A117929 Number of partitions of n into 2 distinct primes.

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, 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, 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

Offset

1,16

Comments

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

Links

T. D. Noe, Table of n, a(n) for n = 1..10000

Formula

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

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

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

a(n) = Sum_{i=2..floor((n-1)/2)} A010051(i) * A010051(n-i). - Wesley Ivan Hurt, Oct 29 2017

Example

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

Maple

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);
# alternative
A117929 := proc(n)
    local a, i, p ;
    a := 0 ;
    p := 2 ;
    for i from 1 do
        if 2*p >= n then
            return a;
        end if;
        if isprime(n-p) then
            a := a+1 ;
        end if;
        p := nextprime(p) ;
    end do:
end proc:
seq(A117929(n), n=1..80) ; # R. J. Mathar, Oct 01 2021

Mathematica

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

Prog

(PARI) a(n)=my(s); forprime(p=2, (n-1)\2, s+=isprime(n-p)); s \\ Charles R Greathouse IV, Feb 26 2014
(Python)
from sympy import sieve
from collections import Counter
from itertools import combinations
def aupton(max):
    sieve.extend(max)
    a = Counter(c[0]+c[1] for c in combinations(sieve._list, 2))
    return [a[n] for n in range(1, max+1)]
print(aupton(105)) # Michael S. Branicky, Feb 16 2024

Crossrefs

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

Column k=2 of A219180. - Alois P. Heinz, Nov 13 2012

Sequence in context: A178687 A325538 A238417 * A306439 A107455 A039701

Adjacent sequences: A117926 A117927 A117928 * A117930 A117931 A117932

Keyword

nonn,easy

Author

Emeric Deutsch, Apr 03 2006

Status

approved