OFFSET
1,1
COMMENTS
The sum of two consecutive sphenic numbers varies as 4n log n/(log log n)^2, so this sequence is not smaller than that. - Charles R Greathouse IV, Jan 21 2025
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
EXAMPLE
385 = 5*7*11 is a member because 385 = 190+195, sum of 18-th and 19-th sphenic number.
555 = 3*5*37 is a member because 555 = 273+292, sum of 28-th and 29-th sphenic number.
MAPLE
issphenic:= proc(n) local F; F:= ifactors(n)[2]; F[.., 2] = [1, 1, 1] end proc:
S:= select(issphenic, [$1..10000]):
select(issphenic, S[1..-2]+S[2..-1]);
# Robert Israel, Jan 20 2025
MATHEMATICA
sphenicQ[n_] := FactorInteger[n][[;; , 2]] == {1, 1, 1}; Select[MovingMap[Total, Select[Range[3200], sphenicQ], 1], sphenicQ] (* Amiram Eldar, Jan 21 2025 *)
PROG
(PARI) issphenic(n) = my(f=factor(n)); (bigomega(f)==3) && (omega(f)==3);
lista(nn) = my(v=select(issphenic, [1..nn])); select(issphenic, vector(#v-1, k, v[k]+v[k+1])); \\ Michel Marcus, Jan 20 2025
(PARI) sphen(lim)=my(v=List(), t); forprime(p=2, sqrtnint(lim\=1, 3), forprime(q=p+1, sqrtint(lim\p), t=p*q; forprime(r=q+1, lim\t, listput(v, t*r)))); Set(v)
has(n, f=factor(n))=f[, 2]==[1, 1, 1]~
list(lim)=my(v=List(), u=sphen(lim\2)); for(i=2, #u, if(has(u[i]+u[i-1]), listput(v, u[i]+u[i-1]))); forfactored(k=lim\2+1, lim\1-u[#u], if(has(0, k[2]), if(has(k[1]+u[#u]), listput(v, k[1]+u[#u])); break)); Vec(v) \\ Charles R Greathouse IV, Jan 21 2025
CROSSREFS
KEYWORD
nonn
AUTHOR
Massimo Kofler, Jan 20 2025
STATUS
approved