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.

Previous Showing 31-39 of 39 results.

A360739 Semiprimes of the form k^2 + 2.

Original entry on oeis.org

6, 38, 51, 123, 146, 291, 326, 731, 843, 1227, 1371, 1766, 1851, 2306, 2603, 2811, 2918, 3027, 3602, 4227, 4358, 4763, 5186, 5331, 5627, 6243, 6891, 7058, 7571, 8102, 8651, 9411, 13227, 14163, 15627, 17426, 17691, 18227, 18771, 19883, 20738, 22502, 23411, 24027
Offset: 1

Views

Author

Elmo R. Oliveira, Feb 18 2023

Keywords

Comments

A242330 gives the corresponding values of k.

Examples

			123 is a term because 11^2 + 2 = 123 = 3*41.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[0, 200]^2 + 2, PrimeOmega[#] == 2 &] (* Amiram Eldar, Feb 18 2023 *)

Formula

a(n) = A242330(n)^2 + 2.

A360740 Semiprimes of the form k^2 + 3.

Original entry on oeis.org

4, 39, 259, 327, 403, 579, 679, 1027, 1159, 1299, 1603, 1939, 2119, 2307, 3139, 3603, 4359, 4627, 6087, 6403, 7747, 9607, 10003, 10407, 10819, 11667, 13459, 13927, 14403, 16387, 18499, 21907, 23107, 26899, 28903, 30279, 30979, 33127, 35347, 36103, 36867, 38419
Offset: 1

Views

Author

Elmo R. Oliveira, Feb 18 2023

Keywords

Comments

A242331 gives the corresponding values of k.

Examples

			259 is a term because 16^2 + 3 = 259 = 7*37.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[0, 200]^2 + 3, PrimeOmega[#] == 2 &] (* Amiram Eldar, Feb 18 2023 *)

Formula

a(n) = A242331(n)^2 + 3.

A360741 Semiprimes of the form k^2 + 4.

Original entry on oeis.org

4, 85, 365, 445, 533, 629, 965, 1685, 1853, 2605, 2813, 3029, 3973, 4765, 5045, 5629, 5933, 6245, 6893, 8285, 8653, 11029, 11453, 11885, 12773, 14165, 15133, 16645, 17165, 17693, 20453, 21029, 22205, 22805, 23413, 24653, 27229, 29245, 29933, 30629, 32765, 34229
Offset: 1

Views

Author

Elmo R. Oliveira, Feb 18 2023

Keywords

Comments

A242332 gives the corresponding values of k.
Except for 4, all terms == 5 (mod 8). - Robert Israel, Feb 18 2023

Examples

			85 is a term because 9^2 + 4 = 85 = 5*17.
		

Crossrefs

Programs

  • Maple
    select(t -> numtheory:-bigomega(t)=2, [seq(i^2+4,i=0..1000)]); # Robert Israel, Feb 18 2023
  • Mathematica
    Select[Range[0, 200]^2 + 4, PrimeOmega[#] == 2 &] (* Amiram Eldar, Feb 18 2023 *)

Formula

a(n) = A242332(n)^2 + 4.

A375391 a(n) is the greatest odd number k such that n^2 + j is a semiprime for all odd numbers j from 1 to k.

Original entry on oeis.org

-1, -1, 1, -1, 1, -1, -1, 1, 1, -1, 1, 1, -1, -1, 1, -1, -1, -1, 1, -1, -1, 1, -1, -1, 1, -1, -1, 1, 1, 1, -1, -1, -1, 3, 1, -1, -1, -1, 1, -1, -1, 1, -1, 9, 1, 3, -1, 3, 1, 1, 1, 1, -1, -1, -1, -1, -1, 1, 1, 3, 1, 1, -1, 1, 1, -1, -1, -1, 1, -1, 1, -1, -1, -1, -1, 1, -1, 3, 1, 3, -1, -1, -1, -1
Offset: 1

Views

Author

Robert Israel, Aug 15 2024

Keywords

Comments

a(n) = -1 if n^2 + 1 is not a semiprime.
a(n) <= 1 if n is odd, since n^2 + 3 is divisible by 4.
a(n) <= 15 since one of n^2 + 1, n^2 + 3, ..., n^2 + 17 is divisible by 9.
First occurrences of values: a(3) = 1, a(34) = 3, a(152) = 5, a(102) = 7, a(44) = 9, a(824264) = 11, a(21394) = 13, a(121364) = 15.

