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.

A249302 Prime-partitionable numbers a(n) for which there exists a 2-partition of the set of primes < a(n) that has a smallest subset containing three primes only.

Original entry on oeis.org

22, 130, 222, 246, 280, 286, 288, 320, 324, 326, 356, 416, 426, 454, 470, 494, 516, 528, 556, 590, 612, 634, 670, 690, 738, 746, 804, 818, 836, 838, 870, 900, 902, 904, 922, 936, 1002, 1026, 1074, 1106, 1116, 1140, 1144, 1150, 1206, 1208, 1262, 1264, 1326, 1338
Offset: 1

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Prime-partitionable numbers are defined in A059756.
To demonstrate that a number is prime-partitionable a suitable 2-partition {P1, P2} of the set of primes < a(n) must be found. In this sequence we are interested in prime-partitionable numbers such that the smallest P1 contains 3 odd primes.
Conjecture:
If P1 = {p1a, p1b, p1c} with p1a, p1b and p1c odd primes and p1a < p1b < p1c then the union of the integer solutions to the three equation groups below, {{m1}, {m2}, {m3}}, contains all even members of {a(n)}:
m1 = v1*p1a + 1 = v2*p1b + p1a = p1c + p1b
m2 = v3*p1a + 1 = p1b + p1a^2 = p1c + p1a
m3 = v4*p1a + p1b = v5*p1b + 1 = p1c + p1a
where v1, v2, v3, v4 and v5 are odd naturals.

Examples

			a(1) = 22 because A059756(2) = 22 and both the 2-partitions {3, 13, 19}, {2, 3, 11, 13, 19} and {5, 7, 17}, {2, 5, 7, 11, 17} of the set of primes < 22 demonstrate it.
		

Crossrefs

Programs

  • PARI
    prime_part(n)=
    {
      my (P = primes(primepi(n-1)));
      for (k1 = 2, #P - 1,
        for (k2 = 1, k1 - 1,
          for (k3 = 1, k2 - 1,
            mask = 2^k1 + 2^k2 + 2^k3;
            P1 = vecextract(P, mask);
            P2 = setminus(P, P1);
            for (n1 = 1, n - 1,
              bittest(n - n1, 0) || next;
              setintersect(P1, factor(n1)[,1]~) && next;
              setintersect(P2, factor(n-n1)[,1]~) && next;
              next(2)
            );
            print1(n, ", ");
          );
        );
      );
    }
    # PP = {{2x, x = 1:1000} - {A245664(n), 1:145}}
    PP=[2, 4, 6, 8, 10, 12, 14, 18, 20, 22, 24, 26, 28, 30, \
        32, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, \
        ...
        1980, 1984, 1986, 1988, 1990, 1994, 1996, 1998, 2000];
    for(m=1,#PP,prime_part(PP[m]));

A245372 Prime-partitionable numbers a(n) for which there exists a 2-partition of the set of primes < a(n) that has a smallest subset containing four primes only.

Original entry on oeis.org

46, 76, 96, 106, 134, 142, 146, 204, 218, 276, 310, 408, 438, 466, 518, 534, 536, 546, 580, 624, 650, 672, 680, 694, 792, 800, 896, 970, 1000, 1016, 1100, 1160, 1170, 1318, 1344, 1358, 1364, 1384, 1470, 1480
Offset: 1

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Prime-partitionable numbers are defined in A059756.
To demonstrate that a number is prime-partitionable a suitable 2-partition {P1, P2} of the set of primes < a(n) must be found. In this sequence we are interested in prime-partitionable numbers such that the smallest P1 contains 4 odd primes.
Conjecture:
If P1 = {p1a, p1b, p1c, p1d} with p1a, p1b, p1c and p1d odd primes and p1a < p1b < p1c < p1d then the union of the integer solutions to the ten equation groups below, {{m1}, {m2}, {m3}, {m4}, {m5}, {m6}, {m7}, {m8}, {m9}, {m10}}, contains all even members of {a(n)}:
m1 = v1*p1a+1 = v2*p1b+p1a = v3*p1c+p1b = v4*p1d+p1c
m2 = v5*p1a+1 = v6*p1b+p1a^2 = v7*p1c+p1b = v8*p1d+p1a
m3 = v9*p1a+1 = v10*p1b+p1a^3 = v11*p1c+p1a^2 = v12*p1d+p1a
m4 = v13*p1a+1 = v14*p1b+p1c = v15*p1c+p1a = v16*p1d+p1b
m5 = v17*p1a+1 = v18*p1b+p1c = v19*p1c+p1a^2 = v20*p1d+p1a
m6 = v21*p1a+p1b = v22*p1b+1 = v23*p1c+p1a = v24*p1d+p1c
m7 = v25*p1a+p1b = v26*p1b+1 = v27*p1c+p1a^2 = v28*p1d+p1a
m8 = v29*p1a+p1b = v30*p1b+p1c = v31*p1c+1 = v32*p1d+p1a
m9 = v33*p1a+p1c = v34*p1b+1 = v35*p1c+p1b = v36*p1d+p1a
m10 = v37*p1a+p1c = v38*p1b+p1a = v39*p1c+1 = v40*p1d+p1b
where the vi, i = 1..40 are constrained odd naturals.

Examples

			a(1) = 46 because A245602(5) = 46 and the 2-partition {3, 19, 37, 43} {2, 5, 7, 11, 13, 17, 23, 29, 31, 41} of the set of primes < 46 demonstrates it.
		

Crossrefs

Programs

  • PARI
    prime_part(n)=
    {
      my (P = primes(primepi(n-1)));
      for (k1 = 2, #P - 1,
        for (k2 = 1, k1 - 1,
          for (k3 = 1, k2 - 1,
            for (k4 = 1, k3 - 1,
              mask = 2^k1 + 2^k2 + 2^k3 + 2^k4;
              P1 = vecextract(P, mask);
              P2 = setminus(P, P1);
              for (n1 = 1, n - 1,
                bittest(n - n1, 0) || next;
                setintersect(P1, factor(n1)[,1]~) && next;
                setintersect(P2, factor(n-n1)[,1]~) && next;
                next(2)
                   );
              print1(n, ", ");
            );
          );
        );
      );
    }
    # PP = {{2x, x = 1:1000} - {A245664(n), n = 1:145}
    #                        - {A249302(n), n = 1:77}}
    PP = [2, 4, 6, 8, 10, 12, 14, 18, 20, 24, 26, 28, 30, 32, \
          38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, \
          ...
          1994, 1996, 1998, 2000];
    for(m=1,#PP,prime_part(PP[m]));

A259301 Taken over all those prime-partitionable numbers m for which there exists a 2-partition of the set of primes < m that has one subset containing two primes only, a(n) is the frequency with which the smaller prime occurs, where n is the prime index.

Original entry on oeis.org

0, 0, 1, 1, 3, 3, 3, 2, 4, 4, 3, 4, 5, 7, 8, 5, 8, 7, 8, 9, 10, 10, 11, 12, 12, 14, 13, 13, 12, 15, 14, 14, 17, 14, 19, 17, 12, 18, 13, 19, 20, 22, 20, 23, 21, 15, 21, 21, 23, 25, 26, 23, 26, 26, 19, 23, 27, 24, 29, 27, 26, 28, 31, 29, 30, 25, 30, 29, 34, 30
Offset: 1

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Comments

A number n is called a prime partitionable number if there is a partition {P1,P2} of the primes less than n such that for any composition n1+n2=n, either there is a prime p in P1 such that p | n1 or there is a prime p in P2 such that p | n2.
To demonstrate that a positive integer m is prime-partitionable, a suitable 2-partition {P1, P2} of the set of primes < m must be found. In this sequence we are interested in prime-partitionable numbers such that P1 contains 2 odd primes.
Conjecture: If P1 = {p1a, p1b} with p1a and p1b odd primes, p1a < p1b and p1b = 2*k*p1a + 1 for some positive integer k such that 2*k <= p1a - 3 and if m = p1a + p1b then m is prime-partitionable.

