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 1-10 of 13 results. Next

A046791 A046790 has several definitions, one of which is: "Numbers i such that there is a smaller positive number j such that (i+j)/2 and sqrt(i*j) are integers". The present sequence gives the smallest choice for j.

Original entry on oeis.org

2, 1, 4, 2, 6, 1, 3, 2, 4, 10, 5, 12, 1, 2, 6, 14, 7, 4, 2, 3, 20, 1, 22, 10, 6, 2, 11, 4, 26, 12, 28, 13, 30, 1, 5, 14, 2, 15, 34, 4, 3, 6, 38, 17, 10, 2, 42, 1, 19, 7, 44, 20, 46, 21, 12, 4, 22, 2, 23, 52, 6, 14, 1, 58, 26, 60, 2, 3, 5, 62, 10, 28, 4, 29, 66, 30, 68, 11, 31, 70, 2, 1, 6, 74, 33
Offset: 1

Views

Author

David W. Wilson, Dec 11 1999

Keywords

Comments

Note that A046790 is the complement of A078779. - Omar E. Pol, Jun 11 2016

Examples

			From _Vladimir Shevelev_, Jun 07 2016: (Start)
A046790(5)=24 with even squarefree part (6), so a(5) = 6;
A046790(12)=48 with odd squarefree part (3), so a(12) = 3*4=12.
(End)
		

Crossrefs

Cf. A046790.

Programs

  • PARI
    a(n) = my(n=A046790(n),f=factor(n),p=n%2);f[,2]=f[,2]%2;r=prod(i=1,matsize(f)[1],f[i,1]^f[i,2]);r*=(4^(n%2==0&&r%2==1)) \\ David A. Corneth, Jun 07 2016

Formula

Let b(n)=A046790(n). Let k=k(n) be the greatest number whose square divides b(n) and is such that b(n) and b(n)/k^2 are of the same parity. Then a(n) = b(n)/k^2. - Vladimir Shevelev, Jun 07 2016
Or, equivalently, a(n) is the squarefree part s(n) of b(n), if either b(n) is odd or s(n) is even. Otherwise, when b(n) is even, but s(n) is odd, a(n)=4*s(n). - David A. Corneth, Jun 07 2016

Extensions

Entry revised by N. J. A. Sloane, with help from Don Reble and several OEIS editors. Jun 07 2016

A078779 Union of S, 2S and 4S, where S = odd squarefree numbers (A056911).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17, 19, 20, 21, 22, 23, 26, 28, 29, 30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 44, 46, 47, 51, 52, 53, 55, 57, 58, 59, 60, 61, 62, 65, 66, 67, 68, 69, 70, 71, 73, 74, 76, 77, 78, 79, 82, 83, 84, 85, 86, 87, 89, 91, 92, 93, 94, 95, 97, 101
Offset: 1

Views

Author

Benoit Cloitre, Jan 11 2003

Keywords

Comments

Numbers n such that the cyclic group Z_n is a DCI-group.
Numbers n such that A008475(n) = A001414(n).
A193551(a(n)) = A000026(a(n)) = a(n). - Reinhard Zumkeller, Aug 27 2011
Union of squarefree numbers and twice the squarefree numbers (A005117). - Reinhard Zumkeller, Feb 11 2012
The complement is A046790. - Omar E. Pol, Jun 11 2016

Crossrefs

Programs

  • Haskell
    a078779 n = a078779_list !! (n-1)
    a078779_list = m a005117_list $ map (* 2) a005117_list where
       m xs'@(x:xs) ys'@(y:ys) | x < y     = x : m xs ys'
                               | x == y    = x : m xs ys
                               | otherwise = y : m xs' ys
    -- Reinhard Zumkeller, Feb 11 2012, Aug 27 2011
    
  • PARI
    is(n)=issquarefree(n/gcd(n,2)) \\ Charles R Greathouse IV, Nov 05 2017

Formula

a(n) = (Pi^2/7)*n + O(sqrt(n)). - Vladimir Shevelev, Jun 08 2016

Extensions

Edited by N. J. A. Sloane, Sep 13 2006

A274141 Positive numbers divisible by 3^3 or by the square of some other prime.

