|
%I #36 Jan 07 2022 19:34:22
%S 3,5,6,8,10,12,13,17,19,20,22,27,32,34,36,37,41,43,44,46,61,67,68,82,
%T 84,91,95,107,119,126,129,131,153,167,204,211,214,252,261,416,452,489,
%U 499,537,6006,6265,6266,6312,190852,207403,208524,208806,211967,213074,213594,213677,214781,215042,215075,215077
%N Positive integers j of the form Sum_{i=1..k} b(i)c(i), i.e., not in A297345 such that there is only one set {c(1),...,c(k)} where the c(i) are drawn with repetition from {b(0),...,b(k)} and b(k+1) is the smallest element of A297345 that is larger than j, where b() is A297345.
%e The first positive integer not in b() is 3. To check if 3 is a(1) we note that the smallest element of b() larger than 3 is b(3)=7, hence k=2. There is only one set of coefficients {c(1),c(2)} that allows 3 to be obtained from Sum_{i=1..k} b(i)c(i). These are c(1)=2 and c(2)=1. So 3 is in fact a(1).
%e The next integer not in b() is 4. To see if it is a(2) we note that k is still 2 in this case. Now there are two possible sets of coefficients that allow the representation of 4: {0,2} and {2,1}, so 4 is not a term.
%o (Python)
%o # generates all elements of the sequence, smaller than 6268
%o import numpy as np
%o import itertools
%o def g(i,s,perms):
%o c = 0
%o for iks in perms:
%o t=np.asarray(iks)
%o if np.dot(t,s) == i:
%o c += 1
%o if c == 2:
%o break
%o if c == 1:
%o print i
%o S=[1, 2, 7,24,85,285,1143]
%o S1=[0,1, 2, 7,24,85,285,1143]
%o perms = [p for p in itertools.product(S1, repeat=len(S))]
%o s=np.asarray(S,dtype=np.int)
%o for i in range(1,6268):
%o if i not in S:
%o g(i,s,perms)
%Y Cf. A297345.
%K nonn
%O 1,1
%A _Luis F.B.A. Alexandre_, Mar 11 2018
|
