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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A218716 a(n) is smallest number such that a(n)^2 + 1 is divisible by 61^n.

Original entry on oeis.org

0, 11, 682, 51412, 6304056, 144762466, 9435321777, 988322434636, 71294762793847, 3138611770750343, 283798117998769727, 15409745938584647495, 320007169218635518122, 45443939732277600209579, 207359227164430355867160, 59053635973003478214807486
Offset: 0

Views

Author

Michel Lagneau, Nov 04 2012

Keywords

Examples

			a(3) = 51412 because 51412^2+1 = 5 * 17 * 61 ^ 3 * 137.
		

Crossrefs

Programs

  • Mathematica
    b=11;n61=61;jo=Join[{0,b},Table[n61=61*n61;b=PowerMod[b,61,n61];b=Min[b,n61-b],{99}]]

A069187 Numbers k such that core(k) = ceiling(sqrt(k)) where core(k) is the squarefree part of k (the smallest integer such that k*core(k) is a square).

Original entry on oeis.org

1, 2, 20, 90, 272, 650, 1332, 4160, 6642, 10100, 14762, 20880, 28730, 38612, 50850, 65792, 83810, 130682, 160400, 194922, 234740, 280370, 332352, 391250, 457652, 532170, 615440, 708122, 810900, 924482, 1187010, 1337492, 1501850, 1680912, 1875530, 2314962, 2561600
Offset: 1

Views

Author

Benoit Cloitre, Apr 14 2002

Keywords

Comments

Conjecture: sequence is A071253 minus those entries of A071253 that have their index in A049532, i.e., a(n) is of form n^2*(n^2+1) for all n not in A049532. - Ralf Stephan, Aug 18 2004

Crossrefs

Programs

  • Mathematica
    core[n_] := Times @@ Apply[ Power, {#[[1]], Mod[#[[2]], 2]}& /@ FactorInteger[n], {1}]; Select[Range[500000], core[#] == Ceiling[Sqrt[#]]&] (* Jean-François Alcover, Jul 26 2011 *)

Extensions

More terms from Amiram Eldar, Sep 10 2020

A124808 Number of numbers k <= n such that k^2 + 1 is squarefree.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 30, 31, 32, 33, 34, 35, 35, 36, 37, 37, 38, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 61, 62, 62, 63, 64
Offset: 0

Views

Author

Reinhard Zumkeller, Nov 08 2006

Keywords

Crossrefs

Programs

  • Maple
    ListTools:-PartialSums([seq(numtheory:-mobius(k^2+1)^2, k=0..100)]); # Robert Israel, Jul 15 2015
  • Mathematica
    Accumulate[Table[If[SquareFreeQ[k^2+1],1,0],{k,0,80}]] (* Harvey P. Dale, Mar 04 2014 *)
    Table[Sum[MoebiusMu[k^2 + 1]^2, {k, 0, n}], {n, 0, 100}] (* Wesley Ivan Hurt, Jul 15 2015 *)
  • PARI
    a(n)={my(k,r=0);for(k=0,n,if(issquarefree(k^2+1),r++));return(r);}
    main(size)=my(n);vector(size,n,a(n-1)) /* Anders Hellström, Jul 15 2015 */

Formula

a(0) = 1, a(n) = a(n-1) + 0^(A059592(n) - 1).
a(n) = Sum_{k=0..n} mu(k^2+1)^2, where mu(n) is the Mobius function (A008683). - Wesley Ivan Hurt, Jul 15 2015
a(n) ~ c*n where c = Product_{p prime, p == 1 (mod 4)} (1 - 2/p^2) = 0.894841... (A335963). - Amiram Eldar, Feb 23 2021

A217798 Numbers n such that n^2 + 1 and (n+1)^2 + 1 are divisible by a square.

Original entry on oeis.org

117, 407, 606, 775, 943, 1193, 1252, 1482, 1743, 1957, 2267, 2563, 3217, 3281, 3309, 3457, 3506, 3618, 3718, 3817, 4007, 4632, 4831, 5168, 5742, 5743, 5845, 6031, 6182, 6492, 6768, 7506, 7843, 8042, 8118, 8331, 8368, 8418, 8707, 8782, 8857, 9056, 9292, 9393
Offset: 1

Views

Author

Michel Lagneau, Oct 12 2012

Keywords

Comments

Also numbers n such that mu(n^2+1) = mu((n+1)^2+1)=0, where mu is the Moebius-function (A008683).

Examples

			117 is in the sequence because 117^2+1 = 2*5*37^2 and 118^2+1 = 5^2*557.
		

Crossrefs

Subsequence of A049532.

