A353986 -id:A353986 - cp's OEIS frontend

A353987 Numbers k such that F(k), F(k+1) and F(k+2) have the same binary weight (A000120), where F(k) is the k-th Fibonacci number (A000045).

1, 51, 130, 996, 3224, 4287, 9951, 12676, 72004, 53812945, 141422620

Offset: 1

Amiram Eldar, May 13 2022

Numbers k such that A011373(k) = A011373(k+1) = A011373(k+2).

The corresponding values of A011373(k) are 1, 17, 42, 354, 1110, 1490, 3451, 4383, 24988, 18678035, ...

			1 is a term since A011373(1) = A011373(2) = A011373(3) = 1.
51 is a term since A011373(51) = A011373(52) = A011373(53) = 17.

Cf. A000045, A000120, A011373.

Subsequence of A353986.

s[n_] := s[n] = DigitCount[Fibonacci[n], 2, 1]; Select[Range[10^4], s[#] == s[# + 1] == s[# + 2] &]

hf(k) = hammingweight(fibonacci(k)); \\ A011373
isok(k) = my(h=hf(k)); (h == hf(k+1)) && (h == hf(k+2)); \\ Michel Marcus, May 13 2022

# if Python version < 3.10, replace c.bit_count() with bin(c).count('1')
from itertools import islice
def A353987_gen(): # generator of terms
        b, c, k, ah, bh = 1, 2, 1, 1, 1
        while True:
            if ah == (ch := c.bit_count()) == bh:
                yield k
            b, c, ah, bh = c, b+c, bh, ch
            k += 1
A353987_list = list(islice(A353987_gen(),7)) # Chai Wah Wu, May 14 2022

a(11) from Dennis Yurichev, Jul 10 2024

A354300 Numbers k such that k! and (k+1)! have the same binary weight (A000120).

Original entry on oeis.org

0, 1, 3, 5, 7, 8, 12, 13, 15, 31, 63, 88, 127, 129, 131, 244, 255, 262, 263, 288, 300, 344, 511, 793, 914, 1012, 1023, 1045, 1116, 1196, 1538, 1549, 1565, 1652, 1817, 1931, 1989, 2047, 2067, 2096, 2459, 2548, 2862, 2918, 2961, 3372, 3478, 3540, 3588, 3673, 3707

Offset: 1

Amiram Eldar, May 23 2022

Numbers k such that A079584(k) = A079584(k+1).
The corresponding values of A079584(k) are 1, 1, 2, 4, 6, 6, 12, 12, 18, 42, ...
This sequence is infinite as it contains A000225. - Rémy Sigrist, May 23 2022

			1 is a term since A079584(1) = A079584(2) = 1.
3 is a term since A079584(3) = A079584(4) = 2.

Amiram Eldar, Table of n, a(n) for n = 1..1000

A354301 is a subsequence.

Cf. A000120, A000142, A000225, A079584, A353986.

s[n_] := s[n] = DigitCount[n!, 2, 1]; Select[Range[0, 4000], s[#] == s[# + 1] &]

isok(k) = hammingweight(k!) == hammingweight((k+1)!); \\ Michel Marcus, May 23 2022

from itertools import count, islice
def wt(n): return bin(n).count("1")
def agen(): # generator of terms
    n, fn, fnplus, wtn, wtnplus = 0, 1, 1, 1, 1
    for n in count(0):
        if wtn == wtnplus: yield n
        fn, fnplus = fnplus, fnplus*(n+2)
        wtn, wtnplus = wtnplus, wt(fnplus)
print(list(islice(agen(), len(data)))) # Michael S. Branicky, May 23 2022

A374960 Numbers k such that 2^k and 2^(k+1) have the same number of terms in their Zeckendorf representation (A007895).

Original entry on oeis.org

0, 5, 6, 7, 11, 18, 20, 25, 39, 52, 61, 96, 104, 157, 176, 199, 206, 210, 279, 326, 333, 339, 369, 380, 397, 411, 426, 473, 542, 576, 743, 860, 898, 921, 961, 970, 993, 1024, 1043, 1049, 1100, 1121, 1176, 1184, 1193, 1199, 1206, 1230, 1253, 1376, 1380, 1387, 1435

Offset: 1

Amiram Eldar, Jul 25 2024

Numbers k such that A020908(k) = A020908(k+1).

The corresponding values of A020908(k) are 1, 3, 3, 3, 6, 7, 8, 9, 18, 20, 28, 44, 37, ... .

			0 is a term since the Zeckendorf representation of 2^0 = 1 is A014417(1) = 1, and the Zeckendorf representation of 2^1 = 2 is A014417(2) = 10, so A020908(0) = A020908(1) = 1.
5 is a term since the Zeckendorf representation of 2^5 = 32 is A014417(32) = 1010100, and the Zeckendorf representation of 2^6 = 64 is A014417(64) = 100010001, so A020908(5) = A020908(6) = 3.

Amiram Eldar, Table of n, a(n) for n = 1..550 (terms below 10^5)

Cf. A007895, A014417, A020908, A353986.

A374961 is a subsequence.

z[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; (* Alonso del Arte at A007895 *)
s[n_] := s[n] = z[2^n]; Select[Range[0, 1500], s[#] == s[# + 1] &]

A007895(n)=if(n<4, n>0, my(k=2, s, t); while(fibonacci(k++)<=n, ); while(k && n, t=fibonacci(k); if(t<=n, n-=t; s++); k--); s); \\ Charles R Greathouse IV at A007895
lista(kmax) = {my(z1 = A007895(1), z2); for(k = 1, kmax, z2 = A007895(2^k); if(z1 == z2, print1(k-1 , ", ")); z1 = z2);}

cp's OEIS Frontend

A353987 Numbers k such that F(k), F(k+1) and F(k+2) have the same binary weight (A000120), where F(k) is the k-th Fibonacci number (A000045).

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A354300 Numbers k such that k! and (k+1)! have the same binary weight (A000120).

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Mathematica

PARI

Python

A374960 Numbers k such that 2^k and 2^(k+1) have the same number of terms in their Zeckendorf representation (A007895).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI