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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A367320 Carmichael numbers k such that (k-1)/lambda(k) > (m-1)/lambda(m) for all Carmichael numbers m < k, where lambda is the Carmichael lambda function (A002322).

Original entry on oeis.org

561, 1105, 1729, 29341, 41041, 63973, 172081, 825265, 852841, 1773289, 5310721, 9890881, 12945745, 18162001, 31146661, 93869665, 133205761, 266003101, 417241045, 496050841, 509033161, 1836304561, 1932608161, 2414829781, 4579461601, 9799928965, 11624584621, 12452890681
Offset: 1

Views

Author

Amiram Eldar, Nov 14 2023

Keywords

Crossrefs

Subsequence of A002997.

Programs

  • Mathematica
    seq[kmax_] := Module[{s = {}, r, rm = 0, lam}, Do[If[CompositeQ[k], lam = CarmichaelLambda[k]; If[Mod[k, lam] == 1, r = (k - 1)/lam; If[r > rm, rm = r; AppendTo[s, k]]]], {k, 9, kmax, 2}]; s]; seq[10^6]
  • PARI
    lista(kmax) = {my(r, rm = 0, lam); forcomposite(k = 4, kmax, if(k % 2, lam = lcm(znstar(k)[2]); if(k % lam == 1, r = (k-1)/lam; if(r > rm, rm = r; print1(k, ", ")))));}

A258839 Carmichael numbers whose prime factors all have the form p=1+x^2+y^2 for some x,y in Z.

Original entry on oeis.org

561, 162401, 410041, 488881, 656601, 2433601, 36765901, 109393201, 171454321, 176659201, 178837201, 189941761, 221884001, 288120421, 600892993, 618068881, 721244161, 931694401, 985052881, 1183104001, 1828377001, 1848112761, 1943951041, 2361232477, 2438403661
Offset: 1

Views

Author

Michel Marcus, Jun 12 2015

Keywords

Comments

Banks & Freiberg show that this sequence is infinite.

Crossrefs

Cf. A002997 (Carmichael numbers), A079545 (primes of the form x^2 + y^2 + 1).

Programs

  • PARI
    has(n)=for(x=sqrtint(n\2),sqrtint(n-1), if(issquare(n-x^2-1), return(1)));0
    Korselt(n,f=factor(n))=for(i=1,#f~,if(f[i, 2]>1||(n-1)%(f[i, 1]-1),return(0))); 1
    is(n)=my(f); if(n%2==0||isprime(n)||!Korselt(n,f=factor(n))||n<9, return(0)); for(i=1,#f~,if(!has(f[i,1]), return(0))); 1 \\ Charles R Greathouse IV, Jun 12 2015

A267462 Carmichael numbers that are not of the form x^2 + y^2 + z^2 where x, y and z are integers.

Original entry on oeis.org

8911, 1152271, 10267951, 14913991, 64377991, 67902031, 139952671, 178482151, 612816751, 652969351, 743404663, 2000436751, 2560600351, 3102234751, 3215031751, 5615659951, 5883081751, 7773873751, 8863329511, 9462932431, 10501586767, 11335174831, 12191597551, 13946829751, 16157879263, 21046047751
Offset: 1

Views

Author

Altug Alkan, Jan 15 2016

Keywords

Comments

Intersection of A002997 and A004215.
Carmichael numbers that are the sum of 4 but no fewer nonzero squares.
Carmichael numbers of the form 8*k + 7.
Subsequence of A185321.
Carmichael numbers of the form x^2 + y^2 + z^2 where x, y and z are integers are 561, 1105, 1729, 2465, 2821, 6601, 10585, 15841, 29341, 41041, 46657, 52633, 62745, 63973, 75361, 101101, 115921, 126217, 162401, 172081, 188461, 252601, 278545, 294409, 314821, 334153, 340561, 399001, 410041, 449065, 488881, 512461, 530881, 552721, ...

Examples

			Carmichael number 561 is not a term of this sequence because 561 = 2^2 + 14^2 + 19^2.
Carmichael number 8911 is a term because there is no integer values of x, y and z for the equation 8911 = x^2 + y^2 + z^2.
Carmichael number 10585 is not a term because 10585 = 0^2 + 37^2 + 96^2.
		

