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 1-5 of 5 results.

A108278 Numbers k such that k^2-1 and k^2+1 are semiprimes.

Original entry on oeis.org

12, 30, 42, 60, 102, 108, 198, 312, 462, 522, 600, 810, 828, 1020, 1050, 1062, 1278, 1452, 1488, 1872, 1950, 2028, 2130, 2142, 2340, 2790, 2802, 2970, 3000, 3120, 3252, 3300, 3330, 3672, 3930, 4020, 4092, 4230, 4548, 4800, 5280, 5640, 5652, 5658, 6198
Offset: 1

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Author

Hugo Pfoertner, May 30 2005

Keywords

Comments

Subsequence of A069062. - Michel Marcus, Jan 22 2016
Subsequence of A014574. - Robert Israel, Jan 24 2016

Examples

			a(1)=12 because 12^2-1=143=11*13 and 12^2+1=145=5*29 are both semiprimes.
		

Crossrefs

Cf. A001358 (semiprimes), A069062 (k^2-1 and k^2+1 have the same number of divisors), A014574 (average of twin prime pairs).

Programs

  • Magma
    IsSemiprime:=func< n | &+[k[2]: k in Factorization(n)] eq 2 >; [ n: n in [4..7000] | IsSemiprime(n^2+1) and IsSemiprime(n^2-1) ]; // Vincenzo Librandi, Jan 22 2016
    
  • Maple
    filter:= n -> isprime(n+1) and isprime(n-1) and numtheory:-bigomega(n^2+1)=2:
    select(filter, [seq(i,i=2..1000, 2)]); # Robert Israel, Jan 24 2016
  • Mathematica
    Select[Range[7000], PrimeOmega[#^2 - 1] == PrimeOmega[#^2 + 1]== 2 &] (* Vincenzo Librandi, Jan 22 2016 *)
  • PARI
    isok(n) = (bigomega(n^2-1) == 2) && (bigomega(n^2+1) == 2); \\ Michel Marcus, Jan 22 2016

A373209 Numbers k such that k^2 - 1 and k^2 + 1 have 8 divisors each.

Original entry on oeis.org

68, 112, 128, 162, 200, 212, 252, 294, 318, 336, 338, 372, 448, 450, 498, 502, 542, 578, 592, 598, 612, 648, 672, 678, 708, 752, 762, 808, 812, 852, 878, 888, 938, 952, 992, 996, 1012, 1038, 1098, 1102, 1116, 1122, 1188, 1202, 1212, 1248, 1258, 1328, 1362, 1380
Offset: 1

Views

Author

Jon E. Schoenfield, Jun 21 2024

Keywords

Comments

Among the first 10000 terms (from a(1) = 68 through a(10000) = 697578), k^2 - 1 and k^2 + 1 are each the product of three distinct primes, except for
125 terms for which k^2 + 1 = 5^3 times a prime
6 terms for which k^2 + 1 = 13^3 times a prime
1 terms for which k^2 + 1 = 17^3 times a prime
1 terms for which k^2 + 1 = 29^3 times a prime, and
4 terms for which k^2 - 1 = p^3 * (p^3 +/- 2) (with p = 19, 29, 37, 83, respectively).
The first term for which both k^2 - 1 and k^2 + 1 are of the form p^3 * q is k = 41457661182: k^2 - 1 = 3461^3 * 41457661183, while k^2 + 1 = 5^3 * 13749901365452077097.

Examples

			68 is a term: both 68^2 - 1 = 4623 = 3 * 23 * 67 and 68^2 + 1 = 4625 = 5^3 * 37 have 8 divisors.
		

Crossrefs

Formula

{ k : tau(k^2 - 1) = tau(k^2 + 1) = 8}, where tau() is the number of divisors function, A000005.

A373903 Numbers k such that k^2 - 1 has fewer divisors than k^2 + 1.

Original entry on oeis.org

18, 72, 132, 138, 182, 192, 228, 242, 268, 278, 282, 327, 348, 360, 378, 382, 408, 418, 432, 438, 618, 632, 642, 660, 682, 684, 693, 718, 772, 788, 798, 822, 843, 858, 882, 918, 948, 957, 1032, 1048, 1068, 1092, 1113, 1118, 1143, 1152, 1227, 1228, 1230, 1282, 1292
Offset: 1

Views

Author

Amiram Eldar, Jun 22 2024

Keywords

Comments

Numbers k such that A347191(k) < A193432(k).

Examples

			18 is a term since 18^2 - 1 = 323 has 4 divisors (1, 17, 19 and 323) while 18^2 + 1 = 325 has 6 divisors (1, 5, 13, 25, 65 and 325).
		

