A001590 -id:A001590 - cp's OEIS frontend

A242576 Prime terms in A214828.

13, 151, 277, 36313, 225949, 7129366889, 933784181621, 19397107178326126131136629644898891137047, 401151570474397232184569825031979125080583558010764826781295643008140597581801

Offset: 1

Robert Price, May 17 2014

nonn

a(10) has 119 digits and thus is too large to display here. It corresponds to A214828(448).

Michel Marcus, Table of n, a(n) for n = 1..17

Cf. A001590, A100683, A231574, A231575, A232543, A214899, A020992, A233554. A214727, A234696, A141523, A235862, A214825, A235873, A242324, A214827, A214828.

a={1,6,6}; For[n=3, n<=1000, n++, sum=Plus@@a; If[PrimeQ[sum], Print[sum]]; a=RotateLeft[a]; a[[3]]=sum]
Select[LinearRecurrence[{1,1,1},{1,6,6},350],PrimeQ] (* Harvey P. Dale, Jul 21 2018 *)

my(x='x+O('x^500)); select(isprime, Vec((1+5*x-x^2)/(1-x-x^2-x^3))) \\ Michel Marcus, Jun 16 2025

A243623 Prime terms in A214829.

Original entry on oeis.org

7, 29, 1087, 1999, 3677, 6763, 5487349608898607, 115507410616162687, 878001744429057971864287, 210582098197038415344728317608265501, 870277059555114378903885645581650740066907

Offset: 1

Robert Price, Jun 07 2014

nonn

a(12) has 114 digits and thus is too large to display here. It corresponds to A214829(426).

Robert Israel, Table of n, a(n) for n = 1..20

Cf. A001590, A100683, A231574, A231575, A232543, A214899, A020992, A233554. A214727, A234696, A141523, A235862,A214825, A235873, A214827, A242324, A214827, A214828, A214829, A242572, A242576, A243622.

f:= gfun:-rectoproc({a(n) = a(n-1) + a(n-2) + a(n-3), a(0) = 1, a(1) = 7, a(2) = 7},a(n),remember):
select(isprime, map(f, [$2..1000])); # Robert Israel, Sep 02 2024

a={1,7,7}; Print["7"]; Print["7"]; For[n=3, n<=1000, n++, sum=Plus@@a; If[PrimeQ[sum], Print[sum]]; a=RotateLeft[a]; a[[3]]=sum]

7 inserted as a(1) by Robert Israel, Sep 02 2024

A247192 Indices of primes in the hexanacci numbers sequence A000383.

Original entry on oeis.org

7, 9, 30, 31, 33, 46, 52, 54, 82, 102, 109, 124, 210, 301, 351, 365, 369, 1045, 2044, 2125, 2143, 2815, 4377, 4754, 4893, 7310, 11558, 17602, 17929, 28389, 32100, 44298, 106725, 151678, 197953

Offset: 1

Robert Price, Dec 03 2014

a(36) > 2*10^5.

Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas n-step Sequences, J. of Integer Sequences, Vol. 8 (2005), Article 05.4.4

Cf. A001590, A001631, A100683, A231574, A231575, A232542, A214899, A230607, A020992, A232498, A214727, A081172, A214752, A141523, A214825, A235862, A000288, A000322, A000383, A249413.

a={1,1,1,1,1}; For[n=5, n<=1000, n++, sum=Plus@@a; If[PrimeQ[sum], Print[n]]; a=RotateLeft[a]; a[[5]]=sum]

A248920 Indices of primes in the pentanacci numbers sequence A000322.

Original entry on oeis.org

5, 7, 13, 18, 19, 34, 35, 38, 43, 48, 188, 286, 450, 501, 759, 1446, 2021, 2419, 2997, 3715, 5677, 13566, 46303, 57174, 108844, 117145, 166683, 178863

Offset: 1

Robert Price, Oct 16 2014

a(29) > 2*10^5.

Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas n-step Sequences, J. of Integer Sequences, Vol. 8 (2005), Article 05.4.4

Cf. A001590, A001631, A100683, A231574, A231575, A232542, A214899, A230607, A020992, A232498, A214727, A081172, A214752, A141523, A214825, A235862, A000288, A000322.

a={1,1,1,1,1}; For[n=5, n<=1000, n++, sum=Plus@@a; If[PrimeQ[sum], Print[n]]; a=RotateLeft[a]; a[[5]]=sum]

A278044 Length of tribonacci representation of n (cf. A278038).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Nov 18 2016

For n>=2, n appears A001590(n+2) times. - John Keith, May 23 2022

N. J. A. Sloane, Table of n, a(n) for n = 0..20000
Wolfdieter Lang, The Tribonacci and ABC Representations of Numbers are Equivalent, arXiv preprint arXiv:1810.09787 [math.NT], 2018.

Cf. A278038, A278042, A278043.
Cf. A001590.
Similar to, but strictly different from, A201052.

t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; a[0] = 1; a[n_] := Module[{k = 1}, While[t[k] <= n, k++]; k - 1]; Array[a, 100, 0] (* Amiram Eldar, Mar 04 2022 *)

a(n) = A278042(n) + A278043(n).

