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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A375390 Numbers k such that k^2 + 1, k^2 + 3 and k^2 + 5 are semiprimes.

Original entry on oeis.org

44, 102, 104, 108, 152, 188, 226, 234, 296, 328, 426, 526, 586, 692, 720, 842, 846, 856, 926, 994, 1076, 1278, 1284, 1386, 1426, 1484, 1498, 1574, 1704, 1746, 1764, 1822, 1826, 1848, 1952, 2058, 2114, 2128, 2142, 2148, 2164, 2186, 2386, 2416, 2442, 2484, 2640, 2704, 2904, 2948, 3108, 3142, 3164
Offset: 1

Views

Author

Zak Seidov and Robert Israel, Aug 15 2024

Keywords

Comments

All terms are even.
a(n)^2 + 3 or a(n)^2 + 5 is 3 times a prime. In the first case, a(n)/3 is in A111051.

Examples

			a(3) = 104 is a term because 104^2 + 1 = 10817 = 29 * 373, 104^2 + 3 = 10819 = 31 * 349 and 104^2 + 5 = 10821 = 3 * 3607 are all semiprimes.
		

Crossrefs

Cf. A001358, A111051. Intersection of A085722, A242331 and A242333.

Programs

  • Maple
    select(t -> andmap(s -> numtheory:-bigomega(t^2+s)=2, [1,3,5]), 2*[$1..2000]);
  • Mathematica
    Select[Range[3000], 2 == PrimeOmega[1 + #^2] == PrimeOmega[3 +
    #^2] ==   PrimeOmega [5 + #^2] &]

A180507 Numbers k such that k^2 + 1 = p*q, p and q prime with p == q (mod k).

Original entry on oeis.org

3, 8, 12, 144, 1020, 8040, 13860, 34840, 729180, 1728240, 3232060, 17576520, 39279240, 85184880, 117649980, 778689840, 884737920, 1225045140, 1771563420, 3723878100, 3869896140, 4574299320, 7762395960, 12487172640, 14348911860, 14886940920, 21484957560, 24137574780
Offset: 1

Views

Author

Michel Lagneau, Jan 20 2011

Keywords

Comments

q - p = k with k = 3, 8, 144.
The next terms with q - p = k are F(432) = 85738...5984 and F(570) where F(n) is the n-th Fibonacci number. All such terms are in A001906; the next such term, if one exists, has more than 25000 decimal digits. - Charles R Greathouse IV, Jan 21 2011

Examples

			a(3) = 12 because 12^2 + 1 = 5*29 and 29 - 5 = 2*12;
a(8) = 34840 because 34840^2 + 1 = 4289 * 283009 and 283009 - 4289 = 278720 = 8*34840.
		

Crossrefs

Programs

  • Maple
    with(numtheory):for k from 1 to 40000 do: x:=k^2+1:y:=factorset(x):yy:=bigomega(x):if
      yy=2 and irem(y[2],k) =y[1] then printf(`%d, `,k):else fi:od:
  • PARI
    w(m, r) = Vec(x*(1-x)/(1-(m^2+2)*x+x^2) + O(x^r));
    isok(s, t) = isprime(s) && isprime(s+t);
    lista(nn) = {my(g, k, m=1, r, u=w(1, nn), v=List([])); for(i=2, r=#u, g=k=(u[i]+sqrtint(5*u[i]^2-4))/2; if(isok(u[i], k), listput(v, k))); while(r>2, u=w(m++, r); for(i=2, #u, k=(m*u[i]+sqrtint((m^2+4)*u[i]^2-4))/2; if(kJinyuan Wang, Mar 29 2020

Extensions

More terms from Charles R Greathouse IV, Jan 24 2011
Missing terms inserted and more terms from Jinyuan Wang, Mar 30 2020

A186689 Numbers n such that n^4 + 1 is a semiprime.

Original entry on oeis.org

3, 5, 7, 8, 10, 11, 12, 13, 14, 17, 18, 21, 22, 23, 26, 29, 30, 32, 35, 36, 38, 39, 40, 42, 50, 52, 57, 58, 61, 62, 65, 68, 71, 72, 73, 78, 81, 84, 86, 92, 94, 98, 100, 102, 103, 105, 108, 112, 113, 114, 115, 116, 119, 120, 122, 124, 128, 129, 130, 138, 146, 148, 152, 153, 158
Offset: 1

Views

Author

Michel Lagneau, Feb 25 2011

Keywords

Comments

Corresponding semiprimes n^4+1 are in A186688.