Original entry on oeis.org

4, 8, 12, 16, 20, 24, 25, 27, 28, 32, 36, 40, 44, 48, 49, 50, 52, 54, 56, 60, 64, 68, 72, 75, 76, 80, 81, 84, 88, 92, 96, 98, 100, 104, 108, 112, 116, 120, 121, 124, 125, 128, 132, 135, 136, 140, 144, 147, 148, 150, 152, 156, 160, 162, 164, 168, 169, 172, 175, 176
Offset: 1

Views

Author

Vladimir Shevelev, Jun 11 2016

Keywords

Comments

Or numbers n>=4 having a divisor k^2>=4 such that n and n/k^2 equal modulo 3.
All positive multiples of 4 are in the sequence.
Or numbers n such that there is a smaller positive number j == n (mod 3) such that sqrt(j*n) is an integer. The smallest such j corresponds to the greatest k; or, the same, j = 3*A007913(n/3), if n is divisible by 3 and otherwise j=A007913(n).
Or complement to the sequence: S, 3*S and 9*S, where S denotes the sequence of the squarefree numbers not divisible by 3.

Crossrefs

Programs

  • Mathematica
    Select[Range[200], (e = IntegerExponent[#, 3]) > 2 || ! SquareFreeQ[#/3^e] &] (* Amiram Eldar, Feb 25 2021 *)
  • PARI
    isok(n) = (((v=valuation(n, 3)) >= 3) || (((m = n/3^v) > 1) && (vecmax((factor(m))[,2]) >=2))); \\ Michel Marcus, Jun 12 2016

Formula

Let A(x) be the number of a(n)<=x. Then A(x)~(1 - 6.5/Pi^2)*x = 0.34141230...*x as x goes to infinity.

A332713 a(n) = Sum_{d|n} phi(d/gcd(d, n/d)).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 7, 8, 10, 11, 12, 13, 14, 15, 13, 17, 16, 19, 20, 21, 22, 23, 21, 22, 26, 22, 28, 29, 30, 31, 24, 33, 34, 35, 32, 37, 38, 39, 35, 41, 42, 43, 44, 40, 46, 47, 39, 44, 44, 51, 52, 53, 44, 55, 49, 57, 58, 59, 60, 61, 62, 56, 46, 65, 66, 67, 68, 69, 70
Offset: 1

Views

Author

Ilya Gutkovskiy, Feb 20 2020

Keywords

Crossrefs

Cf. A000010, A001616, A010052, A046790 (numbers n such that a(n) < n), A055653, A061884, A078779 (fixed points), A332619, A332686, A332712.

Programs

  • Mathematica
    Table[Sum[EulerPhi[d/GCD[d, n/d]], {d, Divisors[n]}], {n, 1, 70}]
    A055653[n_] := Sum[Boole[GCD[d, n/d] == 1] EulerPhi[d], {d, Divisors[n]}]; a[n_] := Sum[Boole[IntegerQ[(n/d)^(1/2)]] A055653[d], {d, Divisors[n]}]; Table[a[n], {n, 1, 70}]
  • PARI
    a(n) = sumdiv(n, d, eulerphi(d/gcd(d, n/d))); \\ Michel Marcus, Feb 20 2020

