cp's OEIS Frontend

Equally, a(n) = minimum number of steps needed to repeat k = A235027(k)(starting from k = n) until A001222(A235027(k)) = A001222(k).

Or in other words, how many times are needed to repeatedly factorize the number, to reverse the bits of each odd prime factor (with A056539) and factorize and bit-reverse the reversed factors again, until the number of prime divisors no more grows, meaning that we have found either a fixed point or entered a cycle of two.

Sequence consists of all the primes in A204219 (after 2), together with all of their multiples.

Note that this is not the same as the numbers that do not occur in A235027 ("Garden of Eden" numbers for A235027), a subsequence of this sequence, which begins as: 19, 38, 57, 59, 76, 79, 89, 95, 103, 109, 114, 118, 133, 137, 139, 149, 152, 157, 158, 171, 177, 178, 179, 190, 191, 206, 211, 218, 228, 236, 237, 239, 241, 266, 267, 271, 274, 278, 281, 285, 293, 298, 304, 309, 311, 314, 316, 317, 327, 342, 347, 354, 356, 358, ...

The first term that occurs in this sequence, but not in the "GoE"-sequence is a(27)=209, as a(139) = 209 = 11*19 and 139 = A235146(2), the least integer which requires two steps to reach a fixed point or 2-cycle.

Both the above "GoE"-sequence, and its differences from this will be submitted later.

Note, as for all composite values A235145(u * v) = max(A235145(u), A235145(v)) which can be further reduced as A235145(n) = Max_{p|n} A235145(p), and because for any odd prime p, lpf(A056539(p)) >= 3 (where lpf = A020639, the least prime dividing n) while 1/2 < A056539(n)/n < 2, it follows that this sequence gives also the positions of the records in A235145, as its new values must appear in order.

Also, because of that multiplicativity criterion, all terms (after zero) must be primes, and specifically, the terms are a subset of A235030 (i.e., of A204219).

Conjecture: additional property is that the primes here belong to that subset of p in A204219 for which A056539(p) > p. The list of such primes begins as: 19, 79, 103, 137, 139, 149, 157, 179, 191, 239, 271, 281, 293, 311, 317, 347, 367, 379, 439, 523, 541, 547, 557, 563, 569, 587, 607, 613, 647, 659, 719, 733, 743, 751, 787, ...

The first 20 terms are equal with A057890, after which a(21)=25, while A057890(21)=27. On the other hand, 33 is the first term which occurs in A057890 but does not occur here.

If terms x and y are included, then also their product x*y is included. If term x is included, then 2^k * x is also included. The sequence contains also all primes in A016041 and their mutual multiples. However, in addition to that, there are also terms like 143 = 11*13, where A235027 will map the factors to each other (as their binary expansions '1011' and '1101' are mirror images of each other), even although neither of them is present in A016041. (These latter kind of primes are in A074832).

Please use the "graph" link to see how the terms get rarer.

Because A056539(n)/n < 2 for all n, and already for the tenth term of this sequence 1349 we have A235027(1349)/1349 = 2.094... it follows that the only primes present are terms a(2) .. a(7): 11, 19, 67, 71, 263, 271. Conjecture: every term after that is a product of some of those six primes. For example: 781 = 11*71, 1273 = 19*67, 1349 = 19*71, 2981 = 11*271, 4757 = 67*71, 5041 = 71*71.

The original name was "Bit-reverse of n, including as many leading as trailing zeros." - Antti Karttunen, Dec 25 2024

A permutation of integers consisting only of fixed points and pairs. a(n)=n when n is a binary palindrome (including as many leading as trailing zeros), otherwise a(n)=A003010(n) (i.e. n has no axis of symmetry). A057890 gives the palindromes (fixed points, akin to A006995) while A057891 gives the "antidromes" (pairs). See also A280505.

This is multiplicative in domain GF(2)[X], i.e. with carryless binary arithmetic. A193231 is another such permutation of natural numbers. - Antti Karttunen, Dec 25 2024

"Encoded in binary representation" means that a polynomial a(n)*X^n+...+a(0)*X^0 over GF(2) is represented by the binary number a(n)*2^n+...+a(0)*2^0 in Z (where each coefficient a(k) = 0 or 1).

To determine whether n belongs to this sequence: first find a unique multiset of terms i, j, ..., k (terms not necessarily distinct) from A014580 for which i x j x ... x k = n, where x stands for the carryless multiplication (A048720). If and only if NONE of those i, j, ..., k is a composite (in other words, if all are primes in N, i.e. terms of A091206), then n is a member.

Equally, numbers which can be constructed as p x q x ... x r, where p, q, ..., r are terms of A091206. (Compare to the definition of A236860.)

Also fixed points of A236851(n). Proof: if k is a term of this sequence, the operation described in A236851 reduces to an identity operation. On the other hand, if k is not a term of this sequence, then it contains at least one irreducible GF(2)[X]-factor which is a composite in N, which is thus "broken" by A236851 to two or more separate GF(2)[X]-factors (either irreducible or not), and because the original factor was irreducible, and GF(2)[X] is a unique factorization domain, the new product computed over the new set of factors (with one or more "broken" pieces) cannot be equal to the original k. (Compare this to how primes are "broken" in a similar way in A235027, also A235145.)

Also by similar to above reasoning, positions where A234742(n) = A236837(n).

This is a subsequence of A236841, from which this differs for the first time at n=43, where A236841(43)=43, while from here 43 is missing, and a(43)=44.

After 0 and 1, positive integers which are products of p * q * ... * r, where p, q, ..., r are terms of A091206.

Also fixed points of A236852(n). Proof: if k is a term of this sequence, the operation described in A236852 reduces to an identity operation. On the other hand, if k is not a term of this sequence, then it has at least one prime divisor which is reducible in polynomial ring GF(2)[X], which is thus "broken" by A236852 (A234742) to two or more separate factors (either prime or not), and because the original factor was prime, and N is a unique factorization domain, the new product computed over the new set of factors (with one or more "broken" pieces) cannot be equal to the original k. (Compare this to how primes are "broken" in a similar way in A235027, also A235145.)

Note: This sequence is not equal to all n for which A234741(n) = A236846(n). The first counterexample occurs at a(325) = 741 (= 3*13*19) for which we have: A236846(741) = 281 (= 3 x 247 = 3 x (13*19)) while A234741(741) = 329 (= 3 x 13 x 19). Contrast this with the behavior of the "dual sequence" A236850, where the corresponding property holds.