A104457 -id:A104457 - cp's OEIS frontend

A257672 Numbers of the form floor(r^i) + floor(s^j), where r = (1 + sqrt(5))/2, s = r^2, i >= 0, j >= 0.

2, 3, 4, 5, 6, 7, 8, 10, 12, 13, 17, 18, 19, 21, 23, 28, 30, 31, 34, 35, 46, 47, 48, 50, 52, 57, 63, 75, 77, 78, 82, 92, 93, 122, 123, 124, 126, 128, 133, 139, 151, 168, 198, 200, 201, 205, 216, 244, 245, 321, 322, 323, 325, 327, 332, 338, 350, 367, 397, 443

Offset: 1

Clark Kimberling, May 03 2015

Cf. A001622, A104457.

Cf. A257671.

r = GoldenRatio; s = r/(r - 1); Take[Union[Flatten[Table[Floor[r^i] + Floor[s^j], {i, 0, 100}, {j, 0, 100}]]], 100]

A309434 a(n) = floor(nIm(2e^(iPi/5))/(Im(2e^(i*Pi/5)) - 1)).

Original entry on oeis.org

6, 13, 20, 26, 33, 40, 46, 53, 60, 66, 73, 80, 87, 93, 100, 107, 113, 120, 127, 133, 140, 147, 154, 160, 167, 174, 180, 187, 194, 200, 207, 214, 220, 227, 234, 241, 247, 254, 261, 267, 274, 281, 287, 294, 301, 308, 314, 321, 328, 334, 341

Offset: 1

Karl V. Keller, Jr., Jun 06 2020

nonn

This is the Beatty sequence for Im(2*e^(i*Pi/5))/(Im(2*e^(i*Pi/5)) - 1).
This is the complement of A335137.
Im(2*e^(i*Pi/5))/(Im(2*e^(i*Pi/5)) - 1) = (5 + sqrt(5))/2 + sqrt(5 + 2*sqrt(5)) = 6.695717525925148250774877410... = 2 + phi + tan(2*Pi/5) = A296184 + A019970.
For n < 10, a(n) = A109235(n).
Re(2*e^(i*Pi/5))/(Re(2*e^(i*Pi/5)) - 1) = (3 + sqrt(5))/2 = 1 + phi = phi^2 = A104457.
Floor(n*Re(2*e^(i*Pi/5))/(Re(2*e^(i*Pi/5)) - 1)) is A001950 (floor(n*phi^2)).

			For n = 3, floor(3*6.69571) = 20.

Karl V. Keller, Jr., Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Beatty Sequence.
Index entries for sequences related to Beatty sequences

Cf. A001950, A019970, A104457, A109235, A296184, A335137.

a[n_] := Floor[n * Im[2 * Exp[I * Pi/5]]/(Im[2 * Exp[I * Pi/5]] - 1)]; Array[a, 100] (* Amiram Eldar, Jul 06 2020 *)

from sympy import floor, im, exp, I, pi
for n in range(1, 101): print(floor(n*im(2*exp(I*pi/5))/(im(2*exp(I*pi/5)) - 1)), end=', ')

from sympy import floor, sqrt
for n in range(1, 101): print(floor(n*((5 + sqrt(5))/2 + sqrt(5 + 2*sqrt(5)))), end=', ')

A337301 Triangle read by rows in which row n lists the closest integers to diagonal lengths of regular n-gon with unit edge length, n >= 4.

Original entry on oeis.org

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

Offset: 4

Mohammed Yaseen, Aug 22 2020

			Triangle begins:
1;
2, 2;
2, 2, 2;
2, 2, 2, 2;
2, 2, 3, 2, 2;
2, 3, 3, 3, 3, 2;
2, 3, 3, 3, 3, 3, 2;
2, 3, 3, 4, 4, 3, 3, 2;
2, 3, 3, 4, 4, 4, 3, 3, 2;
2, 3, 3, 4, 4, 4, 4, 3, 3, 2;
2, 3, 4, 4, 4, 4, 4, 4, 4, 3, 2;
2, 3, 4, 4, 5, 5, 5, 5, 4, 4, 3, 2;
2, 3, 4, 4, 5, 5, 5, 5, 5, 4, 4, 3, 2;
...
Row n lists the closest integers to the length of the diagonals drawn from a fixed vertex of a regular n-gon with unit edge length, n >= 4.
The lengths of the diagonals drawn from vertex A of a regular 8-gon ABCDEFGH with unit edge length are:
AC = 1.84775...
AD = 2.41421...
AE = 2.61312...
AF = 2.41421...
AG = 1.84775...
So the row for n=8 is 2, 2, 3, 2, 2.

Cf. A064313.
Decimal expansion of diagonal lengths of regular n-gons with unit edge length:
n=4  A002193.
n=5  A001622.
n=6  A002194, A000038.
n=7  A160389, A231187.
n=8  A179260, A014176, A121601.
n=9  A332437.
n=10 A188593, A104457, A019970, A134945.
n=11 A231186.
n=12 A188887, A090388, A337402, A019973, A214726.

T[n_,k_]:=Round[Sin[(k+1)*Pi/n]/Sin[Pi/n]]; Flatten[Table[T[n,k],{n,4,16},{k,1,n-3}]] (* Stefano Spezia, Sep 07 2020 *)

T(n,k) = round(sin((k+1)*Pi/n)/sin(Pi/n)), n >= 4, 1 <= k <= n-3.

A351354 Numbers k such that the k-th centered 40-gonal numbers (A195317) is a square.

Original entry on oeis.org

1, 3, 7, 45, 117, 799, 2091, 14329, 37513, 257115, 673135, 4613733, 12078909, 82790071, 216747219, 1485607537, 3889371025, 26658145587, 69791931223, 478361013021, 1252365390981, 8583840088783, 22472785106427, 154030760585065, 403257766524697, 2763969850442379

Offset: 1

Lamine Ngom, Feb 08 2022

nonn

Corresponding square roots are listed in A351353.
3 and 7 are the unique primes in this sequence, a(2*n+1) and a(2*n) always sharing common factors that are closely linked to Fibonacci (A000045) and Lucas (A000032) numbers (detailed in formula section).
In addition, the ratio a(2*n+1)/a(2*n) converges to 2.618033988 ... = golden ratio squared: A104457.

			45 is in the sequence because the 45th centered 40-gonal number is 39601, which is a square: 199^2 = A000032(11)^2.
799 is in the sequence because the 799th centered 40-gonal number is 12752041, which is a square: 3571^2 = A000032(17)^2.

Index entries for linear recurrences with constant coefficients, signature (1,18,-18,-1,1).

Cf. A000032, A000045, A007310, A077259, A104457, A195317, A351353.

a[1] := 1: a[2] := 3: a[3] := 7: a[4] := 45: a[5] := 117:
for n from 6 to 30 do a[n] := a[n - 1] + 18*a[n - 2] - 18*a[n - 3] - a[n - 4] + a[n - 5]: od:
seq(a[n], n = 1 .. 30);

LinearRecurrence[{1, 18, -18, -1, 1}, {1, 3, 7, 45, 117}, 30] (* Amiram Eldar, Feb 08 2022 *)

a(n) = A077259(n-1) + 1.
a(1)=1, a(2)=3, a(3)=7, a(4)=45, a(5)=117 and a(n) = a(n-1) + 18*a(n-2) - 18*a(n-3) - a(n-4) + a(n-5).
gcd(a(2*n+1), a(2*n)) = A000045(n)*(A000032(2*n) - 1)/2, if n is odd.
gcd(a(2*n+1), a(2*n)) = A000032(n)*(A000032(2*n) - 1)/2, if n is even.
A195317(a(n)) = A000032(A007310(n))^2 = A351353(n)^2.

A370944 Decimal expansion of ((1+sqrt(5))/2)sqrt(3) = A001622A002194.

