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 1-4 of 4 results.

A112009 Numbers n with even length such that phi(n)=d_1^d_2*d_3^d_4*...* d_(k-1)^d_k where d_1 d_2 ... d_k is the decimal expansion of n.

Original entry on oeis.org

113724, 116680, 126620, 176453, 236520, 12146841, 12514635, 13334445, 13469331, 13813728, 16473510, 18259344, 20116537, 20119347, 21324832, 23336066, 27923616, 30352728, 34425425, 35424571, 36311184, 37837170, 39171345, 43362816, 45429360, 45449216, 45916416, 46544032, 50713684, 50816880, 61642672, 62193744, 62226711, 62263890, 62288272, 64245272, 64808352, 64832560, 66707233, 66807126, 66827180, 81913446, 84943040
Offset: 1

Views

Author

Farideh Firoozbakht, Aug 26 2005

Keywords

Examples

			27923616 is in the sequence because phi(27923616)=2^7*9^2*3^6*1^6.
11600069 and 23635500 are not members, since 0^0 is undefined.
		

Crossrefs

Programs

  • Mathematica
    Do[h = IntegerDigits[n]; k = Length[h]; If[EvenQ[k] && Select[ Range[k/2], h[[2#-1]] == 0 &] == {} && EulerPhi[n]==Product[ h[[2j-1]]^h[[2j]], {j, k/2}], Print[n]], {n, 30000000}]

Extensions

Edited by N. J. A. Sloane, Apr 02 2009
More terms from Max Alekseyev, Oct 16 2012

A112010 Numbers m with even length such that phi(m)=phi(d_1^d_2*d_3^d_4*...* d_(k-1)^d_k) where d_1 d_2 ... d_k is the decimal expansion of m.

Original entry on oeis.org

24, 1064, 2592, 6520, 106434, 145166, 237165, 262535, 372780, 491520, 531765, 546410, 566250, 636352, 12716544, 12806910, 13666320, 15116832, 15408692, 17473715, 21645616, 23473515, 23726640, 23728264, 26722436, 26757024, 27933192, 30537364, 30869280, 32118177, 33452293, 34114338, 39602752, 42262365, 44373490
Offset: 1

Views

Author

Farideh Firoozbakht, Aug 26 2005

Keywords

Examples

			33452293 is in the sequence because phi(33452293)=phi(3^3*4^5*2^2*9^3).
		

Crossrefs

Programs

  • Mathematica
    Do[h = IntegerDigits[n]; k = Length[h]; If[EvenQ[k] && Select[ Range[k/2], h[[2#-1]] == 0 &] == {} && EulerPhi[n]==EulerPhi [Product[h[[2j-1]]^h[[2j]], {j, k/2}]], Print[n]], {n, 31000000}]
    epQ[n_]:=Module[{idn=IntegerDigits[n]},EvenQ[Length[idn]]&& FreeQ[ Take[ idn, {1,-1,2}],0] && EulerPhi[n] == EulerPhi[Times@@(#[[1]]^#[[2]]&/@ Partition[ idn,2])]]; Join[Select[Range[10,99],epQ],Select[Range[ 1000,9999], epQ], Select[Range[100000,999999],epQ], Select[Range[ 10000000, 44999999], epQ]] (* Harvey P. Dale, Feb 24 2016 *)

Extensions

More terms from Max Alekseyev, Oct 16 2012

A112011 Numbers n with even length such that phi(n)=phi(d_1^d_2)*phi(d_3^d_4) *...*phi(d_(k-1)^d_k) where d_1 d_2 ... d_k is the decimal expansion of n.

Original entry on oeis.org

24, 1064, 2592, 6520, 9234, 145166, 245344, 296480, 372780, 491520, 531765, 546410, 566250, 664062, 12806910, 12826710, 14466530, 15408692, 15621268, 17473715, 19946352, 22297520, 23256720, 30537364, 30869280, 32118177
Offset: 1

Views

Author

Farideh Firoozbakht, Aug 26 2005

Keywords

Comments

For the third term we have the relation 2592=2^5*9^2. So phi(2592)=phi(2^5*9^2)=phi(2^5)*phi(9^2).

Examples

			39602752 is in the sequence because phi(39602752)=
phi(3^9)*phi(6^0)*phi(2^7)*phi(5^2).
		

Crossrefs

Programs

  • Mathematica
    Do[h = IntegerDigits[n]; k = Length[h]; If[EvenQ[k] && Select[ Range[k/2], h[[2#-1]] == 0 &] == {} && EulerPhi[n]== Product[EulerPhi[h[[2j-1]]^h[[2j]]], {j, k/2}], Print[n]], {n, 35000000}]

A099650 Solutions to x+phi(x) = sigma(x)/2.

Original entry on oeis.org

456, 828, 7584, 33462, 1596048, 1964544, 19800384, 26211264, 31451136, 106805184, 156868224, 316113024, 502349274, 503291904, 1557940992, 8024671392, 8052965376, 11697091872, 22149447168, 87877745664, 443520605184, 626058783744
Offset: 1

Views

Author

Labos Elemer, Nov 05 2004

Keywords

Comments

If 5*2^n-1 is prime then m=3*2^(n+1)*(5*2^n-1) is in the sequence because m+phi(m)=2^(n+1)*3*(5*2^n-1)+2^(n+1)*(5*2^n-2)=2^(n+1) *(20*2^n-5)=2^(n+1)*5*(2^(n+2)-1)=1/2*4*(2^(n+2)-1)*(5*2^n)= 1/2*sigma(3)*sigma(2^(n+1))*sigma(5*2^n-1)=1/2*sigma(3*2^(n+1) *(5*2^n-1))=1/2*sigma(m). So 3*2^(A001770+1)*(5*2^A001770-1) is a subsequence of this sequence. A110084 is this subsequence. Next term is greater than 10^8. - Farideh Firoozbakht, Aug 04 2005
a(23) > 10^12. - Donovan Johnson, Feb 29 2012

Examples

			n=456: phi(456) = 144, sigma(456) = 1200.
		

Crossrefs

Programs

  • Mathematica
    Do[If[DivisorSigma[1, m] == 2m + 2 EulerPhi[m], Print[m]], {m, 100000000}] (Firoozbakht)

Extensions

Two more terms from Farideh Firoozbakht, Aug 04 2005
a(10)-a(22) from Donovan Johnson, Feb 29 2012
Showing 1-4 of 4 results.