A179328 -id:A179328 - cp's OEIS frontend

A178942 a(1) = 3; for n >= 2, a(n) is the smallest prime q > a(n-1) such that, for the previous prime p and the following prime r, the fraction (q-p)/(r-q) has denominator equal to A001223(n)/2 (or 0, if no such prime exists).

3, 5, 11, 13, 17, 19, 29, 37, 47, 53, 61, 67, 71, 79, 83, 131, 137, 151, 163, 173, 233, 277, 331, 359, 379, 397, 401, 419, 439, 773, 823, 941, 947, 1021, 1031, 1033, 1063, 1087, 1097, 1117, 1123, 1153, 1187, 1237, 1277, 1709, 1789, 1823

Offset: 1

Vladimir Shevelev, Jan 06 2011

nonn

Conjecture: a(n) > 0 for all n.

The smallest prime(k) > a(n-1) such that the denominator of A001223(k-1)/A001223(k) equals A001223(n)/2. - R. J. Mathar, Jan 07 2011

Cf. A001223, A168253, A179210, A179234, A179240, A179328.

A001223 := proc(n) ithprime(n+1)-ithprime(n) ; end proc:
A178942 := proc(n) option remember; local p,q,r ; if n = 1 then 3; else for q from procname(n-1)+1 do if isprime(q) then p := prevprime(q) ; r := nextprime(q) ; denom((q-p)/(r-q)) ; if % = A001223(n)/2 then return q; end if; end if; end do: end if; end proc: # R. J. Mathar, Jan 07 2011

A001223[n_] := Prime[n + 1] - Prime[n];
a[n_] := a[n] = Module[{p, q, r, d}, If[n == 1, 3, For[q = a[n - 1] + 1, True, q++, If [PrimeQ[q], p = NextPrime[q, -1]; r = NextPrime[q]; d = Denominator[(q - p)/(r - q)]; If[d == A001223[n]/2, Return[q]]]]]];
Array[a, 48] (* Jean-François Alcover, May 21 2020, after Maple *)

More terms from Alois P. Heinz, Jan 06 2011

A179479 a(n) is the smallest prime q > a(n-1) such that, for the previous prime p and the following prime r, the fraction (q-p)/(r-q) has denominator 2, for odd n and 1 for even n (or 0, if such a prime does not exist).

Original entry on oeis.org

3, 5, 7, 11, 13, 17, 19, 29, 43, 53, 67, 71, 79, 97

Offset: 1

Vladimir Shevelev, Jan 08 2011

nonn

Conjecture: a(n) > 0 for all n.

Since, as it is accepted in the OEIS, we consider the uncancelled fractions, then, by the condition, for even n, we have (r-q)|(q-p).

			If n=1, then denominator should be 2. Thus a(1)=3, since (3-2)/(5-3)=1/2. If n=2, then denominator should be 1. Thus a(2)=5, since (5-3)/(7-5)=1/1, etc.

Cf. A168253, A179210, A179234, A179240, A179256, A179328.

A179921 a(n) = prime(n) if n<=3; for n>3, a(n) is the smallest prime >a(n-1), such that the denominator of fraction (a(n-1)-a(n-2))/(a(n)-a(n-1)) did not appear earlier.

Original entry on oeis.org

2, 3, 5, 7, 13, 23, 31, 53, 67, 79, 113, 131, 151, 193, 233, 271, 307, 353, 379, 409, 457, 557, 613, 691, 761, 809, 883, 907, 1013, 1069, 1123, 1181, 1213, 1279, 1361, 1423, 1483, 1571, 1657, 1709, 1811, 1933, 1997, 2087, 2179, 2273, 2341, 2459

Offset: 1

Vladimir Shevelev, Jan 12 2011

nonn

Using Dirichlet's theorem on arithmetic progressions, it is easy to prove that the sequence is infinite. The sequence of the corresponding denominators begins with 2,1,3,5,4,11,7,6,17, ...

			The first four terms 2,3,5,13 give three denominators: 2,1,3. Then a(5) is not in {17, 19}, since (13-5)/(17-13) = 2/1, (13-5)/(19-13) = 4/3 and denominators 1 and 3 already appeared earlier. Since (13-5)/(23-13) = 4/5 and 5 is not yet in the denominator sequence, a(5) = 23.

Cf. A168253, A178942, A179210, A179234, A179256, A179328.

Edited by Alois P. Heinz, Jan 12 2011

cp's OEIS Frontend

A178942 a(1) = 3; for n >= 2, a(n) is the smallest prime q > a(n-1) such that, for the previous prime p and the following prime r, the fraction (q-p)/(r-q) has denominator equal to A001223(n)/2 (or 0, if no such prime exists).

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A179479 a(n) is the smallest prime q > a(n-1) such that, for the previous prime p and the following prime r, the fraction (q-p)/(r-q) has denominator 2, for odd n and 1 for even n (or 0, if such a prime does not exist).

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A179921 a(n) = prime(n) if n<=3; for n>3, a(n) is the smallest prime >a(n-1), such that the denominator of fraction (a(n-1)-a(n-2))/(a(n)-a(n-1)) did not appear earlier.

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