A017581 -id:A017581 - cp's OEIS frontend

A089911 a(n) = Fibonacci(n) mod 12.

0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1

Offset: 0

Casey Mongoven, Nov 14 2003

From Reinhard Zumkeller, Jul 05 2013: (Start)
Sequence has been applied by several composers to 12-tone equal temperament pitch structure. The complete Fibonacci mod 12 system (a set of 10 periodic sequences) exhausts all possible ordered dyads; that is, every possible combination of two pitches is found in these sets.
a(A008594(n)) = 0;
a(A227144(n)) = 1;
a(3*A047522(n)) = 2;
a(A017569(n)) = a(2*A016933(n)) = a(4*A016777(n)) = 3;
a(2*A017629(n)) = a(3*A017137(n)) = a(6*A004767(n)) = 4;
a(A227146(n)) = 5;
a(nonexistent) = 6;
a(2*A017581(n)) = 7;
a(2*A017557(n)) = a(4*A016813(n)) = 8;
a(A017617(n)) = a(2*A016957(n)) = a(4*A016789(n)) = 9;
a(3*A047621(n)) = 10;
a(2*A017653(n)) = 11. (End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..1199
Ron Knott, Fibonacci Numbers and the Golden Section
M. Renault, The Fibonacci Sequence Modulo M
A. P. Shah, Fibonacci Sequence Modulo m, Fibonacci Quarterly, Vol.6, No.2 (1968), 139-141.
D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.
Index entries for linear recurrences with constant coefficients, signature (1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1).

Cf. A000045, A001175, A003893, A010881.

a089911 n = a089911_list !! n
a089911_list = 0 : 1 : zipWith (\u v -> (u + v) `mod` 12)
                       (tail a089911_list) a089911_list
-- Reinhard Zumkeller, Jul 01 2013

[Fibonacci(n) mod 12: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014

with(combinat,fibonacci); A089911 := proc(n) fibonacci(n) mod 12; end;

Table[Mod[Fibonacci[n], 12], {n, 0, 100}] (* Vincenzo Librandi, Feb 04 2014 *)

a(n)=fibonacci(n)%12 \\ Charles R Greathouse IV, Feb 03 2014

Has period of 24, restricted period 12 and multiplier 5.

a(n) = (a(n-1) + a(n-2)) mod 12, a(0) = 0, a(1) = 1.

More terms from Ray Chandler, Nov 15 2003

A126979 a(n) = 24*n + 233.

Original entry on oeis.org

233, 257, 281, 305, 329, 353, 377, 401, 425, 449, 473, 497, 521, 545, 569, 593, 617, 641, 665, 689, 713, 737, 761, 785, 809, 833, 857, 881, 905, 929, 953, 977, 1001, 1025, 1049, 1073, 1097, 1121, 1145, 1169, 1193, 1217, 1241, 1265, 1289, 1313, 1337, 1361

Offset: 0

Robert H Barbour, Mar 20 2007, Jun 12 2007

Superhighway created by 'LQTL Ant' L45R135L45R135 from iteration 233 where the Ant moves in a 'Moore neighborhood' (nine cells), the L indicates a left turn, the R a right turn, and the numerical value is the turn angle in degrees.

P. Sakar, "A Brief History of Cellular Automata," ACM Computing Surveys, vol. 32, 2000.
S. Wolfram, A New Kind of Science, 1st ed. Il.: Wolfram Media Inc., 2002.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A031041, A017581, A126978, A126980. Has many terms in common with A031041.

a:=[233, 257];; for n in [3..60] do a[n]:=2*a[n-1]-a[n-2]; od; a; # G. C. Greubel, May 28 2019

I:=[233, 257]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..60]]; // G. C. Greubel, May 28 2019

