A020908 -id:A020908 - cp's OEIS frontend

A182535 Number of terms in Zeckendorf representation of prime(n).

1, 1, 1, 2, 2, 1, 3, 3, 2, 2, 3, 2, 3, 3, 2, 4, 3, 3, 4, 3, 3, 3, 4, 1, 2, 4, 3, 3, 4, 3, 4, 3, 4, 4, 2, 3, 2, 4, 3, 3, 3, 3, 3, 4, 5, 2, 5, 4, 5, 5, 1, 3, 2, 3, 3, 4, 3, 4, 4, 4, 4, 3, 5, 4, 5, 4, 4, 4, 5, 5, 5, 4, 5, 6, 2, 3, 4, 4, 3, 4, 3, 4, 5, 3, 4, 4, 5, 5, 4, 5, 3, 3, 3, 5, 6, 4, 5, 2, 3, 5, 4, 4, 4, 5, 5

Offset: 1

Alex Ratushnyak, May 05 2012

nonn

Alternately, the minimum number of Fibonacci numbers which sum to prime(n). - Alan Worley, Apr 17 2015

			prime(4)=7, and 7 is represented as 5+2, so a(4)=2.
prime(7)=17, and 17 is represented as 13+3+1, so a(7)=3.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A007895, A020908, A020910, A025496-A025502.

f[n_Integer] := Block[{k = Ceiling[ Log[ GoldenRatio, n*Sqrt[5]]], t = n, fr = {}}, While[k > 1, If[t >= Fibonacci[k], AppendTo[fr, 1]; t = t - Fibonacci[k], AppendTo[fr, 0]]; k--]; Count[fr, 1]]; f@# & /@ Prime@ Range@ 105 (* Robert G. Wilson v, Apr 22 2015 *)

a(n) = A007895(A000040(n)).

A182576 Number of 1's in the Zeckendorf representation of n^2.

Original entry on oeis.org

0, 1, 2, 2, 2, 3, 2, 3, 3, 3, 3, 4, 1, 4, 4, 4, 3, 3, 3, 4, 3, 4, 4, 3, 3, 3, 4, 5, 5, 6, 4, 5, 3, 3, 5, 3, 4, 3, 6, 4, 2, 4, 4, 5, 6, 6, 7, 3, 4, 6, 5, 4, 5, 5, 5, 5, 6, 3, 5, 7, 4, 5, 6, 4, 6, 4, 5, 6, 5, 6, 6, 6, 4, 6, 7, 7, 8, 5, 6, 7, 6, 6, 7, 4, 4, 6, 3

Offset: 0

Alex Ratushnyak, May 05 2012

nonn

Also the minimum number of different Fibonacci numbers that sum up to n^2. - Carmine Suriano, Jul 03 2013

See A088060 for some comments related to the occurrence of 2's. - Peter Munn, Mar 22 2021

			a(11)=4 since 11^2 = 121 = 89 + 21 + 8 + 3 = fib(11) + fib(8) + fib(6) + fib(4). - _Carmine Suriano_, Jul 03 2013

Carmine Suriano, Table of n, a(n) for n = 0..1000

Cf. A007895, A020908, A020910, A025496-A025502, A088060, A182535, A182580.

A182577 Number of ones in Zeckendorf representation of n!

Original entry on oeis.org

1, 1, 1, 2, 2, 4, 3, 5, 6, 9, 8, 11, 11, 11, 16, 17, 17, 18, 23, 23, 28, 31, 33, 27, 33, 29, 40, 37, 42, 42, 41, 44, 47, 44, 53, 56, 57, 50, 64, 55, 59, 68, 63, 72, 70, 61, 69, 85, 80, 83, 87, 97, 98, 101, 87, 91, 100, 102, 114, 108, 116, 109, 117, 117, 113, 124

Offset: 0

Alex Ratushnyak, May 05 2012

nonn

			5! = {1, 0, 0, 1, 0, 1, 0, 0, 1, 0} in the Zeckendorf base.

Chai Wah Wu, Table of n, a(n) for n = 0..10000

Cf. A007895, A020908, A020910, A025496-A025502, A182535, A182576, A000142.

from math import factorial
def A182577(n):
    m, tlist, s = factorial(n), [1,2], 0
    while tlist[-1]+tlist[-2] <= m:
        tlist.append(tlist[-1]+tlist[-2])
    for d in tlist[::-1]:
        if d <= m:
            s += 1
            m -= d
    return s # Chai Wah Wu, Jun 15 2018

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);}

A182578 Number of ones in Zeckendorf representation of n^n.

Original entry on oeis.org

1, 1, 2, 3, 3, 6, 3, 10, 13, 12, 16, 15, 20, 24, 20, 30, 25, 31, 26, 33, 33, 31, 34, 42, 49, 49, 53, 55, 56, 55, 58, 64, 64, 67, 73, 78, 70, 76, 77, 75, 89, 83, 92, 90, 106, 99, 100, 99, 107, 116, 107, 115, 125, 125, 122, 119, 127, 137, 127, 138, 155, 156, 153, 160

Offset: 0

Alex Ratushnyak, May 05 2012

nonn

			5^5 = {1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0} in the Zeckendorf base.

