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User: Bobby Browning

Bobby Browning has authored 2 sequences.

A185718 For positive n with prime decomposition n = Product_{j=1..m} (p_j^k_j) define A_n = Sum_{j=1..m} (p_j*k_j) and B_n = Sum_{j=1..m} (p_j^k_j). This sequence gives those n for which A_n and B_n are both prime and B_n = A_n + 2 (i.e., form a twin prime pair).

40, 88, 184, 424, 808, 1048, 1384, 1528, 1864, 2104, 2184, 3080, 4504, 4744, 5224, 5928, 6440, 6568, 7224, 8104, 8360, 8840, 9784, 10264, 10472, 11480, 11544, 11848, 12808, 12904, 14136, 14840, 14968, 16280, 16648, 18664, 19608, 20344, 21080, 22040, 23240, 23704, 24440, 24648, 24920, 26008, 26584, 27384, 27608, 27688, 28264, 28952, 29240

Offset: 1

Bobby Browning, Feb 10 2011

nonn

Assuming the twin prime conjecture, my advisor and I are able to prove there are infinitely many of these pairs. In other words, there are infinitely many n such that A_n and B_n are prime and B_n = A_n + 2.
From Bobby Browning, Feb 14 2011: (Start)
8*A046138 is a subsequence of A185718 for the following reasons:
i) the n in A185718 for which A_n and B_n form a twin prime pair are of the form n=2^3*p_1*p_2*...p_k.
ii) the A046138 sequence consists of primes p such that p+6 and p+8 form a twin prime pair.
iii) so if p is a prime such that p+6 and p+8 form a twin prime pair and n = 2^3*p then A_n = p+6 and B_n = p+8. Thus, the integers such that n = 2^3*p are a subsequence of A185718. (End)

			a(1) = 40  = 2^3*5^1, with a = 11 and b = 13.
a(2) = 88  = 2^3*11^1 with a = 17 and b = 19.
a(3) = 184 = 2^3*23^1 with a = 29 and b = 31.

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A001414 (A_n), A008475 (B_n).

okQ[n_] := Module[{p, e, a, b}, {p, e} = Transpose[FactorInteger[n]]; a = Plus @@ (p*e); b = Plus @@ (p^e); b == a + 2 && PrimeQ[a] && PrimeQ[b]]; Select[Range[30000], okQ]

sopfr(n)=my(f=factor(n));sum(i=1,#f[,1],f[i,1]*f[i,2]);
forstep(n=8,1e5,16,if(issquarefree(n/8)&&isprime(k=sopfr(n))&isprime(k+2), print1(n", ")))

A179435 For positive n with prime decomposition n = Product_{j=1..m} (p_j^k_j) define a_n = Sum_{j=1..m} (p_j*k_j) and b_n = Sum_{j=1..m} (p_j^k_j). This sequence gives those n for which a_n and b_n are both prime and unequal.

Original entry on oeis.org

40, 48, 54, 88, 108, 184, 250, 384, 424, 432, 448, 808, 864, 1048, 1216, 1384, 1528, 1575, 1680, 1792, 1864, 1890, 2104, 2184, 2457, 2925, 2944, 3080, 3120, 3328, 3510, 3696, 3712, 3915, 4125, 4158, 4288, 4504, 4744, 4950, 5224, 5488, 5632, 5928, 5940, 6240

Offset: 1

Bobby Browning and Rohan Hemasinha (rhemasin(AT)uwf.edu), Jan 07 2011

nonn

Is the sequence infinite?

Odd terms in the sequence are: a(18) = 1575, a(25) = 2457, a(26) = 2925, a(34) = 3915, a(35) = 4125, a(47) = 6345, a(50) = 6669, ...

			a(1) = 40 = 2^3*5^1, with a = 11 and b = 13.
a(2) = 48 = 2^4*3^1, with a = 11 and b = 19.
Notice that a and b are both prime and not equal.

Alois P. Heinz, Table of n, a(n) for n = 1..1000

a:= proc(n) option remember; local an, bn, k, l;
      for k from 1 +`if`(n=1, 0, a(n-1)) do
        l:= ifactors(k)[2];
        an:= add( i[1] * i[2], i=l);
        bn:= add( i[1] ^ i[2], i=l);
        if isprime(an) and isprime(bn) and an<>bn then break fi
      od; k
    end:
seq(a(n), n=1..50);  # Alois P. Heinz, Jan 20 2011

a[n_] := a[n] = Module[{an, bn, k, p, e}, For[k = 1 + If[n==1, 0, a[n-1]], True, k++, {p, e} = Transpose[FactorInteger[k]]; an = p.e; bn = Total[p^e]; If[PrimeQ[an] && PrimeQ[bn] && an != bn, Break[]]]; k];
Array[a, 50] (* Jean-François Alcover, Nov 20 2020 *)

More terms from Alois P. Heinz, Jan 20 2011

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User: Bobby Browning

A185718 For positive n with prime decomposition n = Product_{j=1..m} (p_j^k_j) define A_n = Sum_{j=1..m} (p_j*k_j) and B_n = Sum_{j=1..m} (p_j^k_j). This sequence gives those n for which A_n and B_n are both prime and B_n = A_n + 2 (i.e., form a twin prime pair).

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