Formula

Dirichlet g.f.: zeta(s) * zeta(2*s) * zeta(s - 1) * Product_{p prime} (1 - p^(-s) + p^(-2*s) - p^(1 - 2*s)).
a(n) = Sum_{d|n} phi(lcm(d, n/d)/d).
a(n) = Sum_{d|n} A010052(n/d) * A055653(d).
Sum_{k=1..n} a(k) ~ c * Pi^6 * n^2 / 1080, where c = A330523 = Product_{primes p} (1 - 1/p^2 - 1/p^3 + 1/p^4) = 0.5358961538283379998085... - Vaclav Kotesovec, Feb 22 2020
From Richard L. Ollerton, May 10 2021: (Start)
a(n) = Sum_{k=1..n} phi(gcd(n,k)/gcd(gcd(n,k),n/gcd(n,k)))/phi(n/gcd(n,k)).
a(n) = Sum_{k=1..n} phi(n/gcd(n,k)/gcd(gcd(n,k),n/gcd(n,k)))/phi(n/gcd(n,k)).
a(n) = Sum_{k=1..n} phi(lcm(gcd(n,k),n/gcd(n,k))/gcd(n,k))/phi(n/gcd(n,k)).
a(n) = Sum_{k=1..n} phi(lcm(gcd(n,k),n/gcd(n,k))*gcd(n,k)/n)/phi(n/gcd(n,k)).
a(n) = Sum_{k=1..n} A010052(gcd(n,k))*A055653(n/gcd(n,k))/phi(n/gcd(n,k)).
a(n) = Sum_{k=1..n} A010052(n/gcd(n,k))*A055653(gcd(n,k))/phi(n/gcd(n,k)). (End)

A274188 Number n such that there is a smaller positive number j == n (mod 5) such that sqrt(j*n) is an integer.

Original entry on oeis.org

9, 16, 18, 20, 27, 32, 36, 40, 45, 48, 49, 54, 60, 63, 64, 72, 80, 81, 90, 96, 98, 99, 100, 108, 112, 117, 120, 121, 125, 126, 128, 135, 140, 144, 147, 153, 160, 162, 169, 171, 176, 180, 189, 192, 196, 198, 200, 207, 208, 216, 220, 224, 225, 234, 240, 242, 243, 245, 250
Offset: 1

Views

Author

Vladimir Shevelev, Jun 12 2016

Keywords

Comments

Or numbers n>=9 having a divisor t^2 > 1, where t=k/m, 1<= m < k, such that n == n/t^2 (mod 5).
Or positive numbers n such that if n == 0 (mod 5), then it divisible by 5^3 or by the square of some other prime; otherwise n divisible by k^2, such that there is a k_1, 0< k_1
A generalization see in our seqfan list from Jun 13 (correction Jun 14) 2016.

Examples

			9 is member, since 4 == 9 (mod 5) and 4*9 is a square;
32 is member, since 2 == 32 (mod 5) and 2*32 is a square;
45 is member, since 5 == 45 (mod 5) and 5*45 is a square.
		

Crossrefs

Programs

  • Mathematica
    Select[Range@ 250, Function[n, Count[Sqrt[# n] & /@ Select[Range[n - 1], Mod[#, 5] == Mod[n, 5] &], k_ /; IntegerQ@ k] > 0]] (* Michael De Vlieger, Jun 14 2016 *)
  • PARI
    isok(n)=j = n-5; while (j >0, if (issquare(j*n), return (1)); j -= 5;); 0; \\ Michel Marcus, Jun 14 2016
    
  • PARI
    is(n)=!issquarefree(n/if(n%5, if(n%4, 1, 4), 5)) \\ Charles R Greathouse IV, Jun 15 2016

Formula

Let A(x) be the number of a(n)<=x. Then A(x)/x ~ 1 - 149/(20*Pi^2) = 1 - 7.45/Pi^2 = 0.24515718... as x goes to infinity. - Vladimir Shevelev, Jun 15 2016; corrected by Charles R Greathouse IV, Jun 15 2016

