Freimut Marschner - cp's OEIS frontend

A250623 a(n) = floor(nlog(prime(n))) + ceiling(nlog(n)) - 2*prime(n).

-4, -2, -2, -1, -2, 0, -1, 2, 2, -1, 2, -1, 0, 3, 4, 2, 0, 4, 1, 3, 8, 7, 8, 6, 1, 2, 8, 10, 16, 18, 3, 5, 4, 9, 2, 8, 7, 6, 8, 8, 7, 13, 5, 12, 15, 22, 10, -1, 2, 9, 13, 12, 19, 12, 12, 12, 11, 18, 18, 22, 29, 22, 8, 12, 19, 23, 8, 8, 2, 9, 13, 13, 11, 11, 11

Offset: 1

Freimut Marschner, Dec 02 2014

It is known that n*log(n) < prime(n) < n*prime(n), n >= 4. The arithmetic mean of the limits of this inequality is f(n) = (floor((n*log(n)) + ceiling(n*prime(n))))/2. So a(n) is the difference between twice this quantity and 2*prime(n).

			a(4) = floor(4*log(7)) + ceiling(4*log(4)) - 2*7 = floor(7.78...) + ceiling(5.54...) - 14 = 7 + 6 - 14 = -1;
a(6) = floor(6*log(13)) + ceiling(6*log(6)) - 2*13 = floor(15.38...) + ceiling(10.75..) - 26 = 15 + 11 - 26 = 0.

Freimut Marschner, Table of n, a(n) for n = 1..10000

a250623[n_] :=
Floor[#*Log[Prime[#]]] + Ceiling[#*Log[#]] - 2*Prime[#] & /@ Range[n]; a250623[137] (* Michael De Vlieger, Dec 26 2014 *)

vector(100,n,floor(n*log(prime(n)))+ceil(n*log(n))-2*prime(n)) \\ Derek Orr, Dec 30 2014

a(n) = A250621(n) + A050502(n) - 2*A000040(n).

A251482 a(n) = floor(prime(n)/log(n)) + ceiling(prime(n)/log(prime(n))) - 2*n, n >=2.

Original entry on oeis.org

3, 2, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, -1, 0, -1, 0, 2, 0, 0, -1, -2, -3, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, -1, 1, 0, -1, -3, 0, 3, 2, 1, 0, 0, -2, 0, 1, 1, 1, -1, -1, -2, -3, -2, 2, 1, -1, -1, 2, 1, 3, 2, 1, 1, 2, 1, 1, 1, 1, 2, 0, 2, 3, 1, 3, 1, 1, 0, 0, 1, 0, -2, -3, -1, 0, 0, 0, -1, -1, 1, -1, 3

Offset: 2

Freimut Marschner, Dec 07 2014

The prime number theorem implies prime(n)/log(prime(n)) < n < prime(n)/log(n), n >= 2. From this follows a(n).

			a(4) = floor(5.04...) + ceiling(3.59...) - 2*4 = 5 + 4 - 2*4 = 1.

Michael De Vlieger, Table of n, a(n) for n = 2..10000

Cf. A086861 (floor(prime(n)/log(prime(n)))), A085581 (floor(prime(n)/log(n))).
Cf. A087724 (-PrimePi(n) + floor(prime(n)/log(n)) - 2), A000720 (pi(n)).
Cf. A060715 (Number of primes between n and 2n exclusive).

[Floor(NthPrime(n)/Log(n)) + Ceiling(NthPrime(n)/Log(NthPrime(n))) - 2*n: n in [2..100]]; // Vincenzo Librandi, Mar 25 2015

a251482[n_Integer] :=
Floor[Prime[#]/Log[#]] + Ceiling[Prime[#]/Log[Prime[#]]] - 2 # & /@
Range[2, n]; a251482[100] (* Michael De Vlieger, Dec 15 2014 *)

vector(100,n,floor(prime(n+1)/log(n+1))+ceil(prime(n+1)/log(prime(n+1)))-2*n-2) \\ Derek Orr, Dec 30 2014

a(n) = A085581(n) + (A086861(n) + 1) - 2*n.

A250622 a(n) = floor(n*log(prime(n)))-prime(n), n >= 1.

Original entry on oeis.org

-2, -1, -1, 0, 0, 2, 2, 4, 5, 4, 6, 6, 7, 9, 10, 10, 10, 12, 12, 14, 17, 17, 18, 18, 17, 18, 22, 23, 27, 28, 23, 25, 25, 28, 26, 29, 30, 30, 32, 33, 33, 37, 34, 38, 40, 44, 40, 36, 38, 42, 45, 45, 49, 47, 48, 49, 49, 53, 54, 57, 61, 59, 53, 56, 60, 63, 57, 58, 56, 60, 63, 64, 64, 65, 66, 69, 70, 69

Offset: 1

Freimut Marschner, Dec 02 2014

Since n*log(prime(n)) > prime(n), n >= 4 and ceiling(prime(n) - n*log(n)) < prime(n), then n*log(n) < prime(n) < n*log(prime(n)), n >= 4. This inequality is included in the prime number theorem PNT. Remark: a(n) >= 0 for n >=4 otherwise a(n) < 0.

			n = 1, a(1) = floor(1*0.6931...) - 2 = 0 - 2 = -2;
n = 5, a(5) = floor(5*2.3978...) - 11 = floor( 11.9894...) - 11 = 11 - 11 = 0;
n = 6, a(6) = floor(6*2.5649...) - 13 = floor(15.3896...) - 13 = 15 - 13 = 2.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A000040, A064658 (ceiling(prime(n) - n*log(n))), A250621 (floor(n*log(prime(n)))).

a250622[n_Integer] := Table[Floor[i*Log[Prime[i]]] - Prime[i], {i, n}]; a250622[121] (* Michael De Vlieger, Dec 11 2014 *)

vector(100,n,floor(n*log(prime(n))-prime(n))) \\ Derek Orr, Dec 13 2014

a(n) = floor(n*log(prime(n))) - prime(n) = A250621(n) - A000040(n).

A250621 a(n) = floor(n*log(prime(n))).

Original entry on oeis.org

0, 2, 4, 7, 11, 15, 19, 23, 28, 33, 37, 43, 48, 52, 57, 63, 69, 73, 79, 85, 90, 96, 101, 107, 114, 119, 125, 130, 136, 141, 150, 156, 162, 167, 175, 180, 187, 193, 199, 206, 212, 218, 225, 231, 237, 243, 251, 259, 265, 271, 278, 284, 290, 298, 305, 312, 318, 324, 331

Offset: 1

Freimut Marschner, Nov 26 2014

From n < prime(n), n >= 1 follows that n*log(n) < prime(n) < n*log(prime(n)), n >= 4. This inequality is included in the prime number theorem PNT.

			For n = 1, prime(1) = 2, floor(1*0.69... = 0.69...) = 0 ;
For n = 25, prime(25) = 97, floor(25*4.57... = 114.36...) = 114.

Cf. A050504 (floor(n*log(n))), A086861 (floor(prime(n)/log(prime(n)))), A085581 (floor(prime(n)/log(n))), A050504 (integer part of n*log(n)), A050503 (nearest integer to n*log(n)), A050502 (ceiling of n*log(n)).

Table[Floor[n Log[Prime[n]]],{n,60}] (* Harvey P. Dale, Aug 13 2019 *)

vector(100,n,floor(n*log(prime(n)))) \\ Derek Orr, Nov 28 2014

A247583 Primes extracted from a pseudo-Collatz cycle '3*n-1' by consecutive arithmetic derivatives, here with starting point prime(99147) = 1287511.

Original entry on oeis.org

1287511, 1448449, 2172673, 37122139, 44596859, 91644073, 28996757, 3440533, 3870599, 4354423, 3265817, 7348087, 8266597, 9299921, 20924821, 31387231, 17655317, 19862231, 22345009, 33517513, 50276269, 75414403, 21499669, 34438309, 55163509, 9817919

Offset: 1

Freimut Marschner, Sep 21 2014

sign

a(n) is defined as a sequence of subsequences of prime numbers extracted from the pseudo-Collatz cycle '3*n-1' , C = c(z) by consecutive arithmetic derivatives AD(i) of C. The starting point here is c(1) = prime(99147) = 1287511, the length is z = 560. The arithmetic derivative AD(i), i >=0 is a tool to select prime numbers out of a given sequence of integers, because the AD of prime numbers is 1.
Let AD(i,C(k)) be the i-th AD of the AD of C(k), then AD(1,C(k)) is the first AD of C(k) with AD(0,C(k)) = C(k). So a(n) = AD(i,C(k)) is a sequence of consecutive values of AD(i) of C(k).
The selection of the prime numbers can be made under the conditions:
(1) If AD(i+1,C(k)) = 1 then AD(i,C(k)) is prime.
(2) If AD(i,C(k)) mod 2 = 1 and AD(i,C(k)) > AD(i+1,C(k)) then AD(i,C(k)) is uneven and is (probably) convergent to a prime number.
(3) If AD(i,C(k)) mod 2 = 0 and AD(i,C(k)) < AD(i+1,C(k)) then AD(i) is even and (probably) divergent.
If any of the conditions 1 - 3 are not satisfied then the search for primes by AD in that sequence is hopeless.
In Tables 1 and 3, i is the number of the AD, np the counting number of primes of the AD and a(n) the last prime number of the i'th AD.
Table 1
i      0    1    2      3      4     5       6       7        8             9        10   ...
np     65   33   27     19     10    10      1       3        4             2        0    ...
n      65   98   125    144    154   164     165     168      172           174
a(n)   17   19   103    71     5     7       101     271      967721        5

			Example for starting point prime(7) = 17. This pseudo-Collatz cycle is repetitive (see A246007).
Table 2
Number         1    2  3   4   5     6   7     8   9  10  11 12   13 14   15   16 17  18  19
Sequence      17   50 25  74  37   110  55   164  82  41 122 61  182 91  272  136 68  34  17
Primes( AD)   17   37 41  61  17    43 131    19   7
Table 3
i        0    1  2  3 ...
np       5    3  1  0 ...
n        5    8  9
a(n)    17   19  7

Freimut Marschner, Table of n, a(n) for n = 1..174

Cf. A246007 (length of pseudo-Collatz cycles '3*n - 1' of prime numbers).

A246010 a(n) = floor(5*prime(n)^2 / 4).

Original entry on oeis.org

5, 11, 31, 61, 151, 211, 361, 451, 661, 1051, 1201, 1711, 2101, 2311, 2761, 3511, 4351, 4651, 5611, 6301, 6661, 7801, 8611, 9901, 11761, 12751, 13261, 14311, 14851, 15961, 20161, 21451, 23461, 24151, 27751, 28501, 30811, 33211

Offset: 1

Freimut Marschner, Sep 28 2014

Let f(x) = -x^2 + b*x + b^2 be a polynomial function with b = prime(n), n >= 1, then the vertex of the graph of f(x) is at the point (vx;f(vx)) = (b/2;5*b^2/4) with f’(vx) = -2*vx + b = 0. If b = n, n >= 0, then the sequence of the vertex of this polynomial is A032527, the concentric pentagonal numbers: floor( 5*n^2 / 4). So a(n) = floor( 5*prime(n)^2 / 4), n >= 1 is a subsequence of A032527.

			a(4) = floor(5*7^2 / 4) = floor(61.25) = 61.

Freimut Marschner, Table of n, a(n) for n = 1..1044

Cf. A032527 (the concentric pentagonal numbers: floor( 5*n^2 / 4)).

[Floor(5*NthPrime(n)^2 / 4): n in [1..40]]; // Vincenzo Librandi, Oct 21 2014

Floor[(5*Prime[Range[40]]^2)/4] (* Harvey P. Dale, Sep 15 2019 *)

vector(100,n,floor(5*prime(n)^2/4)) \\ Derek Orr, Sep 30 2014

a(n) = A032527(A000040(n)). - Michel Marcus, Sep 30 2014

A246009 Length of Collatz cycles '3*n + 1' of prime numbers.

Original entry on oeis.org

2, 8, 6, 17, 15, 10, 13, 21, 16, 19, 107, 22, 110, 30, 105, 12, 33, 20, 28, 103, 116, 36, 111, 31, 119, 26, 88, 101, 114, 13, 47, 29, 91, 42, 24, 16, 37, 24, 68, 32, 32, 19, 45, 120, 27, 120, 40, 71, 14, 35, 84, 53, 22, 66, 123, 79, 30, 43, 17, 43, 61, 118, 38, 87, 131, 38, 25, 113, 126, 33, 126, 51, 46, 20, 59

Offset: 1

Freimut Marschner, Aug 12 2014

nonn

Define a Collatz cycle C(prime(n)) = {c(z+1) = c(z)/2 if c(z) mod 2 = 0, otherwise c(z+1) = 3*c(z) + 1}, z >= 1, c(1) = prime(n), n >= 1}; then the length of C(prime(n)) depends only on the starting point c(1) if C ends with c(z) = 1. The length of C(prime(n)) is z, so a(n) = z.
The longest C(prime(n)) out of 10^5 prime numbers is C(prime(96648)) = C(1252663) with a(96648) = 510.
Until now C is not proved mathematically. So if the ending point c(z) is not equal to 1 then C(prime(n)) is not a 'true' Collatz cycle or does not exist.

			a(1) = {c(1) = prime(1) = 2, 2 mod 2 = 0, c(2) = 2/2 = 1, z=2} = 2;
a(3) = {c(1) = prime(3) = 5, 5 mod 2 = 1, c(2) = 3*5 + 1 = 16; 16 mod 2 = 0, c(3) = 16/2 = 8; 8 mod 2 = 0, c(4) = 8/2 = 4; 4 mod 2 = 0, c(5) = 4/2 = 2; 2 mod 2 = 0, c(6) = 2/2 = 1, z=6} = 6.

Freimut Marschner, Table of n, a(n) for n = 1..100000

A006577 (Number of halving and tripling steps to reach 1 in '3x+1' problem), A070975 (Number of steps to reach 1 in '3x+1' (or Collatz) problem starting with prime(n)).

a(n)=n=prime(n);A=List;while(n != 1,listput(A,n);if(n%2==0,n=n/2,n=3*n+1));listput(A,1);return(#Vec(A)) \\ Edward Jiang, Sep 06 2014

a(n) = z where {c(z+1) = c(z)/2 if c(z) mod 2 = 0, otherwise c(z+1) = 3*c(z) + 1}, z >= 1, c(1) = prime(n), n >= 1}, a(n) = A006577(prime(n)) + 1 = A070975(n) + 1.

A246007 Length of pseudo-Collatz cycles '3*n - 1' of prime numbers.

Original entry on oeis.org

2, 5, 3, 6, 7, 7, 19, 5, 4, 11, 7, 15, 10, 9, 14, 17, 12, 8, 21, 20, 16, 15, 11, 33, 36, 36, 18, 10, 14, 31, 26, 22, 21, 13, 26, 34, 16, 12, 21, 42, 25, 16, 16, 37, 20, 29, 19, 24, 32, 90, 28, 28, 19, 19, 85, 23, 40, 14, 36, 27, 22, 49, 17, 31, 31, 40, 13, 44, 43, 26, 66, 43, 25, 25, 25, 30, 21, 30, 30, 51, 20, 25, 25, 33, 47, 16, 47, 91, 46, 46, 29, 46, 28

