

A370521


The smallest number k that can be partitioned in n ways as the sum of two Blum numbers (A016105).


1



1, 42, 90, 162, 234, 474, 270, 378, 558, 594, 774, 846, 970, 810, 1050, 630, 1370, 1134, 990, 1170, 1470, 1730, 1530, 2054, 1970, 1386, 1638, 1710, 2178, 2070, 2630, 2250, 1890, 2730, 2394, 2310, 3234, 3230, 3530, 2790, 2898, 3650, 3470, 4010, 3570, 3654, 2970, 3150
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,2


LINKS



EXAMPLE

1 cannot be written as the sum of two Blum numbers, so a(0) = 1.
Since A016105(k) >= 21, for k >= 1, the numbers 2 through 41 cannot be written as the sum of two Blum numbers. 42 = 21 + 21 = A016105(1) + A016105(1), so a(1) = 42.
90 = 21 + 69 = A016105(1) + A016105(4), 90 = 33 + 57 = A016105(2) + A016105(3), and the numbers 1 to 89 cannot be written in two ways as the sum of two Blum numbers. Thus a(2) = 90.


PROG

(Magma) pp:=PrimeDivisors; blum:=func<n#Divisors(n) eq 4 and #pp(n) eq 2 and pp(n)[1] mod 4 eq 3 and pp(n)[2] mod 4 eq 3>; b:=[n: n in [1..5000]blum(n)]; a:=[]; for n in [0..47] do k:=1; while #RestrictedPartitions(k, 2, Set(b)) ne n do k:=k+1; end while; Append(~a, k); end for; a;
(Python)
from sympy import factorint
from itertools import takewhile
from collections import Counter
def okA016105(n):
f = factorint(n)
return len(f)==2 and sum(f.values())==2 and all(p%4==3 for p in f)
def aupto(N): # N is limit of terms considered; use 2*10**6 for bfile
s = [k for k in range(1, N+1) if okA016105(k)]
c = Counter(x+y for i, x in enumerate(s) if 2*i<=N for y in s[i:] if x+y<=N)
adict = {0: 1}
for k in sorted(c):
v = c[k]
if v not in adict: adict[v] = k
adict_rev = (adict.get(i) for i in range(max(adict)+1))
return list(takewhile(lambda v:v!=None, adict_rev))


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