Original entry on oeis.org

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

Offset: 1

Paolo Xausa, Mar 07 2024

Long space diagonal of a regular dodecahedron with unit edges.

			2.80251707688814708935335587064413598888786347955...

Paolo Xausa, Table of n, a(n) for n = 1..10000
Matt Parker, The Five Compound Platonic Solids, YouTube video, 2024.

Cf. A001622, A002194.

Cf. A094887 (short diagonal), A104457 (medium diagonal).

(sqrt(3) + sqrt(15))/2: evalf(%, 86);  # Peter Luschny, Mar 07 2024

First[RealDigits[GoldenRatio*Sqrt[3], 10, 100]]

Equals 2*A179296. - Hugo Pfoertner, Mar 07 2024

A131309 Rabbit-like sequence for phi^2.

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0

Offset: 0

Gary W. Adamson, Jun 27 2007

The terms given can be computed as the iterates of the morphism 1 -> 110, 0 -> 10, with axiom 1, concatenated. - Joerg Arndt, Mar 01 2022
Ratio of 1's to 0's tends to phi^2, by way of example, in the subset of 8 terms (1, 1, 0, 1, 1, 0, 1, 0), there are five 1's and three 0's.
Subsets have A001906: (1, 3, 8, 21, ...) terms, being partial sums of A027941: (1, 4, 12, 33, ...). After 33 total terms, there are (1 + 3 + 8) zeros and (1 + 2 + 5 + 13) = 21 ones; with the ratio of ones to zeros tending to phi^2 = 2.618...

			By rows, we get:
  1;
  1, 1, 0;
  1, 1, 0, 1, 1, 0, 1, 0;
  ...
Then append n-th row to the end of (n-1)-th row, forming a continuous string.

Cf. A027941, A001906, A104457 (phi^2).

Substitution rules T => 2T = t; t => T + t; are derived directly from the matrix generator [2,1; 1,0] (eigenvalue phi^2). Then substitute 1 for T and 0 for t.

A173428 The smallest prime appearing in the truncated version of the decimal expansion of (Golden Ratio)^n shifted iteratively left.

Original entry on oeis.org

1618033, 2618033988749, 42360679, 6854101, 1109, 179, 2903, 469787, 760131556174964248389559523684316960024905121133959373, 1229, 19900502499874064149, 32199689437, 5210019193, 8429

Offset: 1

Michel Lagneau, Feb 18 2010

The n-th power of the golden ratio A001622 is successively shifted left, building floor(A001622^n *10^k) for k = 0, 1, 2, 3,...

As soon as this becomes a prime, we let a(n) be this prime.

			1618033 is the first prime found in the decimal expansion of Golden Ratio A001622, after 6 shifts to the left.
2618033988749 is the first prime found in the decimal expansion of (Golden Ratio)^2, A104457.
42360679 is the first prime found in the decimal expansion of (Golden Ratio)^3, A098317.

Eric Weisstein's World of Mathematics, Phi-Prime

Digits := 200:for n from 1 to 50 do: n0 := evalf(((sqrt(5)+1)/2)^n): for p from 1 to 100 while (type(trunc(10^p*n0),prime)= false) do:od: n2:= trunc(10^p*n0): print (n2): od:

Edited by R. J. Mathar, Feb 24 2010

A281387 Pairs (x, y) of relatively prime positive integers such that (x^2 - 5)/y and (y^2 - 5)/x are both positive integers.

Original entry on oeis.org

4, 11, 11, 29, 29, 76, 76, 199, 199, 521, 521, 1364, 1364, 3571, 3571, 9349, 9349, 24476, 24476, 64079, 64079, 167761, 167761, 439204, 439204, 1149851, 1149851, 3010349, 3010349, 7881196, 7881196, 20633239, 20633239, 54018521, 54018521, 141422324