Examples

			The table below shows all p1a and p1b pairs for p1a <= 29 that demonstrate that m is prime-partitionable.
. n    p1a    p1b     2k      m
. 3      5     11      2     16
. 4      7     29      4     36
. 5     11     23      2     34
.       11     67      6     78
.       11     89      8    100
. 6     13     53      4     66
.       13     79      6     92
.       13    131     10    144
. 7     17    103      6    120
.       17    137      8    154
.       17    239     14    256
. 8     19    191     10    210
.       19    229     12    248
. 9     23     47      2     70
.       23    139      6    162
.       23    277     12    300
.       23    461     20    484
.10     29     59      2     88
.       29    233      8    262
.       29    349     12    378
.       29    523     18    552
By examining the p1a column it can be seen that
a(1) = 0, a(2) = 0, a(3) = 1, a(4) = 1, a(5) = 3, a(6) = 3,
a(7) = 3, a(8) = 2, a(9) = 4, a(10) = 4.
		

Crossrefs

Programs

  • Maple
    # Makes use of conjecture in COMMENTS section.
    ppgen := proc (ub)
      local freq_p1a, i, j, k, nprimes, p1a, p1b, pless;
      # Construct set of primes < ub in pless.
      pless := {};
      for i from 3 to ub do
        if isprime(i) then
          pless := `union`(pless, {i});
        end if
      end do;
      nprimes := numelems(pless);
      # Determine frequency of each p1a.
      printf("0, ");    # For prime 2.
      for j to nprimes do
        p1a := pless[j];
        freq_p1a := 0;
        for k to (p1a-3)/2 do
          p1b := 2*k*p1a+1;
          if isprime(p1b) then
            freq_p1a := freq_p1a+1;
          end if;
        end do;
        printf("%d, ", freq_p1a);
      end do;
    end proc:
    ub := 1000:
    ppgen(ub):

A259560 Primes p such that p = 2kq + 1 for k a positive integer, q an odd prime and 2k <= q - 3.

Original entry on oeis.org

11, 23, 29, 47, 53, 59, 67, 79, 83, 89, 103, 107, 131, 137, 139, 149, 167, 173, 179, 191, 223, 227, 229, 233, 239, 263, 269, 277, 283, 293, 311, 317, 347, 349, 359, 367, 373, 383, 389, 431, 439, 461, 467, 479, 499, 503, 509, 523, 557, 563, 569, 587, 593, 607
Offset: 1

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Keywords

Comments

This sequence is associated with the conjecture in A245664 that p + q is prime-partitionable.
There are 138438 values of p in the first 216816 primes, i.e., 63.85%, all of which are distinct.

Examples

			The table lists values of n, q, 2k and p for 1 <= n <= 20.
.n      q     2k      p (a(n))
.1      5      2     11
.2     11      2     23
.3      7      4     29
.4     23      2     47
.5     13      4     53
.6     29      2     59
.7     11      6     67
.8     13      6     79
.9     41      2     83
10     11      8     89
11     17      6    103
12     53      2    107
13     13     10    131
14     17      8    137
15     23      6    139
16     37      4    149
17     83      2    167
18     43      4    173
19     89      2    179
20     19     10    191
		

Crossrefs

Programs

  • Maple
    ppgen := proc (n)
      local i, j, k, nprimes, p1a, p1b, p1b_ind, pless, pless_idx, p1b_ind_num_0, p1b_ind_num_1;
      pless := {};
      for i from 3 to n do
        if isprime(i) then
          pless := `union`(pless, {i})
        end if
      end do;
      nprimes := numelems(pless);
      p1b_ind := Vector(nprimes);
      for j to nprimes do
        p1a := pless[j];
        if (1/2)*pless[-1]+1/2 < p1a then
          break
        end if;
        for k to (1/2)*p1a-3/2 do
          p1b := 2*k*p1a+1;
          if member(p1b, pless, 'pless_idx') then
            p1b_ind[pless_idx] := 1
          elif pless[-1] < p1b then
            break
          end if
        end do
      end do;
      p1b_ind_num_0 := 1;
      p1b_ind_num_1 := 0;
      for i to nprimes do
        if p1b_ind[i] = 0 then
          p1b_ind_num_0 := p1b_ind_num_0+1
        else
          p1b_ind_num_1 := p1b_ind_num_1+1;
          fprintf(fop, "%d %d\n", p1b_ind_num_1, pless[i])
        end if
      end do
    end proc;
    n := 376200;
    ppgen(n);
  • PARI
    is(n)=my(f=factor(n\2)[,1]); for(i=1,#f, if(n\2/f[i]*2<=f[i]-3, return(isprime(n)))); 0 \\ Charles R Greathouse IV, Jul 15 2015
Showing 1-4 of 4 results.