A124129 -id:A124129 - cp's OEIS frontend

A349660 Numbers which are the sum of a prime and the square of the next prime.

11, 28, 54, 128, 180, 302, 378, 548, 864, 990, 1400, 1718, 1890, 2252, 2856, 3534, 3780, 4550, 5108, 5400, 6314, 6968, 8004, 9498, 10298, 10710, 11552, 11988, 12878, 16242, 17288, 18900, 19458, 22340, 22950, 24800, 26726, 28052, 30096, 32214, 32940, 36662

Offset: 1

Karl-Heinz Hofmann, Nov 24 2021

			a(2) = 3 + 5^2 = 28; a(3) = 5 + 7^2 = 54.

Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000

Cf. A140511, A000040, A001248, A036690, A001223, A001043, A036689, A349661, A124129.

nterms=100;Table[Prime[n]+Prime[n+1]^2,{n,nterms}] (* Paolo Xausa, Nov 24 2021 *)

a(n) = prime(n) + prime(n+1)^2; \\ Michel Marcus, Nov 24 2021

from sympy import sieve;
for n in range(1,10001): print(sieve[n] + sieve[n+1]**2)

a(n) = prime(n) + prime(n+1)^2.
a(n) = A000040(n) + A001248(n+1).
a(n) = A036690(n+1) - A001223(n).
a(n) = A001043(n) + A036689(n+1). - Michel Marcus, Nov 24 2021

A200474 a(n) = floor(10*(prime(n+1)-prime(n))/log(prime(n))^2).

Original entry on oeis.org

3, 6, 2, 4, 6, 1, 5, 3, 1, 2, 4, 3, 1, 3, 2, 1, 3, 2, 3, 3, 1, 0, 1, 0, 1, 6, 1, 2, 0, 4, 0, 2, 2, 1, 2, 2, 0, 3, 0, 1, 0, 4, 4, 1, 0, 1, 2, 0, 3, 1, 1, 1, 0, 1, 1, 0, 3, 4, 1, 0, 1, 4, 1, 2, 0, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 2, 0, 2, 0, 1, 1, 1, 2, 1, 0, 1, 3

Offset: 5

Arkadiusz Wesolowski, Nov 18 2011

nonn

Cramer's conjecture is true if, for every n >= 5, a(n) is smaller than 10.
If Cramer's conjecture is true, then Andrica's conjecture is true. [John W. Nicholson, Feb 06 2012]
Some mathematicians are trying to prove: if Andrica's conjecture is true, then Cramer's conjecture is true. [Arkadiusz Wesolowski, Feb 22 2012]

			a(9) = 6 because 10*(29-23)/log(23)^2 = 6.1029419977....

Arkadiusz Wesolowski, Table of n, a(n) for n = 5..10000
Carlos Rivera, Conjecture 7. The Cramer's Conjecture, The Prime Puzzles and Problems Connection.
Eric Weisstein's World of Mathematics, Cramer Conjecture

Cf. A000040, A001223, A086142, A124129, A200324.

Table[Floor[10*(Prime[n + 1] - Prime[n])/Log[Prime[n]]^2], {n, 5, 100}]

a(n) = floor(10*A001223(n)/log(A000040(n))^2), n >= 5.

A133697 a(n) = smallest number k such that P(k)/P(k+1) > P(k+1)/P(k+2) > ... > P(k+n+1)/P(k+n+2), where P(k) = k-th prime = A000040(k).

Original entry on oeis.org

1, 7, 69, 420, 1796, 12073, 101397, 1139211, 5440508, 320620306, 2058187481, 36451609409, 54594153615, 4100904808215

