A361505 -id:A361505 - cp's OEIS frontend

A359804 a(1) = 1, a(2) = 2; thereafter let p be the smallest prime that does not divide a(n-2)*a(n-1), then a(n) is the smallest multiple of p that is not yet in the sequence.

1, 2, 3, 5, 4, 6, 10, 7, 9, 8, 15, 14, 11, 12, 20, 21, 22, 25, 18, 28, 30, 33, 35, 16, 24, 40, 42, 44, 45, 49, 26, 27, 50, 56, 36, 55, 63, 32, 60, 70, 66, 13, 65, 34, 39, 75, 38, 77, 48, 80, 84, 88, 85, 51, 46, 90, 91, 99, 52, 95, 54, 98, 100, 57, 105, 58, 110, 69, 112, 115, 72, 119, 120, 121

Offset: 1

David James Sycamore, Mar 08 2023

nonn

Let i = a(n-2), j = a(n-1). For k > 1, m >= 1, a(n) = m*prime(k) iff rad(i*j) = primorial(k-1), and this is the m-th such occurrence. This suggests the late appearance of most primes (namely those >= 7), apparent in the lowest part of scatterplot, where for example a(717126), a(63056215) = 31, 37 respectively.
As Scott R. Shannon has just observed, the following proof is incomplete, since it requires a proof that every even number appears. Even the induction step seems a little dubious. - N. J. A. Sloane, Mar 18 2023
All multiples of all primes appear in the sequence, for if not there is a least prime p such that m*p is not a term for any [some?] m >= 1. Choose any prime q < p; then every multiple of q must appear, so then p*q must be a term; contradiction since this is a multiple of p. [But what if p = 2?]
Corollary: This sequence is a permutation of the positive integers. [This question appears to be still open. - N. J. A. Sloane, Mar 18 2023]
Conjecture: The primes appear in their natural order.

			a(3) must be 3 because a(1,2) = 1,2 and 3 is the least prime which does not divide 2.
a(4) = 5 since this is the least multiple of the smallest prime which does not divide 2*3 = 6.
a(8) = 7 because a(6,7) = 6,10 and 7 is the smallest prime which does not divide 60, rad(60) = 2*3*5 = 30.
a(19,20) = 18,28, and 5 is the smallest prime not dividing rad(18*28) = 42. Since multiples of 5 have appeared 5 times already, a(20) = 6*5 = 30.

Winston de Greef, Table of n, a(n) for n = 1..10000
Michael De Vlieger, Log log scatterplot of a million terms showing primes in red.
Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^16, showing primes in red, composite prime powers in gold, squarefree composites in green, and numbers neither prime power nor squarefree in blue.

Cf. A002110, A007947, A053669.
See also A361502, A361503, A361504, A361505.
A351495 has a very similar definition.

R:= 1,2: S:= {1,2}:
for i from 3 to 100 do
  s:= R[i-2]*R[i-1]:
  p:= 2;
  while s mod p = 0 do p:= nextprime(p) od:
  for r from p by p while member(r,S) do od:
  R:= R,r; S:= S union {r}
od:
R; # Robert Israel, Mar 08 2023

nn = 2^10; c[] = False; q[] = 1;
Array[Set[{a[#], c[#]}, {#, True}] &, 2];
Set[{i, j}, {a[1], a[2]}]; u = 3;
Do[(k = q[#];
      While[c[k #], k++]; k *= #;
    While[c[# q[#]], q[#]++]) &[(p = 2;
    While[Divisible[i j, p], p = NextPrime[p]]; p)];
  Set[{a[n], c[k], i, j}, {k, True, j, k}];
  If[k == u, While[c[u], u++]], {n, 3, nn}];
Array[a, nn] (* Michael De Vlieger, Mar 08 2023 *)

findp(n) = forprime(p=2, , if (n%p, return(p)));
lista(nn) = my(va = vector(nn, k, if (k<=2, k))); for (n=3, nn, my(vsa = vecsort(va), p=findp(va[n-1]*va[n-2]), k=p); while (vecsearch(vsa, k), k+=p); va[n] = k;); va; \\ Michel Marcus, Mar 09 2023

from itertools import count, islice
from sympy import prime, primefactors, primepi
def A359804_gen(): # generator of terms
    aset, bset, cset = set(), {1}, {1,2}
    yield from (1,2)
    while True:
        for i in count(1):
            if not (i in aset or i in bset):
                p = prime(i)
                for j in count(1):
                    if (m:=j*p) not in cset:
                        yield m
                        cset.add(m)
                        break
                break
        aset, bset = bset, set(map(primepi,primefactors(m)))
A359804_list = list(islice(A359804_gen(),30)) # Chai Wah Wu, Mar 18 2023

A363593 Numbers k such that both A359804(k) and A359804(k+1) are odd.

