Patrick D McLean - cp's OEIS frontend

A302126 Interleaved Fibonacci and Lucas numbers.

0, 2, 1, 1, 1, 3, 2, 4, 3, 7, 5, 11, 8, 18, 13, 29, 21, 47, 34, 76, 55, 123, 89, 199, 144, 322, 233, 521, 377, 843, 610, 1364, 987, 2207, 1597, 3571, 2584, 5778, 4181, 9349, 6765, 15127, 10946, 24476, 17711, 39603, 28657, 64079, 46368, 103682, 75025, 167761

Offset: 0

Patrick D McLean, Apr 01 2018

			a(10) = Fibonacci(5) = 5;
a(11) = Lucas(5) = 11.

Colin Barker, Table of n, a(n) for n = 0..1000
T. Crilly, Interleaving Integer Sequences, The Mathematical Gazette, Vol. 91, No. 520 (Mar., 2007), pp. 27-33.
Wikipedia, Interleave sequence
Index entries for linear recurrences with constant coefficients, signature (0,1,0,1).

Interleaves A000045 and A000032.

Flat(List([1..25],n->[Fibonacci(n),Lucas(1,-1,n)[2]])); # Muniru A Asiru, Apr 02 2018

a:= n-> (<<0|1>, <1|1>>^iquo(n, 2, 'r'). <<2*r, 1>>)[1, 1]:
seq(a(n), n=0..60);  # Alois P. Heinz, Apr 23 2018

Table[{Fibonacci[n], LucasL[n]}, {n, 0, 25}] // Flatten
LinearRecurrence[{0, 1, 0, 1}, {0, 2, 1, 1}, 52]
Flatten@ Array[{LucasL@#, Fibonacci@#} &, 26, 0] (* or *)
CoefficientList[Series[(x^3 - x^2 - 2x)/(x^4 + x^2 - 1), {x, 0, 51}], x] (* Robert G. Wilson v, Apr 02 2018 *)

concat(0, Vec(x*(2 - x)*(1 + x) / (1 - x^2 - x^4) + O(x^60))) \\ Colin Barker, Apr 02 2018

a(0) = 0; a(1) = 2; a(2) = 1; a(3) = 1; a(n) = a(n-2) + a(n-4), n >= 4.
G.f.: x*(2 - x)*(1 + x) / (1 - x^2 - x^4). - Colin Barker, Apr 02 2018
a(0) = 0; a(1) = 2; a(2n) = (a(2n-1) + a(2n-2))/2; a(2n+1) = a(2n) + 2*a(2n-2), n >= 1. - Daniel Forgues, Jul 29 2018

More terms from Colin Barker, Apr 02 2018

A284786 Pisano period of sequence A006054 modulo n.

Original entry on oeis.org

1, 7, 26, 14, 62, 182, 42, 28, 78, 434, 266, 182, 12, 42, 806, 56, 614, 546, 254, 434, 546, 266, 1106, 364, 310, 84, 234, 42, 28, 5642, 1986, 112, 3458, 4298, 1302, 546, 2814, 1778, 156, 868, 40, 546, 42, 266, 2418, 1106, 4514, 728, 294, 2170, 7982, 84, 5726, 1638, 8246, 84, 3302, 28, 7082, 5642, 582

Offset: 1

Patrick D McLean, Apr 02 2017

nonn

Robert Israel, Mathematics StackExchange, When does x^3-x^2-2x+1 split mod p.
Wikipedia, Pisano period.

Cf. A001175 Pisano periods of Fibonacci numbers mod n.

Cf. A045472.

f:= proc(n) option remember; local F, t, k, a;
F:= ifactors(n)[2];
if nops(F) > 1 then
  return(ilcm(seq(procname(t[1]^t[2]),t=F)))
fi;
a:= [0,0,1];
for k from 1 do
  a:= [a[2],a[3],2*a[3]+a[2]-a[1] mod n];
  if  a = [0,0,1] then return k fi;
od:
end proc:
f(1):= 1:
map(f, [$1..100]); # Robert Israel, Jun 14 2017

Table[FindTransientRepeat[
    Mod[LinearRecurrence[{2, 1, -1}, {0, 0, 1}, 100000], n], 2] //
   Last // Length, {n, 1, 20}]

From Robert Israel, Jun 14 2017: (Start)
If m and n are coprime, a(m*n) = lcm(a(m),a(n)).
If p is a prime such that the polynomial x^3-x^2-2*x+1 splits into distinct factors mod p, then a(p) divides p-1. These primes are A045472. (End)

More terms from Robert Israel, Jun 14 2017

A262773 A Beatty sequence: a(n)=floor(q*n) where q=A231187.

Original entry on oeis.org

0, 2, 4, 6, 8, 11, 13, 15, 17, 20, 22, 24, 26, 29, 31, 33, 35, 38, 40, 42, 44, 47, 49, 51, 53, 56, 58, 60, 62, 65, 67, 69, 71, 74, 76, 78, 80, 83, 85, 87, 89, 92, 94, 96, 98, 101, 103, 105, 107, 110, 112, 114, 116, 119, 121, 123, 125, 128, 130, 132, 134, 137, 139

Offset: 0

Patrick D McLean, Sep 30 2015

nonn

Beatty sequence of the longer diagonal (A231187) in a regular heptagon with sidelength 1.

