A242208 -id:A242208 - cp's OEIS frontend

A328912 Continued fraction expansion of log_2((sqrt(5)+1)/2) = 0.6942419... = A242208.

0, 1, 2, 3, 1, 2, 3, 2, 4, 2, 1, 2, 11, 2, 1, 11, 1, 1, 134, 2, 2, 2, 1, 4, 1, 1, 3, 1, 7, 1, 13, 1, 3, 5, 1, 1, 1, 8, 1, 3, 4, 1, 1, 1, 3, 4, 1, 3, 1, 4, 1, 4, 1, 3, 40, 1, 1, 5, 4, 3, 3, 1, 3, 1, 2, 6, 1, 1, 2, 28, 11, 1, 71, 2, 1, 4, 8, 5, 1, 2, 1, 1, 14

Offset: 0

M. F. Hasler, Oct 31 2019

This number is also the solution to 1 + 2^x = 4^x, or 1 + 1/2^x = 2^x, which clarifies the relation to Phi = (sqrt(5)+1)/2, solution to 1 + 1/x = x.

			log_2((sqrt(5)+1)/2) = 0.6942419... = 0 + 1/(1 + 1/(2 + 1/(3 + 1/(1 + ...))))

Paolo Xausa, Table of n, a(n) for n = 0..10000

Cf. A242208, A001622 (decimals of Phi), A000012 (cont. frac. of Phi).

ContinuedFraction[Log2[GoldenRatio], 100] (* Paolo Xausa, Mar 07 2024 *)

localprec(1000); contfrac(log(sqrt(5)+1)/log(2)-1)

Some terms corrected by Paolo Xausa, Mar 07 2024

A011373 Number of 1's in binary expansion of Fibonacci(n).

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 1, 3, 3, 2, 5, 4, 2, 5, 6, 4, 8, 7, 4, 5, 8, 6, 8, 11, 6, 6, 9, 11, 11, 12, 8, 11, 9, 13, 12, 11, 12, 14, 10, 12, 16, 17, 14, 16, 18, 15, 21, 13, 12, 18, 18, 17, 17, 17, 16, 22, 21, 16, 24, 20, 16, 19, 26, 23, 20, 25, 19, 26, 15, 23, 23, 22, 25, 27, 24, 23, 23, 22

Offset: 0

N. J. A. Sloane

			a(8) = 3 because Fibonacci(8) = 21, which in binary is 11001 and that has 3 on bits.
a(9) = 2 because Fibonacci(9) = 34, which in binary is 100010 and that only has 2 on bits.

Alois P. Heinz, Table of n, a(n) for n = 0..10000 (first 1001 terms from T. D. Noe)
C. L. Stewart, On the representation of an integer in two different bases, Journal für die reine und angewandte Mathematik 319 (1980): 63-72.
Index entries for sequences related to binary expansion of n.

Cf. A000045, A000120, A059016, A242208.

A000120 := proc(n) add(d,d=convert(n,base,2)) ; end proc:
A011373 := proc(n) A000120(combinat[fibonacci](n)) ; end proc:
seq(A011373(n),n=0..50) ; # R. J. Mathar, Mar 22 2011

DigitCount[#, 2, 1]&/@Fibonacci[Range[0, 79]] (* Harvey P. Dale, Mar 14 2011 *)
Table[Plus@@IntegerDigits[Fibonacci[n], 2], {n, 0, 79}]

a(n)=hammingweight(fibonacci(n)) \\ Charles R Greathouse IV, Mar 02 2014

from sympy import fibonacci
def a(n): return int(fibonacci(n)).bit_count() # David Radcliffe, Jul 03 2025

def fibonacci(n: BigInt): BigInt = {
  val zero = BigInt(0)
  def fibTail(n: BigInt, a: BigInt, b: BigInt): BigInt = n match {
    case `zero` => a
    case _ => fibTail(n - 1, b, a + b)
  }
  fibTail(n, 0, 1)
} // Based on tail recursion by Dario Carrasquel
(0 to 79).map(fibonacci().bitCount) // _Alonso del Arte, Apr 13 2019

a(n) = A000120(A000045(n)). - Michel Marcus, Dec 27 2014
a(n) = [x^Fibonacci(n)] (1/(1 - x))*Sum_{k>=0} x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Mar 27 2018
Conjecture: Limit_{n->oo} a(n)/n = log_2(phi)/2 = A242208/2 = 0.3471209568... . - Amiram Eldar, May 13 2022
Limit_{n->oo} a(n) = oo by a result of C. L. Stewart, linked above. - David Radcliffe, Jul 03 2025

A328900 Decimal expansion of s = 1.507126591638653..., solution to 2^s + 3^s = 4^s.

