A297345 a(0)=0; for n>0, a(n) is the least positive integer that cannot be represented as Sum_{k=1..n-1} a(i_k)*a(k), with 0 <= i_k < n.

0, 1, 2, 7, 24, 85, 285, 1143, 6268, 216784, 1059813, 6100794, 226303113

Offset

0,3

Links

Table of n, a(n) for n=0..12.

Example

a(1)= 1 since it is not possible to write 1 using only a(0). a(2)=2, since it is not possible to obtain 2 using only a(0) and a(1). The following numbers up to 6 can be represented using these first 3 elements of the sequence: 3 = 1*1 + 1*2, 4 = 0*1 + 2*2, 5 = 1*1 + 2*2, 6 = 2*1 + 2*2. Again we reach a number that cannot be represented as defined above, so that number is appended to the sequence. It happens here when we try to represent 7 using only a(0)=0, a(1)=1, and a(2)=2. So 7 becomes a(3).
A larger example: 216752 = 1*1 + 1*2 + 85*7 + 285*24 + 85*85 + 85*285 + 24*1143 + 24*6268

Mathematica

Nest[Function[a, Append[a, 1 + LengthWhile[Differences@ #, # == 1 &] &@ Union[Total /@ Map[a # &, Tuples[a, Length@ a]]]]], {0}, 8] (* Michael De Vlieger, Jan 09 2018 *)

Prog

(Python)
# Generate all the elements in the sequence, S, necessary to represent all
# numbers until the integer 'last'. It also shows how each integer is
# represented by showing the sequence elements and the respective
# multiplicative factors.
import numpy as np
import itertools
last=100
def generate(i, S):
    n=len(S)
    s=np.asarray(S, dtype=np.int)
    perms = [p for p in itertools.product(S, repeat=n)]
    for iks in perms:
        t=np.asarray(iks)
        if np.dot(t, s) == i:
            print('%d=' %i, end=', ')
            print(t, 'x', s)
            return 0
    return -1
S=[0]
for i in range(1, last+1):
        if generate(i, S) == -1:
            S.append(i)
            generate(i, S)

Crossrefs

Sequence in context: A369266 A144170 A369296 * A052986 A053368 A141753

Adjacent sequences: A297342 A297343 A297344 * A297346 A297347 A297348

Keyword

nonn,more,hard

Author

Luis F.B.A. Alexandre, Dec 28 2017

Extensions

a(9) from Robert G. Wilson v, Jan 09 2018

a(10)-a(11) from Jon E. Schoenfield, Jan 16 2018

a(12) from Giovanni Resta, Jan 22 2018

Status

approved