Programs

  • Magma
    A002522:=func; [n: n in [1..10^4]| not IsSquarefree(A002522(n)) and not IsSquarefree(A002522(n+1))]; // Bruno Berselli, Oct 15 2012
  • Maple
    with(numtheory):for n from 1 to 10000 do :x:=n^2+1:y:=(n+1)^2+1:if issqrfree(x)=false and issqrfree(y)=false then printf(`%d, `,n):else fi:od:
  • Mathematica
    Select[ Range[2, 10000], Max[ Transpose[ FactorInteger[ #^2+1 ]] [[2]]] > 1 && Max[ Transpose[ FactorInteger[ (#+1)^2 + 1]] [[2]]] > 1 &]

A218564 Numbers n such that n^2 + 1 is divisible by a 5th power.

Original entry on oeis.org

1068, 2057, 4193, 5182, 7318, 8307, 10443, 11432, 13568, 14557, 16693, 17682, 19818, 20807, 22943, 23932, 26068, 27057, 29193, 30182, 32318, 33307, 35443, 36432, 38568, 39557, 41693, 42682, 44818, 45807, 47943, 48932, 51068, 52057, 54193, 55182, 57318, 58307
Offset: 1

Views

Author

Michel Lagneau, Nov 02 2012

Keywords

Comments

For each prime p == 1 (mod 4), there are two values of x (mod p^5) that solve x^2 + 1 == 0 (mod p^5), and then x + k*p^5 is in the sequence for every k. Thus the asymptotic density of this sequence should be 1 - Product_p (1 - 2/p^5), where the product is over all primes p == 1 (mod 4). - Robert Israel, Sep 04 2018

Examples

			1068 is in the sequence because 1068^2+1 = 1140625 = 5^6*73;
143044 is in the sequence because 143044^2+1 = 20461585937 = 13^5*55109;
390112 is in the sequence because 390112^2+1 = 152187372545 = 5*13*17^6*97.
		

Crossrefs

Programs

  • Maple
    N:= 10^5: # to get all terms <= N
    P:= select(isprime,[seq(i,i=5..floor((N^2+1)^(1/5)),4)]):
    g:= proc(x,r,N) local t; t:= rhs(op(x)); seq(t+r*k,k=0..(N-t)/r) end proc:
    R:= `union`(seq(map(g, {msolve(n^2+1,p^5)},p^5,N),p=P)):
    sort(convert(R,list)); # Robert Israel, Sep 04 2018
  • Mathematica
    Select[Range[2,20000],Max[Transpose[FactorInteger[#^2+1]][[2]]]>4&]
  • PARI
    isok(n) = vecmax(factor(n^2+1)[,2]) >= 5; \\ Michel Marcus, Sep 04 2018

A218717 a(n) is smallest number such that a(n)^2 + 1 is divisible by 73^n.

Original entry on oeis.org

0, 27, 776, 153765, 6459524, 404034898, 41865466758, 3219884218827, 239822883201307, 9110883894036198, 991706090146518323, 142813358470363920740, 8641533837443707913816, 586811715371303018585730, 2756887299416274753296336, 729513196939063257288876118
Offset: 0

Views

Author

Michel Lagneau, Nov 04 2012

Keywords

Examples

			a(3) = 153765 because 153765^2+1 = 2 * 73 ^ 3 * 30389.
		

Crossrefs

Programs

  • Mathematica
    b=27;n73=73;jo=Join[{0,b},Table[n73=73*n73;b=PowerMod[b,73,n73];b=Min[b,n73-b],{99}]]

A124895 Triangle read by rows, 1 <= k <= n: T(n,k) = mu(n^2 + k^2) with mu = A008683.

Original entry on oeis.org

-1, -1, 0, 1, -1, 0, -1, 0, 0, 0, 1, -1, 1, -1, 0, -1, 0, 0, 0, -1, 0, 0, -1, 1, 1, 1, 1, 0, 1, 0, -1, 0, -1, 0, -1, 0, 1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 0, -1, 0, 0, 0, -1, 0, -1, 0, 1, 0, -1, -1, 1, -1, -1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, -1, 0, 0, 0, 1, 0, -1, -1, 1, 1, 1, 1, 1, -1, 0, -1, -1, -1, 0, -1, 0, 1, 0, 1, 0, 0, 0, -1, 0, -1, 0, 1, 0, 1, -1, 0, -1, 0, 0, 1, 0, 0, 0, 1, 0, 1, -1, 0
Offset: 1

Views

Author

Reinhard Zumkeller, Nov 12 2006

Keywords

Examples

			Triangle begins:
 -1
 -1,  0
  1, -1,  0
 -1,  0,  0,  0
  1, -1,  1, -1,  0
 -1,  0,  0,  0, -1, 0
  0, -1,  1,  1,  1, 1,  0
  1,  0, -1,  0, -1, 0, -1, 0
  1,  1,  0, -1,  1, 0, -1, 1,  0
 -1,  0, -1,  0,  0, 0, -1, 0, -1, 0
		

