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

Author

Antti Karttunen (after Mikhail Kurkov's A325803), Dec 10 2019 [verification needed]

Keywords

• sign

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) [verification needed]

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.

Crossrefs

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

Cf. also A284005, A293233.

Programs

Mathematica

f[n_?EvenQ] := f[n] = -f[n/2]; f[0] = 0; f[n_] := f[n] = f[(n - 1)/2] + 1; Table[Product[1 + f[Floor[n/(2^k)]], {k, 0, Floor[Log2[n]]}], {n, 0, 120}]  (* Michael De Vlieger, Apr 22 2024, after Jean-François Alcover at A004718 *)

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

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).