Examples

			a(44) = 9 since 44^2 + 1 = 1937 = 13 * 149, 44^2 + 3 = 1939 = 7 * 277, 44^2 + 5 = 1941 = 3 * 647, 44^2 + 7 = 1943 = 29 * 67 and 44^2 + 9 = 1945 = 5 * 389 are all semiprimes but 44^2 + 11 = 1947 = 3 * 11 * 59 is not a semiprime.
		

Crossrefs

Programs

  • Maple
    f:= proc(n) local i;
      for i from 1 by 2 while numtheory:-bigomega(n^2+i) = 2 do od:
      i-2
    end proc:
    map(f, [$1..100]);

A115398 Numbers k such that both k^2+1 and 2^k + 1 are semiprimes.

Original entry on oeis.org

3, 5, 11, 12, 19, 28, 61, 64, 79, 92, 101, 104, 199, 356, 596, 692, 1709, 3539, 3824
Offset: 1

Views

Author

Zak Seidov, Mar 08 2006

Keywords

Comments

Intersection of A085722 and A092559.

Examples

			11 is a term because 11^2 + 1 = 122 = 2*61 (semiprime) and 2^11 + 1 = 2049 = 3*683 (semiprime).
		

Crossrefs

Programs

  • Magma
    IsSemiprime:=func; [n: n in [2..700] | IsSemiprime(n^2+1) and IsSemiprime(2^n+1)]; // Vincenzo Librandi, Oct 10 2013
  • Mathematica
    Select[Range[700],PrimeOmega[#^2+1]==PrimeOmega[2^#+1]==2&] (* Harvey P. Dale, Apr 14 2019 *)
  • PARI
    isok(n) = (bigomega(n^2+1) == 2) && (bigomega(2^n+1) == 2); \\ Michel Marcus, Oct 10 2013
    

Extensions

a(17)-a(19) from Robert Israel, Nov 27 2023

A258780 a(n) is the least k such that k^2 + 1 is a semiprime p*q, p < q, and (q - p)/2^n is prime.

Original entry on oeis.org

8, 12, 140, 64, 2236, 196, 1300, 1600, 6256, 5084, 248756, 246196, 484400, 36680, 887884, 821836, 1559116, 104120, 126072244, 9586736, 4156840, 542759984, 1017981724, 2744780140, 405793096, 148647496, 1671024916
Offset: 2

Views

Author

Michel Lagneau, Jun 10 2015

Keywords

Comments

The corresponding primes are 2, 3, 71, 7, 1069, 7, 5, 5, 59, 2, 368471, 180463, 12421, 2, 29, 125683, 226169, 5, 369704891, 197, 5, 263, 7444559, 239621423, 594271, 2, 474359, ...
All terms are even, in order for k^2+1 to be odd. Otherwise, with k^2+1 being even, p-q would be odd and hence not a multiple of 2^n. - Michel Marcus, Apr 13 2019

Examples

			a(2)=8 because 8^2+1 = 5*13 and (13-5)/2^2 = 2 is prime. The number 8 is the first term of the sequence 8, 22, 34, 46, 50, 58, ...
a(3)=12 because 12^2+1 = 5*29 and (29-5)/2^3 = 3 is prime. The number 12 is the first term of the sequence 12, 28, 44, 52, 76, 80, ...
a(4)=140 because 140^2+1 = 17*1153 and (1153-17)/2^4 = 71 is prime. The number 140 is the first term of the sequence 140, 296, 404, 604, ...
		

