A117929 - OEIS

%I #47 Feb 16 2024 11:27:19

%S 0,0,0,0,1,0,1,1,1,1,0,1,1,1,1,2,0,2,1,2,1,2,0,3,1,2,0,2,0,3,1,2,1,3,

%T 0,4,0,1,1,3,0,4,1,3,1,3,0,5,1,4,0,3,0,5,1,3,0,3,0,6,1,2,1,5,0,6,0,2,

%U 1,5,0,6,1,4,1,5,0,7,0,4,1,4,0,8,1,4,0,4,0,9,1,4,0,4,0,7,0,3,1,6,0,8,1,5,1

%N Number of partitions of n into 2 distinct primes.

%C Number of distinct rectangles with prime length and width such that L + W = n, W < L. For example, a(16) = 2; the two rectangles are 3 X 13 and 5 X 11. - _Wesley Ivan Hurt_, Oct 29 2017

%H T. D. Noe, <a href="/A117929/b117929.txt">Table of n, a(n) for n = 1..10000</a>

%F G.f.: Sum_{j>0} Sum_{i=1..j-1} x^(p(i)+p(j)), where p(k) is the k-th prime.

%F G.f.: A(x)^2/2 - A(x^2)/2 where A(x) = Sum_{p in primes} x^p. - _Geoffrey Critzer_, Nov 21 2012

%F a(n) = [x^n*y^2] Product_{i>=1} (1+x^prime(i)*y). - _Alois P. Heinz_, Nov 22 2012

%F a(n) = Sum_{i=2..floor((n-1)/2)} A010051(i) * A010051(n-i). - _Wesley Ivan Hurt_, Oct 29 2017

%e a(24) = 3 because we have [19,5], [17,7] and [13,11].

%p g:=sum(sum(x^(ithprime(i)+ithprime(j)),i=1..j-1),j=1..35): gser:=series(g,x=0,130): seq(coeff(gser,x,n),n=1..125);

%p # alternative

%p A117929 := proc(n)

%p     local a,i,p ;

%p     a := 0 ;

%p     p := 2 ;

%p     for i from 1 do

%p         if 2*p >= n then

%p             return a;

%p         end if;

%p         if isprime(n-p) then

%p             a := a+1 ;

%p         end if;

%p         p := nextprime(p) ;

%p     end do:

%p end proc:

%p seq(A117929(n),n=1..80) ; # _R. J. Mathar_, Oct 01 2021

%t l = {}; For[n = 1, n <= 1000, n++, c = 0; For[k = 1, Prime[k] < n/2, k++, If[PrimeQ[n - Prime[k]], c = c + 1] ]; AppendTo[l, c] ] l (* _Jake Foster_, Oct 27 2008 *)

%t Table[Count[IntegerPartitions[n,{2}],_?(AllTrue[#,PrimeQ]&&#[[1]]!= #[[2]] &)],{n,120}] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Jul 26 2020 *)

%o (PARI) a(n)=my(s);forprime(p=2,(n-1)\2,s+=isprime(n-p));s \\ _Charles R Greathouse IV_, Feb 26 2014

%o (Python)

%o from sympy import sieve

%o from collections import Counter

%o from itertools import combinations

%o def aupton(max):

%o     sieve.extend(max)

%o     a = Counter(c[0]+c[1] for c in combinations(sieve._list, 2))

%o     return [a[n] for n in range(1, max+1)]

%o print(aupton(105)) # _Michael S. Branicky_, Feb 16 2024

%Y Cf. A010051, A045917, A061358, A073610, A166081 (positions of 0), A077914 (positions of 2), A080862 (positions of 6).

%Y Column k=2 of A219180. - _Alois P. Heinz_, Nov 13 2012

%K nonn,easy

%O 1,16

%A _Emeric Deutsch_, Apr 03 2006