Table[24*n + 233, {n, 0, 60}] (* Stefan Steinerberger, Jun 17 2007 *)
LinearRecurrence[{2,-1}, {233,257}, 60] (* G. C. Greubel, May 28 2019 *)

my(x='x+O('x^60)); Vec((233-209*x)/(1-x)^2) \\ G. C. Greubel, May 28 2019

((233-209*x)/(1-x)^2).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, May 28 2019

From Chai Wah Wu, May 30 2016: (Start)
a(n) = 2*a(n-1) - a(n-2) for n > 1.
G.f.: (233 - 209*x)/(1 - x)^2. (End)
E.g.f.: (233 + 24*x)*exp(x). - G. C. Greubel, May 28 2019

More terms from Stefan Steinerberger, Jun 17 2007

A351974 a(n) is the first maximum reached by iterating the reduced Collatz function R on 4n-1: a(n) = R^s(4n-1), where R(k) = A139391(k) and s the number of iterations required.

Original entry on oeis.org

5, 17, 17, 53, 29, 53, 41, 161, 53, 89, 65, 161, 77, 125, 89, 485, 101, 161, 113, 269, 125, 197, 137, 485, 149, 233, 161, 377, 173, 269, 185, 1457, 197, 305, 209, 485, 221, 341, 233, 809, 245, 377, 257, 593, 269, 413, 281, 1457, 293, 449, 305, 701, 317, 485

Offset: 1

Ya-Ping Lu, Feb 26 2022

Iterating R on 4n-1 (n>=1) starts with an increasing trajectory before reaching the first maximum. However, iterating R on 4n-3 (n>=1) starts with a decreasing trajectory before reaching 1 or the first minimum.

			For n = 1, iterating R on 4n-1=3 gives 3->5->1, in which the first maximum is 5, and thus a(0) = 5.
For n = 8, iterating R on 4n-1=31 gives 31->47->71->107->161->121->91->137->103->155->233->175...->23->35->53->5->1, in which the first maximum is 161, and thus a(8) = 161.

Cf. A004767, A017581, A001511, A139391, A075684.

a(n) = my(s=valuation(n,2)); n>>(s-1)*3^(s+1) - 1; \\ Kevin Ryde, Feb 28 2022

def A351974(n): s = (n&-n).bit_length(); return 4*n*3**s//2**s - 1

a(n) = 4*n*(3/2)^s - 1, where s = A001511(n).
a(n) == 5 (mod 12).
For s >= 1 and m >= 0, a(2^s*m+2^(s-1)) = 2*(3^s)*(2m+1) - 1. For example, a(2m+1) = 12m+5 = A017581(m); a(4m+2) = 36m+17; and a(8m+4) = 108m+53.

A360962 Square array T(n,k) = k((3+6n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

Original entry on oeis.org

0, 0, 1, 0, 4, 5, 0, 7, 17, 12, 0, 10, 29, 39, 22, 0, 13, 41, 66, 70, 35, 0, 16, 53, 93, 118, 110, 51, 0, 19, 65, 120, 166, 185, 159, 70, 0, 22, 77, 147, 214, 260, 267, 217, 92, 0, 25, 89, 174, 262, 335, 375, 364, 284, 117, 0, 28, 101, 201, 310, 410, 483, 511, 476, 360, 145

Offset: 0

Paul Curtz, Feb 27 2023

The main diagonal is A024394.

The antidiagonals sums are A000537.

			The rows are:
  0  1  5  12  22  35  51  70 ... = A000326
  0  4 17  39  70 110 159 217 ... = A022266
  0  7 29  66 118 185 267 364 ... = A022272
  0 10 41  93 166 260 375 511 ... = A022278
  0 13 53 120 214 335 483 658 ... = A022284
  ... .
Columns: A000004, A016777, A017581, A154266=3*A017209, 2*A348845, 5*A161447, 3*A158057(n+1), ... (coefficients from A026741).
Difference between two consecutive rows are: A033428.
This square array read by antidiagonals leads to the triangle
  0
  0  1
  0  4  5
  0  7 17 12
  0 10 29 39  22
  0 13 41 66  70  35
  0 16 53 93 118 110 51
  ... .

