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.

Previous Showing 41-48 of 48 results.

A271817 Smallest k > 1 such that none of k^2 + 0, k^2 + 1, k^2 + 2,..., k^2 + n are squarefree.

Original entry on oeis.org

2, 7, 32, 41, 239, 13982, 240761, 240761, 26220368, 93499279
Offset: 0

Views

Author

Juri-Stepan Gerasimov, Apr 14 2016

Keywords

Comments

a(10) > 1.2*10^8. - Giovanni Resta, Apr 15 2016

Examples

			a(0) = 2 because 2 > 1 and none of 2^2 + 0 = 4 = 2^2 is squarefree,
a(1) = 7 because none of 7^2 + 0 = 49 = (7^2), 7^2 + 1 = 50 =(5^2)*2 are squarefree,
a(2) = 32 because none of 32^2 + 0 = 1024 = (2^2)*156, 32^2 + 1 = 1025 = (5^2)*41, 32^2 + 2 = 1026 = (3^2)*114 are squarefree.
		

Crossrefs

Programs

  • PARI
    isok(k, n) = {for (j=0, n, if (issquarefree(k^2+j), return (0));); 1;}
    a(n) = {my(k = 2); while (! isok(k, n), k++); k;} \\ Michel Marcus, Apr 14 2016

Extensions

a(8)-a(9) from Giovanni Resta, Apr 15 2016

A374536 a(n) is the least exponentially odd number that is nonsquarefree and is followed by exactly n successive exponentially odd numbers that are squarefree, or -1 if no such number exists.

Original entry on oeis.org

135, 24, 120, 27, 96, 88, 32, 40, 328, 168, 136, 104, 1288, 1161, 352, 488, 8, 783, 189, 952, 4520, 56, 11576, 67384, 5088, 1336, 35768, 16173, 53768, 80328, 128169, 28576, 247375, 208552, 2556192, 1486568, 3099368, 1653032, 910568, 7864008, 34242976, 14484152
Offset: 0

Views

Author

Amiram Eldar, Jul 11 2024

Keywords

Examples

			a(0) = 135 because 135 and 136 are successive nonsquarefree exponentially odd numbers with no squarefree number between them.
a(1) = 24 because 24 and 27 are successive nonsquarefree exponentially odd numbers with one squarefree number between them, 26.
a(2) = 120 because 120 and 125 are successive nonsquarefree exponentially odd numbers with two squarefree number between them, 122 and 123.
		

Crossrefs

Programs

  • Mathematica
    sq[k_] := Module[{e = FactorInteger[k][[;;, 2]]}, If[AnyTrue[e, EvenQ], 0, If[k == 1 || Max[e] == 1, 2, 1]]]; seq[len_, kmax_ : Infinity] := Module[{v = Table[0, {len}], c = 0, k = 1, k0 = 0, m, i = 1}, While[c < len && k < kmax, m = sq[k]; If[m > 0, If[m == 2, i++, If[k0 > 0, If[i <= len && v[[i]] == 0, c++; v[[i]] = k0]; i = 1];	k0 = k]]; k++]; v]; seq[10]
  • PARI
    issq(k) = {my(e = factor(k)[, 2]); for(i = 1, #e, if(!(e[i] % 2), return(0))); if(k == 1 || vecmax(e) == 1, 2, 1);}
    lista(len, kmax = oo) = {my(v = vector(len), c = 0, k = 1, k0 = 0, m, i = 1); while(c < len && k < kmax, m = issq(k); if(m > 0, if(m == 2, i++, if(k0 > 0, if(i <= len && v[i] == 0, c++; v[i] = k0); i = 1); k0 = k)); k++); v; }

Formula

a(n) = A268335(A374460(k)), where k is the least number such that A374459(k+1) - A374459(k) = n + 1.

A380595 a(n) is the first nonsquarefree number k such that the n consecutive nonsquarefree numbers starting with k are in arithmetic progression.

Original entry on oeis.org

4, 4, 16, 28, 28, 5050, 6348, 144946, 3348550, 221167422, 221167422, 47255689915, 82462576220, 1043460553364, 79180770078548, 3215226335143218, 23742453640900972, 125781000834058568
Offset: 1

Views

Author

Robert Israel, Jan 27 2025

Keywords

Comments

Since multiples of 4 are not squarefree, the common difference of the arithmetic progression will be 1, 2 or 4 in each case.
For an arithmetic progression of length 10 or more with initial term even and common difference 2 or 4, there would be an odd multiple of 9 between the first and last term. Since multiples of 9 are not squarefree, these could not be consecutive nonsquarefree numbers. Thus for n >= 10, the common difference must be 1, and a(n) = A045882(n).

