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.

Showing 1-3 of 3 results.

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

A145236 a(n) is the least positive integer such that if p_n is the n-th prime then (ceiling(sqrt(a(n)*p_n)))^2 - a(n)*p_n is a perfect square.

Original entry on oeis.org

2, 1, 1, 3, 5, 5, 9, 9, 13, 17, 19, 23, 25, 27, 31, 35, 41, 41, 47, 51, 51, 57, 61, 65, 73, 75, 77, 81, 83, 85, 99, 101, 107, 109, 117, 119, 125, 129, 133, 139, 145, 145, 155, 157, 161, 163, 173, 183, 187, 189, 193, 199, 201, 209, 215, 221, 225, 227, 233, 237, 239, 247
Offset: 1

Views

Author

Vladimir Shevelev, Oct 05 2008, Oct 07 2008

Keywords

Comments

Conjectures: 1) for n >= 2, the sequence does not decrease; 2) for n > 1, a(n) is odd; 3) a(n) can be equal to a(n+1) only for twins: p_(n+1) - p_n = 2 (although there also exist twins for which a(n) < a(n+1)).
All these conjectures are proved using the formula a(n) = p_n - 2*floor(sqrt(2p_n)) + 2, n > 1. See also A145701 and A145714. - Vladimir Shevelev, Oct 18 2008

Crossrefs

Cf. A145016, A145022, A145023, A145047, A145048, A145049, A145050, A145215.

Programs

  • Maple
    A145236 := proc(n) local p,k,a ; p := ithprime(n) ; for k from 1 do ceil(sqrt(ceil(k*p))) ; a := %^2-k*p ; if issqr(a) then return k ; end if; end do: end proc:
for n from 1 do printf("%d,\n",A145236(n)) ; end do: # R. J. Mathar, Aug 02 2010

Extensions

a(12)=23 (not 21). - Vladimir Shevelev, Oct 16 2008
Extended by R. J. Mathar, Aug 02 2010

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
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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