Chai Wah Wu, Table of n, a(n) for n = 0..10000

Cf. A007895, A020908, A020910, A025496-A025502, A182535, A182576, A182577, A000312.

def A182578(n):
    m, tlist, s = n**n, [1,2], 0
    while tlist[-1]+tlist[-2] <= m:
        tlist.append(tlist[-1]+tlist[-2])
    for d in tlist[::-1]:
        if d <= m:
            s += 1
            m -= d
    return s # Chai Wah Wu, Jun 14 2018

A374961 Numbers k such that 2^k, 2^(k+1) and 2^(k+2) have the same number of terms in their Zeckendorf representation (A007895).

Original entry on oeis.org

5, 6, 1931, 4127, 26584

Offset: 1

Amiram Eldar, Jul 25 2024

Numbers k such that A020908(k) = A020908(k+1) = A020908(k+2).
The corresponding values of A020908(k) are 3, 3, 763, 1660, 10596, ... .
a(6) > 10^5, if it exists.

			5 is a term since A020908(5) = A020908(6) = A020908(7) = 3.
763 is a term since A020908(1931) = A020908(1932) = A020908(1933) = 763.

Cf. A007895, A020908, A353987.

Subsequence of A374960.

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, 4200], s[#] == s[# + 1] == s[# + 2] &]

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 = A007895(2), z3); for(k = 2, kmax, z3 = A007895(2^k); if(z1 == z2 && z2 == z3, print1(k-2 , ", ")); z1 = z2; z2 = z3);}

A374962 Numbers k such that the number of terms in the Zeckendorf representation of 2^k equals the binary weight of Fibonacci(k).

Original entry on oeis.org

1, 3, 4, 7, 8, 13, 14, 20, 26, 50, 55, 58, 90, 140, 270, 314, 603

Offset: 1

Amiram Eldar, Jul 25 2024

Numbers k such that A007895(A000079(k)) = A000120(A000045(k)), or equivalently A020908(k) = A011373(k).
The corresponding values of A020908(k) = A011373(k) are 1, 1, 2, 3, 3, 5, 6, 8, 9, 18, 22, 24, 33, 53, 106, 122, 232, ... .
a(18) > 63000, if it exists.
a(18) > 333333, if it exists. - Lucas A. Brown, Aug 13 2024

			  n | k = a(n) | 2^k | A014417(2^k) | F(k) | A007088(F(k)) | Number of 1's
  --+----------+-----+--------------+------+---------------+--------------
  1 |        1 |   2 |           10 |    1 |             1 |             1
  2 |        3 |   8 |        10000 |    2 |            10 |             1
  3 |        4 |  16 |       100100 |    3 |            11 |             2
  4 |        7 | 128 |   1010001000 |   13 |          1101 |             3
  5 |        8 | 256 | 100001000010 |   21 |         10101 |             3

Lucas A. Brown, Python program.

Cf. A000045, A000079, A000120, A007088, A007895, A011373, A014417, A020908.

z[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; (* Alonso del Arte at A007895 *)
Select[Range[700], z[2^#] == DigitCount[Fibonacci[#], 2, 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
is(k) = A007895(2^k) == hammingweight(fibonacci(k));

A356977 a(n) is the number of solutions, j >= 0 and 2 <= m_1 <= ... <= m_n, of the equation Sum_{k=1..n} F(m_k) = 2^j where F(i) is the i-th Fibonacci number.

Original entry on oeis.org

0, 3, 6, 10, 36, 66

Offset: 0

Peter Munn and Jon E. Schoenfield, Sep 07 2022

The difficulty of this sequence comes in determining which is the largest j in a solution for a(n), equivalently the last nonzero term in each sum from the A319394-based formula in the formula section.
a(2) derives from Bravo and Luca, a(3) from Bravo and Bravo, a(4) from Pagdame Tiebekabe and Diouf. Pagdame has indicated A356928(5), from which a(5) is derived, has been determined.
a(6) >= 178, a(7) >= 478.

			For n = 2, the a(2) = 6 solutions are j = 1 with (2,2), j = 2 with (2,4) and (3,3), j = 3 with (4,5), j = 4 with (4,7) and (6,6) according to the paper of Bravo and Luca. [That is, 2 = 1+1, 4 = 1+3 = 2+2, 8 = 3+5, 16 = 3+13 = 8+8.]

J. J. Bravo, and F. Luca, On the Diophantine equation F_n+F_m=2^a, Quaest. Math. 39 (2016) 391-400.
P. Tiebekabe and I. Diouf, On solutions of Diophantine equation F_{n_1}+F_{n_2}+F_{n_3}+F_{n_4}=2^a, Journal of Algebra and Related Topics, Volume 9, Issue 2 (2021), 131-148.

E. F. Bravo and J. J. Bravo, Powers of two as sums of three Fibonacci numbers, Lithuanian Mathematical Journal, 55, pp. 301-311 (2015).

Cf. A020908, A319394, A356928.

a(n) = Sum_{i >= 0} A319394(2^i, k).

cp's OEIS Frontend

A182535 Number of terms in Zeckendorf representation of prime(n).

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A182576 Number of 1's in the Zeckendorf representation of n^2.

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A182577 Number of ones in Zeckendorf representation of n!

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

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A182578 Number of ones in Zeckendorf representation of n^n.

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A374961 Numbers k such that 2^k, 2^(k+1) and 2^(k+2) have the same number of terms in their Zeckendorf representation (A007895).

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A374962 Numbers k such that the number of terms in the Zeckendorf representation of 2^k equals the binary weight of Fibonacci(k).

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A356977 a(n) is the number of solutions, j >= 0 and 2 <= m_1 <= ... <= m_n, of the equation Sum_{k=1..n} F(m_k) = 2^j where F(i) is the i-th Fibonacci number.

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