%I
%S 5,13,19,43,103,113,229,293,349,463,739,773,859,1171,1429,1483,3079,
%T 3229,3319,3823,4003,4273,5449,6781,6899,7129,7369,7499,7873,7993,
%U 10729,11173,11321,11779,12241,12553,13523,13693,14533,14699,17203,17389
%N Numbers that are the first of two consecutive primes having a sum that is the product of two consecutive numbers.
%C 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.
%H Klaus Brockhaus, <a href="/A154634/b154634.txt">Table of n, a(n) for n=1..1000</a> [From _Klaus Brockhaus_, Jan 15 2009]
%F {A000040(i): A001043(i) in A002378}.  _R. J. Mathar_, Jan 15 2009
%e For the pair of consecutive primes 1429 and 1433, their sum is 2862=53*54.
%e 773 and 787 are consecutive primes. 773+787 = 1560 = 39*40, hence 773 is in the sequence.  _Klaus Brockhaus_, Jan 15 2009
%p 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
%p 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
%o (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
%K easy,nonn
%O 1,1
%A _J. M. Bergot_, Jan 13 2009
%E Corrected and extended by several correspondents, Jan 15 2009