Crossrefs

Programs

  • Maple
    filter:= proc(n)
      local q;
      if isprime(n) then return false fi;
      if 2 &^ (n-1) mod n <> 1 then return false fi;
      for q in ifactors(n)[2] do
        if q[2] > 1 or (n-1) mod (q[1]-1) <> 0 then return false fi
        od;
        true
    end proc:
    select(filter, [seq(8*k+7, k=0..10^7)]); # Robert Israel, Jan 18 2016
  • Mathematica
    Select[8*Range[1,8000000]+7, CompositeQ[#] && Divisible[#-1, CarmichaelLambda[#]] &] (* Amiram Eldar, Jun 26 2019 *)
  • PARI
    isA004215(n) = { my(fouri, j) ; fouri=1 ; while( n >=7*fouri, if( n % fouri ==0, j= n/fouri -7 ; if( j % 8 ==0, return(1) ) ; ) ; fouri *= 4 ; ) ; return(0) ; } { for(n=1, 400, if(isA004215(n), print1(n, ", ") ; ) ; ) ; }
    isA002997(n) = { my(f); bittest(n, 0) && !for(i=1, #f=factor(n)~, (f[2, i]==1 && n%(f[1, i]-1)==1)||return) && #f>1 }
    for(n=0, 1e10, if(isA002997(n) && isA004215(n), print1(n, ", ")));
    
  • PARI
    isA002997(n) = { my(f); bittest(n, 0) && !for(i=1, #f=factor(n)~, (f[2, i]==1 && n%(f[1, i]-1)==1)||return) && #f>1 }
    for(n=0, 1e10, if(isA002997(k=8*n+7), print1(k, ", ")));

A277366 Composite numbers k such that phi(k)*lambda(k) divides (k-1)^2, where phi(k) = A000010(k) and lambda(k) = A002322(k).

Original entry on oeis.org

1729, 670033, 6840001, 83099521, 193708801, 321197185, 367804801, 484662529, 1752710401, 2320690177, 5064928705, 12820178449, 32220147601, 257124585601, 270177600001, 301036080385, 7043394657601, 13237329899521, 14276860416001, 85661522006401, 119377939968001
Offset: 1

Views

Author

Thomas Ordowski, Oct 11 2016

Keywords

Comments

Are there infinitely many such numbers?
Such k must be a Carmichael number since phi(k)*lambda(k) = m*lambda(k)^2 for some integer m. - Nathan McNew, Oct 11 2016

Crossrefs

Subsequence of A002997 and of A173703.

Programs

  • Mathematica
    Select[Range[10^8], CompositeQ[#] && Divisible[(# - 1)^2, EulerPhi[#] * CarmichaelLambda[#]] &] (* Amiram Eldar, Feb 02 2019 *)
  • PARI
    lista(nn) = forcomposite(n=4, nn, if (((n-1)^2 % (eulerphi(n)*lcm(znstar(n)[2]))) == 0, print1(n, ", "));); \\ Michel Marcus, Oct 11 2016
    
  • PARI
    is(n,f=factor(n))=(n-1)^2%(eulerphi(f)*lcm(znstar(f)[2])) == 0 && !isprime(n) && n>1 \\ Charles R Greathouse IV, Oct 11 2016

Extensions

a(2)-a(3) from Michel Marcus, Oct 11 2016
a(4)-a(8) from Charles R Greathouse IV, Oct 11 2016
a(9)-a(13) from David A. Corneth, Oct 11 2016
More terms from Amiram Eldar, Feb 02 2019

A277720 Numbers k > 2 such that lambda(k)^2 divides k-1, where lambda(k) = A002322(k).

Original entry on oeis.org

2320690177, 17069520863233, 42182344790209, 65465530560001, 3432376805760001, 13322002122777601, 20388795375960001, 129009714848870401, 580007888606160001, 1096591987029196801, 3029756968906340401, 5806765663003468801, 6213994663149504001, 6367205158826803201, 7802569551798000001, 10319507991273499201
Offset: 1

Views

Author

Keywords

Comments

Squarefree numbers k > 2 such that (p-1)^2 | k-1 for every prime p|k.
For the first five terms, lambda(k)^2 | phi(k). - Thomas Ordowski, Apr 11 2017

Crossrefs

Subsequence of A002997 and of A277389.
Cf. A002322.