Offset: 1

Michel Lagneau, Jan 21 2017

nonn

For x, y > 2, the solutions start with (4,11) -> (11, 29) -> (29, 76) -> ...
The sequence is infinite (see the proof in the second reference).
Consider the pairs of the form (a(2n-1), a(2n)). Limit_{n->oo} a(2n)/a(2n-1) = phi^2 = 2.618033988749894... (A104457).
Property: a(2n-1)^2 + a(2n)^2 = 3*a(2n-1)*a(2n) + 5.
Apparently a(2*n) = a(2*n+1) = A002878(n) for n >= 1. - Georg Fischer, Dec 05 2022

Art of Problem Solving, Problem A112
Peter Vandendriessche and Hojoo Lee, Problems in Elementary Number Theory (see problem A112, p. 15). [Via Wayback Machine]

Cf. A001622, A002878, A104457.

nn:=10^6:a:=4:
for b from a+1 to nn do:
x:=(a^2-5)/b:y:=(b^2-5)/a:
if x>0 and y>0 and gcd(a,b)=1 and x=floor(x) and y=floor(y)
then
printf(`%d, `,a): printf(`%d, `,b):a:=b:
else fi:
od:

nn = 10^6; a = 4; Reap[For[b = a+1, b <= nn, b++, x = (a^2-5)/b; y = (b^2-5)/a; If[x>0 && y>0 && GCD[a, b] == 1 && x == Floor[x] && y == Floor[y], Print[a, " ", b]; Sow[a]; Sow[b]; a = b]]][[2, 1]] (* adapted from Maple *)
(* Second program: *)
Clear[a]; a[n_] := 2^(-n-2)*((7-3*Sqrt[5])*(1-Sqrt[5])^n-(-Sqrt[5]-1)^(n+1) - (Sqrt[5]-1)^(n+1) + (3*Sqrt[5]+7)*(Sqrt[5]+1)^n); Table[a[n] // Simplify, {n, 1, 36}] (* Jean-François Alcover, Jan 25 2017 *)

Conjectures from Chai Wah Wu, Jan 28 2024: (Start)
a(n) = 3*a(n-2) - a(n-4) for n > 4.
G.f.: x*(-4*x^3 - x^2 + 11*x + 4)/(x^4 - 3*x^2 + 1). (End)

A341616 Table read by ascending antidiagonals: T(n,j) = Fibonacci(n)*Lucas(n+j), product of the n-th term in the Fibonacci sequence (with F(1)=1 and F(2)=1) and the (n+j)-th term in the Lucas sequence (with L(1)=1 and L(2)=3 and j=0,1,2,...).

Original entry on oeis.org

1, 3, 3, 8, 4, 4, 21, 14, 7, 7, 55, 33, 22, 11, 11, 144, 90, 54, 36, 18, 18, 377, 232, 145, 87, 58, 29, 29, 987, 611, 376, 235, 141, 94, 47, 47, 2584, 1596, 988, 608, 380, 228, 152, 76, 76, 6765, 4182, 2583, 1599, 984, 615, 369, 246, 123, 123

Offset: 1

Jens Rasmussen, Feb 16 2021

j is the offset when combining terms from the two initial sequences.

			T(4,3) = Fibonacci(4)*Lucas(4+3) = 3*29 = 87.
Square array showing T(n,j) begins:
      j=0 j=1 j=2 j=3 j=4  ..
  n=1   1   3   4   7  11  ..
  n=2   3   4   7  11  18  ..
  n=3   8  14  22  36  58  ..
  n=4  21  33  54  87 141  ..
  ...  ..  ..  ..  ..  ..  ..

For j=0 the resulting sequence is used as input in A341414.

Cf. A000045, A000032, A001622, A094214, A104457.

T(n,j) = fibonacci(2*n+j) - (-1)^n*fibonacci(j);
matrix(7,7,n,k, T(n,k-1)) \\ Michel Marcus, Mar 02 2021

For phi=(1+sqrt(5))/2 and tau=(1-sqrt(5))/2:
T(n,j) = Fibonacci(n)*Lucas(n+j).
T(n,j) = (phi^n - tau^n)*(phi^(n+j) + tau^(n+j))/sqrt(5).
T(n,j) = Fibonacci(2n+j) - (-1)^n*Fibonacci(j).
Lim_{n, j -> oo} T(n+1,j)/T(n,j) = phi^2 (A104457).
Lim_{n, j -> oo} T(n,j+1)/T(n,j) = phi (A001622).