Examples

			3 is in the sequence because 3^4 + 1 = 82 = 2*41 is semiprime.
		

Crossrefs

Programs

  • Mathematica
    SemiPrimeQ[ n_] := (n > 1) && (2 == Plus @@ (Transpose[FactorInteger[n]][[2]]));
      Select[Range[300], SemiPrimeQ[#^4 + 1] &]
    Select[Range[200],PrimeOmega[#^4+1]==2&] (* Harvey P. Dale, Jan 27 2013 *)

A234693 Primes of the form n^2 + 1 such that (n - 1)^2 + 1 and (n + 1)^2 + 1 are semiprimes.

Original entry on oeis.org

17, 101, 28901, 324901, 608401, 902501, 2016401, 5664401, 7452901, 14822501, 16974401, 18490001, 34222501, 40449601, 41731601, 46240001, 48580901, 50410001, 52417601, 76038401, 92736901, 103022501, 111936401, 121220101, 124768901, 139948901, 151290001
Offset: 1

Views

Author

Michel Lagneau, Dec 29 2013

Keywords

Comments

The corresponding n are 4, 10, 170, 570, 780, 950, 1420, 2380...
Property: n^2 + 1 = p + q - 1 and for a(n) > 17, a(n) == 1 mod 100.

Examples

			101 = 10^2 + 1 is in the sequence because 9^2 + 1 = 2*41 and 11^2 + 1 = 2*61.
		

Crossrefs

Programs

  • Maple
    with(numtheory):for n from 1 to 10^5 do:n1:=n^2+1:n2:=(n+1)^2+1:n3:=(n+2)^2+1: if type(n2,prime)=true and bigomega(n1)=2 and bigomega(n3)=2 then printf(`%d, `,n2):else fi:od:
  • PARI
    forstep(n=4,1e5,2,if(isprime(n^2+1) && isprime(n^2/2-n+1) && isprime(n^2/2+n+1), print1(n^2+1", "))) \\ Charles R Greathouse IV, Dec 29 2013

A238947 Numbers k such that k^2 + 1 = p*q, p < q primes and q-p is a power of 2.

Original entry on oeis.org

8, 100, 3524, 5084, 36680, 77980, 21474824, 134201344, 148647496, 300741464, 73851531256, 153122539756, 778318386944, 6669171349484, 16526971109344, 596403262068016, 9376599920124524, 26698166963373164, 140144514160214876, 1613032378604451500
Offset: 1

Views

Author

Michel Lagneau, Mar 07 2014

Keywords

Comments

Note that if n^2+1 = p*q, then p+q cannot be a power of 2. Proof by contradiction: There are two cases: p an odd prime and p=2. Case 1: suppose p and q are odd primes and q = 2^m-p. Then p(2^m-p) = n^2+1 for some even n. Rearranging terms, we obtain p*2^m-1 = p^2+n^2. Looking at this equation modulo 4, we obtain -1 = 1, a contradiction. Case 2: Let p=2. Then we obtain 2^(m+1)-n^2 = 5, which has no solutions in integers.

Examples

			8^2+1 = 65 = 5*13 and 13-5 = 2^3;
100^2+1 = 10001 = 73*137 and 137-73 = 2^6;
3524^2+1 = 12418577 = 3049*4073 and 4073-3049 = 2^10.
		

Crossrefs

Subsequence of A085722.

Programs

  • Maple
    with(numtheory):for a from 1 to 200000 do:p:=ithprime(a):for i from 1 to 50 do:q:=p+2^i:if type(q,prime)=true then x:=sqrt(p*q-1):if x=floor(x) then print(x):else fi:fi:od:od:
  • Mathematica
    Select[Range[10^5],!PrimeQ[#^2+1]&&Plus@@Last/@FactorInteger[#^2+1]==2&&PrimeNu[#^2+1]==2&&IntegerQ[Log[2,FactorInteger[#^2+1][[2]][[1]]-FactorInteger[#^2+1][[1]][[1]]]]&]
  • PARI
    isok(n) = (bigomega(n^2+1) == 2) && (f = factor(n^2+1)) && ((f[2, 1] - f[1, 1])== 2^(valuation(f[2, 1] - f[1, 1], 2))); \\ Michel Marcus, Mar 07 2014

Extensions

a(7)-a(20) from Giovanni Resta, Mar 07 2014

A241975 Numbers n such that n^4 - n^3 - n - 1 is a semiprime.