Offset: 0

Philippe LALLOUET (philip.lallouet(AT)orange.fr), Jan 04 2008

In other words, the rank of the smallest prime number such that the ratio between each prime and the following one is decreasing for at least n+2 consecutive ratios.
The sequence of primes P[a(n)] begins 2,17,347,2903,15373,128981,... - Robert G. Wilson v, Mar 01 2008
a(9) > 120000000. - Robert G. Wilson v, Mar 01 2008
If 113 is, as conjectured, the last term of A124129, then P(a(n)) = A158939(n+2). Proof: Let x and y be the prime gaps following the prime p = P(j) > 113, so that P(j+1) = P(j) + x and P(j+2) = P(j) + x + y. The inequality P(j)/P(j+1) > P(j+1)/P(j+2) can be written as p/(p+x) > (p+x)/(p+x+y), which simplifies to y > x+x^2/p. By assumption, x^2 < p, so this holds if and only if y > x. So the condition P(j)/P(j+1) > P(j+1)/P(j+2) is equivalent to increasing prime gaps, P(j+2) - P(j+1) > P(j+1) - P(j). (In fact, since all prime gaps except the first are even, it is enough to assume the weaker conjecture that 7 is the only prime P(j) such that (P(j+1)-P(j))^2 >= 2*P(j).) - Pontus von Brömssen, Nov 19 2021

			P(1)=2, P(2)=3, P(3)=5; 2/3 > 3/5, hence a(0)=1.
17/19 > 19/23 > 23/29 is the first double inequality satisfied by consecutive primes, hence a(1)=7 as 17=P(7).
347/349 > 349/353 > 353/359 > 359/367 is the first triple inequality satisfied by consecutive primes, hence a(2)=69 as 347=P(69).

Cf. A000040, A124129, A158939.

(* for the 6th term *) n = 12000; While[ Prime[n]/Prime[n + 1] < Prime[n + 1]/Prime[n + 2] || Prime[n + 1]/Prime[n + 2] < Prime[n + 2]/Prime[n + 3] || Prime[n + 2]/Prime[n + 3] < Prime[n + 3]/Prime[n + 4] || Prime[n + 3]/Prime[n + 4] < Prime[n + 4]/Prime[n + 5] || Prime[n + 4]/Prime[n + 5] < Prime[n + 5]/Prime[n + 6] || Prime[n + 5]/Prime[n + 6] < Prime[n + 6]/Prime[n + 7] || Prime[n + 6]/Prime[n + 7] < Prime[n + 7]/Prime[n + 8], n++ ]; Print[n] (* Robert G. Wilson v, Mar 01 2008 *)

a(6)-a(8) from Robert G. Wilson v, Mar 01 2008
a(9)-a(13) (based on data for A158939) from Pontus von Brömssen, Nov 19 2021
Edited to make name and offset consistent by Pontus von Brömssen, Nov 19 2021

A283225 Primes prime(k) such that prime(k)^2 mod prime(k+2) is different from prime(k+2)^2 mod prime(k).

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 43, 47, 53, 59, 61, 73, 79, 83, 89, 109, 113, 137, 139, 199, 211, 241, 283, 293, 313, 317, 523, 1321, 1327

Offset: 1

Arnaud Vernier, Mar 03 2017

I conjecture that there are no other terms in this sequence.
A124129 is constructed in a similar way: by comparing the values of prime(k)^2 mod prime(k+1) and prime(k+1)^2 mod prime(k).
If it exists, then a(35) > 10^12. - Lucas A. Brown, Feb 11 2021

			a(10) = prime(10) = 29 is in the sequence because the remainder of the division of 29^2 = 841 by prime(12) = 37 is 27, which is different from the remainder of the division of 37^2 = 1369 by prime(10) = 29, which is 6.

Cf. A124129.

Select[Prime[Range[250]],PowerMod[#,2,NextPrime[#,2]] != PowerMod[ NextPrime[ #,2],2,#]&]  (* Harvey P. Dale, Nov 17 2020 *)

cp's OEIS Frontend

A349660 Numbers which are the sum of a prime and the square of the next prime.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Python

Formula

A200474 a(n) = floor(10*(prime(n+1)-prime(n))/log(prime(n))^2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A133697 a(n) = smallest number k such that P(k)/P(k+1) > P(k+1)/P(k+2) > ... > P(k+n+1)/P(k+n+2), where P(k) = k-th prime = A000040(k).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Extensions

A283225 Primes prime(k) such that prime(k)^2 mod prime(k+2) is different from prime(k+2)^2 mod prime(k).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica