A260850 -id:A260850 - cp's OEIS frontend

A370974 A260850 sorted into increasing order and duplicates omitted.

1, 2, 6, 20, 24, 120, 140, 858, 924, 1008, 1120, 1430, 10080, 11088, 12012, 22880, 176358, 388960, 1662804, 1750320, 3879876, 4056234, 9694845, 10029150, 10400600, 10816624, 33256080, 270415600, 280816200, 290845350, 300540195, 1037158320, 1452021648, 3181073742, 3267048708, 9617286240, 13784652882, 20583576819, 35263382880, 120880802196

Offset: 1

N. J. A. Sloane, Apr 15 2024

nonn

It seems very likely that there are no duplicates in A260850. Compare the proof of the analogous property of A008336.

Michael De Vlieger, Table of n, a(n) for n = 1..3365
Michael De Vlieger, Plot p(i)^m(i) | a(n) at (x,y) = (n,i), n = 1..2048, with a color function where black indicates m(i) = 1, red indicates m(i) = 2, ..., magenta indicates the largest m(i) for n <= 2048.

Cf. A008336, A065422, A260850, A370968.

A008336 a(n+1) = a(n)/n if n|a(n) else a(n)*n, a(1) = 1.

Original entry on oeis.org

1, 1, 2, 6, 24, 120, 20, 140, 1120, 10080, 1008, 11088, 924, 12012, 858, 12870, 205920, 3500640, 194480, 3695120, 184756, 3879876, 176358, 4056234, 97349616, 2433740400, 93605400, 2527345800, 90262350, 2617608150, 87253605, 2704861755, 86555576160, 2856334013280

Offset: 1

Bernardo Recamán

The graph of log_10(a(n)+1) seems to suggest that log(a(n)) is asymptotic to C*n where C is approximately 0.8. - Daniel Forgues, Sep 18 2011
Comments from N. J. A. Sloane, Apr 14 2024: (Start)
See A370968 for the terms in increasing order with duplicates omitted.
See A337486 and A195504 for the n such that a(n+1) = a(n)/n.
Guy and Nowakowski give bounds on a(n).
Theorem: 1 is the only repeated term.
Proof: Write a(n) for A008336(n).
Suppose, seeking a contradiction, that for 1 < r < s we have a(r) = a(s).
This means that a(r)*r^e_0*(r+1)^e_1*(r+2)^e_2*...(s-1)^e_t = a(s) = a(r),
where the exponents e_* are +1 or -1. The product (P1, say) of the terms with exponent +1 must equal the product (P2, say) of the terms with exponent -1. Since r>1, we need s >= r+2.
The product P1*P2 = P1^2 of all these terms is (s-1)!/(r-1)!.
But this contradicts Erdos's theorem (Erdos 1939) that the product of two or more consecutive integers is never a square.  QED.
(End)

P. Erdos, On the product of consecutive integers, J. London Math. Soc., 14 (1939), 194-198.

Indranil Ghosh, Table of n, a(n) for n = 1..2732 (terms 1..1000 from T. D. Noe)
R. K. Guy and R. Nowakowski, Unsolved Problems, Amer. Math. Monthly, vol. 102 (1995), 921-926; circa page 924.
R. K. Guy and R. Nowakowski, Annotated extract from previous link
Nick Hobson, Python program for this sequence
N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
Index entries for sequences related to Recamán's sequence

Cf. A005132 (the original Recaman sequence).
A065422 and A260850 are variants of the present sequence.
Cf. also A195504 = Product of numbers up to n-1 used as divisors in A008336(n), n >= 2; a(1) = 1.
Cf. also A337486, A370968.

a008336 n = a008336_list !! (n-1)
a008336_list = 1 : zipWith (/*) a008336_list [1..] where
    x /* y = if x `mod` y == 0 then x `div` y else x*y
-- Reinhard Zumkeller, Feb 22 2012, Oct 25 2010