Examples

			a(2) = 4 because the 2 nonsquarefree numbers starting with 4 are 4, 6, forming an arithmetic progression with difference 2.
a(3) = 16 because the 3 nonsquarefree numbers starting with 16 are 16, 18, 20, forming an arithmetic progression with difference 2.
a(4) = a(5) = 28 because the 5 nonsquarefree numbers starting with 28 are 28, 32, 36, 40, 44, forming an arithmetic progression with difference 4.
a(6) = 5050 because the 6 nonsquarefree numbers starting with 5050 are 5050, 5052, 5054, 5056, 5058, 5060, forming an arithmetic progression with difference 2.
a(7) = 6348 because the 7 nonsquarefree numbers starting with 6348 are 6348, 6350, 6352, 6354, 6356, 6358, 6360, forming an arithmetic progression with difference 2.
a(8) = 144946, because the 8 nonsquarefree numbers starting with 144946 are 144946, 144948, 144950, 144952, 144954, 144956, 144958, 144960, forming an arithmetic progression with difference 2.
a(9) = 3348550, because the 9 nonsquarefree numbers starting with 3348550 are 3348550, 3348552, 3348554, 3348556, 3348558, 3348560, 3348562, 3348564, 3348566, forming an arithmetic progression with difference 2.
		

Crossrefs

Programs

  • Maple
    nsf:= remove(numtheory:-issqrfree, [$4..4*10^6]):
    S:= nsf[2..-1]-nsf[1..-2]: nS:= nops(S):
    R:= NULL:
    for m from 1 do
      found:= false;
      for t from 1 to nS +1-m do
        if nops(convert(S[t..t+m-1],set))=1 then R:= R,nsf[t]; found:= true; break fi
      od;
      if not found then break fi;
    od:
    R;

Extensions

a(1) = 4 prepended by David A. Corneth, Jan 28 2025

A055554 An arithmetic progression each term of which is followed by at least 4 nonsquarefree consecutive integers.

Original entry on oeis.org

242, 1757711582, 3515422922, 5273134262, 7030845602, 8788556942, 10546268282, 12303979622, 14061690962, 15819402302, 17577113642, 19334824982, 21092536322, 22850247662, 24607959002, 26365670342
Offset: 1

Views

Author

Labos Elemer, Jul 10 2000

Keywords

Comments

From the terms of A051681 or A045882 similar progressions for longer chains can be constructed.

Crossrefs

Formula

a(n) = d*n + 242, where d = LCM(242, 243, 244, 245) = 1757711340 = 2*2*3*3*3*3*3*5*7*11*11*61.

A172436 Smallest m such that the Moebius function takes successively, from m, n values 1,0,1,0,... ending with 1 or 0.

Original entry on oeis.org

1, 15, 55, 159, 411, 411, 411, 3647, 15243, 15243, 15243, 113343, 1133759, 1133759, 1133759, 29149139
Offset: 1

Views

Author

Michel Lagneau, Feb 02 2010

Keywords

Comments

It's easy to prove that a(n) for n >= 17 does not exist, because in all sequences of 17 consecutive numbers such that the first is squarefree, there are necessarily two numbers r, s where 9 divides r and s, so Moebius(r) = Moebius(s) = 0 with r - s odd.

Examples

			a(3) = 55 since Moebius(55) = 1, Moebius(56) = 0, Moebius(57) = 1, and this pattern does not occur for any smaller value of n.
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 826.
  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 24.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 161, #16.
  • Deleglise, Marc and Rivat, Joel, Computing the summation of the Mobius function. Experiment. Math. 5 (1996), no. 4, 291-295.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, 1979, th. 262 and 287.

Crossrefs

Moebius (or Mobius) function mu(n): A008683, A002321, A045882.

Programs

  • PARI
    a(n)=local(ok,m);m=1;while(1,ok=1;for(k=1,n,if(moebius(m+k-1)!=k%2,ok=0;break));if(ok,return(m));m++)

Extensions

Edited by Franklin T. Adams-Watters, May 15 2010

A268270 Least prime that is at distance > n from the nearest squarefree number.

