cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A145281 a(n) is the least prime such that (ceiling(sqrt(a(n)*p_n)))^2 - a(n)*p_n is a perfect square, where p_n is the n-th prime.

Original entry on oeis.org

2, 3, 3, 3, 5, 5, 11, 11, 13, 17, 19, 23, 29, 29, 31, 37, 41, 41, 47, 53, 53, 59, 61, 67, 73, 79, 79, 83, 83, 89, 101, 101, 107, 109, 127, 127, 127, 131, 137, 139, 149, 149, 157, 157, 163, 163, 173, 191, 191, 191, 193, 199, 211, 211, 223, 223, 227, 227, 233, 239
Offset: 1

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Author

Vladimir Shevelev, Oct 06 2008, Oct 07 2008

Keywords

Comments

Theorem. p_n - 2*sqrt(2p_n) + 2= A145236(n).
Or a(n) is the least prime q_n <= p_n such that sqrt(p_n) - sqrt(q_n) < sqrt(2) [or (p_n + q_n)/2 < sqrt(p_n*q_n) + 1]. See also our comment to A145300. - Vladimir Shevelev, Oct 09 2008
The above conjecture is true. This means that a(n) is the nearest prime p > p_n - 2*floor(sqrt(2*p_n)) + 2. A considerably more important and deep question is whether p < p_n. The answer does not follow even from the Riemann conjecture about zeros of the zeta function. - Vladimir Shevelev, Oct 17 2008

Crossrefs

Cf. A145236, A000040, A145016, A145022, A145023, A145047, A145048, A145049, A045050.

Programs

  • PARI
    a(n) = {my(p = prime(n)); my(q = 2); while (! issquare(ceil(sqrt(q*p))^2 - q*p), q = nextprime(q+1)); q;} \\ Michel Marcus, Jul 06 2015

Extensions

More terms from Michel Marcus, Jul 06 2015

A249298 Smallest positive integer k, such that s-k*n is a square where s is the smallest square >= k*n.

Original entry on oeis.org

1, 2, 1, 1, 1, 2, 3, 1, 1, 4, 5, 1, 5, 6, 1, 1, 9, 2, 9, 2, 1, 12, 13, 1, 1, 14, 1, 3, 17, 2, 19, 1, 3, 20, 1, 1, 23, 24, 3, 1, 25, 2, 27, 6, 1, 30, 31, 1, 1, 2, 3, 7, 35, 4, 1, 2, 3, 40, 41, 1, 41, 42, 1, 1, 1, 4, 47, 10, 5, 2, 51, 1, 51, 52, 3, 12, 1, 6, 57, 1, 1, 60, 61, 1, 3, 62, 7, 3, 65, 2
Offset: 1

Views

Author

Valtteri Raiko, Oct 24 2014

Keywords

Comments

For any n>=3, there exists at least one positive integer k, 1 <= k <= n-1 such that the difference between the smallest square >= k*n and k*n is a square. To prove this, consider the multiplier k = n-2. Then (n-2)*n = (n-1)^2-1, thus the difference from the next square is 1, which is a square. If n = 1, k = 1 and if n = 2, k = 2.
Smallest positive integer k such that ceiling(sqrt(k*n))^2-k*n is a square.

Examples

			a(10) = 4, for ceiling(sqrt(10))^2-10 = 6, ceiling(sqrt(2*10))^2-2*10 = 5, ceiling(sqrt(3*10))^2-3*10 = 6 and ceiling(sqrt(4*10))^2-4*10 = 9 = 3^2.

Crossrefs

Cf. A000290, A145236 (equals a(A000040)), A068527 (difference for k=1).
Cf. A145016, A145022, A145023, A145047, A145048, A145049, A145050, A145215.

Programs

  • Mathematica
    dif[n_] := Ceiling[Sqrt[n]]^2 - n;a[k_] := Module[{n = 1}, While[dif[dif[n*k]] != 0, n++]; Return[n]];Table[a[k], {k, 1, 90}]
  • PARI
    a(n) = {k=1; while(!issquare(ceil(sqrt(k*n))^2-k*n), k++); k;} \\ Michel Marcus, Oct 24 2014

A145230 Numbers of different values of the minimal factors s for primes of the form 4k+1 not exceeding 10^n (see A145215).

Original entry on oeis.org

1, 2, 3, 7, 17, 38
Offset: 1

Views

Author

Vladimir Shevelev, Oct 05 2008

Keywords

Crossrefs

A145016 A145022 A145023 A145047 A145048 A145049 A145050 A145215
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