Offset: 1

Freimut Marschner, Aug 10 2014

nonn

Define a pseudo-Collatz cycle C(prime(n)) = {c(z+1) = c(z)/2 if c(z) mod 2 = 0, otherwise c(z+1) = 3*c(z) - 1}, z >= 1, c(1) = prime(n), n >= 1} depending of the starting point c(1). If c(1) = prime(n) then c(z) might be
  (1) finite convergent to c(z) = 1 or
  (2) infinite periodic from c(z) = 7 or from c(z) = 17 or
  (3) no cycle if c(z) = -1.
The case (3) is not observed out of 10^5 prime numbers. So a(n) = z is the length of the C(prime(n)) up to the stoppping point, where c(z) = 1 or up to the periodical point, where c(z) = 7 or c(z) = 17 or c(z) = c(1). See Table for examples of cases (1) and (2). The longest sequence here is a(99147) = 560 with starting point c(1) = prime(99147) = 1287511 up to the periodical point c(560) = 17.

			a(1) = {c(1) = prime(1) = 2, 2 mod 2 = 0, c(2) = 2/2 = 1, z=2} = 2.
Table for cases (1) and (2):
case (1)
c(1) = prime(2) = 3
z    1 2 3 4 5
c(z) 3 8 4 2 1
a(2) = 5
c(1) = prime(3) = 5
z    1  2 3
c(z) 5 14 7
a(3) = 3
c(1) = prime(10) = 29
z     1  2  3   4  5  6  7 8 9 10 11
c(z) 29 86 43 128 64 32 16 8 4  2  1
a(10) = 11
case (2)
c(1) = prime(4) = 7
z    1  2  3 4  5 6  7 ...
c(z) 7 20 10 5 14 7 20 ...
a(4) = 6
c(1) = prime(7) = 17
z     1  2  3  4  5   6  7   8  9 10  11 12  13
c(z) 17 50 25 74 37 110 55 164 82 41 122 61 182
z    14  15  16 17 18 19 20 ...
c(z) 91 272 136 68 34 17 50 ...
a(7) = 19

Freimut Marschner, Table of n, a(n) for n = 1..100000

A003627 (Primes of form 3n-1), A006370 (Image of n under the '3x+1' map), A014682 (The Collatz or 3x+1 function: a(n) = n/2 if n is even, otherwise (3n+1)/2), A006577(Number of halving and tripling steps to reach 1 in '3x+1' problem), A016789({3n+2, n >=0} = {3n-1, n >= 1}).

a(n) = z where {c(z+1) = c(z)/2 if c(z) mod 2 = 0, otherwise c(z+1) = 3*c(z) - 1}, z >= 1, c(1) = prime(n), n>= 1}.

A245838 Arithmetic derivative of (3*n + 1), n >= 1, (A016777)'.

Original entry on oeis.org

4, 1, 7, 1, 32, 1, 13, 10, 32, 1, 19, 1, 68, 1, 25, 14, 56, 16, 31, 1, 192, 1, 59, 1, 80, 1, 43, 22, 140, 20, 49, 1, 140, 1, 55, 1, 240, 28, 61, 22, 128, 1, 101, 26, 212, 1, 73, 34, 152, 1, 113, 1, 432, 1, 85, 26, 176, 95, 91, 1, 284, 28, 143, 1, 252, 1, 103

Offset: 1

Freimut Marschner, Aug 06 2014

nonn

Comparing A016777(n) = 3n+1 and its arithmetic derivative a(n) = (A016777(n))' allows us to select the primes in A016777, because (prime(n))' = 1.

			a(2) = (3*2 + 1)' = (7)' = 1, a(5846) = (A016777(5846))' = (3*5846 + 1)' = (17539)' = 1.