Original entry on oeis.org

1, 5, 0, 7, 1, 2, 6, 5, 9, 1, 6, 3, 8, 6, 5, 3, 1, 3, 3, 9, 8, 6, 8, 8, 3, 3, 6, 0, 8, 3, 8, 6, 3, 1, 1, 6, 4, 3, 7, 3, 9, 9, 4, 0, 9, 4, 4, 8, 5, 6, 5, 6, 8, 9, 6, 6, 7, 5, 3, 6, 4, 3, 5, 9, 4, 4, 3, 8, 1, 4, 7, 3, 3, 8, 0, 4, 8, 5, 1, 5, 7, 2, 5, 9, 2, 2, 8

Offset: 1

M. F. Hasler, on suggestion from Artur Jasinski, Oct 30 2019

Equivalently, solution to 1/(1 - 2^-s) = 1 + 2^-s + 3^-s, related to partial sums and Euler product approximating zeta(s).
When a + b = c, then the only solution to a^x + b^x = c^x is trivially x = 1. The solution to 1 + 2^x = 4^x is log_2(Phi) = A242208.
See A328904, A328905 for (a, b, c) = (1, 3, 5) and (1, 2, 5).

			1.5071265916386531339868833608386311643739940944856568966753643594438147338...

M. F. Hasler, Table of n, a(n) for n = 1..1000
M. F. Hasler, Solutions to a^x + b^x = c^x, OEIS wiki, Nov. 2019

Cf. A328913 (continued fraction).
Cf. A242208 (1 + 2^x = 4^x), A328912 (continued fraction thereof).
Cf. A328904 (1 + 3^x = 5^x), A328905 (1 + 2^x = 5^x), A328906 (1 + 2^x = 6^x), A328907 (1 + 3^x = 6^x).

RealDigits[ x /. FindRoot[2^x + 3^x == 4^x, {x, 1.5}, WorkingPrecision -> 100]][[1]] (* Artur Jasinski, Oct 30 2019 *)

solve(s=1,2,2^s+3^s-4^s) \\ use e.g. \p200 to get more digits

A328904 Decimal expansion of x = 0.7271601514124259243... solution to 1 + 3^x = 5^x.

Original entry on oeis.org

7, 2, 7, 1, 6, 0, 1, 5, 1, 4, 1, 2, 4, 2, 5, 9, 2, 4, 3, 0, 4, 4, 7, 0, 8, 4, 4, 0, 0, 9, 5, 2, 1, 7, 6, 9, 3, 5, 4, 5, 8, 9, 0, 4, 5, 5, 6, 4, 5, 8, 3, 3, 0, 4, 1, 4, 2, 5, 7, 7, 7, 6, 4, 1, 7, 5, 2, 9, 0, 8, 6, 8, 4, 3, 2, 3, 0, 5, 7, 7, 3, 3, 5, 5, 0

Offset: 0

M. F. Hasler, Oct 31 2019

			0.7271601514124259243044708440095217693545890455645833041425777641752908684323...

Cf. A329334 (continued fraction).

Cf. A242208 (1 + 2^x = 4^x), A328900 (2^x + 3^x = 4^x), A328905 (1 + 2^x = 5^x).

RealDigits[x /. FindRoot[1 + 3^x == 5^x, {x, 1}, WorkingPrecision -> 120]][[1]] (* Amiram Eldar, Jun 28 2023 *)

print(c=solve(x=0,1, 1+3^x-5^x)); digits(c\.1^default(realprecision))[^-1] \\ [^-1] to discard possibly incorrect last digit. Use e.g. \p999 to get more digits. - M. F. Hasler, Oct 31 2019

A328905 Decimal expansion of the solution x = 0.56389552425993647949... to 1 + 2^x = 5^x.