Cf. A000326, A000537, A022266, A022272, A022278, A022284, A024394.
Cf. A033428.
Cf. A000004, A016777, A017209, A017581, A026741, A154266, A158057, A161447, A348845.

T:= (n,k)-> k*(k*(3+6*n)-1)/2:
seq(seq(T(d-k,k), k=0..d), d=0..10);  # Alois P. Heinz, Feb 28 2023

T[n_, k_] := ((6*n + 3)*k - 1)*k/2; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Feb 27 2023 *)

T(n,k) = k*((3+6*n)*k-1)/2; \\ Michel Marcus, Feb 27 2023

Take successively sequences n*(3*n-1)/2, n*(9*n-1)/2, n*(15*n-1)/2, n*(21*n-1)/2, ... listed in the EXAMPLE section.
From Stefano Spezia, Feb 21 2024: (Start)
G.f.: y*(1 + 2*y + x*(2 + y))/((1 - x)^2*(1 - y)^3).
E.g.f.: exp(x+y)*y*(2 + 3*y + 6*x*(1 + y))/2. (End)

A017582 a(n) = (12n + 5)^2.

Original entry on oeis.org

25, 289, 841, 1681, 2809, 4225, 5929, 7921, 10201, 12769, 15625, 18769, 22201, 25921, 29929, 34225, 38809, 43681, 48841, 54289, 60025, 66049, 72361, 78961, 85849, 93025, 100489, 108241, 116281

Offset: 0

N. J. A. Sloane

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

lst={};Do[AppendTo[lst,(12*n+5)^2],{n,0,5!}];lst (* Vladimir Joseph Stephan Orlovsky, Jun 29 2009 *)
CoefficientList[Series[(25 + 214 x + 49 x^2) / (1 - x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 28 2014 *)
(12*Range[0,30]+5)^2 (* or *) LinearRecurrence[{3,-3,1},{25,289,841},30] (* Harvey P. Dale, Jul 19 2019 *)

a(n)=(12*n+5)^2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = A017581(n)^2. - Michel Marcus, Dec 28 2014

G.f.: (25 + 214*x + 49*x^2)/(1-x)^3. - Vincenzo Librandi, Dec 28 2014

A017590 a(n) = (12*n+5)^10.

Original entry on oeis.org

9765625, 2015993900449, 420707233300201, 13422659310152401, 174887470365513049, 1346274334462890625, 7326680472586200649, 31181719929966183601, 110462212541120451001, 339456738992222314849

Offset: 0

N. J. A. Sloane

Index entries for linear recurrences with constant coefficients, signature (11, -55, 165, -330, 462, -462, 330, -165, 55, -11, 1).

Cf. A008454, A017581.

A017590:=(12*n+5)^10: seq(A017590(n), n=0..20); # Wesley Ivan Hurt, Apr 03 2017

(12*Range[0,20]+5)^10 (* Harvey P. Dale, Feb 19 2015 *)

A110598 Balanced numbers k such that k mod 12 = 5.

Original entry on oeis.org

137885, 145145, 3501605, 6605945, 6953765, 8409305, 10055045, 11413205, 11569805, 16540205, 18545285, 19648805, 21902705, 22806905, 25965005, 26655005, 29811665, 45680921, 71569745, 79989845, 91681289, 196492205, 214218389, 223086125, 229554941, 233601641

Offset: 1

Walter Kehowski, Sep 13 2005

nonn

For the first 26 terms, the quotient sigma(k)/phi(k) is 2 or 3.

Amiram Eldar, Table of n, a(n) for n = 1..7013 (terms below 6.5*10^14, calculated using data from Jud McCranie)

Intersection of A017581 and A020492.