A008336 := proc(n) option remember; if n = 1 then 1 elif A008336(n-1) mod (n-1) = 0 then A008336(n-1)/(n-1) else A008336(n-1)*(n-1); fi; end;

a[n_] := a[n] = If[ Divisible[ a[n-1], n-1], a[n-1]/(n-1), a[n-1]*(n-1)]; a[1] = 1; Table[a[n], {n, 1, 28}] (* Jean-François Alcover, Dec 02 2011 *)
nxt[{n_,a_}]:={n+1,If[Divisible[a,n],a/n,n*a]}; Transpose[ NestList[ nxt,{1,1},30]][[2]] (* Harvey P. Dale, May 09 2016 *)

from functools import lru_cache
@lru_cache(maxsize=None)
def A008336(n):
    if n == 1: return 1
    a, b = divmod(c:=A008336(n-1),n-1)
    return c*(n-1) if b else a # Chai Wah Wu, Apr 11 2024

A371906 a(n) = sum of 2^(k-1) such that floor(n/prime(k)) is odd.

Original entry on oeis.org

0, 1, 3, 2, 6, 5, 13, 12, 14, 11, 27, 24, 56, 49, 55, 54, 118, 117, 245, 240, 250, 235, 491, 488, 492, 461, 463, 454, 966, 961, 1985, 1984, 2002, 1939, 1951, 1948, 3996, 3869, 3903, 3898, 7994, 7985, 16177, 16160, 16166, 15911, 32295, 32292, 32300, 32297, 32363

Offset: 1

Michael De Vlieger, Apr 15 2024

The only powers of 2 in the sequence are likely 1 and 2.

			a(1) = 0 since n = 1 is the empty product.
a(2) = 1 since for n = prime(1) = 2, floor(2/2) = 1 is odd. Therefore a(2) = 2^(1-1) = 1.
a(3) = 3 since for n = 3 and prime(1) = 2, floor(3/2) = 1 is odd, and for prime(2) = 3, floor(3/3) = 1 is odd. Hence a(3) = 2^(1-1) + 2^(2-1) = 1 + 2 = 3.
a(4) = 2 since for n = 4 and prime(1) = 2, floor(4/2) = 2 is even, but for prime(2) = 3, floor(4/3) = 1 is odd. Therefore, a(n) = 2^(2-1) = 2.
a(5) = 6 since for n = 5, though floor(5/2) = 2 is even, floor(5/3) and floor(5/5) are both odd. Therefore, a(n) = 2^(2-1) + 2^(3-1) = 2 + 4 = 6, etc.
Table relating a(n) with b(n), diagramming powers of 2 with "x" that sum to a(n), or prime factors with "x" that produce b(n), where b(n) = A372000(n).
             Power of 2
   n   a(n)  01234567      b(n)
  ----------------------------
   1     0   .               1
   2     1   x               2
   3     3   xx              6
   4     2   .x              3
   5     6   .xx            15
   6     5   x.x            10
   7    13   x.xx           70
   8    12   ..xx           35
   9    14   .xxx          105
  10    11   xx.x           42
  11    27   xx.xx         462
  12    24   ...xx          77
  13    56   ...xxx       1001
  14    49   x...xx        286
  15    55   xxx.xx       4290
  16    54   .xx.xx       2145
  17   118   .xx.xxx     36465
  18   117   x.x.xxx     24310
  19   245   x.x.xxxx   461890
  20   240   ....xxxx    46189
  ----------------------------
                 1111
             23571379
             Prime factor

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Michael De Vlieger, Plot powers 2^(i-1) that sum to a(n) at (x,y) = (n,i) for n = 1..2048.

Cf. A008336, A260850, A372000.

Table[Total[2^(-1 + Select[Range@ PrimePi[n], OddQ@ Quotient[n, Prime[#]] &])], {n, 50}]

a(n) = sum(k=1, primepi(n), if (n\prime(k) % 2, 2^(k-1))); \\ Michel Marcus, Apr 16 2024

A372000 a(n) = product of primes p such that floor(n/p) is odd.

