A154634 Numbers that are the first of two consecutive primes having a sum that is the product of two consecutive numbers.

5, 13, 19, 43, 103, 113, 229, 293, 349, 463, 739, 773, 859, 1171, 1429, 1483, 3079, 3229, 3319, 3823, 4003, 4273, 5449, 6781, 6899, 7129, 7369, 7499, 7873, 7993, 10729, 11173, 11321, 11779, 12241, 12553, 13523, 13693, 14533, 14699, 17203, 17389

Offset: 1

Author

J. M. Bergot, Jan 13 2009

Keywords

• easy
• nonn

Comments

Is the sequence mostly uniformly distributed or do clusters occur for the products? One could also find sums of 2n consecutive primes equaling the product of 2n numbers.

Examples

			For the pair of consecutive primes 1429 and 1433, their sum is 2862=53*54.
773 and 787 are consecutive primes. 773+787 = 1560 = 39*40, hence 773 is in the sequence. - _Klaus Brockhaus_, Jan 15 2009
		

Links

• Klaus Brockhaus, Table of n, a(n) for n=1..1000 [From _Klaus Brockhaus_, Jan 15 2009]

Programs

Magma

[ p: p in PrimesUpTo(18000) | r*(r+1) eq s where r is Iroot(s, 2) where s is p+NextPrime(p) ]; // Klaus Brockhaus, Jan 15 2009

Maple

isA002378 := proc(n) local a; a := floor(sqrt(n)) ; RETURN( a*(a+1) = n ) ; end: for i from 1 to 5000 do p := ithprime(i) ; a001043 := p+nextprime(p) ; if isA002378(a001043) then printf("%d,",p) ; fi; od: # R. J. Mathar, Jan 15 2009
a := proc (n) local p, s: p := ithprime(n): s := p+nextprime(p): if type((1/2)*sqrt(1+4*s)-1/2, integer) = true then p else end if end proc: seq(a(n), n = 1 .. 3000); # Emeric Deutsch, Jan 15 2009

Mathematica

sp2Q[{a_,b_}]:=Module[{s=Floor[Sqrt[a+b]]},a+b==s(s+1)]; Select[Partition[ Prime[ Range[2100]],2,1],sp2Q][[All,1]] (* Harvey P. Dale, Jun 28 2020 *)

Formula

{A000040(i): A001043(i) in A002378}. - R. J. Mathar, Jan 15 2009

Extensions

Corrected and extended by several correspondents, Jan 15 2009