Crossrefs

Programs

  • Mathematica
    Select[Range[2, 1300], DivisorSigma[0, #^2 - 1] < DivisorSigma[0, #^2 + 1] &]
  • PARI
    is(k) = k > 1 && numdiv(k^2 - 1) < numdiv(k^2 + 1);

A373213 Numbers k such that k^2 - 1 and k^2 + 1 have 6 divisors each.

Original entry on oeis.org

168, 1368, 97968, 10374840, 16104168, 44049768, 68674368, 100741368, 281803368, 486775968, 1177381968, 1262878368, 1336852968, 2321986968, 2404627368, 3476635368, 4374102768, 5102102040, 5142754368, 5182128168, 5385651768, 6035269968, 9218496168, 10657878168
Offset: 1

Views

Author

Jon E. Schoenfield, Jun 21 2024

Keywords

Comments

Each term is a number of the form k = sqrt(p^2 * q + 1) such that q = p^2 - 2 and k^2 + 1 = r^2 * s, where p, q, r, and s are distinct primes.

Examples

			168 is a term: both 168^2 - 1 = 28223 = 13^2 * 167 and 168^2 + 1 = 28225 = 5^2 * 1129 have 6 divisors.
		

Crossrefs

Formula

{ k : tau(k^2 - 1) = tau(k^2 + 1) = 6}, where tau() is the number of divisors function, A000005.

A373756 Table read by antidiagonals: T(n,k) is the smallest m > 1 such that m^2 - 1 and m^2 + 1 have 2n and 2k divisors, respectively, or -1 if no such m exists.

Original entry on oeis.org

2, 4, -1, 10, 3, -1, 14, 8, 18, -1, 28560, 5, 168, 72, -1, 26, 9, 32, 360, 16068, -1, 25071688922457240, 15, 7, 68, 369465818568, 1620, -1, 56, 728, 332, 28398240, 182, 744768, 1407318, -1, 170, 11, 161245807967271241368, 98, 248872305817685706212070112080, 132, 4175536688568, 642, -1
Offset: 1

Views

Author

Jon E. Schoenfield, Jun 16 2024

Keywords

Comments

m=1 is excluded because m^2 - 1 would be 0.
For all m > 1, both m^2 - 1 and m^2 + 1 are nonsquares, so each has an even number of divisors.
For k=1, m^2 + 1 is a prime, so T(n,1) == 0 (mod 2) for all n.
For n=1, m^2 - 1 = (m-1)*(m+1) is a prime, which occurs only at m=2; 2^2 + 1 = 5 is also a prime, so T(1,1) = 2 and T(1,k) = -1 for k > 1.
For n=2, m^2 - 1 = (m-1)*(m+1) has 4 divisors, so (except for T(2,2) = 3) T(2,k) is the average of a twin prime pair (A014574).
Is T(n,k) > 0 for all n > 1?

Examples

			T(5,1) is the smallest integer m > 1 such that m^2 - 1 and m^2 + 1 have 10 and 2 divisors, respectively; since m^2 - 1 cannot be the 9th power of a prime, this requires that p^4 * q + 1 = m^2 = r - 1, where p, q, and r are distinct primes. The smallest such m is 28560, which gives a solution with p = 13, q = 28559, r = 815673601.
T(5,5) is the smallest integer m > 1 such that m^2 - 1 and m^2 + 1 each have 10 divisors; since neither m^2 - 1 nor m^2 + 1 can be the 9th power of a prime, this is the smallest m such that p^4 * q + 1 = m^2 = r^4 * s - 1, where p, q, r, and s are distinct primes: 22335421^4 * 248872305817685706212070112079 + 1 = 248872305817685706212070112080^2 = 13^4 * 2168601400616633822685176617536070987718973054081571441 - 1.
The first eight antidiagonals of the table are shown below.
.
  n\k|                 1   2   3        4            5      6       7  8
  ---+------------------------------------------------------------------
   1 |                 2  -1  -1       -1           -1     -1      -1 -1
   2 |                 4   3  18       72        16068   1620 1407318
   3 |                10   8 168      360 369465818568 744768
   4 |                14   5  32       68          182
   5 |             28560   9   7 28398240
   6 |                26  15 332
   7 | 25071688922457240 728
   8 |                56
		

Crossrefs

Formula

Define f(m) = tau(m^2 - 1) and g(m) = tau(m^2 + 1), where tau is the number of divisors function (A000005). Then
T(n,k) = min_{ m : f(m) = 2n and g(m) = 2k },
or -1 if no such m exists.
Showing 1-5 of 5 results.