A329893 a(n) = Product_{k=0..floor(log_2(n))} (1 + A004718(floor(n/(2^k)))), where A004718 is Per Nørgård's "infinity sequence".

1, 2, 0, 6, 0, 0, -6, 24, 0, 0, 0, 0, -18, 0, -48, 120, 0, 0, 0, 0, 0, 0, 0, 0, 18, -72, 0, 0, -192, 48, -360, 720, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 54, 0, 144, -360, 0, 0, 0, 0, 384, -960, 144, 0, -1800, 720, -2880, 5040, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -54, 216

Offset

0,2

Comments

The composer Per Nørgård's name is also written in the OEIS as Per Noergaard.

From Mikhail Kurkov, May 22 2019 (comments originally given in A325803): (Start)

Number of positive terms on the interval 2^m <= n < 2^(m+1) for m > 0 equals f(m-1,2,1) (and f(m-1,4,3) for negative) with f(m,g,h) = binomial(m, floor(m/2) + floor((m+g)/4) - floor((m+h)/4)), so total number of nonzero terms equals binomial(m, floor(m/2)) = A001405(m).

Sum_{n=0..2^m-1} a(n) = 3^m, m >= 0.

More generally, if we define a(n,k) = (-1)^(n+1)*a(floor(n/k),k) + n mod k, a(0,k) = 0, so Sum_{n=0..k^m-1} Product_{i=0..floor(log_k(n))} (1 + a(floor(n/(k^i)),k)) = binomial(k+1, 2)^m for any k = 2p, p > 0.

(End)

Links

Antti Karttunen, Table of n, a(n) for n = 0..16383

Antti Karttunen, Data supplement: n, a(n) computed for n = 0..65537

Will Sawin, Voyage into the golden screen (sequence defined by recurrence relation), Answer to question 333031 on Mathoverflow, June 1, 2019.

Formula

a(0) = 1; for n > 1, a(n) = (1+A004718(n)) * a(floor(n/2)).

a(n) = Product_{k=0..floor(log_2(n))} (1 + A004718(floor(n/(2^k)))).

a(A325804(n)) = A325803(n).

Prog

(PARI)
up_to = 65537;
A004718list(up_to) = { my(v=vector(up_to)); v[1]=1; v[2]=-1; for(n=3, up_to, v[n] = if(n%2, 1+v[n>>1], -v[n/2])); (v); }; \\ After code in A004718.
v004718 = A004718list(up_to);
A004718(n) = if(!n, n, v004718[n]);
A329893(n) = { my(m=1); while(n, m *= 1+A004718(n); n >>= 1); (m); };
(Python)
from math import prod
def A329893(n):
    c, s = [0]*(m:=n.bit_length()), bin(n)[2:]
    for i in range(m):
        if s[i]=='1':
            for j in range(m-i):
                c[j] = c[j]+1
        else:
            for j in range(m-i):
                c[j] = -c[j]
    return prod(1+d for d in c) # Chai Wah Wu, Mar 03 2023

Crossrefs

Cf. A001405, A004718, A325803 (nonzero terms), A325804 (their positions).

Cf. also A284005, A293233.

Sequence in context: A151336 A346092 A180491 * A047918 A321981 A322481

Adjacent sequences: A329890 A329891 A329892 * A329894 A329895 A329896

Keyword

sign

Author

Antti Karttunen (after Mikhail Kurkov's A325803), Dec 10 2019

Status

approved