Crossrefs

Programs

  • Mathematica
    lst={};Do[k=2;While[!(Plus@@Last/@FactorInteger[k^2+1]==2&&PrimeQ[(FactorInteger[k^2+1][[-1,1]]-FactorInteger[k^2+1][[1,1]])/2^n]),k=k+2];Print[n," ",k],{n,2,19}];lst
  • PARI
    isok(k, n) = my(kk=k^2+1, f=factor(kk)[,1]~); (bigomega(kk) == 2) && (#f == 2) && (p=f[1]) && (q=f[2]) && (qq=(q-p)/2^n) && !frac(qq) && isprime(qq);
    a(n) = my(k=2); while (!isok(k,n), k+=2); k; \\ Michel Marcus, Apr 13 2019

Extensions

Name edited by Jon E. Schoenfield, Sep 12 2017
a(20)-a(22) from Daniel Suteu, Apr 13 2019
a(23)-a(28) from Daniel Suteu, Nov 09 2019

A321519 Let d(n,i), i = 1..k be the k divisors of n^2 + 1 (the number 1 is not counted). a(n) is the number of ordered pairs d(n,i) < d(n,j) such that gcd(d(n,i), d(n,j)) = 1.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 2, 1, 1, 0, 1, 1, 6, 0, 1, 0, 6, 2, 1, 0, 6, 1, 6, 0, 1, 0, 6, 1, 1, 1, 6, 2, 6, 1, 1, 0, 6, 2, 1, 0, 2, 1, 11, 1, 1, 1, 25, 1, 1, 1, 1, 1, 6, 0, 6, 0, 16, 1, 1, 1, 1, 1, 6, 1, 1, 0, 6, 3, 1, 2, 1, 6, 25, 0, 6, 1, 6, 1, 1, 1, 6, 2, 25, 0, 1, 1
Offset: 1

Views

Author

Michel Lagneau, Nov 12 2018

Keywords

Comments

Terms only depends on prime signature of n^2+1. - David A. Corneth, Nov 14 2018
We observe an interesting statistic for n <= 10^5: the four values of a(n) = 0, 1, 6, 25 represent more than 82% (see the table below).
a(A005574(n)) = 0, a(A085722(n)) = 1, a(A272078(n)) = 6, a(A316351(n)) = 25.
In the general case, a(k) = m if k^2+1 = p*q^m, m = 1, 2, 3, ... with p, q primes.
+--------------+-----------------------+------------+
| | number of occurrences | |
| a(n) | for n <= 10^5 | percentage |
+--------------+-----------------------+------------+
| 0 | 6656 | 6.656% |
| 1 | 23255 | 23.255% |
| 6 | 31947 | 31.947% |
| 25 | 20461 | 20.461% |
| other values | 17681 | 17.681% |
+--------------+-----------------------+------------+

Examples

			a(13) = 6 because the divisors {d(i)} of 13^2 + 1 = 170 (without the number 1)  are  {2, 5, 10, 17, 34, 85, 170}, and gcd(d(i), d(j)) = 1 for the 6 following pairs of elements of {d(i)}: (2, 5), (2, 17), (2, 85), (5, 17), (5, 34) and (10, 17).
		

Crossrefs

Programs

  • Maple
    with(numtheory):nn:=10^3:
    for n from 1 to nn do:
      it:=0:d:=divisors(n^2+1):n0:=nops(d):
       for k from 2 to n0-1 do:
        for l from k+1 to n0 do:
         if gcd(d[k],d[l])= 1
          then
          it:=it+1
          else
         fi:
       od:
      od:
      printf(`%d, `,it):
    od:
  • Mathematica
    f[n_] := (DivisorSigma[0, n^2] - 1)/2 - DivisorSigma[0, n] + 1; Map[f, Range[0,100]^2+1] (* Amiram Eldar, Nov 14 2018 after Robert G. Wilson v at A089233 *)
  • PARI
    a(n) = {my(d=divisors(n^2+1)); sum(k=2, #d, sum(j=2, k-1, gcd(d[k], d[j]) == 1));} \\ Michel Marcus, Nov 12 2018

Formula

a(n) = A089233(n^2+1). - Michel Marcus, Nov 13 2018

A321985 Numbers m such that m^2+1 is semiprime with (m-1)^2+1 and (m+1)^2+1 primes.