Cf. A000010, A000203, A062699, A068391.

with(numtheory); BNM5:=[]: for z from 1 to 1 do for m from 1 to 1000000 do n:=12*m+5; if sigma(n) mod phi(n) = 0 then BNM5:=[op(BNM5),n] fi; od; od; BNM5;

Select[Range[5,12000000,12],MemberQ[{2,3},DivisorSigma[1,#]/EulerPhi[#]]&]  (* Harvey P. Dale, May 06 2012 *)

a(10)-a(26) from Donovan Johnson, Aug 30 2012

A160080 Lodumo_4 of Fibonacci numbers.

Original entry on oeis.org

0, 1, 5, 2, 3, 9, 4, 13, 17, 6, 7, 21, 8, 25, 29, 10, 11, 33, 12, 37, 41, 14, 15, 45, 16, 49, 53, 18, 19, 57, 20, 61, 65, 22, 23, 69, 24, 73, 77, 26, 27, 81, 28, 85, 89, 30, 31, 93, 32, 97, 101, 34, 35, 105, 36, 109, 113, 38, 39, 117, 40, 121, 125, 42, 43, 129, 44, 133, 137, 46

Offset: 0

Philippe Deléham, May 01 2009

Permutation of nonnegative numbers.

OEIS wiki, Lodumo transform
Index entries for sequences that are permutations of the nonnegative integers
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,2,0,0,0,0,0,-1).

Cf. A004767, A008586, A016825, A017533, A017581, A017629, A079343.

a(n) = lod_4(A000045(n)).
From Philippe Deléham, Nov 30 2023: (Start)
a(n) = 2*a(n-6) - a(n-12) for n >= 12.
a(6*n) = 4*n, a(6*n+1) = 12*n+1, a(6*n+2) = 12*n+5, a(6*n+3) = 4*n+2, a(6*n+4) = 4*n+3, a(6*n+5) = 12*n+9.
G.f.: (x + 5*x^2 + 2*x^3 + 3*x^4 + 9*x^5 + 4*x^6 + 11*x^7 + 7*x^8 + 2*x^9 + x^10 + 3*x^11) / ((1-x)^2*(1+x+x^2)^2*(1+x^3)^2). (End)

A361226 Square array T(n,k) = k((1+2n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 0, 2, 5, 3, 0, 3, 9, 12, 6, 0, 4, 13, 21, 22, 10, 0, 5, 17, 30, 38, 35, 15, 0, 6, 21, 39, 54, 60, 51, 21, 0, 7, 25, 48, 70, 85, 87, 70, 28, 0, 8, 29, 57, 86, 110, 123, 119, 92, 36, 0, 9, 33, 66, 102, 135, 159, 168, 156, 117, 45

Offset: 0

Paul Curtz, Mar 05 2023

The main diagonal is A002414.
The first upper diagonal is A160378(n+1).
The antidiagonals sums are A034827(n+2).
b(n) = (A034827(n+3) = 0, 2, 10, 30, 70, ...) - (A002414(n) = 0, 1, 9, 30, 70, ...) = 0, 1, 1, 0, 0, 5, 21, 56, ... .
b(n+2) = A299120(n). b(n+4) = A033275(n). b(n+4) - b(n) = A002492(n).

			The rows are
  0, 0,  1,  3,  6,  10,  15,  21, ...   = A161680
  0, 1,  5, 12, 22,  35,  51,  70, ...   = A000326
  0, 2,  9, 21, 38,  60,  87, 119, ...   = A005476
  0, 3, 13, 30, 54,  85, 123, 168, ...   = A022264
  0, 4, 17, 39, 70, 110, 159, 217, ...   = A022266
  ... .
Columns: A000004, A001477, A016813, A017197=3*A016777, 2*A017101, 5*A016873, 3*A017581, 7*A017017, ... (coefficients from A026741).
Difference between two consecutive rows: A000290. Hence A143844.
This square array read by antidiagonals leads to the triangle
  0
  0   0
  0   1   1
  0   2   5   3
  0   3   9  12   6
  0   4  13  21  22  10
  0   5  17  30  38  35  15
  ... .