A000308 a(n) = a(n-1)a(n-2)a(n-3) with a(1)=1, a(2)=2 and a(3)=3.

Original entry on oeis.org

1, 2, 3, 6, 36, 648, 139968, 3265173504, 296148833645101056, 135345882205792807436868315512832, 130876399105969522361889021452224949874232743897657526714368

Offset: 1

N. J. A. Sloane

nonn

			a(6)=36*6*3=648.

John Cerkan, Table of n, a(n) for n = 1..15

Cf. A000304, A000336, A063401.

A000308 := proc(n) option remember; if n <=3 then n else A000308(n-1)*A000308(n-2)*A000308(n-3); fi; end;

a(n) = 2^A001590(n-1)*3^A000073(n-1). - Henry Bottomley, Jul 16 2001

A061282 Minimal number of steps to get from 0 to n by (a) adding 1 or (b) multiplying by 3. A stopping problem: begin with n and at each stage if a multiple of 3 divide by 3, otherwise subtract 1.

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 3, 4, 5, 3, 4, 5, 4, 5, 6, 5, 6, 7, 4, 5, 6, 5, 6, 7, 6, 7, 8, 4, 5, 6, 5, 6, 7, 6, 7, 8, 5, 6, 7, 6, 7, 8, 7, 8, 9, 6, 7, 8, 7, 8, 9, 8, 9, 10, 5, 6, 7, 6, 7, 8, 7, 8, 9, 6, 7, 8, 7, 8, 9, 8, 9, 10, 7, 8, 9, 8, 9, 10, 9, 10, 11, 5, 6, 7, 6, 7, 8, 7, 8, 9, 6, 7, 8, 7, 8, 9, 8, 9, 10, 7, 8

Offset: 0

Henry Bottomley, Jun 06 2001

n > 0 occurs A001590(n+2) times in this sequence. - Peter Kagey, Jul 19 2015

a(n) gives the number of iterations of A260316 to reach 0. - Peter Kagey, Jul 22 2015

			a(25)=7 since 25=((0+1+1)*3+1+1)*3+1.

Alois P. Heinz, Table of n, a(n) for n = 0..10000
Index to sequences related to the complexity of n

Cf. A001590, A007089, A053735, A056792, A062153, A260316.

Analogous sequences with a different multiplier k: A056792 (k=2), A260112 (k=4).

c i = if i `mod` 3 == 0 then i `div` 3 else i - 1
b 0 foldCount = foldCount
b sheetCount foldCount = b (c sheetCount) (foldCount + 1)
a061282 n = b n 0 -- Peter Kagey, Sep 02 2015

a:= n-> (l-> nops(l)+add(i, i=l)-1)(convert(n, base, 3)):
seq(a(n), n=0..105);  # Alois P. Heinz, Jul 16 2015

a(n)=sumdigits(n,3)+#digits(n,3)-1 \\ Charles R Greathouse IV, Jul 16 2015

a(n) = A062153(n) + A053735(n) = (number of base 3 digits of n) + (sum of base 3 digits of n)-1. a(3n) = a(n)+1, a(3n+1) = a(n)+2, a(3n+2) = a(n)+3; a(0)=0, a(1)=1, a(2)=2.

A077988 Expansion of 1/(1+2x-2x^3).

Original entry on oeis.org

1, -2, 4, -6, 8, -8, 4, 8, -32, 72, -128, 192, -240, 224, -64, -352, 1152, -2432, 4160, -6016, 7168, -6016, 0, 14336, -40704, 81408, -134144, 186880, -210944, 153600, 66560, -555008, 1417216, -2701312, 4292608, -5750784, 6098944, -3612672, -4276224, 20750336, -48726016, 88899584

Offset: 0

N. J. A. Sloane, Nov 17 2002

G. C. Greubel, Table of n, a(n) for n = 0..1000
J. Pan, Multiple Binomial Transforms and Families of Integer Sequences , J. Int. Seq. 13 (2010), 10.4.2, T^(-1).
Index entries for linear recurrences with constant coefficients, signature (-2, 0, 2).

Cf. A000073, A001590, A000213, A077940, A135491.

a:=[1,-2,4];; for n in [4..50] do a[n]:=-2*a[n-1]+2*a[n-3]; od; a; # G. C. Greubel, Jun 25 2019

R:=PowerSeriesRing(Integers(), 50); Coefficients(R!( 1/(1+2*x-2*x^3) )); // G. C. Greubel, Jun 25 2019

LinearRecurrence[{-2, 0, 2}, {1, -2, 4}, 50] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)
CoefficientList[Series[1/(1+2x-2x^3),{x,0,50}],x] (* Harvey P. Dale, May 21 2024 *)

my(x='x+O('x^50)); Vec(1/(1+2*x-2*x^3)) \\ G. C. Greubel, Jun 25 2019

(1/(1+2*x-2*x^3)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Jun 25 2019

a(n) = (-1)^n * A077940(n). - G. C. Greubel, Jun 25 2019

A078046 Expansion of (1-x)/(1 + x + x^2 - x^3).

