Werner D. Sand - cp's OEIS frontend

A249241 a(n) = p - prime(n)!/prime(n)#, where p is the smallest prime number > prime(n)!/prime(n)#+1.

2, 2, 3, 5, 11, 7, 29, 17, 17, 397, 47, 67, 23, 41, 31, 157, 409, 31, 151, 109, 199, 191, 131, 61, 103, 547, 179, 269, 389, 317, 181, 331, 307, 173, 1259, 1289, 619, 131, 223, 683, 139, 241, 191, 101, 1039, 1367, 1153, 241, 1187, 479, 149, 181, 487, 1093, 571, 1151, 809, 199, 823, 491, 191, 151, 1321, 197, 163, 337, 467, 659, 673, 877, 487, 743, 313, 673, 857, 677, 1021

Offset: 1

Werner D. Sand, Oct 23 2014

nonn

Conjecture: All terms are prime.

While Fortune's conjecture (A005235) uses products of primes, this sequence uses products of composite numbers (more exactly: of nonprimes, because 1 belongs to them). It looks like all multiples of prime(n)# (except some powers) lead to a sequence which contains only prime numbers.

			n = 1; prime(1)!/prime(1)# = 2/2 = 1; p = nextprime(1+1) = 3; a(1) = 3-1 = 2.

Cf. A092435.

q:=1; p:=1; for i from 1 to 100 do q:=nextprime(q+1); p:=p*q; N:=nextprime((fact(q)/p)+2)-fact(q)/p; print(i,N); end_for:

A092435(n)=prime(n)!/prod(i=1,n,prime(i))
a(n)=my(t=A092435(n)); nextprime(t+2)-t \\ Charles R Greathouse IV, Oct 23 2014

A248714 a(n) = p - prime(n)#^2, where prime(n)# is the product of the first n primes and p is the smallest prime > prime(n)#^2 + 1.

Original entry on oeis.org

3, 5, 7, 11, 17, 29, 23, 41, 29, 37, 89, 79, 89, 71, 439, 389, 163, 79, 151, 73, 89, 211, 113, 113, 419, 167, 139, 199, 173, 137, 487, 197, 401, 167, 739, 641, 461, 199, 223, 331, 379, 401, 293, 223, 251, 647, 593, 613, 317, 271, 257, 947, 331, 347, 593, 433

Offset: 1

Werner D. Sand, Oct 12 2014

nonn

Conjecture: Analogous to Fortune's Conjecture (A005235) all a(n) are prime, so are all members of a(n)=p-k*prime(n)#, k=natural number.

Besides, many powers p-prime(n)#^m, m=natural number, behave as well, e.g. p-prime(n)#^29 does, p-prime(n)#^30 does not.

Cf. A002110, A005235, A037153.

q:=1;p:=1;for i from 1 to 100 do q:=nextprime(q+1);p:=p*q;N:=nextprime(p^2+2)-p^2;print(i,N);end_for: \\ Werner D. Sand, Oct 13 2014

a(n) = {hp = prod(ip=1, n, prime(ip)); nextprime(hp^2+2) - hp^2;} \\ Michel Marcus, Oct 12 2014

A135097 Decimal expansion of a certain constant U (see A135096 for further information).

Original entry on oeis.org

1, 3, 6, 5, 2, 1, 2, 3, 8, 8, 9, 6, 8, 5, 1, 8, 7, 2, 9, 7, 2, 9, 3, 3, 8, 6, 6, 7, 6, 7, 7, 7, 5, 1, 5, 7, 3, 1, 3, 7, 3, 5, 6, 8, 9, 9, 6, 7, 8, 7

Offset: 1

Werner D. Sand, Nov 25 2007, Dec 01 2007

			For example, take c=log 2. Each n>2 has the form n=floor(m^(log 2)+p), where m is natural number and p prime.

Cf. A135096.

A135096 Decimal expansion of certain constant L.

Original entry on oeis.org

5, 7, 8, 1, 2, 9, 6, 5, 2, 6, 3, 0, 5, 6, 2, 8, 1, 3, 7, 2, 4, 0, 5, 7, 7, 5, 7, 9, 8, 0, 2, 9, 0, 2, 6, 1, 9, 1, 2, 1, 5, 9, 6, 2, 1, 7, 1, 5, 6

Offset: 0

Werner D. Sand, Nov 25 2007, Dec 01 2007

Conjecture: With any real number c in the interval [L,U] there exists for each natural number n>2 a natural number m and a prime number p so that n=floor(m^c)+p. L is the lower bound of c. U is the upper bound of c. L = 0.5781296526305628137240577579803... U = 1.3652123889685187297293386676777... (A135097).

