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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A115910 Numbers k such that phi(k)*k is a triangular number.

Original entry on oeis.org

1, 3, 15, 84, 217, 255, 1710, 2967, 5363, 11242, 24066, 27370, 29697, 29953, 60977, 65535, 69324, 103257, 159761, 209305, 228333, 446516, 598559, 615410, 691410, 1231305, 1238358, 1365175, 1467732, 1841154, 1966220, 2081070, 2562370
Offset: 1

Views

Author

Giovanni Resta, Feb 06 2006

Keywords

Examples

			phi(24066)*24066 = 164611440 = T(18144).
		

Crossrefs

Programs

  • Mathematica
    Select[Range[2.6*10^6],OddQ[Sqrt[8#*EulerPhi[#]+1]]&] (* Harvey P. Dale, Jul 11 2017 *)
  • PARI
    isok(n) = ispolygonal(n*eulerphi(n), 3); \\ Michel Marcus, Jan 25 2014

A189393 a(n) = phi(n^4).

Original entry on oeis.org

1, 8, 54, 128, 500, 432, 2058, 2048, 4374, 4000, 13310, 6912, 26364, 16464, 27000, 32768, 78608, 34992, 123462, 64000, 111132, 106480, 267674, 110592, 312500, 210912, 354294, 263424, 682892, 216000, 893730, 524288, 718740, 628864, 1029000, 559872
Offset: 1

Views

Author

Vincenzo Librandi, Apr 21 2011

Keywords

Crossrefs

Cf. A002618 (phi(n^2)), A053191 (phi(n^3)), A238533 (phi(n^5)), A239442 (phi(n^7)), A239443 (phi(n^9)).

Programs

  • Magma
    [ n^3*EulerPhi(n) : n in [1..100] ]
    
  • Mathematica
    EulerPhi[Range[100]^4] (* T. D. Noe, Dec 27 2011 *)
  • PARI
    vector(66,n,n^3*eulerphi(n))  /* Joerg Arndt, Apr 22 2011 */

Formula

a(n) = n^3*phi(n).
Dirichlet g.f.: zeta(s - 4) / zeta(s - 3). The n-th term of the Dirichlet inverse is n^3 * A023900(n) = (-1)^omega(n) * a(n) / A003557(n), where omega=A001221. - Álvar Ibeas, Nov 24 2017
Sum_{k=1..n} a(k) ~ 6*n^5 / (5*Pi^2). - Vaclav Kotesovec, Feb 02 2019
Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + p/(p^5 - p^4 - p + 1)) = 1.15762316629211803144... - Amiram Eldar, Dec 06 2020

A374456 The Euler phi function values of the exponentially odd numbers (A268335).

Original entry on oeis.org

1, 1, 2, 4, 2, 6, 4, 4, 10, 12, 6, 8, 16, 18, 12, 10, 22, 8, 12, 18, 28, 8, 30, 16, 20, 16, 24, 36, 18, 24, 16, 40, 12, 42, 22, 46, 32, 52, 18, 40, 24, 36, 28, 58, 60, 30, 48, 20, 66, 44, 24, 70, 72, 36, 60, 24, 78, 40, 82, 64, 42, 56, 40, 88, 72, 60, 46, 72, 32, 96
Offset: 1

Views

Author

Amiram Eldar, Jul 09 2024

Keywords

Crossrefs

Similar sequences related to phi: A002618, A049200, A323333, A358039.
Similar sequences related to exponentially odd numbers: A366438, A366439, A366534, A366535, A367417, A368711, A374457.

Programs

  • Mathematica
    f[p_, e_] := If[OddQ[e], (p-1) * p^(e-1), 0]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; Select[Array[s, 100], # > 0 &]
  • PARI
    s(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i, 2] % 2, (f[i, 1]-1) * f[i, 1]^(f[i, 2] - 1), 0));}
    lista(kmax) = {my(s1); for(k = 1, kmax, s1 = s(k); if(s1 > 0, print1(s1, ", ")));}

Formula

a(n) = A000010(A268335(n)).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = A307868 / A065463^2 = 0.95051132596733153581... .