Original entry on oeis.org

1, -2, 1, 2, -5, 4, 3, -12, 13, 2, -27, 38, -9, -56, 103, -56, -103, 262, -215, -150, 627, -692, -85, 1404, -2011, 522, 2893, -5426, 3055, 5264, -13745, 11536, 7473, -32754, 36817, 3410, -72981, 106388, -29997, -149372, 285757, -166382, -268747, 720886, -618521, -371112, 1710519, -1957928

Offset: 0

N. J. A. Sloane, Nov 17 2002

The root of the denominator [1 + x + x^2 - x^3] is the tribonacci constant.

This is the negative of the tribonacci numbers, signature (0, 1, 0), in reverse order, starting from A001590(-1), going backwards A001590(-2), A001590(-3), ... - Peter M. Chema, Dec 31 2016

			G.f. = 1 - 2*x + x^2 + 2*x^3 - 5*x^4 + 4*x^5 + 3*x^6 - 12*x^7 + 13*x^8 + ...

Benjamin Braun, W. K. Hough, Matching and Independence Complexes Related to Small Grids, arXiv preprint arXiv:1606.01204 [math.CO], 2016.
Wesley K. Hough, On Independence, Matching, and Homomorphism Complexes, (2017), Theses and Dissertations--Mathematics, 42.
Index entries for linear recurrences with constant coefficients, signature (-1,-1,1).

First differences of A057597.

Cf. A001590.

a[ n_] := If[ n >= 0, SeriesCoefficient[ (1 - x) / (1 + x + x^2 - x^3), {x, 0, n}], SeriesCoefficient [ -x^2 (1 - x) / (1 - x - x^2 - x^3),{x, 0, -n}]]; (* Michael Somos, Jun 01 2014 *)

Vec((1-x)/(1+x+x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

{a(n) = if( n>=0, polcoeff( (1 - x) / (1 + x + x^2 - x^3) + x * O(x^n), n), polcoeff( -x^2 * (1 - x) / (1 - x - x^2 - x^3) + x * O(x^-n), -n))}; /* Michael Somos, Jun 01 2014 */

G.f.: (1-x)/(1+x+x^2-x^3).
Recurrence: a(n) = a(n-3) - a(n-2) - a(n-1) for n > 2.
a(-1 - n) = - A001590(n). - Michael Somos, Jun 01 2014

A099328 Number of Catalan knight paths from (0,0) to (n,0) in the region between and on the lines y=0 and y=3. (See A096587 for the definition of Catalan knight.).

Original entry on oeis.org

1, 0, 1, 0, 2, 2, 8, 8, 21, 28, 69, 108, 226, 370, 736, 1280, 2473, 4392, 8281, 14920, 27874, 50706, 94088, 171880, 317693, 582116, 1073853, 1970836, 3630914, 6669730, 12279296, 22568896, 41533777, 76360464, 140493041, 258344528, 475256898

Offset: 1

Clark Kimberling, Oct 12 2004

			a(6) counts 8 paths from (0,0) to (6,0); the final move in 5 of the paths is from the point (5,2) and the final move in the other 3 paths is from (4,1).

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

Cf. A099329, A099330, A099331.

LinearRecurrence[{1,1,-1,3,1,1,-1},{1,0,1,0,2,2,8},40] (* Harvey P. Dale, Aug 11 2017 *)

Taking A099328 to A099331 as the rows of an array T, the recurrences for these row sequences are given for n>=2 by T(n, 0) = T(n-1, 2) + T(n-2, 1), T(n, 1) = T(n-1, 3) + T(n-2, 0) + T(n-2, 2), T(n, 2) = T(n-1, 0) + T(n-2, 1) + T(n-2, 3), T(n, 3) = T(n-1, 1) + T(n-2, 2), with initial values T(0, 0)=1, T(1, 2)=1.
From Chai Wah Wu, Aug 09 2016: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3) + 3*a(n-4) + a(n-5) + a(n-6) - a(n-7) for n > 7.
G.f.: x*(1 - x - 2*x^4)/((x^4 - 2*x^3 - 1)*(x^3 + x^2 + x - 1)). (End)
2*a(n) = A001590(n)-(-1)^n*( A052922(n-1)+A052922(n-3)) . - R. J. Mathar, Nov 22 2024

cp's OEIS Frontend

A242576 Prime terms in A214828.

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A243623 Prime terms in A214829.

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A247192 Indices of primes in the hexanacci numbers sequence A000383.

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A248920 Indices of primes in the pentanacci numbers sequence A000322.

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Mathematica

A278044 Length of tribonacci representation of n (cf. A278038).

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A000308 a(n) = a(n-1)*a(n-2)*a(n-3) with a(1)=1, a(2)=2 and a(3)=3.

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A061282 Minimal number of steps to get from 0 to n by (a) adding 1 or (b) multiplying by 3. A stopping problem: begin with n and at each stage if a multiple of 3 divide by 3, otherwise subtract 1.

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A077988 Expansion of 1/(1+2*x-2*x^3).

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A000308 a(n) = a(n-1)a(n-2)a(n-3) with a(1)=1, a(2)=2 and a(3)=3.

A077988 Expansion of 1/(1+2x-2x^3).