Complement of Beatty sequence A262770 of the longer diagonal (A160389) in a regular heptagon with sidelength 1.

Peter Steinbach, Golden Fields: A Case for the Heptagon, Mathematics Magazine, Vol. 70, No. 1, Feb. 1997
Index entries for sequences related to Beatty sequences

Complement of A262770.

Table[Floor[n/(2 Cos[3 Pi/7])], {n, 0, 106}] (* Michael De Vlieger, Oct 05 2015 *)

q=roots([1,-2,-1,1])(1); a(n)=floor(q*n)

a(n) = floor(n/(2*cos(3*Pi/7))) \\ Michel Marcus, Oct 05 2015

A262770 A Beatty sequence: a(n)=floor(np) where p=2cos(Pi/7)=A160389.

Original entry on oeis.org

0, 1, 3, 5, 7, 9, 10, 12, 14, 16, 18, 19, 21, 23, 25, 27, 28, 30, 32, 34, 36, 37, 39, 41, 43, 45, 46, 48, 50, 52, 54, 55, 57, 59, 61, 63, 64, 66, 68, 70, 72, 73, 75, 77, 79, 81, 82, 84, 86, 88, 90, 91, 93, 95, 97, 99, 100, 102, 104, 106, 108, 109, 111, 113, 115, 117, 118, 120, 122, 124, 126, 127, 129, 131, 133, 135, 136, 138, 140, 142, 144, 145, 147, 149, 151, 153, 154, 156, 158, 160, 162, 163, 165, 167, 169, 171, 172, 174, 176, 178, 180, 181, 183, 185, 187, 189, 191

Offset: 0

Patrick D McLean, Sep 30 2015

nonn

Beatty sequence of the shorter diagonal (A160389) in a regular heptagon with sidelength 1.
Complement of Beatty sequence A262773 of the longer diagonal (A231187) in a regular heptagon with sidelength 1.
First 106 terms agree with A187318, but A187318(106)=190 while A262770(106)=191.

Peter Steinbach, Golden Fields: A Case for the Heptagon, Mathematics Magazine, Vol. 70, No. 1, Feb. 1997.
Index entries for sequences related to Beatty sequences

Complement of A262773.

Initially agrees with A187318 (because 2*cos(Pi/7) is close to 9/5).

Table[Floor[2 n Cos[Pi/7]], {n, 0, 106}] (* Michael De Vlieger, Oct 05 2015 *)

p=roots([1,-1,-2,1])(1); a(n)=floor(p*n)

a(n) = floor(n*2*cos(Pi/7)); \\ Michel Marcus, Oct 05 2015

A239526 For 0<=n<=100, a(n) is the number of positive responses x such that x/N rounds to n%, minimized over sample size N.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 2, 3, 1, 3, 2, 3, 4, 1, 5, 3, 5, 2, 3, 4, 6, 1, 10, 6, 4, 3, 3, 7, 2, 7, 5, 3, 4, 5, 6, 7, 10, 17, 1, 18, 11, 8, 7, 6, 5, 4, 7, 10, 3, 11, 5, 5, 7, 11, 19, 2, 13, 9, 7, 5, 13, 8, 14, 3, 13, 10, 7, 11, 4, 13, 9, 5, 16, 11, 6, 7, 7, 8, 9, 10, 11, 13, 15, 18, 22, 28, 39, 66, 1

Offset: 0

Patrick D McLean, Mar 21 2014

			a(31)=4 because 4/13=0.31 (2DP).

Cf. A239525 (Minimal sample sizes).

Table[LinearProgramming[{1, 0}, {{-n/100 + 0.005, 1}, {n/100 + 0.005, -1}}, {0, 0}, {1, 1}, Integers], {n, 0, 100}] // Transpose // Last

from itertools import count
def A239526(n):
    for y in count(1):
        x, z = divmod(y*((n<<1)+1),200)
        if not z: return x
        x, z = divmod(y*((n<<1)-1),200)
        if (x:=x+bool(z)) and (200*x+y)//(y<<1) == n:
            return x # Chai Wah Wu, Jun 28 2025

A239525 For 0 <= n <= 100, a(n) is smallest integer N such that some positive x/N rounds to n%, with x > 0.