Original entry on oeis.org

3, 8, 22, 29, 36, 42, 45, 53, 57, 64, 82, 85, 88, 94, 110, 119, 124, 132, 135, 141, 144, 152, 159, 165, 170, 177, 183, 190, 195, 201, 214, 220, 224, 231, 239, 246, 252, 264, 270, 281, 287, 292, 299, 302, 306, 309, 323, 328, 334, 341, 347, 350, 356, 361, 372, 378, 381, 386, 397, 402, 411, 418, 424, 431

Offset: 1

Michael De Vlieger and David James Sycamore, Jun 12 2023

nonn

Odd numbers may occur no more than twice in a row in A359804 as consequence of definition of that sequence.
Let b(n) = A359804(n). Let D(n) = b(a(n)..a(n)+1).
Since the product of 2 odd numbers b(n-2) and b(n-1) is odd, and since b(n) = mp, where p = A053669(b(n-2)*b(n-1)) = 2, D(n) implies b(a(n)+2) = 2m.
b(a(n)+2) = 2k and b(a(n+j)+2) = 2m, j >= 1 imply m > k as consequence of definition of A359804.
Perfect powers 2^k = b(j) occur such that j = a(n)+2 for some n. Therefore, A361505 is a subset of { a(n) + 2 }. Generally, perfect powers p^e in A246547 follow b(n-2) and b(n-1) such that b(n-2)*b(n-1) mod p != 0.
Conjecture: for prime q > 11, even squarefree semiprimes 2q follow D(n) for some n. Consider that primes in A359804 appear late for q > 11, yet pairs of successive odd numbers in that sequence occur rather often.
Conjectured to be an infinite sequence, meaning that consecutive odd terms appear infinitely many times in A359804. - David James Sycamore, Jun 21 2023

			Table of a(n) showing i = b(n) = p(i)*m(i), j = b(n+1) = p(j)*m(j), and k = b(n+2), where p(n) = A361503(n) and m(n) = A359804(n)/A361503(n):
   n  a(n)    i     j    k   p(i) p(j) m(i) m(j)
  ----------------------------------------------
   1    3     3     5    4     3    5    1    1
   2    8     7     9    8     7    3    1    3
   3   22    33    35   16    11    7    3    5
   4   29    45    49   26     5    7    9    7
   5   36    55    63   32     5    7   11    9
   6   42    13    65   34    13    5    1   13
   7   45    39    75   38     3    5   13   15
   8   53    85    51   46     5    3   17   17
   9   57    91    99   52     7   11   13    9
  10   64    57   105   58     3    7   19   15
  11   82   143    81   62    11    3   13   27
  12   85   135   147   64     5    7   27   21
  ...

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Michael De Vlieger, Plot A359804(n) mod 2 at (x,y) = (n mod 256, -floor(n/256)), 8X magnification, where white represents even terms, and odd terms are shown in color. Singleton odd numbers are shown in dark blue, while red indicates two odd terms in a row. Shows A359804(n) mod 2 for n = 1..2^16.
Michael De Vlieger, Plot A359804(n) mod 2 at (x,y) = (n mod 2^10, -floor(n/2^10)), where white represents even terms, and odd terms are shown in color. Singleton odd numbers are shown in dark blue, while red indicates two odd terms in a row. Shows A359804(n) mod 2 for n = 1..2^20.

Cf. A000079, A053669, A100484, A359804, A361503, A361505, A363594.

nn = 432; c[] = False; q[] = 1;
Set[{i, j}, {1, 2}]; c[1] = c[2] = True; q[2] = 2; u = 3;
Reap[Do[
    (k = q[#]; While[c[k #], k++]; k *= #;
       While[c[# q[#]], q[#]++]) &[(p = 2;
      While[Divisible[i j, p], p = NextPrime[p]]; p)];
    If[OddQ[j k], Sow[n - 1]];
    Set[{c[k], i, j}, {True, j, k}];
    If[k == u, While[c[u], u++]], {n, 3, nn}] ][[-1, -1]]

A361503(a(n)+1) = 2, consequence of definition of A359804.

cp's OEIS Frontend

A359804 a(1) = 1, a(2) = 2; thereafter let p be the smallest prime that does not divide a(n-2)*a(n-1), then a(n) is the smallest multiple of p that is not yet in the sequence.

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