Crossrefs

Programs

  • Mathematica
    row[n_] := Table[MoebiusMu[n^2 + k^2], {k, 1, n}]; Array[row, 15] // Flatten (* Amiram Eldar, May 12 2025 *)
  • PARI
    row(n) = vector(n, k, moebius(n^2 + k^2)); \\ Amiram Eldar, May 12 2025

Formula

T(n,k) = A008683(A070216(n,k)).
T(n,1) = A124897(1); T(A049533(n),1) <> 0; T(A049532(n),1) = 0.
T(n,n) = -A000007(n-1).
A124896(n) = Sum_{k=1..n} abs(T(n,k)), row sums of absolute values.

A218565 Numbers k such that k^2 + 1 is divisible by a 6th power.

Original entry on oeis.org

1068, 14557, 16693, 30182, 32318, 45807, 47943, 61432, 63568, 77057, 79193, 92682, 94818, 108307, 110443, 123932, 126068, 139557, 141693, 155182, 157318, 170807, 172943, 186432, 188568, 202057, 204193, 217682, 219818, 233307, 235443, 248932, 251068, 264557
Offset: 1

Views

Author

Michel Lagneau, Nov 02 2012

Keywords

Examples

			1068 is in the sequence because 1068^2 + 1 = 5^6 * 73.
390112 is in the sequence because 390112^2 + 1 = 5 * 13 * 17 ^ 6 * 97.
1999509 is in the sequence because 1999509^2 + 1 = 2 * 13 ^ 6 * 29 * 14281.
		

Crossrefs

Cf. A001014 (6th powers).

Programs

  • Mathematica
    Select[Range[2,27000],Max[Transpose[FactorInteger[#^2+1]][[2]]]>5&]

A218718 a(n) is smallest number such that a(n)^2 + 1 is divisible by 89^n.

Original entry on oeis.org

0, 34, 3861, 344464, 20099637, 2153335831, 102666405913, 4867146503697, 923990886302412, 50251663587824641, 5655954122907587985, 909925832091926912414, 85120439454684773642745, 2631773999763198769695986, 41332517834853462204330752
Offset: 0

Views

Author

Michel Lagneau, Nov 04 2012

Keywords

Examples

			a(3) = 344464 because 344464^2+1 = 37 * 89 ^ 3 * 4549.
		

Crossrefs

Programs

  • Mathematica
    b=34;n89=89;jo=Join[{0,b},Table[n89=89*n89;b=PowerMod[b, 89,n89];b=Min[b,n89-b],{99}]]

A260824 Least positive integer b such that b^(2^n)+1 is not squarefree.

Original entry on oeis.org

3, 7, 110, 40, 392, 894, 315, 48
Offset: 0

Views

Author

Jeppe Stig Nielsen, Aug 04 2015

Keywords

Comments

For any n, a(n) <= A261117(n).
The smallest square in the factors of b^(2^n)+1 are 2^2, 5^2, 17^2, 17^2, 769^2. - Robert Price, Mar 07 2017; edited by Jeffrey Shallit, May 10 2017
a(8) <= 50104 (corresponding square 10753^2). - Jeffrey Shallit, May 10 2017
Some better bounds than A261117(n): a(9) <= 65863 (factor 13313^2), a(12) <= 265801 (factor 65537^2), a(16) <= 1493667 (factor 1179649^2), a(18) <= 15834352 (factor 7340033^2), a(19) <= 15786037 (factor 23068673^2), a(21) <= 78597313 (factor 230686721^2), a(22) <= 13753565041 (factor 469762049^2), a(23) <= 6276931961 (factor 469762049^2). - Max Alekseyev, Feb 20 2018

Examples

			a(1) = A049532(1) = 7.
For n=4, we consider b^16+1. The first time it is not squarefree is for b=392, where 392^16+1 is divisible by 769^2. So a(4)=392.
		

Crossrefs

Subsequence of A248214.

Programs

  • PARI
    a(n) = for(b=1,10^42, !issquarefree(b^(2^n)+1) & return(b) );
    
  • Python
    from sympy.ntheory.factor_ import core
    def a(n):
      b, pow2, t = 1, 2**n, 2
      while core(t, 2) == t:
        b += 1
        t = b**(pow2) + 1
      return b
    print([a(n) for n in range(4)]) # Michael S. Branicky, Mar 07 2021

Formula

a(n) = A248214(2^n).

Extensions

Edited and a(5)-a(7) added by Max Alekseyev, Feb 20 2018
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