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

1,3

Links

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

Example

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 ...
...

Mathematica

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

Prog

(Python)
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)

Crossrefs

Cf. A332343, A058027, A058299, A071316.

Sequence in context: A143517 A145352 A086834 * A173539 A292250 A245492

Adjacent sequences: A332339 A332340 A332341 * A332343 A332344 A332345

Keyword

nonn,tabl

Author

Andrey Zabolotskiy, Feb 10 2020

Status

approved