Original entry on oeis.org

1, 2, 6, 3, 15, 10, 70, 35, 105, 42, 462, 77, 1001, 286, 4290, 2145, 36465, 24310, 461890, 46189, 969969, 176358, 4056234, 676039, 3380195, 520030, 1560090, 111435, 3231615, 430882, 13357342, 6678671, 220396143, 25928958, 907513530, 151252255, 5596333435, 589087730, 22974421470, 2297442147

Offset: 1

Michael De Vlieger, Apr 15 2024

The only primes in the sequence are 2 and 3.
We can approach the sequence in a manner akin to A260850, a variant of A008336. Set k = 1. Then for all prime factors p | n, if p | k, divide k by p, otherwise multiply k by p. Then we set a(n) = k. This accounts for the "toggling on or off" of prime factors as n increases.
For n >= 1, A055773(n) | a(n), where A055773(n) = A034386(n) / A034386(floor(n/2)).

			a(1) = 1 since n = 1 is the empty product.
a(2) = 2 since for n = 2, floor(n/p) = floor(2/2) = 1 is odd.
a(3) = 6 since for n = 3 and p = 2, floor(3/2) = 1 is odd, and for p = 3, floor(3/3) = 1 is odd. Hence a(3) = 2*3 = 6.
a(4) = 3 since for n = 4 and p = 2, floor(4/2) = 2 is even, but for p = 3, floor(4/3) = 1 is odd. Therefore, a(n) = 3.
a(5) = 15 since for n = 5, though floor(5/2) = 2 is even, floor(5/3) and floor(5/5) are both odd. Therefore, a(n) = 3*5 = 15, etc.
Table relating a(n) with b(n), diagramming prime factors with "x" that produce a(n), or powers of 2 with "x" that sum to b(n), where b(n) = A371906(n).
                Prime factor
                    1111
   n      b(n)  23571379   b(n)
  ----------------------------
   1        1   .            0
   2        2   x            1
   3        6   xx           3
   4        3   .x           2
   5       15   .xx          6
   6       10   x.x          5
   7       70   x.xx        13
   8       35   ..xx        12
   9      105   .xxx        14
  10       42   xx.x        11
  11      462   xx.xx       27
  12       77   ...xx       24
  13     1001   ...xxx      56
  14      286   x...xx      49
  15     4290   xxx.xx      55
  16     2145   .xx.xx      54
  17    36465   .xx.xxx    118
  18    24310   x.x.xxx    117
  19   461890   x.x.xxxx   245
  20    46189   ....xxxx   240
  ----------------------------
                01234567
                Power of 2

Michael De Vlieger, Table of n, a(n) for n = 1..3384
Michael De Vlieger, Plot prime(i) | a(n) at (x,y) = (n,i) for n = 1..2048.

Cf. A005117, A008336, A034386, A055773, A260850, A371906.

Table[Times @@ Select[Prime@ Range@ PrimePi[n], OddQ@ Quotient[n, #] &], {n, 40}] (* or *)
Table[Product[Prime[i], {j, 1 + Floor[PrimePi[n]/2]}, {i, 1 + PrimePi[Floor[n/(2 j)]], PrimePi[Floor[n/(2 j - 1)]]}], {n, 40}]

a(n) = vecprod(select(x->((n\x) % 2), primes([1, n]))); \\ Michel Marcus, Apr 16 2024

print([prod(p for p in prime_range(n + 1) if is_odd(n//p)) for n in range(1, 41)])
# Peter Luschny, Apr 16 2024

a(n) = Product_{k = 1..floor(pi(n)/2)+1} Product_{j = 1+floor(n/(2*k))..floor(n/(2*k-1))} prime(j), where pi(x) = A000720(n).

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A370974 A260850 sorted into increasing order and duplicates omitted.

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A008336 a(n+1) = a(n)/n if n|a(n) else a(n)*n, a(1) = 1.

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A371906 a(n) = sum of 2^(k-1) such that floor(n/prime(k)) is odd.

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A372000 a(n) = product of primes p such that floor(n/p) is odd.

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