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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A045882 Smallest term of first run of (at least) n consecutive integers which are not squarefree.

Original entry on oeis.org

4, 8, 48, 242, 844, 22020, 217070, 1092747, 8870024, 221167422, 221167422, 47255689915, 82462576220, 1043460553364, 79180770078548, 3215226335143218, 23742453640900972, 125781000834058568
Offset: 1

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Author

Erich Friedman

Keywords

Comments

Solution for n=10 is same as for n=11.
This sequence is infinite and each term initiates a suitable arithmetic progression with large differences like squares of primorials or other suitable products of primes from prime factors being on power 2 in terms and in chains after. Proof includes solution of linear Diophantine equations and math. induction. See also A068781, A070258, A070284, A078144, A049535, A077640, A077647, A078143 of which first terms are recollected here. - Labos Elemer, Nov 25 2002

Examples

			a(3) = 48 as 48, 49 and 50 are divisible by squares.
n=5 -> {844=2^2*211; 845=5*13^2; 846=2*3^2*47; 847=7*11^2; 848=2^4*53}.

References

  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 242, p. 67, Ellipses, Paris 2008.

Links

Crossrefs

Cf. A013929, A053806, A049535, A077647, A078143. Also A069021 and A051681 are different versions.

Programs

  • Mathematica
    cnt = 0; k = 0; Table[While[cnt < n, k++; If[! SquareFreeQ[k], cnt++, cnt = 0]]; k - n + 1, {n, 7}]
  • PARI
    a(n)=my(s);for(k=1,9^99,if(issquarefree(k),s=0,if(s++==n,return(k-n+1)))) \\ Charles R Greathouse IV, May 29 2013

Formula

a(n) = 1 + A020754(n+1). - R. J. Mathar, Jun 25 2010
Correction from Jeppe Stig Nielsen, Mar 05 2022: (Start)
a(n) = 1 + A020754(n+1) for 1 <= n < 11.
a(n) = 1 + A020754(n) for 11 <= n < N where N is unknown.
Possibly a(n) = 1 + A020754(n-d) for some higher n, depending on how many repeated terms the sequence has. (End)
a(n) <= A061742(n) = A002110(n)^2 is the trivial bound obtained from the CRT. - Charles R Greathouse IV, Sep 06 2022

Extensions

a(9)-a(11) from Patrick De Geest, Nov 15 1998, Jan 15 1999
a(12)-a(15) from Louis Marmet (louis(AT)marmet.org) and David Bernier (ezcos(AT)yahoo.com), Nov 15 1999
a(16) was obtained as a result of a team effort by Z. McGregor-Dorsey et al. [Louis Marmet (louis(AT)marmet.org), Jul 27 2000]
a(17) was obtained as a result of a team effort by E. Wong et al. [Louis Marmet (louis(AT)marmet.org), Jul 13 2001]
a(18) was obtained as a result of a team effort by L. Marmet et al.

A182433 Smallest number such that the next n integers each have the square of one of the first n primes as a factor in order.

Original entry on oeis.org

7, 547, 29347, 1308247, 652312447, 180110691547, 65335225716547, 38733853511213647, 4368761145612023947, 1804216772228848838647, 14884872991210984993091647, 9816873967836575781598117447, 143397994078495393809327283088347
Offset: 2

Views

Author

Alonso del Arte, Apr 28 2012

Keywords

Comments

These are found by an application of the Chinese remainder theorem. The remainders are the numbers prime(n)^2 - n (A182174), and the moduli are the squares of primes (A001248).
This guarantees a run of at least n nonsquarefree numbers. But just as n! + 1 guarantees a run of at least n - 1 composite numbers, this might not be the smallest run of n nonsquarefree numbers (for that, see A045882).
Marmet credits Erick Bryce Wong with the idea of applying the Chinese remainder theorem and a sieving process to obtain upper limits for squarefree gaps. From this it occurred to me to just apply the Chinese remainder theorem to find these squarefree gaps exhibiting the squares of primes in order.
Also, beyond a(4), that is n > 4, we will observe that some of the numbers in the run of nonsquarefree numbers are divisible by more than one prime power, e.g., a(n) + 5 is divisible both by 49 (the square of the fourth prime) and 4.

Examples

			a(3) = 547 as that is the solution to the simultaneous congruences x = 3 mod 4 = 7 mod 9 = 22 mod 25. We verify that the next 3 integers meet the requirement: 548 = 4 * 137, 549 = 9 * 61, 550 = 25 * 2 * 11.
a(4) = 29347 as that is the solution to the simultaneous congruences x = 3 mod 4 = 7 mod 9 = 22 mod 25 = 45 mod 49. We verify that the next 4 integers meet the requirement: 29348 = 4 * 11 * 23 * 29, 29349 = 9 * 3 * 1087, 29350 = 25 * 2 * 587, 29351 = 49 * 599.

Links

Crossrefs

Cf. A069021, A051681.

Programs

  • Mathematica
    Table[ChineseRemainder[Prime[Range[n]]^2 - Range[n], Prime[Range[n]]^2], {n, 2, 14}]

A069022 Start of the first occurrence of two consecutive numbers divisible by an n-th power.

Original entry on oeis.org

1, 8, 80, 80, 1215, 16767, 76544, 636416, 3995648, 24151040, 36315135, 689278976, 1487503359, 1487503359, 155240824832, 785129144319, 4857090670592, 45922887663615, 157197025673216, 1375916505694208, 2280241934368767
Offset: 1

Views

Author

Amarnath Murthy, Apr 02 2002

Keywords

Comments

Essentially the same as A063528. [R. J. Mathar, Sep 29 2008]

Examples

			(8, 9) is the first occurrence of two consecutive numbers divisible by a square, so a(2) = 8.

Crossrefs

Cf. A069021.

Programs

  • Mathematica
    a = {1, 1}; k = 4; Do[ While[Min[a] < n, k++; a = Drop[a, 1]; a = Append[ a, Max[ Transpose[ FactorInteger[k]] [[2]] ]]]; Print[k - 1], {n, 2, 9}] (* Robert G. Wilson v *)

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Oct 12 2003
Showing 1-3 of 3 results.

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