L is close to the Euler-Mascheroni constant Gamma = 0.5772... I am continuing calculating further digits.

			Let c=1.23456789 (arbitrary). Each natural number n>2 has the form n=floor(m^1.23456789+p), where m is natural number and p prime.

Cf. A135097.

Leading zero removed by R. J. Mathar, Feb 05 2009

A109756 If you sum 3 consecutive odd prime numbers p,q,r, you get a number s which is either prime or not: p+q+r=s. If s is prime, you call it p and repeat the game. If s is not prime, you call the largest prime factor p and repeat the game. Finally, you get into an infinite cycle, which is one of the above 3 sequences, no matter what initial prime numbers you choose.

Original entry on oeis.org

7, 31, 109, 349, 1061, 103, 29, 97, 43, 13, 11, 41, 131, 37, 17, 59

Offset: 1

Werner D. Sand, Aug 12 2005

You can invent numerous variations which generate other cycles, but always you end in 2 or 3 cycles.

			p=7
7+11+13=31
31+37+41=109
109+113+127=349
349+353+359=1061
1061+1063+1069=3193=31*103
103+107+109=319=11*29
29+31+37=97
97+101+103=301=7*43
43+47+53=143=11*13
13+17+19=49=7*7
7+11+13...

Cf. A117631.

p+q+r=s, prime. s=p. repeat. p+q+r=s=f1*f2*f3..., fi prime. Largest f=p. repeat.

A075765 a(n) = floor(prime(n)/n) + (prime(n) mod n).

Original entry on oeis.org

2, 2, 3, 4, 3, 3, 5, 5, 7, 11, 11, 4, 5, 4, 5, 8, 11, 10, 13, 14, 13, 16, 17, 20, 25, 26, 25, 26, 25, 26, 7, 7, 9, 7, 13, 11, 13, 15, 15, 17, 19, 17, 23, 21, 21, 19, 27, 35, 35, 33, 33, 35, 33, 39, 41, 43, 45, 43, 45, 45, 43, 49, 59, 59, 57, 57, 67, 69, 7, 73, 73, 75, 7, 8, 9, 8, 9, 12

Offset: 1

Werner D. Sand, Oct 09 2002

nonn

The digital sum base n of the n-th prime. - Hieronymus Fischer, Dec 24 2007

			p(9)/9=23/9=2+5/9; a(9)=2+5=7

Cf. A007953.

fmQ[n_]:=Module[{pn=Prime[n]},Floor[pn/n]+Mod[pn,n]]
fmQ/@Range[90]  (* Harvey P. Dale, Feb 25 2011 *)

p(n)/n = k + r; r

a(n) = ds_n(prime(n)), where ds_n = digital sum base n. - Hieronymus Fischer, Oct 09 2007

a(n) = prime(n) - (n-1)*sum_{k>0} floor(prime(n)/n^k) = prime(n) - (n-1)*floor(prime(n)/n). - Hieronymus Fischer, Oct 09 2007

Corrected by Hieronymus Fischer, Oct 09 2007

A074500 Difference between n*sqrt(n)+1 and prime(n), rounded to nearest integer.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 3, 5, 5, 4, 6, 6, 7, 10, 12, 12, 12, 16, 17, 19, 24, 25, 28, 30, 29, 33, 38, 42, 48, 52, 47, 51, 54, 60, 59, 66, 69, 72, 78, 81, 85, 92, 92, 100, 106, 114, 112, 111, 117, 126, 132, 137, 146, 147, 152, 157, 162, 172, 177, 185, 194, 196, 194, 202, 212

Offset: 1

Werner D. Sand, Sep 26 2002

nonn

			a(3) = round(3*sqrt(3) + 1 - prime(3)) = round(5.196... + 1 - 5) = round(1.196...) = 1.

a(n) = round(n*sqrt(n) + 1 - prime(n)).

A074766 a(n) = prime(2n) - 2*prime(n) - n.

Original entry on oeis.org

-2, -1, 0, 1, 2, 5, 2, 7, 6, 3, 6, 3, 6, 7, 4, 9, 4, 11, 10, 11, 14, 13, 10, 21, 10, 11, 18, 21, 24, 25, 8, 17, 10, 25, 16, 21, 22, 19, 24, 23, 22, 29, 18, 27, 24, 35, 22, 9, 18, 33, 40, 39, 42, 37, 32, 31, 24, 41, 34, 37, 46, 35, 24, 33, 42, 43, 28, 27, 24, 41, 44, 37, 32, 37, 30, 39, 32

Offset: 1

Werner D. Sand, Sep 29 2002

sign

Conjecture: p(2n) - 2p(n) < 2n for all n.

			a(9) = p(18) - 2p(9) - 9 = 61 - 2*23 - 9 = 6.