Original entry on oeis.org

2, 17, 727, 47527, 29002021, 494501773, 44633461423, 7489886099881
Offset: 0

Views

Author

Keywords

Examples

			a(0)=2 is the least prime and it is at distance 1 from the nearest squarefree numbers (1 and/or 3).
a(1)=17 is the least prime that has no squarefree neighbor: both 16 and 18 are divisible by a square; the nearest squarefree numbers, 15 and 19, are both at distance 2.
a(2)=727 is the least prime p such that p-2 and p+1 are (two consecutive terms) in A068781, namely A068781(75..76).
a(3)=47527 is the least prime p such that p-3 and p+1 are (two consecutive terms) in A070258, namely A070258(878..879).
a(4)=29002021 is the least prime p such that p-4 and p+1 are (two consecutive terms) in A070284.
a(5)=494501773 is the least prime p such that p-5 and p+1 are (two consecutive terms) in A078144.
Similarly, for n = 6, 7, 8 and 9, a(n) is the least prime p such that p-n and p+1 are (two consecutive terms) in A049535, A077640, A077647 and A078143, respectively.
		

Crossrefs

Programs

  • PARI
    a(n)=forprime(p=n,,for(s=1,n,(issquarefree(p-s)||issquarefree(p+s)) && next(2)); return(p))

Extensions

a(4) corrected and a(5) computed by Christopher E. Thompson, Jan 20 2016
a(6)-a(7) from Bert Dobbelaere, Jan 28 2019

A319502 Indices of Mertens's function M(n) (A002321) whose nearest neighbors have value 0.

Original entry on oeis.org

165, 237, 330, 354, 357, 365, 402, 406, 421, 426, 794, 797, 813, 885, 894, 897, 905, 914, 1257, 1281, 1290, 1298, 1301, 1337, 1522, 1526, 1545, 1842, 1865, 2094, 2098, 2118, 2121, 2137, 4602, 4609, 4621, 4629, 4726, 4729, 4738, 5106, 5109, 5198, 5206, 5221
Offset: 1

Views

Author

Torlach Rush, Sep 20 2018

Keywords

Comments

This sequence records the shortest intervals where M(n) leaves 0 before returning to 0.
a(n) - 1 and a(n) + 1 are both terms of A028442.
Both A045882 and A028442 are infinite and this allows for the possibility that this sequence is also infinite (for A028442 see comment of A002321).

Examples

			165 is a term because A002321(164) = A002321(166) = 0.
237 is a term because A002321(236) = A002321(238) = 0.
		

Crossrefs

Programs

  • Maple
    with(numtheory): a:=n->add(mobius(k),k=1..n): select(n->a(n-1)=0 and a(n+1)=0,[$2..2200]); # Muniru A Asiru, Sep 20 2018
  • Mathematica
    With[{s = Partition[Accumulate@ Array[MoebiusMu, 5300], 3, 1]}, 1 + First /@ Position[s, {0, k_, 0} /; k != 0]] (* Michael De Vlieger, Sep 24 2018 *)
  • PARI
    isok(n) = {if (n > 1, x = sum(k=1, n-1, moebius(k)); if (x == 0, if (x + moebius(n) + moebius(n+1) == 0, return (1)););); return (0);} \\ Michel Marcus, Sep 27 2018

Formula

(A002321(a(n)) - A008683(a(n))) = (A002321(a(n)) + A008683(a(n+1))) = (A008683(a(n)) + A008683(a(n+1))) = 0.

A319520 Starts of strictly increasing runs of 0's in Mertens's function A002321.

Original entry on oeis.org

2, 39, 331, 422, 45371, 22898822, 871469945, 1319506393071, 26296710367071
Offset: 1

Views

Author

Torlach Rush, Sep 21 2018

Keywords

Comments

a(n) is always squarefree.
It is not known whether this sequence is infinite. Sequence A045882 is infinite but it appears that increasing runs of consecutive nonsquarefree numbers thin out very quickly. The requirement that the runs consist of 0's is much stronger and makes it uncertain whether this sequence is also infinite.

Examples

			2 is a term because M(2) = 0.
39 is a term because M(39) = M(40) = 0.
331 is a term because M(331) = M(332) = M(333) = 0.
422 is a term because M(422) = ... = M(425) = 0.
45371 is a term because M(45371) = ... = M(45376) = 0.
		

Crossrefs

Programs

  • Mathematica
    With[{s = Map[Boole[# == 0] &, Accumulate@ Array[MoebiusMu, 10^5]]}, Union@ Array[SequencePosition[s, ConstantArray[1, #]][[1, 1]] &, 5]] (* Michael De Vlieger, Sep 26 2018 *)
  • PARI
    M=S=R=0;for(n=1,oo,if(!M+=moebius(n),S||S=n,S,n-S>R&&print1(S",")+R=n-S;S=0)) \\ M. F. Hasler, Nov 23 2018

Extensions

a(6)-a(7) from Amiram Eldar, Sep 26 2018
a(8)-a(9) from Henri Lifchitz, Mar 09 2025
Previous Showing 41-48 of 48 results.