Cf. A000004, A000290, A000326, A001477, A002414, A005476, A016777, A016813, A016873, A017017, A017101, A017197, A017581, A022264, A022266, A026741, A034827, A160378, A161680, A360962.

Cf. A002492, A033275, A143844, A299120.

T[n_, k_] := k*((2*n + 1)*k - 1)/2; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Mar 05 2023 *)

a(n) = { my(row = (sqrtint(8*n+1)-1)\2, column = n - binomial(row + 1, 2)); binomial(column, 2) + column^2 * (row - column) } \\ David A. Corneth, Mar 05 2023

# Seen as a triangle:
from functools import cache
@cache
def Trow(n: int) -> list[int]:
    if n == 0: return [0]
    r = Trow(n - 1)
    return [r[k] + k * k if k < n else r[n - 1] + n - 1 for k in range(n + 1)]
for n in range(7): print(Trow(n)) # Peter Luschny, Mar 05 2023

Take successively sequences n*(n-1)/2, n*(3*n-1)/2, n*(5*n-1)/2, ... listed in the EXAMPLE section.
G.f.: y*(x + y)/((1 - y)^3*(1 - x)^2). - Stefano Spezia, Mar 06 2023
E.g.f.: exp(x+y)*y*(2*x + y + 2*x*y)/2. - Stefano Spezia, Feb 21 2024

A367705 Coefficients of expansion of (1 + 5x + 11x^2 + 5x^3 + 7x^4 + x^5)/(1 - x^3)^2 in powers of x.

Original entry on oeis.org

1, 5, 11, 7, 17, 23, 13, 29, 35, 19, 41, 47, 25, 53, 59, 31, 65, 71, 37, 77, 83, 43, 89, 95, 49, 101, 107, 55, 113, 119, 61, 125, 131, 67, 137, 143, 73, 149, 155, 79, 161, 167, 85, 173, 179, 91, 185, 191, 97, 197, 203, 103, 209, 215, 109, 221, 227, 115, 233, 239

Offset: 0

Philippe Deléham, Nov 27 2023

Based on an idea of Pierre CAMI.

Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A006369, A007310, A016921, A017581, A017653, A130196.

CoefficientList[Series[(1 + 5*x + 11*x^2 + 5*x^3 + 7*x^4 + x^5)/(1 - x^3)^2, {x, 0, 60}], x] (* or *)
LinearRecurrence[{0, 0, 2, 0, 0, -1}, {1, 5, 11, 7, 17, 23}, 60] (* Amiram Eldar, Nov 28 2023 *)

a(n) = 3*A006369(n) + A130196(n).
a(n) = A007310(A006369(n) + 1).
a(n) = 2*a(n-3) - a(n-6) for n >= 6.
a(3*n) = 6*n+1, a(3*n+1) = 12*n+5, a(3*n+2) = 12*n+11.
Sum_{n>=0} (-1)^n/a(n) = ((2+sqrt(2))*Pi + sqrt(3)*log(7+4*sqrt(3)) + sqrt(6)*log(5-2*sqrt(6)))/12. - Amiram Eldar, Nov 28 2023

cp's OEIS Frontend

A089911 a(n) = Fibonacci(n) mod 12.

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A126979 a(n) = 24*n + 233.

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A351974 a(n) is the first maximum reached by iterating the reduced Collatz function R on 4n-1: a(n) = R^s(4n-1), where R(k) = A139391(k) and s the number of iterations required.

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A360962 Square array T(n,k) = k*((3+6*n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

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A017582 a(n) = (12n + 5)^2.

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A017590 a(n) = (12*n+5)^10.

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A110598 Balanced numbers k such that k mod 12 = 5.

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A360962 Square array T(n,k) = k((3+6n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

A361226 Square array T(n,k) = k((1+2n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

A367705 Coefficients of expansion of (1 + 5x + 11x^2 + 5x^3 + 7x^4 + x^5)/(1 - x^3)^2 in powers of x.