Original entry on oeis.org

4, 6, 10, 14, 16, 20, 36, 40, 54, 56, 66, 84, 90, 94, 116, 126, 146, 150, 156, 160, 170, 204, 210, 260, 264, 306, 340, 350, 386, 396, 406, 420, 464, 474, 496, 570, 634, 674, 696, 700, 716, 740, 764, 780, 816, 826, 864, 890, 966, 1054, 1070, 1094, 1106, 1144
Offset: 1

Views

Author

Vincenzo Librandi, Aug 10 2014

Keywords

Comments

Since n^4 - n^3 - n - 1 = (n^2 + 1)*(n^2 - n - 1), these are also numbers n such that n^2 + 1 and n^2 - n - 1 are both prime. Numbers in the intersection of A005574 and A002328. - Derek Orr, Aug 10 2014 [Sequence numbers corrected by Jens Kruse Andersen, Aug 11 2014]

Crossrefs

Programs

  • Magma
    IsSemiprime:=func; [ n: n in [2..1500] | IsSemiprime(n^4 - n^3 - n - 1)];
    
  • Mathematica
    Select[Range[2000], PrimeOmega[#^4 - #^3 - # - 1]==2 &]
  • PARI
    for(n=1,10^4,if(isprime(n^2+1)&&isprime(n^2-n-1),print1(n,", "))) \\ Derek Orr, Aug 10 2014

A259036 Smallest divisor of n^2+1 >= sqrt(n^2+1).

Original entry on oeis.org

1, 2, 5, 5, 17, 13, 37, 10, 13, 41, 101, 61, 29, 17, 197, 113, 257, 29, 25, 181, 401, 26, 97, 53, 577, 313, 677, 73, 157, 421, 53, 37, 41, 109, 89, 613, 1297, 137, 85, 761, 1601, 58, 353, 50, 149, 1013, 73, 65, 461, 1201, 61, 1301, 541, 281, 2917, 89, 3137, 65
Offset: 0

Views

Author

Michel Lagneau, Jun 17 2015

Keywords

Comments

Subsequence of A033677.
a(n) = n^2+1 if n^2+1 is prime (see A005574) or n=0. - Michel Marcus, Jul 01 2015
If n^2+1=p*q for primes p,q with pA085722), then a(n)=q. - Robert Israel, Dec 03 2019

Examples

			a(7) = 10 because 7^2+1 = 2*5*5 and 2*5 = 10 is the smallest divisor >=sqrt(7^2+1) = 7.0710678118...
		

Crossrefs

Programs

  • Magma
    [Min([d:d in Divisors(k^2+1)|d ge Sqrt(k^2+1) ]):k in [0..60]]; // Marius A. Burtea, Dec 03 2019
  • Maple
    f:= proc(n) local m,k;
      m:= n^2+1;
      min(select(t -> t^2 >= m, numtheory:-divisors(m)))
    end proc:
    map(f, [$0..100]); # Robert Israel, Dec 03 2019
  • Mathematica
    Table[Select[Divisors[n^2+1], # >= Sqrt[n^2+1] &, 1] // First, {n, 80}]
  • PARI
    concat(1,vector(100,n,d=divisors(n^2+1);k=1;while(d[k]Derek Orr, Jun 27 2015
    

Formula

a(n) = A033677(A002522(n)).

A261546 Numbers k such that the five numbers k^2+1, (k+1)^2+1, ..., (k+4)^2+1 are all semiprime.