Programs

  • PARI
    isok(n) = (n % lcm(znstar(n)[2])^2)  == 1; \\ Michel Marcus, Apr 22 2017

Extensions

a(7)-a(16) from Max Alekseyev, Apr 23 2017

A290281 Numbers k such that (k-1) mod phi(k) = lambda(k), where phi = A000010 and lambda = A002322.

Original entry on oeis.org

6601, 11972017, 34657141, 67902031, 139952671, 258634741, 2000436751, 8801128801, 9116583841, 9462932431, 38069223721, 326170416001, 359316634951, 1860929324101, 2022188518351, 2283475947391, 2648686458601, 2697891108151, 4513362899761, 5020030521001, 5472940991761, 6163867710001, 7507903975951, 19288340548471
Offset: 1

Views

Author

Robert Israel and Thomas Ordowski, Jul 25 2017

Keywords

Comments

Numbers k such that A215486(k) = A002322(k).
Subsequence of the Carmichael numbers (A002997).
Composite numbers k such that (k-1) == lambda(k) (mod phi(k)).
Composite numbers k such that A277127(k) == 1 (mod A000010(k)).
Problem: are there infinitely many such numbers?
Conjecture: these are numbers k such that phi(k) + lambda(k) = k - 1. Checked up to 2^64. - Amiram Eldar and Thomas Ordowski, Dec 06 2019

Crossrefs

Subsequence of A264012.

Programs

  • Maple
    # Using data files for A002997
    count:= 0:
    for cfile in ["carmichael-16","carmichael17","carmichael18"] do
    do
        S:= readline(cfile);
        if S = 0 then break fi;
        L:= map(parse, StringTools:-Split(S));
        n:= L[1]; pm:= map(`-`,L[2..-1],1);
        phin:= convert(pm,`*`);
        lambdan:= ilcm(op(pm));
        if n-1 - lambdan mod phin = 0 then
          count:= count+1; A[count]:= n;
        fi
    od:
       fclose(cfile);
    od:
    seq(A[i],i=1..count); # Robert Israel, Jul 26 2017
  • Mathematica
    Select[Range[10^8], Divisible[# - 1, (lam = CarmichaelLambda[#])] && Mod[# - 1, EulerPhi[#]] == lam &] (* Amiram Eldar, Dec 06 2019 *)

A290497 Carmichael numbers with a record number of aliquot divisors that are also Carmichael numbers.

Original entry on oeis.org

561, 63973, 31146661, 509033161, 84127131361, 11985185775745, 712484043821641, 24349841028259201, 53545320695780641, 141125066711098561, 16223841675726285601, 562477984940049379201
Offset: 1

Views

Author

Amiram Eldar, Aug 04 2017

Keywords

Comments

The number of aliquot divisors is 0, 1, 2, 5, 7, 8, 10, 11, 17, 20, 26, 27, ...
The terms were calculated using Pinch's tables of Carmichael numbers (see link below).

Examples

			509033161 is in the sequence since it is a Carmichael number, and 5 of its divisors are also Carmichael numbers (1729, 63973, 126217, 188461 and 294409), more than for any smaller Carmichael number.
		

Crossrefs

Cf. A002997.

Programs

  • Mathematica
    A002997 =  Cases[Range[1, 100000, 2], n_ /; Mod[n, CarmichaelLambda[n]] == 1 && ! PrimeQ[n]]; carmichaelQ[n_] := Not[PrimeQ[n]] && Divisible[n - 1, CarmichaelLambda[n]]; numSol[n_] := Module[{m = 0}, ds = Divisors[n];   Do[d = ds[[k]]; If[! carmichaelQ[d], Continue[]]; m++, {k, 2, Length[ds] - 1}]; m]; numSolmax = -1; seq = {}; Do[n = A002997[[j]]; m = numSol[n]; If[m > numSolmax, AppendTo[seq, n]; numSolmax = m], {j, 1, Length[A002997]}];seq

Extensions

a(11) from Amiram Eldar, Jun 29 2019
a(12) calculated using data from Claude Goutier and added by Amiram Eldar, Apr 24 2024

A291616 Carmichael numbers k such that 2^d == 2^(k/d) (mod k) for all d|k.