Extensions

Corrected and extended by Michel Marcus, Jun 14 2016

A078771 a(n) = A008475(n) - A001414(n).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 2, 3, 0, 0, 0, 0, 0, 0, 8, 0, 3, 0, 0, 0, 0, 0, 2, 15, 0, 18, 0, 0, 0, 0, 22, 0, 0, 0, 3, 0, 0, 0, 2, 0, 0, 0, 0, 3, 0, 0, 8, 35, 15, 0, 0, 0, 18, 0, 2, 0, 0, 0, 0, 0, 0, 3, 52, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 15, 0, 0, 0, 0, 8, 69, 0, 0, 0, 0, 0, 0, 2, 0, 3, 0, 0, 0, 0, 0, 22, 0, 35, 3, 15
Offset: 1

Author

Benoit Cloitre, Jan 11 2003

Keywords

Comments

a(n) is not zero if n is in A046790.

Crossrefs

Programs

Formula

Additive with a(p^e) = p^e - p*e. - Amiram Eldar, May 03 2025

A057111 Numbers n which are a factor of (LPP(n)-1)!, where LPP(n) is the largest prime power factor of n.

Original entry on oeis.org

8, 9, 16, 18, 24, 25, 27, 32, 36, 40, 45, 48, 49, 50, 54, 56, 63, 64, 72, 75, 80, 81, 90, 96, 98, 100, 108, 112, 120, 121, 125, 126, 128, 135, 144, 147, 150, 160, 162, 168, 169, 175, 176, 180, 189, 192, 196, 200, 208, 216, 224, 225, 240, 242, 243, 245, 250, 252
Offset: 1

Author

Henry Bottomley, Aug 08 2000

Keywords

Examples

			18 is in the sequence since 9 is the largest prime power factor of 18 and 18 is a factor of (9-1)!=8!=40320=18*2240.
		

Crossrefs

Cf. A002034, A034699, A057109, A057110. Distinct from A046790 although similar initial terms.

Extensions

More terms from James Sellers, Aug 22 2000

A186129 Numbers that can be partitioned into four parts s, t, u, v such that s+k = t-k = u*k = v/k for some k > 1.

Original entry on oeis.org

18, 27, 32, 36, 45, 48, 50, 54, 63, 64, 72, 75, 80, 81, 90, 96, 98, 99, 100, 108, 112, 117, 125, 126, 128, 135, 144, 147, 150, 153, 160, 162, 171, 175, 176, 180, 189, 192, 196, 198, 200, 207, 208, 216, 224, 225, 234, 240, 242, 243, 245, 250, 252, 256, 261
Offset: 1

Author

Manuel Valdivia, Feb 13 2011

Keywords

Comments

Equivalently, solutions n to a*(b+1)^2 = b*n with a > b >= 2.
The general rule to obtain such a partition is to start with any number b > 1 and one of its multiples a = k*b (k > 1 and a < n) and let s = a-b, t = a+b, u = a/b and v = a*b.
Sequence appears to be a subsequence of A013929, of A046790, and of A072903.

Examples

			18 = 2+6+2+8; for k=2 we have 2+2 = 6-2 = 2*2 = 8/2 = 4, hence 18 is a term.
45 = 8+12+5+20; for k=2 we have 8+2 = 12-2 = 5*2 = 20/2 = 10, hence 45 is a term.
		

References

  • José Estalella, Ciencia Recreativa. Gustavo Gili - Editor. Barcelona, 1918, pp. 5-6.

Crossrefs

Programs

  • Magma
    [ n: n in [1..300] | exists{ b: b in [2..n] | exists{ a: a in [b+1..n div 4] | n*b eq a*(b+1)^2 } } ]; // Klaus Brockhaus, Feb 15 2011

A274240 Numbers n such that there is a smaller positive number j == n (mod 7) such that j*n is a square.