Original entry on oeis.org

5, 6, 3, 8, 9, 5, 5, 2, 4, 2, 5, 9, 9, 3, 6, 4, 7, 9, 4, 9, 0, 3, 9, 2, 9, 4, 5, 9, 3, 7, 9, 5, 6, 5, 6, 5, 5, 1, 5, 2, 1, 1, 7, 3, 0, 5, 0, 9, 9, 5, 5, 2, 9, 8, 5, 9, 2, 8, 0, 8, 3, 8, 0, 1, 2, 0, 4, 6, 6, 2, 0, 0, 5, 2, 2, 8, 1, 9, 7, 3, 5, 5, 0, 4, 2

Offset: 0

M. F. Hasler, Oct 31 2019

			0.5638955242599364794903929459379565655152117305099552985928083801204662005228...

Cf. A329334 (continued fraction).

Cf. A242208 (1 + 2^x = 4^x), A328900 (2^x + 3^x = 4^x), A328904 (1 + 3^x = 5^x).

RealDigits[x /. FindRoot[1 + 2^x == 5^x, {x, 1}, WorkingPrecision -> 120]][[1]] (* Amiram Eldar, Jun 28 2023 *)

print(c=solve(x=0,1, 1+2^x-5^x)); digits(c\.1^default(realprecision))[^-1] \\ [^-1] to discard possibly incorrect last digit. Use e.g. \p999 to get more digits. - M. F. Hasler, Oct 31 2019

A328907 Decimal expansion of the solution x = 0.6009668516... to 1 + 3^x = 6^x.

Original entry on oeis.org

6, 0, 0, 9, 6, 6, 8, 5, 1, 6, 1, 3, 6, 7, 5, 4, 8, 5, 7, 1, 5, 7, 0, 5, 2, 6, 4, 6, 3, 1, 8, 3, 8, 1, 2, 0, 6, 7, 7, 2, 2, 7, 9, 9, 2, 1, 3, 3, 0, 5, 1, 3, 5, 8, 8, 5, 0, 2, 6, 3, 9, 4, 0, 1, 9, 1, 6, 9, 2, 1, 2, 0, 4, 0, 9, 8, 0, 5, 1, 3, 9, 9, 6, 8, 5, 2, 3, 4, 8, 3, 7, 0, 2, 5, 3, 1, 3, 9, 8

Offset: 0

M. F. Hasler, Nov 11 2019

			0.6009668516136754857157052646318381206772279921330513588502639401916921204...

Cf. A329337 (continued fraction).

Cf. A242208 (1 + 2^x = 4^x), A328900 (2^x + 3^x = 4^x), A328904 (1 + 3^x = 5^x), A328905 (1 + 2^x = 5^x), A328906 (1 + 2^x = 6^x).

RealDigits[x /. FindRoot[1 + 3^x == 6^x, {x, 1}, WorkingPrecision -> 120]][[1]] (* Amiram Eldar, Jun 28 2023 *)

print(c=solve(x=0,1, 1+3^x-6^x)); digits(c\.1^default(realprecision))[^-1] \\ [^-1] to discard possibly incorrect last digit. Use e.g. \p999 to get more digits. - M. F. Hasler, Oct 31 2019

A174868 Partial sums of Stern's diatomic series A002487.

Original entry on oeis.org

0, 1, 2, 4, 5, 8, 10, 13, 14, 18, 21, 26, 28, 33, 36, 40, 41, 46, 50, 57, 60, 68, 73, 80, 82, 89, 94, 102, 105, 112, 116, 121, 122, 128, 133, 142, 146, 157, 164, 174, 177, 188, 196, 209, 214, 226, 233, 242, 244, 253, 260, 272, 277, 290, 298, 309, 312, 322, 329, 340, 344, 353, 358, 364, 365, 372, 378, 389, 394, 408, 417, 430, 434, 449, 460, 478, 485, 502, 512, 525, 528, 542, 553, 572, 580, 601, 614, 632, 637, 654, 666, 685

Offset: 0

Jonathan Vos Post, Dec 01 2010

After the initial 0, identical to A007729.

			a(16) = 0 + 1 + 1 + 2 + 1 + 3 + 2 + 3 + 1 + 4 + 3 + 5 + 2 + 5 + 3 + 4 + 1 = 41.

