A325803 -id:A325803 - cp's OEIS frontend

A325804 Positions of nonzero terms of Product_{k=0..floor(log_2(n))} (1 + A004718(floor(n/(2^k)))).

0, 1, 3, 6, 7, 12, 14, 15, 24, 25, 28, 29, 30, 31, 48, 50, 51, 56, 57, 58, 60, 61, 62, 63, 96, 97, 100, 101, 102, 103, 112, 113, 114, 115, 116, 117, 120, 121, 122, 123, 124, 125, 126, 127, 192, 194, 195, 200, 201, 202, 204, 205, 206, 207, 224, 225, 226, 228

Offset: 1

Mikhail Kurkov, May 22 2019

nonn

Mikhail Kurkov, Table of n, a(n) for n = 1..13495

Cf. A004718, A325803.

a[n_?EvenQ] := a[n] = -a[n/2]; a[0] = 0; a[n_] := a[n] = a[(n - 1)/2] + 1; -1 + Position[Table[Product[ 1 + a[Floor[n/(2^k)]], {k, 0, Floor[Log2[n]]}], {n, 0, 500}], ?(# != 0 &)][[All, 1]] (* _Michael De Vlieger, Apr 22 2024, after Jean-François Alcover at A004718 *)

b(n) = if(n==0, 0, (-1)^(n+1)*b(n\2) + n%2); \\ A004718
f(n) = if(n==0, 1, prod(k=0, logint(n,2), 1+b(n\2^k)));
isok(n) = f(n)!=0; \\ Michel Marcus, May 24 2019

from itertools import count, islice
def A325804_gen(startvalue=0): # generator of terms >= startvalue
    for n in count(max(startvalue,0)):
        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]
        if all(1+d for d in c): yield n
A325804_list = list(islice(A325804_gen(),20)) # Chai Wah Wu, Mar 03 2023

Conjecture: a(n) - a(n-1) belongs to A094373. - Mikhail Kurkov, Feb 20 2021

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

Original entry on oeis.org

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

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

sign

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]

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.

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

Cf. also A284005, A293233.

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 *)

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); };

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

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).

cp's OEIS Frontend

A325804 Positions of nonzero terms of Product_{k=0..floor(log_2(n))} (1 + A004718(floor(n/(2^k)))).

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