Original entry on oeis.org

3, 5, 15, 25, 205, 715, 1095, 1315, 1615, 2055, 2405, 2925, 3755, 4615, 4795, 5015, 5055, 5475, 6785, 7855, 8115, 8175, 9425, 9475, 10415, 10845, 11025, 11245, 12335, 12765, 15225, 16225, 16395, 16405, 18145, 18175, 18275, 21345, 21915, 22905, 23165, 23815
Offset: 1

Views

Author

Michel Lagneau, Nov 23 2018

Keywords

Comments

Subsequence of A085722.
For n>1, a(n) == 5 (mod 10).
The corresponding pairs of primes (p, q) = ((m-1)^2+1, (m+1)^2+1) are congruent to 7 (mod 10), and the semiprimes are of the form m^2+1 = 2r where r is congruent to 3 (mod 10). So, a(n) = (q - 2r - 1)/2 = (2r - p + 1)/2 = (q - p)/4.

Examples

			15 is in the sequence because 15^2 + 1 = 2*113 is semiprime, and 14^2 + 1 = 197, 16^2 + 1 = 257 are prime numbers.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[50000],PrimeQ[(#-1)^2+1]&&PrimeOmega [#^2+1]==2&&PrimeQ[(#+1)^2+1]&]
  • PARI
    isok(m) = (bigomega(m^2+1) == 2) && isprime((m-1)^2+1) && isprime((m+1)^2+1); \\ Michel Marcus, Nov 23 2018

A348594 Numbers m such that m^2 + 1 = p*q with p, q primes and m = (p + q)/2 - 1.

Original entry on oeis.org

8, 50, 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, 2576450, 2761250, 2832200
Offset: 1

Views

Author

Michel Lagneau, Jan 26 2022

Keywords

Comments

Subsequence of A085722.
The corresponding pairs (p, q) of the sequence are (5, 13), (41, 61), (1201, 1301), (1741, 1861), (2381, 2521), (9661, 9941), (14281, 14621), (19801, 20201), (23981, 24421), (100801, 101701), ...
Property:
a(n) = 2* A109306(n)^2 and a(n) == 0 (mod 50) for n > 1. Proof:
From the relations:
(1) m^2 + 1 = p*q
(2) (p + q)/2 = m + 1
We obtain:
(3) p = m + 1 - sqrt(8*m)/2
(4) q = m + 1 + sqrt(8*m)/2
with m = 2*k^2 we obtain:
(5) p = k^2 + (k-1)^2
(6) q = k^2 + (k+1)^2
For n > 1, A109306(n) == 0 (mod 5) => 2*A109306(n)^2 == 0 (mod 50).

Examples

			50 = 2*5^2 is in the sequence because 50^2 + 1 = 41*61 with 50 = (41 + 61)/2 - 1.
		

Crossrefs

Programs

  • Maple
    with(numtheory):nn:=250:printf(`%d, `,8):
    for k from 0 to nn do:
    n:=50*k^2:d:=factorset(n^2+1):
      if bigomega(n^2+1)=2 and (d[1]+d[2])/2 - 1 = n
       then
        printf(`%d, `,n):
        else
      fi:
    od:
  • Mathematica
    q[n_] := Module[{f = FactorInteger[n^2 + 1]}, f[[;; , 2]] == {1, 1} && f[[1, 1]] + f[[2, 1]] == 2*n + 2]; Select[Range[3*10^5], q] (* Amiram Eldar, Jan 26 2022 *)
  • PARI
    isok(m) = my(x); if (bigomega(x=m^2+1)==2, my(f=factor(x)); (f[1,1]+f[2,1] == 2*(m+1))); \\ Michel Marcus, Jan 26 2022
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