A071316 -id:A071316 - cp's OEIS frontend

A332342 Table T(n, k) read by antidiagonals upwards: sum of the terms of the continued fraction for the fractional part of n/k (n>=1, k>=1).

0, 0, 2, 0, 0, 3, 0, 2, 3, 4, 0, 0, 0, 2, 5, 0, 2, 3, 4, 4, 6, 0, 0, 3, 0, 4, 3, 7, 0, 2, 0, 4, 5, 2, 5, 8, 0, 0, 3, 2, 0, 3, 5, 4, 9, 0, 2, 3, 4, 5, 6, 5, 5, 6, 10, 0, 0, 0, 0, 4, 0, 5, 2, 3, 5, 11, 0, 2, 3, 4, 4, 6, 7, 5, 6, 6, 7, 12, 0, 0, 3, 2, 5, 3, 0, 4, 6, 4, 6, 6, 13

Offset: 1

Andrey Zabolotskiy, Feb 10 2020

			2/7 = 1/(3+1/2), so T(2, 7) = 3 + 2 = 5.
The table begins:
0 2 3 4 5 6 7 8 9 ...
0 0 3 2 4 3 5 4 6 ...
0 2 0 4 4 2 5 5 3 ...
0 0 3 0 5 3 5 2 6 ...
0 2 3 4 0 6 5 5 6 ...
0 0 0 2 5 0 7 4 3 ...
...

Andrey Zabolotskiy, Table of n, a(n) for n = 1..4950 (antidiagonals 1..99)

Cf. A332343, A058027, A058299, A071316.

t[n_,k_] := Total@ ContinuedFraction@ FractionalPart[n/k];
Flatten[Table[t[nk+k-1,k], {nk,10}, {k,nk}]]

def cofr(p, q):
    return [] if q == 0 else [p // q] + cofr(q, p % q)
def t(n, k):
    return sum(cofr(n, k)[1:])
tr = []
for nk in range(1, 20):
    for k in range(1, nk+1):
        tr.append(t(nk+1-k, k))
print(tr)

A071915 Number of 1's in the continued fraction expansion of (3/2)^n.

Original entry on oeis.org

0, 0, 1, 0, 2, 3, 3, 6, 3, 5, 1, 2, 8, 2, 3, 5, 2, 3, 3, 6, 10, 8, 6, 4, 2, 3, 6, 5, 2, 9, 12, 7, 17, 10, 7, 9, 8, 10, 13, 13, 10, 12, 14, 9, 11, 10, 11, 6, 9, 5, 3, 13, 13, 19, 18, 13, 8, 12, 15, 14, 18, 7, 19, 19, 17, 15, 13, 14, 16, 13, 20, 16, 10, 20, 25, 17, 19, 14, 19, 14, 18, 22

Offset: 1

Benoit Cloitre, Jun 13 2002

It seems that lim n ->infinity a(n)/n = 0.2... << (log(4)-log(3))/log(2) = 0.415... the expected density of 1's (cf. measure theory of continued fraction).

			The continued fraction of (3/2)^24 is [16834, 8, 1, 10, 2, 25, 1, 3, 1, 1, 57, 6] which contains 4 "1's", hence a(24)=4.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A071337, A071316, A071315, A071529.

a[1] = 0; a[n_] := Count[ContinuedFraction[(3/2)^n], 1]; Array[a, 100] (* Amiram Eldar, Sep 05 2020 *)

for(n=1,200,s=contfrac(frac((3/2)^n)); print1(sum(i=1,length(s),if(1-component(s,i),0,1)),","))

cp's OEIS Frontend

A332342 Table T(n, k) read by antidiagonals upwards: sum of the terms of the continued fraction for the fractional part of n/k (n>=1, k>=1).

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