Amiram Eldar, Table of n, a(n) for n = 0..10000
Michael J. Collins and David Wilson, Equivalence of OEIS A007729 and A174868, arXiv:1812.11174 [math.CO], 2018.
Clemens Heuberger, Daniel Krenn and Gabriel F. Lipnik, Asymptotic Analysis of q-Recursive Sequences, Algorithmica, Vol. 84 (2022), pp. 2480-2532; arXiv preprint, arXiv:2105.04334 [math.CO], 2021-2022.

Cf. A002487, A007729, A084091, A049456, A020946, A020950, A046815, A070871, A070872, A071883, A064881-A064886, A072170, A001045, A002083, A073459, A000123, A126606, A049455.

Cf. A001622, A242208.

a[n_] := a[n] = If[EvenQ[n], 2*a[n/2] + a[n/2 - 1], 2*a[(n - 1)/2] + a[(n + 1)/2]]; a[0] = 0; a[1] = 1; Array[a, 100, 0] (* Amiram Eldar, May 18 2023 *)

from itertools import accumulate, count, islice
from functools import reduce
def A174868_gen(): # generator of terms
    return accumulate((sum(reduce(lambda x,y:(x[0],x[0]+x[1]) if int(y) else (x[0]+x[1],x[1]),bin(n)[-1:2:-1],(1,0))) for n in count(1)),initial=0)
A174868_list = list(islice(A174868_gen(),30)) # Chai Wah Wu, May 07 2023

a(n) = Sum_{i=0..n} A002487(i).
G.f.: (x/(1 - x))*Product_{k>=0} (1 + x^(2^k) + x^(2^(k+1))). - Ilya Gutkovskiy, Feb 27 2017
a(2k) = 2*a(k) + a(k-1); a(2k+1) = 2*a(k) + a(k+1). - Michael J. Collins, Dec 25 2018
a(n) = n^log_2(3) + Psi_D(log_2(n)) + O(n^log_2(phi)), where phi is the golden ratio (A001622) and Psi_D is a 1-periodic continuous function which is Hölder continuous with any exponent smaller than log_2(3/phi) (Heuberger et al., 2022). - Amiram Eldar, May 18 2023

A241002 Decimal expansion of the asymptotic growth rate of the number of odd coefficients in Pascal trinomial triangle mod 2, where coefficients are from (1+x+x^4)^n.

Original entry on oeis.org

7, 3, 6, 2, 1, 1, 5, 5, 5, 7, 3, 9, 3, 0, 7, 9, 3, 1, 6, 5, 4, 9, 2, 0, 9, 3, 8, 9, 2, 4, 5, 8, 0, 9, 8, 3, 1, 8, 5, 0, 0, 5, 7, 7, 6, 4, 8, 4, 5, 9, 3, 6, 7, 7, 3, 9, 7, 9, 4, 6, 9, 1, 6, 8, 5, 7, 9, 4, 3, 9, 4, 2, 9, 8, 1, 1, 4, 3, 2, 3, 5, 8, 1, 2, 9, 4, 4, 6, 8, 2, 4, 4, 2, 9, 0, 1, 1, 1, 9, 8, 2, 2, 8, 9

Offset: 0

Jean-François Alcover, Aug 13 2014

			0.7362115557393079316549209389245809831850057764845936773979469...

Jean-Francois Alcover, Table of n, a(n) for n = 0..103
Steven Finch, Pascal Sebah and Zai-Qiao Bai, Odd Entries in Pascal's Trinomial Triangle (arXiv:0802.2654) p. 11.

Cf. A242208 (1+x+x^2)^n, A242021 (1+x+x^3)^n, A242022 (1+x+x^2+x^3+x^4)^n.

mu = Sort[Table[Root[x^5 - 3*x^4 - 2*x^2 - 8*x + 8, x, n], {n, 1, 5}], N[Abs[#1]] < N[Abs[#2]] &] // Last; RealDigits[Log[mu]/Log[2] - 1, 10, 104] // First

log(abs(mu))/log(2) - 1, where mu is the root of x^5 - 3*x^4 - 2*x^2 - 8*x + 8 with maximum modulus.

