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Revision History for A347436 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

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a(n) is the least odd number that has exactly n decompositions as the sum of three primes, or 0 if there is no such odd number.
(history; published version)
#8 by N. J. A. Sloane at Fri Sep 03 20:41:30 EDT 2021
STATUS

reviewed

approved

#7 by Peter Luschny at Fri Sep 03 15:01:49 EDT 2021
STATUS

proposed

reviewed

#6 by Robert Israel at Fri Sep 03 12:13:25 EDT 2021
STATUS

editing

proposed

#5 by Robert Israel at Fri Sep 03 12:13:19 EDT 2021
COMMENTS

Assuming Goldbach's conjecture, if k is odd then A068307(k) >= pi(k-4)-pi((k-1)/2). Using Pierre Dusart's bounds on pi(x), this implies that, for example, A068307(k) >= 4292 for odd k >= 10^5. Thus (on the assumption of Goldbach's conjecture) the given entries of 0 are correct.

CROSSREFS
STATUS

approved

editing

#4 by N. J. A. Sloane at Thu Sep 02 17:36:37 EDT 2021
STATUS

proposed

approved

#3 by Robert Israel at Thu Sep 02 17:23:00 EDT 2021
STATUS

editing

proposed

#2 by Robert Israel at Thu Sep 02 17:22:53 EDT 2021
NAME

allocated for Robert Israela(n) is the least odd number that has exactly n decompositions as the sum of three primes, or 0 if there is no such odd number.

DATA

1, 7, 9, 15, 17, 21, 31, 27, 35, 33, 39, 41, 45, 47, 55, 51, 53, 57, 0, 63, 67, 65, 71, 0, 79, 81, 0, 85, 77, 83, 99, 0, 0, 89, 97, 95, 103, 111, 101, 0, 0, 0, 115, 107, 0, 129, 121, 113, 0, 141, 119, 0, 0, 125, 133, 147, 0, 131, 159, 145, 153, 151, 137, 0, 0, 143, 0, 0, 149, 155, 0, 0, 0, 163, 189

OFFSET

0,2

COMMENTS

Entries of 0 are conjectural. If nonzero they are greater than 10^5.

EXAMPLE

a(3) = 15 because 15 has exactly 3 decompositions as the sum of 3 primes: 2+2+11 = 3+5+7 = 5+5+5, and it is the smallest odd number that does.

MAPLE

N:= 10^5:

P:= select(isprime, [2, seq(i, i=3..N, 2)]):

nP:=nops(P):

V:= Vector(N):

for i from 1 to nP do

for j from i to nP while P[i]+P[j] <= N do

for k from j to nP do

n:= P[i]+P[j]+P[k];

if n > N then break fi;

V[n]:= V[n]+1;

od od od:

R:= Vector(300):

for i from 1 to N by 2 do

if V[i] <= 300 and V[i] > 0 and R[V[i]] = 0 then R[V[i]]:= i fi

od:

convert(R, list);

CROSSREFS

Cf. A139321.

KEYWORD

allocated

nonn

AUTHOR

J. M. Bergot and Robert Israel, Sep 02 2021

STATUS

approved

editing

#1 by Robert Israel at Thu Sep 02 17:22:53 EDT 2021
NAME

allocated for Robert Israel

KEYWORD

allocated

STATUS

approved