply that result back "upward", from GF(2)[X] to N: a(n) = A234742(A234741(n)).

Original entry on oeis.org

0, 1, 2, 3, 4, 9, 6, 7, 8, 9, 18, 11, 12, 13, 14, 27, 16, 81, 18, 19, 36, 21, 22, 39, 24, 81, 26, 27, 28, 33, 54, 31, 32, 33, 162, 63, 36, 37, 38, 39, 72, 41, 42, 75, 44, 81, 78, 47, 48, 49, 162, 243, 52, 57, 54, 99, 56, 57, 66, 59, 108, 61, 62, 63, 64, 117, 66, 67, 324

Offset: 0

Antti Karttunen, Feb 02 2014

This sequence appears to be completely multiplicative with a(p) = A234742(p) although neither A234741 or A234742 are even multiplicative. Terms tested up to n = 10^7. - Andrew Howroyd, Aug 01 2018

Yes, this is true. Consider for example n = p*q*r*r, where p, q, r are primes in N. Then A234741(n) = p1 x q1 x q2 x q3 x r1 x r2 x r1 x r2, where p1, q1, r1, ..., are the irreducible factors of p, q, r when factored over GF(2), and x stands for multiplication in ring GF(2)[X] (A048720). [Note that these irreducible factors are not necessarily primes in N, except p1 (= p), which must be a term of A091206. Also, A234741(p) = p for any prime p.] Next, a(n) = A234742(p1 x q1 x q2 x q3 x r1 x r2 x r1 x r2) = p1 * q1 * q2 * q3 * r1 * r2 * r1 * r2, which can be obtained also as a(p)*a(q)*a(r)*a(r), thus proving the multiplicativity. - Antti Karttunen, Aug 02 2018

			From _Antti Karttunen_, Aug 02 2018: (Start)
For n = 3, we have A234741(3) = 3 = 11 in binary, which encodes a (0,1)-polynomial x+1, which is irreducible over GF(2) thus A234742(3) = 3 and  a(3) = 3.
For n = 5, we have A234741(5) = 5 = 101 in binary, which encodes a (0,1)-polynomial x^2 + 1, which factorizes as (x+1)(x+1) when factored over GF(2), that is 5 = A048720(3,3), thus it follows that A234742(5) = 3*3 = 9, and a(5) = 9.
For n = 9 = 3*3, we have A234741(9) = A048720(3,3) = 5, and A234742(5) = 9 as shown above. Also by multiplicativity, we have a(3*3) = a(3)*a(3) = 3*3 = 9.
(End)

Antti Karttunen, Table of n, a(n) for n = 0..8192
Index entries for sequences related to polynomials in ring GF(2)[X]

Fixed points: A236860.

Cf. A048720, A234741, A234742, A236851, A236836-A236837, A236846-A236847.

(define (A236852 n) (A234742 (A234741 n)))

a(n) = A234742(A234741(n)).

Keyword mult added after Andrew Howroyd's observation. - Antti Karttunen, Aug 02 2018

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A234742 Product of the binary encodings of the irreducible factors (with multiplicity) of the polynomial over GF(2) whose encoding is n.

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A236844 Numbers that do not occur as results of "upward" remultiplication (GF(2)[X] -> N) of any number; numbers not present in A234742.

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A236846 Least inverse of A234742: a(n) = minimal k such that when it is remultiplied "upwards", from GF(2)[X] to N, the result is n, and 0 if no such k exists.

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A236851 Remultiply n first "upward", from GF(2)[X] to N, and then remultiply that result back "downward", from N to GF(2)[X]: a(n) = A234741(A234742(n)).

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A236852 Remultiply n first "downward", from N to GF(2)[X], and then remultiply that result back "upward", from GF(2)[X] to N: a(n) = A234742(A234741(n)).

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