A127466 Triangle read by rows: A054525 * A127481 as infinite lower triangular matrices.

Original entry on oeis.org

1, 2, 2, 3, 0, 6, 4, 4, 0, 8, 5, 0, 0, 0, 20, 6, 6, 12, 0, 0, 12, 7, 0, 0, 0, 0, 0, 42, 8, 8, 0, 16, 0, 0, 0, 32, 9, 0, 18, 0, 0, 0, 0, 0, 54, 10, 10, 0, 0, 40, 0, 0, 0, 0, 40
Offset: 1

Views

Author

Gary W. Adamson, Jan 15 2007

Keywords

Comments

Mobius transform of A127481.

Examples

			First few rows of the triangle are:
1;
2, 2;
3, 0, 6;
4, 4, 0, 8;
5, 0, 0, 0, 20;
6, 6, 12, 0, 0, 12;
7, 0, 0, 0, 0, 0, 42;
8, 8, 0, 16, 0, 0, 0, 32;
...
		

Crossrefs

Programs

Formula

Sum_{k=1..n} T(n,k) = n^2.
T(n,n) = A002618(n) = n*phi(n).

Extensions

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Aug 23 2007

A127649 A127648 * A054523 as infinite lower triangular matrices.

Original entry on oeis.org

1, 2, 2, 6, 0, 3, 8, 4, 0, 4, 20, 0, 0, 0, 5, 12, 12, 6, 0, 0, 6, 42, 0, 0, 0, 0, 0, 7, 32, 16, 0, 8, 0, 0, 0, 8, 54, 0, 18, 0, 0, 0, 0, 0, 9, 40, 40, 0, 0, 10, 0, 0, 0, 0, 10, 110, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 48, 24, 24, 24, 0, 12, 0, 0, 0, 0, 0, 12, 156, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 13, 84
Offset: 1

Views

Author

Gary W. Adamson, Jan 22 2007

Keywords

Comments

Natural number transform of A054523.
Row sums = n^2, left column = A002618

Examples

			First few rows of the triangle are:
1;
2, 2;
6, 0, 3;
8, 4, 0, 4;
20, 0, 0, 0, 5;
12, 12, 6, 0, 0, 6;
42, 0, 0, 0, 0, 0, 7;
...
		

Crossrefs

Programs

  • Maple
    A054523 := proc(n,k) if n mod k = 0 then numtheory[phi](n/k) ; else 0 ; fi ; end: A127649 := proc(n,k) A054523(n,k)*n ; end: for n from 1 to 20 do for k from 1 to n do printf("%d,",A127649(n,k)) ; od: od: # R. J. Mathar, Nov 01 2007

Formula

T(n,k)=n*A054523(n,k). - R. J. Mathar, Nov 01 2007
T(n,k) = Sum_{y=1..n} Sum_{x=1..n} [GCD(f(x,y), n) = k], where f(x,y) = x - y. - Mats Granvik, Oct 08 2023

Extensions

More terms from R. J. Mathar, Nov 01 2007

A194507 a(n) = y is the unique solution to y*phi(y) = A082473(n).

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 8, 10, 7, 12, 9, 14, 18, 11, 15, 16, 13, 20, 24, 22, 30, 21, 17, 26, 28, 19, 36, 27, 25, 42, 23, 32, 34, 40, 33, 38, 48, 29, 35, 44, 31, 39, 60, 54, 50, 46, 45, 52, 66, 37, 56, 58, 51, 41, 70, 72, 43, 62, 78, 84, 64, 57, 49, 90, 47, 68, 55, 63, 80
Offset: 1

Views

Author

Franz Vrabec, Aug 27 2011

Keywords

Comments

The permutation which rearranges the terms of A002618 into ascending order. - Antti Karttunen, Sep 28 2019

Examples

			a(6) = 5 because 5*phi(5) = 20 = A082473(6).
		