Original entry on oeis.org

200, 67, 40, 29, 23, 19, 16, 14, 12, 11, 10, 9, 8, 8, 7, 13, 19, 6, 11, 16, 5, 14, 9, 13, 17, 4, 19, 11, 18, 7, 10, 13, 19, 3, 29, 17, 11, 8, 8, 18, 5, 17, 12, 7, 9, 11, 13, 15, 21, 35, 2, 35, 21, 15, 13, 11, 9, 7, 12, 17, 5, 18, 8, 8, 11, 17, 29, 3, 19, 13, 10, 7, 18, 11, 19, 4, 17, 13, 9, 14, 5, 16, 11, 6, 19, 13, 7, 8, 8, 9, 10, 11, 12, 14, 16, 19, 23, 29, 40, 67, 1

Offset: 0

Patrick D McLean, Mar 21 2014

When the fractional part is 0.5, rounding in either direction is allowed.
From Jonathan Dushoff, Feb 11 2025: (Start)
The sequence arises when considering how to "reverse-engineer" published percentages: A quoted n percent means that proportion was within at least a total of a(n) things.
Having a(n) based on rounding either way when mid-way between two integer percentages makes the terms agnostic as to what rounding was actually used in the published percentage.
(End)

			a(31)=13 because 4/13 = 0.31 (to two digits after the decimal point).

Cf. A239526 (corresponding x).

Table[LinearProgramming[{1, 0}, {{-n/100 + 0.005, 1}, {n/100 + 0.005, -1}}, {0, 0}, {1, 1}, Integers], {n, 0, 100}] // Transpose // First

from itertools import count
def A239525(n):
    for y in count(1):
        if not y*((n<<1)+1)%200: return y
        x, z = divmod(y*((n<<1)-1),200)
        if (a:=x+bool(z)) and (200*a+y)//(y<<1) == n:
            return y # Chai Wah Wu, Jun 28 2025

Find the smallest N such that there is some x > 0 with abs(100*x/N - n) <= 0.5.

Edited by Jonathan Dushoff, Apr 22 2022

A033917 Coefficients of iterated exponential function defined by y(x) = x^y(x) for e^-e < x < e^(1/e), expanded about x=1.

Original entry on oeis.org

1, 1, 2, 9, 56, 480, 5094, 65534, 984808, 16992144, 330667680, 7170714672, 171438170232, 4480972742064, 127115833240200, 3889913061111240, 127729720697035584, 4479821940873927168, 167143865005981109952, 6610411989494027218368, 276242547290322145178880

Offset: 0

Patrick D McLean

a(n) is the n-th derivative of x^(x^...(x^(x^x))) with n x's evaluated at x=1. - Alois P. Heinz, Oct 14 2016

Alois P. Heinz, Table of n, a(n) for n = 0..400
R. Arthur Knoebel, Exponentials Reiterated, Amer. Math. Monthly 88 (1981), pp. 235-252.
Eric Weisstein's World of Mathematics, Power Tower
Wikipedia, Knuth's up-arrow notation
Wikipedia, Tetration

Cf. A052880, A215703, A216349, A216350.
Row sums of A277536.
Main diagonal of A277537.

a:= n-> add(Stirling1(n, k)*(k+1)^(k-1), k=0..n):
seq(a(n), n=0..25);  # Alois P. Heinz, Aug 31 2012

mx = 20; Table[ i! SeriesCoefficient[ InverseSeries[ Series[ y^(1/y), {y, 1, mx}]], i], {i, 0, n}] (* modified by Robert G. Wilson v, Feb 03 2013 *)
CoefficientList[Series[-LambertW[-Log[1+x]]/Log[1+x], {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Nov 27 2012 *)

Stirling1(n, k)=n!*polcoeff(binomial(x, n), k)
a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, sum(k=0, m, Stirling1(m, k)*(A+x*O(x^n))^k)*x^m/m!)); n!*polcoeff(A, n)
for(n=0,20,print1(a(n),", ")) \\ Paul D. Hanna, Mar 09 2013

E.g.f.: -LambertW(-log(1+x))/log(1+x). a(n) = Sum_{k=0..n} Stirling1(n, k)*(k+1)^(k-1). - Vladeta Jovovic, Nov 12 2003
a(n) ~ n^(n-1) / ( exp(n -3/2 + exp(-1)/2) * (exp(exp(-1))-1)^(n-1/2) ). - Vaclav Kotesovec, Nov 27 2012
E.g.f.: A(x) satisfies A(x) = Sum_{n>=0} x^n/n! * Sum_{k=0..n} Stirling1(n,k) * A(x)^k. - Paul D. Hanna, Mar 09 2013
a(n) = n! * [x^n] (x+1)^^n. - Alois P. Heinz, Oct 19 2016

cp's OEIS Frontend

User: Patrick D McLean

A302126 Interleaved Fibonacci and Lucas numbers.

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A284786 Pisano period of sequence A006054 modulo n.

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A262773 A Beatty sequence: a(n)=floor(q*n) where q=A231187.

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A262770 A Beatty sequence: a(n)=floor(n*p) where p=2*cos(Pi/7)=A160389.

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A239526 For 0<=n<=100, a(n) is the number of positive responses x such that x/N rounds to n%, minimized over sample size N.

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A239525 For 0 <= n <= 100, a(n) is smallest integer N such that some positive x/N rounds to n%, with x > 0.

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A033917 Coefficients of iterated exponential function defined by y(x) = x^y(x) for e^-e < x < e^(1/e), expanded about x=1.

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A262770 A Beatty sequence: a(n)=floor(np) where p=2cos(Pi/7)=A160389.