A359727 Beattific 'primes': numbers n > 1 not equal to floor(kmphi) or floor(kmphi^2) for any smaller element k in this sequence and any positive integer m.

Original entry on oeis.org

2, 4, 7, 8, 13, 14, 17, 23, 24, 28, 30, 39, 40, 43, 46, 49, 50, 53, 59, 65, 66, 70, 72, 75, 76, 81, 86, 88, 92, 96, 98, 107, 108, 114, 117, 118, 123, 127, 134, 140, 143, 144, 149, 150, 153, 156, 159, 160, 163, 166, 175, 176, 179, 182, 185, 191, 195

Offset: 1

James Propp, Jan 11 2023

nonn

Given r = (1+sqrt(5))/2 and s = r^2, we sieve the set {2,3,4,...}, where each time we discover a new "prime" p, we sieve out the numbers floor(p*r), floor(2p*r), floor(3p*r), ... and floor(p*s), floor(2p*s), floor(3p*s), ... It appears that significantly more than half the terms are even.

			The Beattific prime 2 causes us to sieve out 3, 6, 9, ... and 5, 10, ...; then the next Beattific prime, 4, doesn't cause us to throw out anything new; then the next Beattific prime is 7.

Index entries for sequences generated by sieves

Cf. A001622, A104457, A000201, A001950.

bp[limit_] := (*Find all the Beattific primes up to limit*)
  Module[{r = (1 + Sqrt[5])/2, sieve = ConstantArray[1, limit]},
   Do[If[sieve[[n]] == 1,
     sieve[[Table[Floor[k n r], {k, (limit + 1)/(n r)}]]] = 0;
     sieve[[Table[Floor[k n r r], {k, (limit + 1)/(n r r)}]]] = 0],
    {n, 2, limit}];
   Rest@Flatten@Position[sieve, 1]];

h(n)=(n+sqrtint(5*n^2))\2
list(lim)=my(v=vectorsmall(lim\=1,i,1),u=List()); for(n=2,#v, if(v[n]==0, next); listput(u,n); forstep(k=n,h(lim+1)-lim-1,n, v[h(k)]=0); forstep(k=n,2*lim+1-h(lim+1),n, v[h(k)+k]=0)); Vec(u) \\ Charles R Greathouse IV, Jan 25 2023

cp's OEIS Frontend

A257672 Numbers of the form floor(r^i) + floor(s^j), where r = (1 + sqrt(5))/2, s = r^2, i >= 0, j >= 0.

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A309434 a(n) = floor(n*Im(2*e^(i*Pi/5))/(Im(2*e^(i*Pi/5)) - 1)).

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A337301 Triangle read by rows in which row n lists the closest integers to diagonal lengths of regular n-gon with unit edge length, n >= 4.

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Formula

A351354 Numbers k such that the k-th centered 40-gonal numbers (A195317) is a square.

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Maple

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A370944 Decimal expansion of ((1+sqrt(5))/2)*sqrt(3) = A001622*A002194.

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A131309 Rabbit-like sequence for phi^2.

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A173428 The smallest prime appearing in the truncated version of the decimal expansion of (Golden Ratio)^n shifted iteratively left.

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Extensions

A281387 Pairs (x, y) of relatively prime positive integers such that (x^2 - 5)/y and (y^2 - 5)/x are both positive integers.

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Maple

A309434 a(n) = floor(nIm(2e^(iPi/5))/(Im(2e^(i*Pi/5)) - 1)).

A370944 Decimal expansion of ((1+sqrt(5))/2)sqrt(3) = A001622A002194.

A359727 Beattific 'primes': numbers n > 1 not equal to floor(kmphi) or floor(kmphi^2) for any smaller element k in this sequence and any positive integer m.