Freimut Marschner, Table of n, a(n) for n = 1..100000

Cf. A003415 (arithmetic derivative), A016777 (3n+1), A002476 (Primes of form 6m + 1).

ad:= n -> n*add(f[2]/f[1], f=ifactors(n)[2]):
seq(ad(3*n+1), n=1..100); # Robert Israel, Aug 25 2014

a(n) = my(n=3*n+1);sum(i=1, #f=factor(n)~, n/f[1, i]*f[2, i]); \\ Michel Marcus, Aug 12 2014

a(n) = (3*n + 1)' = (A016777(n))' = A003415(A016777(n)).

Edited: in name n>=0 replaced by n>=1. Example corrected. - Wolfdieter Lang, Oct 14 2014

Missing comma in data at a(63) inserted by Andrew Howroyd, Feb 22 2018

A245071 a(n) = 12n - prime(n).

Original entry on oeis.org

10, 21, 31, 41, 49, 59, 67, 77, 85, 91, 101, 107, 115, 125, 133, 139, 145, 155, 161, 169, 179, 185, 193, 199, 203, 211, 221, 229, 239, 247, 245, 253, 259, 269, 271, 281, 287, 293, 301, 307, 313, 323, 325, 335, 343, 353, 353, 353, 361, 371, 379, 385, 395, 397, 403, 409, 415, 425

Offset: 1

Freimut Marschner, Jul 21 2014

Prime(n) > n for n > 0. Let prime(n) = k*n with k as an even integer constant, for example, k = 12; then a(n) = k*n - prime(n) is a sequence of odd integers that are positive as long as k*n > prime(n). This is the case up to a(40072) = 11. If k*n < prime(n) then a(n) < 0, a(40073) = -5 up to a(40083) = -5. From a(40084) = 5 up to a(40121) = 5, a(n) > 0 again, but a(n) < 0 for n >= 40122. For k = 12 the table shows this result compared with floor(prime(n)/n) and (prime(n) mod n) <= (prime(n+1) mod (n+1)) for n >= 1. Observations:
(1) If k > floor(prime(n)/n) then a(n) is positive.
(2) If k <= floor(prime(n)/n) and (prime(n) mod n) < (prime(n+1) mod (n+1)) and n > 1 then a(n) is negative.
(3) If k <= floor(prime(n)/n) and (prime(n) mod n) > (prime(n+1) mod (n+1)) then a(n) is positive.
.
n     prime(n) floor(prime(n)/n) (prime(n) mod n)  a(n)
40072 480853        12                 5            11
40073 480881        12                23            -5
40083 481001        11             40079            -5
40084 481003        11             40074             5
40121 481447        12                 5             5
40122 481469        12                13            -5

			a(3) = 12*3 - prime(3) = 36 - 5 = 31.

Freimut Marschner, Table of n, a(n) for n = 1..100000

A000040 (prime(n)), A038605 (floor(prime(n)/n)), A004648 (prime(n) mod n), A038606 (Least k such that k-th prime > n * k), A038607 (the smallest prime number k such that k > n*pi(k)), A102281 (the largest number m such that m = pi(n*m)).

Table[12n - Prime[n], {n, 60}] (* Alonso del Arte, Jul 27 2014 *)

PARI
```
vector(133, n, 12*n-prime(n) )
```

a(n) = 12*n - prime(n).

cp's OEIS Frontend

User: Freimut Marschner

A250623 a(n) = floor(n*log(prime(n))) + ceiling(n*log(n)) - 2*prime(n).

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A251482 a(n) = floor(prime(n)/log(n)) + ceiling(prime(n)/log(prime(n))) - 2*n, n >=2.

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A250622 a(n) = floor(n*log(prime(n)))-prime(n), n >= 1.

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A250621 a(n) = floor(n*log(prime(n))).

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A247583 Primes extracted from a pseudo-Collatz cycle '3*n-1' by consecutive arithmetic derivatives, here with starting point prime(99147) = 1287511.

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A246010 a(n) = floor(5*prime(n)^2 / 4).

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A246009 Length of Collatz cycles '3*n + 1' of prime numbers.

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A250623 a(n) = floor(nlog(prime(n))) + ceiling(nlog(n)) - 2*prime(n).