Crossrefs

Cf. A002618, A082473, A327173 (inverse permutation).
Nonzero terms in A327172, in the order of appearance.

Programs

  • Mathematica
    Block[{nn = 3000, s, t}, s = Array[EulerPhi[#] (Sqrt@ # /. (c_: 1) a_^(b_: 0) :> (c a^b)^2) &, nn]; t = TakeWhile[Union@ s, # <= nn &]; Map[Block[{y = 1}, While[y EulerPhi@ y != #, y++]; y] &, t]] (* Michael De Vlieger, Sep 29 2019, after Bill Gosper at A007913 *)
  • PARI
    up_to = 105;
    A327172(n) = { fordiv(n,d,if(eulerphi(d)*d == n, return(d))); (0); };
    A194507list(up_to) = { my(v=vector(up_to),k=1); for(n=1,oo,if((v[k]=A327172(n))>0,k++); if(k>up_to, return(v))); };
    v194507 = A194507list(up_to);
    A194507(n) = v194507[n]; \\ Antti Karttunen, Sep 28 2019

Formula

From Antti Karttunen, Sep 28 2019: (Start)
a(n) = A327172(A082473(n)).
A002618(a(n)) = A082473(n).
(End)

A262747 Number of ordered ways to write n as x^2 + y^2 + phi(z^2) with 0 <= x <= y, z > 0, 2 | x*y*z, and phi(k^2) < n for all 0 < k < z, where phi(.) is Euler's totient function given by A000010.

Original entry on oeis.org

1, 2, 1, 1, 1, 3, 2, 1, 2, 4, 2, 2, 1, 2, 2, 1, 2, 3, 2, 3, 4, 3, 1, 2, 3, 4, 2, 4, 2, 2, 3, 1, 4, 2, 1, 2, 5, 4, 1, 2, 2, 6, 3, 2, 4, 4, 3, 3, 4, 3, 3, 5, 3, 3, 4, 2, 6, 7, 3, 4, 4, 5, 2, 2, 5, 6, 6, 1, 5, 4, 4, 4, 6, 6, 1, 4, 4, 2, 4, 3, 5, 6, 3, 4, 5, 5, 4, 2, 2, 6, 5, 4, 6, 3, 3, 1, 5, 3, 2, 3
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 30 2015

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 4, 5, 8, 13, 16, 23, 32, 35, 39, 68, 75, 96, 215, 219, 243, 363, 471, 723, 759, 923, 1443, 1551, 1839, 2739, 2883.
It is easy to see that all the numbers phi(n^2) = n*phi(n) (n = 1,2,3,...) are pairwise distinct. We have verified that a(n) > 0 for all n = 1,...,10^7.
See also A262311 for a related conjecture.

Examples

			a(5) = 1 since 5 = 0^2 + 2^2 + phi(1^2).
a(8) = 1 since 8 = 0^2 + 0^2 + phi(4^2) with 0*0*4 even, and phi(1^2) = 1, phi(2^2) = 2, phi(3^2) = 6 all smaller than 8.
a(13) = 1 since 13 = 1^2 + 2^2 + phi(4^2).
a(32) = 1 since 32 = 2^2 + 4^2 + phi(6^2).
a(68) = 1 since 68 = 0^2 + 6^2 + phi(8^2).
a(96) = 1 since 96 = 0^2 + 8^2 + phi(8^2).
a(363) = 1 since 363 = 0^2 + 19^2 + phi(2^2).
a(471) = 1 since 471 = 0^2 + 19^2 + phi(11^2).
a(723) = 1 since 723 = 17^2 + 18^2 + phi(11^2).
a(759) = 1 since 759 = 9^2 + 26^2 + phi(2^2).
a(923) = 1 since 923 = 16^2 + 25^2 + phi(7^2).
a(1443) = 1 since 1443 = 19^2 + 24^2 +phi(23^2).
a(1551) = 1 since 1551 = 18^2 + 35^2 + phi(2^2).
a(1839) = 1 since 1839 = 3^2 + 30^2 + phi(31^2).
a(2739) = 1 since 2739 = 1^2 + 24^2 + phi(47^2).
a(2883) = 1 since 2883 = 21^2 + 44^2 + phi(23^2).
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=EulerPhi[n^2]
    SQ[n_]:=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[f[x]>n,Goto[aa]];Do[If[SQ[n-f[x]-y^2]&&(Mod[x*y,2]==0||Mod[Sqrt[n-f[x]-y^2],2]==0),r=r+1],{y,0,Sqrt[(n-f[x])/2]}];Continue,{x,1,n}];Label[aa];Print[n," ",r];Continue,{n,1,100}]