Original entry on oeis.org

16, 25, 28, 32, 36, 48, 50, 56, 63, 64, 72, 75, 80, 81, 84, 96, 100, 108, 112, 121, 125, 126, 128, 140, 144, 150, 160, 162, 168, 169, 175, 176, 180, 189, 192, 196, 200, 208, 216, 224, 225, 240, 242, 243, 250, 252, 256, 272, 275, 280, 288, 289, 300, 304, 308
Offset: 1

Author

Vladimir Shevelev, Jun 15 2016

Keywords

Comments

Or numbers n >= 16 having a divisor t^2 > 1, where t=k/m, 1 <= m < k, such that n == n/t^2 (mod 7).
Or positive numbers n such that if n == 0 (mod 7), then n is divisible by 7^3 or by the square of some other prime; otherwise n is divisible by k^2, such that there is a k_1, 0 < k_1 < k such that k_1^2 == k^2 (mod 7) (or, according to the comment in A130290, n is divisible by some k^2 >= 16).
For a generalization, see the seqfan list from Jun 13 (correction Jun 14) 2016.

Examples

			25 is a member, since 4 == 25 (mod 7) and 4*25 is a square.
32 is a member, since 18 == 32 (mod 7) and 18*32 is a square.
		

Crossrefs

Programs

  • PARI
    is(n) = for(j=1, n-1, if(Mod(j, 7)==n && issquare(j*n), return(1))); return(0) \\ Felix Fröhlich, Jun 15 2016

Extensions

More terms from Felix Fröhlich, Jun 15 2016

A274241 Numbers n such that there is a smaller positive number j == n (mod 11) such that sqrt(j*n) is an integer.

Original entry on oeis.org

36, 44, 49, 64, 72, 81, 88, 98, 99, 100, 108, 128, 132, 144, 147, 162, 169, 176, 180, 192, 196, 198, 200, 216, 220, 225, 243, 245, 252, 256, 264, 275, 288, 289, 294, 297, 300, 308, 320, 324, 338, 343, 352, 360, 361, 384, 392, 396, 400, 405, 432, 440, 441, 448
Offset: 1

Author

Vladimir Shevelev, Jun 15 2016

Keywords

Comments

Or numbers n >= 36 having a divisor t^2 > 1, where t=k/m, 1 <= m < k, such that n == n/t^2 (mod 11).
Or positive numbers n such that if n == 0 (mod 11), then n is divisible by 11^3 or by the square of some other prime; otherwise n is divisible by k^2, such that there is a k_1, 0 < k_1 < k with k_1^2 == k^2 (mod 11) (or, according to the comment in A130290, n is divisible by some k^2 >= 36).
For a generalization, see the Sequence Fans mailing list for Jun 13 2016 (correction Jun 14 2016).
From David A. Corneth, Jun 26 2016: (Start)
If k is a term then m * k is a term for m > 0. Hence closed under multiplication. For k > 11, k^2 is in the sequence. So k^t is as well for t > 2.
Summarizing, k is a term iff
- k is of the form k^2 for floor(11/2) < k except k = 11.
- k is of the form 11 * p^2 for p < floor(11/2)
- of the form k * t for k of one of the forms above and integer t > 0. (End)

Examples

			49 is member, since 16 == 49 (mod 11) and 16*49 is a square.
108 is member, since 75 == 108 (mod 11) and 75*108 is a square.
		

Crossrefs

Programs

  • Mathematica
    Select[Range@500, Function[n, AnyTrue[Range[n - 1], And[Mod[#, 11] == Mod[n, 11], IntegerQ@ Sqrt[# n]] &]]] (* Michael De Vlieger, Jun 23 2016, Version 10 *)
  • PARI
    is(n) = for(j=1, n-1, if(Mod(j, 11)==n && issquare(j*n), return(1))); return(0) \\ Felix Fröhlich, Jun 15 2016
    
  • PARI
    is(n)=my(f=factor(n)); f[,2]=f[,2]%2; t=prod(i=1,matsize(f)[1], f[i,1] ^ f[i,2]); for(i=1,sqrtint((n-1)\t), if(Mod(t*i^2, 11)==n,return(1))); 0 \\ David A. Corneth, Jun 26 2016

Extensions

More terms from Felix Fröhlich, Jun 15 2016
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