A328906 Decimal expansion of the solution x = 0.4895363211996... to 1 + 2^x = 6^x.

Original entry on oeis.org

4, 8, 9, 5, 3, 6, 3, 2, 1, 1, 9, 9, 6, 4, 9, 4, 8, 8, 6, 8, 9, 8, 7, 5, 3, 1, 6, 8, 2, 2, 6, 5, 0, 1, 8, 9, 4, 0, 3, 5, 8, 6, 5, 1, 5, 7, 7, 1, 9, 1, 2, 1, 2, 7, 8, 4, 6, 4, 3, 6, 6, 7, 8, 6, 1, 9, 2, 5, 5, 6, 2, 8, 2, 5, 5, 8, 6, 6, 8, 4, 4, 8, 2, 3, 5, 0, 9, 7, 2, 4, 0, 4, 3, 3, 9, 0, 0, 5, 0

Offset: 0

M. F. Hasler, Nov 11 2019

			0.489536321199649488689875316822650189403586515771912127846436678619255628...

Cf. A329336 (continued fraction).

Cf. A242208 (1 + 2^x = 4^x), A328900 (2^x + 3^x = 4^x), A328904 (1 + 3^x = 5^x), A328905 (1 + 2^x = 5^x), A328907 (1 + 3^x = 6^x).

RealDigits[x /. FindRoot[1 + 2^x == 6^x, {x, 1}, WorkingPrecision -> 120]][[1]] (* Amiram Eldar, Jun 28 2023 *)

print(c=solve(x=0,1, 1+2^x-6^x)); digits(c\.1^default(realprecision))[^-1] \\ [^-1] to discard possibly incorrect last digit. Use e.g. \p999 to get more digits. - M. F. Hasler, Oct 31 2019

A329334 Continued fraction of A328904 = 0.7271601514124259..., solution to 1 + 3^x = 5^x.

Original entry on oeis.org

0, 1, 2, 1, 1, 1, 72, 1, 3, 2, 6, 1, 1, 2, 45, 1, 5, 13, 73, 1, 2, 1, 9, 1, 1, 1, 3, 2, 3, 3, 2, 1, 2, 2, 1, 1, 19, 1, 1, 1, 1, 5, 1, 5, 2, 4, 3, 1, 6, 1, 1, 2, 1, 9, 8, 1, 4, 1, 1, 20, 1, 1, 2, 1, 5, 2, 2, 1, 2, 5, 1, 56, 1, 1, 1, 6, 127, 1, 1, 7, 2, 7, 1, 6, 1, 1, 3, 1, 54, 1, 1, 3, 2, 1, 1, 3

Offset: 0

M. F. Hasler, Nov 11 2019

			0.7271601514124259... = 0 + 1/(1 + 1/(2 + 1/(1 + 1/(1 + 1/(1  + 1/(72 + 1/...))))))

Cf. A328912 (cont. frac. of A242208: 1 + 2^x = 4^x), A328913 (cont. frac. of A328900: 2^x + 3^x = 4^x), A329337 (cont. frac. of A328907: 1 + 3^x = 6^x).

contfrac(c=solve(x=0,1, 1+3^x-5^x))[^-1] \\ discarding possibly incorrect last term. Use e.g. \p999 to get more terms. - M. F. Hasler, Oct 31 2019

cp's OEIS Frontend

A328912 Continued fraction expansion of log_2((sqrt(5)+1)/2) = 0.6942419... = A242208.

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A011373 Number of 1's in binary expansion of Fibonacci(n).

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A328900 Decimal expansion of s = 1.507126591638653..., solution to 2^s + 3^s = 4^s.

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A328904 Decimal expansion of x = 0.7271601514124259243... solution to 1 + 3^x = 5^x.

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A328905 Decimal expansion of the solution x = 0.56389552425993647949... to 1 + 2^x = 5^x.

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A328907 Decimal expansion of the solution x = 0.6009668516... to 1 + 3^x = 6^x.

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A174868 Partial sums of Stern's diatomic series A002487.

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A241002 Decimal expansion of the asymptotic growth rate of the number of odd coefficients in Pascal trinomial triangle mod 2, where coefficients are from (1+x+x^4)^n.

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