A262781 Number of ordered ways to write n as x^2 + phi(y^2) + phi(z^2) (x >= 0 and 0 < y <= z) with y or z prime, where phi(.) is Euler's totient function given by A000010.

Original entry on oeis.org

0, 0, 1, 2, 1, 0, 2, 3, 1, 1, 2, 3, 2, 3, 2, 2, 1, 3, 3, 1, 2, 3, 4, 1, 1, 3, 2, 3, 2, 4, 1, 3, 2, 2, 3, 1, 3, 3, 4, 2, 2, 3, 5, 5, 1, 4, 4, 4, 2, 6, 4, 4, 4, 6, 3, 4, 5, 4, 5, 4, 4, 3, 6, 4, 2, 3, 3, 5, 4, 4, 4, 3, 1, 4, 5, 4, 3, 6, 3, 1, 2, 3, 4, 4, 5, 5, 3, 3, 2, 8, 5, 3, 4, 2, 4, 4, 2, 3, 7, 2
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 01 2015

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 6, and a(n) = 1 only for n = 3, 5, 9, 10, 17, 20, 24, 25, 31, 36, 45, 73, 80, 101, 136, 145, 388, 649.
(ii) For any integer n > 4, we can write 2*n as phi(p^2) + phi(x^2) + phi(y^2) with p prime and p <= x <= y.
See also A262311 for a similar conjecture.

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
    f[n_]:=EulerPhi[n^2]
    Do[r=0;Do[If[f[z]>n,Goto[aa]];Do[If[SQ[n-f[z]-f[y]]&&(PrimeQ[y]||PrimeQ[z]),r=r+1],{y,1,z}];Label[aa];Continue,{z,1,n}];Print[n," ",r];Continue,{n,1,100}]

Formula

a(3) = 1 since 3 = 0^2 + phi(1^2) + phi(2^2) with 2 prime.
a(5) = 1 since 5 = 1^2 + phi(2^2) + phi(2^2) with 2 prime.
a(9) = 1 since 9 = 1^2 + phi(2^2) + phi(3^2) with 2 and 3 both prime.
a(10) = 1 since 10 = 0^2 + phi(2^2) + phi(4^2) with 2 prime.
a(17) = 1 since 17 = 3^2 + phi(2^2) + phi(3^2) with 2 and 3 both prime.
a(20) = 1 since 20 = 4^2 + phi(2^2) + phi(2^2) with 2 prime.
a(24) = 1 since 24 = 4^2 + phi(2^2) + phi(3^2) with 2 and 3 both prime.
a(25) = 1 since 25 = 2^2 + phi(1^2) + phi(5^2) with 5 prime.
a(31) = 1 since 31 = 3^2 + phi(2^2) + phi(5^2) with 2 and 5 both prime.
a(36) = 1 since 36 = 2^2 + phi(5^2) + phi(6^2) with 5 prime.
a(45) = 1 since 45 = 1^2 + phi(2^2) + phi(7^2) with 2 and 7 both prime.
a(73) = 1 since 73 = 5^2 + phi(3^2) + phi(7^2) with 3 and 7 both prime.
a(80) = 1 since 80 = 6^2 + phi(2^2) + phi(7^2) with 2 and 7 both prime.
a(101) = 1 since 101 = 7^2 + phi(5^2) + phi(8^2) with 5 prime.
a(136) = 1 since 136 = 5^2 + phi(1^2) + phi(11^2) with 11 prime.
a(145) = 1 since 145 = 7^2 + phi(7^2) + phi(9^2) with 7 prime.
a(388) = 1 since 388 = 2^2 + phi(7^2) + phi(19^2) with 7 and 19 both prime.
a(649) = 1 since 649 = 11^2 + phi(7^2) + phi(27^2) with 7 prime.