Original entry on oeis.org

48, 58, 1688, 2948, 28338, 36998, 38648, 96248, 100308, 133458, 136798, 187538, 207088, 224508, 253808, 309738, 375348, 545048, 598348, 607688, 659548, 672398, 793958, 1055648, 1055688, 1140008, 1270408, 1317808, 1388398, 1399098, 1529488, 1597008, 1655338
Offset: 1

Views

Author

Michel Lagneau, Aug 24 2015

Keywords

Comments

a(n) == 8 (mod 10).
a(15017) > 10^10. - Hiroaki Yamanouchi, Oct 03 2015

Examples

			48 is in the sequence because of these five semiprimes:
48^2+1 = 2305 = 5*461;
49^2+1 = 2402 = 2*1201;
50^2+1 = 2501 = 41*61;
51^2+1 = 2602 = 2*1301;
52^2+1 = 2705 = 5*541.
		

Crossrefs

Subsequence of A085722.

Programs

  • Magma
    IsSemiprime:=func< n | &+[k[2]: k in Factorization(n)] eq 2 >; [ n: n in [1..3*10^5] | IsSemiprime(n^2+1) and IsSemiprime(n^2+2*n+2)and IsSemiprime(n^2+4*n+5)and IsSemiprime(n^2+6*n+10)and IsSemiprime(n^2+8*n+17)]; // Vincenzo Librandi, Aug 24 2015
  • Maple
    with(numtheory):
      n:=5:
      for k from 1 to 10^6 do:
        jj:=0:
        for m from 0 to n-1 do:
           x:=(k+m)^2+1:d0:=bigomega(x):
           if d0=2
           then
           jj:=jj+1:
           else
           fi:
         od:
            if jj=n
            then
            printf(`%d, `,k):
            else
            fi:
        od:
  • Mathematica
    PrimeFactorExponentsAdded[n_]:=Plus@@Flatten[Table[#[[2]], {1}]&/@FactorInteger[n]]; Select[Range[2 10^5], PrimeFactorExponentsAdded[#^2+1] == PrimeFactorExponentsAdded[#^2 + 2 # + 2]== PrimeFactorExponentsAdded[#^2 + 4 # + 5]== PrimeFactorExponentsAdded[#^2 + 6 # + 10]== PrimeFactorExponentsAdded[#^2 + 8 # + 17] == 2 &] (* Vincenzo Librandi, Aug 24 2015 *)
  • PARI
    has(n) = bigomega(n^2+1)==2;
    isok(n) = has(n) && has(n+1) && has(n+2) && has(n+3) && has(n+4); \\ Michel Marcus, Aug 24 2015
    
  • PARI
    a261546() = {
      nterm = 0;
      for (i = 0, 10^9,
        if (isprime(20*i*i + 32*i + 13) &&
          isprime(50*i*i + 90*i + 41) &&
          isprime(50*i*i + 110*i + 61) &&
          isprime(20*i*i + 48*i + 29) &&
          bigomega(100*i*i + 200*i + 101) == 2,
          nterm += 1;
          print(nterm, " ", 10 * i + 8);
        );
      );
    } \\ - Hiroaki Yamanouchi, Oct 03 2015
    
  • PARI
    issemi(n)=forprime(p=2,97, if(n%p==0, return(isprime(n/p)))); bigomega(n)==2
    list(lim)=my(v=List()); forstep(k=48,lim,[10,30,10], if(issemi(k^2+1) && issemi((k+1)^2+1) && issemi((k+3)^2+1) && issemi((k+4)^2+1) && issemi((k+2)^2+1), listput(v,k))); Vec(v) \\ Charles R Greathouse IV, Jul 06 2017
    

Extensions

a(18)-a(33) from Hiroaki Yamanouchi, Oct 03 2015

A261803 a(n) is the smallest number satisfying a(n)^2+1 = p(n)*q(n), p(n) < q(n) both prime, such that q(n+1)/p(n+1) < q(n)/p(n) with the initial condition q(1)/p(1) < 3/2.

Original entry on oeis.org

50, 334, 516, 670, 844, 1164, 1250, 1800, 2450, 9800, 14450, 20000, 24200, 101250, 105800, 135200, 162450, 168200, 204800, 304200, 336200, 451250, 480200, 490050, 530450, 696200, 924800, 966050, 1008200, 1125000, 1155200, 1428050, 1805000, 2332800, 2420000
Offset: 1

Views

Author

Michel Lagneau, Sep 01 2015

Keywords

Comments

The sequence is probably infinite. [This follows from Schinzel's hypothesis H, for example. - Charles R Greathouse IV, Aug 31 2021]
A majority of numbers in the sequence are of the form 2*q^2 with q = 5, 25, 30, 35, 70, 85, 100, 110, 225, 230, 260, 285, 290, 320, 390, 410, ... So, it seems that {a(n)} = {334, 516, 670, 844, 1164} union {2*A109306(n)^2} where A109306 are the numbers k such that k^2 + (k-1)^2 and k^2 + (k+1)^2 are both primes.