Cf. A066066.

A074766:=n->ithprime(2*n)-2*ithprime(n)-n: seq(A074766(n), n=1..100); # Wesley Ivan Hurt, Jan 16 2017

Table[Prime[2 n] - 2 Prime[n] - n, {n, 80}] (* Harvey P. Dale, Mar 04 2017 *)

a(n) = A066066(n) - n.

A073046 Write 2n = p+q (p,q prime), pq minimal; then a(n) = p*q.

Original entry on oeis.org

4, 9, 15, 21, 35, 33, 39, 65, 51, 57, 95, 69, 115, 161, 87, 93, 155, 217, 111, 185, 123, 129, 215, 141, 235, 329, 159, 265, 371, 177, 183, 305, 427, 201, 335, 213, 219, 365, 511, 237, 395, 249, 415, 581, 267, 445, 623, 1501, 291, 485, 303, 309, 515, 321, 327

Offset: 2

Werner D. Sand, Aug 31 2002

nonn

Least semiprime whose sum of prime factors equals 2*n.

Assuming Goldbach's conjecture, a(n) exists for all n >= 2. - David James Sycamore, Jan 08 2019

			n=13: 2n=26; 26 = 23 + 3 = 19 + 7 = 13 + 13; 23*3 = minimal => p*q = 23*3 = 69.

Reinhard Zumkeller, Table of n, a(n) for n = 2..10000

Cf. A000040, A061397.

Cf. A102084, A193315, A288814.

a073046 n = head $ dropWhile (== 0) $
                   zipWith (*) prims $ map (a061397 . (2*n -)) prims
   where prims = takeWhile (<= n) a000040_list
-- Reinhard Zumkeller, Aug 28 2011

Array[Block[{p = 2, q}, While[! PrimeQ@ Set[q, 2 # - p], p = NextPrime[p]]; p q] &, 55, 2] (* Michael De Vlieger, Aug 02 2020 *)

For all n except 3, a(n) = A288814(2*n). - David James Sycamore, Jan 08 2019

Corrected by Ray Chandler, Jun 11 2005

A073497 a(n) = n^2 - prime(n).

Original entry on oeis.org

-1, 1, 4, 9, 14, 23, 32, 45, 58, 71, 90, 107, 128, 153, 178, 203, 230, 263, 294, 329, 368, 405, 446, 487, 528, 575, 626, 677, 732, 787, 834, 893, 952, 1017, 1076, 1145, 1212, 1281, 1354, 1427, 1502, 1583, 1658, 1743, 1828, 1917, 1998, 2081, 2174, 2271, 2368

Offset: 1

Werner D. Sand, Aug 27 2002

a(n) is never a perfect square for n>=5. [Proof: assume on the contrary n^2 - prime(n) = k^2, equivalent to (n+k)*(n-k) = prime(n). Since prime(n) cannot be the product of two nontrivial factors, this equation can only hold for k=n-1, i.e., prime(n)=2n-1. This contradicts the assumption and completes the proof.] - Alexander R. Povolotsky, Oct 01 2008

G. C. Greubel, Table of n, a(n) for n = 1..1000

[n^2 - NthPrime(n): n in [1..100] ]; // Vincenzo Librandi, Apr 12 2011

Mathematica
```
Table[n^2 - Prime[n], {n, 1, 55}]
```
PARI
```
for(n=1,51,print1(n*n-prime(n),","))
```

More terms from Klaus Brockhaus and Robert G. Wilson v, Aug 28 2002

cp's OEIS Frontend

User: Werner D. Sand

A249241 a(n) = p - prime(n)!/prime(n)#, where p is the smallest prime number > prime(n)!/prime(n)#+1.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

MuPAD

PARI

A248714 a(n) = p - prime(n)#^2, where prime(n)# is the product of the first n primes and p is the smallest prime > prime(n)#^2 + 1.

Views

Author

Keywords

Comments

Crossrefs

Programs

MuPAD

PARI

A135097 Decimal expansion of a certain constant U (see A135096 for further information).

Views

Author

Keywords

Examples

Crossrefs

A135096 Decimal expansion of certain constant L.

Views

Author

Keywords

Comments

Examples

Crossrefs

Extensions

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A075765 a(n) = floor(prime(n)/n) + (prime(n) mod n).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

Extensions

A074500 Difference between n*sqrt(n)+1 and prime(n), rounded to nearest integer.

Views

Author

Keywords

Examples

Formula

A074766 a(n) = prime(2n) - 2*prime(n) - n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

Formula

A073046 Write 2*n = p+q (p,q prime), p*q minimal; then a(n) = p*q.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

A073046 Write 2n = p+q (p,q prime), pq minimal; then a(n) = p*q.