A327172 If there is a divisor d of n such that phi(d)*d = n, then a(n) = d, otherwise a(n) = 0.

Original entry on oeis.org

1, 2, 0, 0, 0, 3, 0, 4, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0, 0, 10, 0, 7, 0, 0, 0, 0, 0, 12, 0, 0, 0, 0, 0, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 14, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 18, 0, 11, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15
Offset: 1

Views

Author

Antti Karttunen, Sep 28 2019

Keywords

Comments

If such a divisor exists, it is necessarily unique. See Franz Vrabec's Dec 12 2012 comment in A002618.
Each natural number n > 0 occurs exactly once in this sequence, at position A002618(n).

Crossrefs

Left inverse of A002618.
Cf. A000010.
Cf. A082473 (the indices of nonzero terms), A194507 (nonzero terms in the order of appearance).

Programs

  • Mathematica
    With[{s = EulerPhi /@ Range@ 120}, Table[DivisorSum[n, # &, # s[[#]] == n &], {n, Length@ s}]] (* Michael De Vlieger, Sep 29 2019 *)
  • PARI
    A327172(n) = { fordiv(n,d,if(eulerphi(d)*d == n, return(d))); (0); };

Formula

a(A002618(n)) = n.
a(A082473(n)) = A194507(n).

A174857 The minimum distance k > 0 such that A020639(n+k) = A020639(n).

Original entry on oeis.org

2, 6, 2, 20, 2, 42, 2, 6, 2, 110, 2, 156, 2, 6, 2, 272, 2, 342, 2, 6, 2, 506, 2, 10, 2, 6, 2, 812, 2, 930, 2, 6, 2, 20, 2, 1332, 2, 6, 2, 1640, 2, 1806, 2, 6, 2, 2162, 2, 28, 2, 6, 2, 2756, 2, 10, 2, 6, 2, 3422, 2, 3660, 2, 6, 2, 20, 2, 4422, 2, 6, 2, 4970, 2, 5256, 2, 6, 2, 14, 2, 6162
Offset: 2

Views

Author

Vladimir Shevelev, Mar 31 2010

Keywords

Comments

The sequence has the same records as A002618.

Crossrefs

Programs

  • Maple
    A174857 := proc(n) local k,aref ; aref := A020639(n) ; for k from 1 do if A020639(n+k) = aref then return k; end if; end do: end proc:
    seq(A174857(n),n=2..80) ; # R. J. Mathar, Dec 07 2010
  • Mathematica
    Block[{s = Array[FactorInteger[#][[1, 1]] &, 10^4]}, Array[If[EvenQ[#], 2, Block[{k = 1, n = s[[#]]}, While[n != s[[# + k]], k++; If[# + k > Length[s], AppendTo[s, FactorInteger[# + k][[1, 1]] ]] ]; k]] &, 78, 2]] (* Michael De Vlieger, Apr 06 2021 *)
  • PARI
    A020639(n) = if(1==n, n, factor(n)[1, 1]);
    A174857(n) = if(isprime(n), (n-1)*n, my(spf=A020639(n)); for(k=1,oo,if(A020639(n+k)==spf,return(k)))); \\ Antti Karttunen, Apr 06 2021

Formula

If n is even, then a(n) = 2.
If n = 3k and A020639(k) >= 3, then a(n) = 6.
If n is prime, then a(n) = A036689(n).
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