Examples

			a(1) = 50 because 50^2+1 = 41*61 => 61/41 = 1.4878... < 1.5
a(2) = 334 because 334^2+1 = 281*397 => 397/281 = 1.4128... < 1.4878...
a(3) = 516 because 516^2+1 = 449*593 => 593/449 = 1.3207... < 1.4128...
a(4) = 670 because 670^2+1 = 593*757 => 757/593 = 1.2765... < 1.3207...
		

Crossrefs

Subsequence of A085722.

Programs

  • Maple
    with(numtheory):nn:=100:d:=1.5:
    for n from 1 to nn do:
      x:=factorset(n^2+1):n0:=bigomega(n^2+1):
       if n0=2
       then
       q:=evalf(x[2]/x[1]):
       if q
    				
  • Mathematica
    (* Assumption: n>7 ==> a(n)=0 mod 50 *) a[n_] := a[n] = For[k = Which[n==1, 0, n <= 7, a[n-1]+1, True, a[n-1] + 50], True, k = Which[n <= 7, k+1, k == a[n-1]+1, k+49, True, k+50], f = FactorInteger[k^2+1]; If[Length[f] == 2, If[f[[All, 2]] == {1, 1}, {p1, q1} = f[[All, 1]]; If[q1/p1 < If[n == 1, 3/2, q[n-1]/p[n-1]], p[n] = p1; q[n] = q1; Return[k]]]]]; Table[Print["a(", n, ") = ", a[n], "  p = ", p[n], "  q = ", q[n],  "  q/p = ", N[q[n]/p[n], 10], "  q-p = ", q[n]-p[n]]; a[n], {n, 1, 100}] (* Jean-François Alcover, Sep 28 2015 *)

A333635 Numbers m such that m^2 + 1 has at most 2 prime factors.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 14, 15, 16, 19, 20, 22, 24, 25, 26, 28, 29, 30, 34, 35, 36, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 56, 58, 59, 60, 61, 62, 64, 65, 66, 69, 71, 74, 76, 78, 79, 80, 84, 85, 86, 88, 90, 92, 94, 95, 96, 100
Offset: 1

Views

Author

Bernard Schott, Mar 30 2020

Keywords

Comments

Equivalently, numbers m such that m^2 + 1 is prime or semiprime.
Henryk Iwaniec proved in 1978 that this sequence is infinite (see link). By contrast, it is not known whether there are infinitely many primes of the form m^2 + 1 (or infinitely many semiprimes of that form).
The integers that have at most 2 prime factors counted with multiplicity are called almost-primes of order 2 and they are in A037143. Here, as m^2 + 1 is not a square for m > 0, all the semiprimes of this form have two distinct prime factors (A144255), and with the primes of the form m^2 + 1 (A002496), they constitute A248742.

Examples

			10^2 + 1 = 101, which is prime, so 10 is in the sequence.
11^2 + 1 = 122 = 2 * 61, so 11 is in the sequence.
12^2 + 1 = 145 = 5 * 29, so 12 is in the sequence.
13^2 + 1 = 170 = 2 * 5 * 17, so 13 is not in the sequence.
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, A1.

Crossrefs

Union of A005574 and A085722.
Cf. A002496 (m^2 + 1 is prime), A005574 (corresponding m).
Cf. A144255 (m^2 + 1 is semiprime), A085722 (corresponding m).
Cf. A248742 (m^2 + 1 is prime or semiprime), this sequence (corresponding m).
Cf. A037143 (numbers with at most 2 prime factors counted with multiplicity).

Programs

  • Mathematica
    Select[Range[100], PrimeQ[(k = #^2 + 1)] || PrimeOmega[k] == 2 &]  (* Amiram Eldar, Mar 30 2020 *)
    Select[Range[100],PrimeOmega[#^2+1]<3&] (* Harvey P. Dale, Aug 08 2025 *)
Previous Showing 21-30 of 39 results. Next