Original entry on oeis.org

1105, 294409, 852841, 3828001, 17098369, 118901521, 150846961, 172947529, 186393481, 200753281, 686059921, 771043201, 1001152801, 1207252621, 1269295201, 1299963601, 1632785701, 1772267281, 2301745249, 4215885697, 4562359201, 4765950001, 4897161361
Offset: 1

Views

Author

Keywords

Comments

Intersection of A002997 and A291601.

Examples

			Carmichael number 294409 = 37*73*109 is a term because 2^37 == 2^(73*109) (mod 294409), 2^73 == 2^(37*109) (mod 294409), 2^109 == 2^(37*73) (mod 294409).
		

Crossrefs

A300949 Carmichael numbers whose prime factors form an arithmetic progression.

Original entry on oeis.org

1729, 2465, 29341, 294409, 1152271, 1857241, 6189121, 19384289, 56052361, 64377991, 118901521, 172947529, 216821881, 228842209, 625482001, 775368901, 1213619761, 1299963601, 2301745249, 4562359201, 8346731851, 9293756581, 9624742921, 9701285761, 11346205609, 13079177569, 13946829751, 14386156093
Offset: 1

Views

Author

Robert Israel and Thomas Ordowski, Mar 16 2018

Keywords

Comments

All terms < 10^18 have three prime factors. There are terms with more, e.g., 97888020200929464481 = 34471 * 91921 * 149371 * 206821, 147681255946700149193521 = 214831 * 572881 * 930931 * 1288981, and 2393527068197020059464161 = 431047 * 1149457 * 1867867 * 2586277.
A term with 3 prime factors is of the form (p-d)p(p+d), where p-d, p and p+d are prime, and p-d-1 | d(2d+3), p-1 | d^2, and p+d-1 | d(2d-3). Thus for each d there are only finitely many possible p that make this work. Note that 6|d, see A262723.
Conjecture: if n is a Carmichael number and lpf(n)gpf(n)(lpf(n)+gpf(n))/2 = n, then (lpf(n)+gpf(n))/2 is prime; and thus n has exactly three prime factors. Such numbers n form a proper subsequence of this sequence, also subsequence of A262723. - Charles R Greathouse IV and Thomas Ordowski, Mar 17 2018. Edited by Max Alekseyev, Mar 17 2018
Proof of the above conjecture: Say n = paq with 2 < p < q being primes and a = (p+q)/2, with (a,p!)=1. If n is a Carmichael number, then pa == 1 (mod q-1), so p^2 + pq == 2 (mod q-1), so p^2 + p == 2 (mod q-1). In particular, p^2 + p - 2 >= q-1, which implies that (p+1)^2 > q. Say a has k prime factors, so that a >= (p+2)^k. But a < q, so q > (p+2)^k. Thus, (p+1)^2 > q > (p+2)^k. This implies k=1. - Carl Pomerance (in a letter to the second author), Mar 18 2018
Note: this does not exclude the existence of the Carmichael numbers m = pq(p+q)/2 with more than three prime factors, where p and q are prime. - Thomas Ordowski, Mar 19 2018

Examples

			29341 = 13*37*61 is a Carmichael number, and [13, 37, 61] is an arithmetic progression of length 3 and with common difference of 37 - 13 = 61 - 37 = 24. We have 37 = (13 + 61)/2.
		

Crossrefs

Programs

A303791 Carmichael numbers (A002997) that are central polygonal numbers (A002061).

Original entry on oeis.org

5310721, 2278677961, 9593125081, 29859667201, 467593730289953281, 98538479002618905601, 146842414757227736821
Offset: 1

Views

Author

Max Alekseyev, Apr 30 2018

Keywords

Comments

Also, Carmichael numbers of the form k^2 + k + 1.
Also, Carmichael numbers of the form k^2 - k + 1.
There are no other terms below 10^22.
Carmichael numbers m such that 4m - 3 is square. - Thomas Ordowski, Apr 30 2018

Crossrefs

Intersection of A002997 and A002061.

Extensions

a(6)-a(7) calculated using data from Claude Goutier and added by Amiram Eldar, Apr 20 2024
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