A001644 -id:A001644 - cp's OEIS frontend

A074475 a(n) = Sum_{j=0..floor(n/2)} T(2j + q), where T(n) are generalized tribonacci numbers (A001644) and q = n - 2floor(n/2).

3, 1, 6, 8, 17, 29, 56, 100, 187, 341, 630, 1156, 2129, 3913, 7200, 13240, 24355, 44793, 82390, 151536, 278721, 512645, 942904, 1734268, 3189819, 5866989, 10791078, 19847884, 36505953, 67144913, 123498752, 227149616, 417793283

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 23 2002

a(n) is the convolution of T(n) with the sequence (1,0,1,0,1,0,...).

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

Cf. A001644, A074331.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (3+x)/(1-2*x^2-2*x^3-x^4) )); // G. C. Greubel, Apr 21 2019

CoefficientList[Series[(3+x)/(1-2*x^2-2*x^3-x^4), {x, 0, 40}], x]
LinearRecurrence[{0,2,2,1},{3,1,6,8},40] (* Harvey P. Dale, Jul 08 2017 *)

my(x='x+O('x^40)); Vec((3+x)/(1-2*x^2-2*x^3-x^4)) \\ G. C. Greubel, Apr 21 2019

((3+x)/(1-2*x^2-2*x^3-x^4)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 21 2019

a(n) = 2*a(n-2) + 2*a(n-3) + a(n-4), a(0)=3, a(1)=1, a(2)=6, a(3)=8.

G.f.: (3+x)/(1-2*x^2-2*x^3-x^4).

A075092 Sum of generalized tribonacci numbers (A001644) and reflected generalized tribonacci numbers (A073145).

Original entry on oeis.org

6, 0, 2, 12, 6, 20, 50, 56, 134, 264, 402, 836, 1542, 2652, 5154, 9392, 16902, 31824, 58082, 106172, 197126, 360932, 662994, 1223784, 2245766, 4130520, 7606770, 13976436, 25711622, 47310252, 86978370, 160002656, 294324230, 541249952

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 31 2002

Conjecture: a(n) >= 0.

For n > 2, a(n) is the number of cyclic sequences (q1, q2, ..., qn) consisting of zeros, ones and twos such that each triple contains 0 and 1 at least once, provided the positions of the zeros and ones are fixed on a circle. For example, a(5)=20 because only the sequences (00101), (01001), (01010), (01011), (01012), (01021), (01101), (01201), (02101), (20101) and those obtained from them by exchanging 0 and 1 contain 0 and 1 in each triple (including triples q4, q5, q1 and q5, q1, q2). For n = 1, 2 the statement is still true provided we allow the sequence to wrap around itself on a circle. E.g., a(2) = 2 since only sequences 01 and 10 can be wrapped so one obtains (010) and (101), respectively. - Wojciech Florek, Nov 25 2021

G. C. Greubel, Table of n, a(n) for n = 0..1000
Wojciech Florek, A class of generalized Tribonacci sequences applied to counting problems, Appl. Math. Comput., 338 (2018), 809-821.
W. O. J. Moser, Cyclic binary strings without long runs of like (alternating) bits, Fibonacci Quart. 31(1) (1993), 2-6.
Index entries for linear recurrences with constant coefficients, signature (0,1,4,1,0,-1).

Cf. A001644, A073145, A075091, A294627.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (6-4*x^2-12*x^3-2*x^4)/(1-x^2-4*x^3-x^4+x^6) ));  // G. C. Greubel, Apr 13 2019

CoefficientList[Series[(6-4*x^2-12*x^3-2*x^4)/(1-x^2-4*x^3-x^4+x^6), {x, 0, 40}], x]

my(x='x+O('x^40)); Vec((6-4*x^2-12*x^3-2*x^4)/(1-x^2-4*x^3-x^4+x^6)) \\ G. C. Greubel, Apr 13 2019

((6-4*x^2-12*x^3-2*x^4)/(1-x^2-4*x^3-x^4+x^6)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 13 2019

a(n) = a(n-2) + 4*a(n-3) + a(n-4) - a(n-6), a(0)=6, a(1)=0, a(2)=2, a(3)=12, a(4)=6, a(5)=20.

G.f.: (6 - 4*x^2 - 12*x^3 - 2*x^4)/(1 - x^2 - 4*x^3 - x^4 + x^6).

A075536 a(n) = ((1+(-1)^n)T(n+1) + (1-(-1)^n)S(n))/2, where T(n) = tribonacci numbers A000073, S(n) = generalized tribonacci numbers A001644.

Original entry on oeis.org

0, 1, 1, 7, 4, 21, 13, 71, 44, 241, 149, 815, 504, 2757, 1705, 9327, 5768, 31553, 19513, 106743, 66012, 361109, 223317, 1221623, 755476, 4132721, 2555757, 13980895, 8646064, 47297029, 29249425, 160004703, 98950096, 541292033, 334745777

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Sep 23 2002

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0, 3, 0, 1, 0, 1).

Cf. A000073, A001644, A005013, A005247.

R:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x*(1+x+4*x^2+x^3-x^4)/(1-3*x^2-x^4-x^6) )); // G. C. Greubel, Apr 21 2019

A075536 := proc(n)
    if type(n,'even') then
        A000073(n+1) ;
    else
        A001644(n) ;
    end if;
end proc:
seq(A075536(n),n=0..80) ; # R. J. Mathar, Aug 05 2021

CoefficientList[Series[(x+x^2+4x^3+x^4-x^5)/(1-3x^2-x^4-x^6), {x, 0, 40}], x]
LinearRecurrence[{0,3,0,1,0,1},{0,1,1,7,4,21},40] (* Harvey P. Dale, Jul 10 2012 *)

my(x='x+O('x^40)); concat([0], Vec(x*(1+x+4*x^2+x^3-x^4)/(1-3*x^2-x^4-x^6))) \\ G. C. Greubel, Apr 21 2019

(x*(1+x+4*x^2+x^3-x^4)/(1-3*x^2-x^4-x^6)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 21 2019

a(2n) = A073717(n) = A000073(2n+1).
a(2n+1) = A001644(2n+1).
a(n) = 3*a(n-2) + a(n-4) + a(n-6), a(0)=0, a(1)=1, a(2)=1, a(3)=7, a(4)=4, a(5)=21.
O.g.f.: x*(1 + x + 4*x^2 + x^3 - x^4)/(1 - 3*x^2 - x^4 - x^6).

Index in definition corrected. - R. J. Mathar, Aug 05 2021

A105285 Indices of Lucas 3-step numbers A001644 which have a nontrivial divisor in common with index.

Original entry on oeis.org

6, 15, 18, 21, 35, 39, 44, 45, 54, 55, 57, 78, 80, 84, 90, 93, 96, 117, 120, 123, 132, 133, 135, 140, 147, 154, 156, 162, 171, 174, 195, 201, 210, 213, 234, 235, 240, 245, 247, 249, 252, 259, 264, 273, 275, 279, 286, 288, 290, 291, 295, 299, 312, 318, 323, 327

Offset: 1

Jonathan Vos Post, Apr 25 2005

Extension by T. D. Noe. Wanted: closed-form formula for this as exists for Fibonacci and Lucas numbers. See also A105762 (prime Lucas 3-step numbers).

			gcd(6, A001644(6)) = gcd(6,39) = 3,
gcd(21, A001644(21)) = gcd(21,361109) = 7.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001644, A104714, A105762.

m=300; s = LinearRecurrence[{1, 1, 1}, {3, 1, 3}, m+1]; Select[Range[m], !CoprimeQ[#, s[[#+1]]] &] (* Amiram Eldar, Sep 05 2019 *)

gcd(a(n), A001644(a(n))) > 1.

More terms from Amiram Eldar, Sep 05 2019

A105762 Primes in A001644 (the Lucas 3-step numbers).

Original entry on oeis.org

3, 7, 11, 71, 131, 241, 443, 1499, 196331, 86992799, 541292033, 292997064989357251, 129824812729295169371, 238785058551151434437, 5026368970977284897651, 105803877284856287511991

Offset: 1

T. D. Noe, Apr 22 2005

nonn

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
Eric Weisstein's World of Mathematics, Lucas n-Step Number

Cf. A104576 (indices of prime Lucas 3-step numbers).

a={-1, -1, 3}; lst={}; Do[s=Plus@@a; a=RotateLeft[a]; a[[ -1]]=s; If[PrimeQ[s], AppendTo[lst, s]], {n, 1000}]; lst

a(n) = A001644(A104576(n)). - Arthur O'Dwyer, Jul 31 2024

Name clarified by Arthur O'Dwyer, Jul 31 2024

A106789 Sum of two consecutive squares of Lucas 3-step numbers (A001644).

Original entry on oeis.org

10, 10, 58, 170, 562, 1962, 6562, 22202, 75242, 254330, 860474, 2911226, 9848050, 33316090, 112707970, 381286954, 1289885834, 4363653034, 14762129274, 49939929610, 168945571442, 571538767370, 1933501811618, 6540989771354

Offset: 0

Jonathan Vos Post, May 16 2005

A106729 is sum of two consecutive squares of Lucas numbers (A001254), for which L(n)^2 + L(n+1)^2 = 5*{F(n)^2 + F(n+1)^2} = 5*A001519(n). Sum of two consecutive squares of Lucas 3-step numbers can be expressed in terms of tribonacci numbers, but not quite as neatly, as derived from the identity A001644(n) = T(n) + 2*T(n-1) + 3*T(n-2) = 3*T(n+1) - 2*T(n) - T(n-1) where the tribonacci numbers T(n) = A000073(n).

			a(0) = A001644(0)^2 + A001644(1)^2 = 3^2 + 1^2 = 9 + 1 = 10.
a(1) = A001644(1)^2 + A001644(2)^2 = 1^2 + 3^2 = 1 + 9 = 10.
a(2) = A001644(2)^2 + A001644(3)^2 = 3^2 + 7^2 = 9 + 49 = 58.
a(3) = A001644(3)^2 + A001644(4)^2 = 7^2 + 11^2 = 49 + 121 = 170 = 13^2 + 1.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000073, A001644.

Cf. A001254, A001519, A106729.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 2*(5-5*x+4*x^2-18*x^3-x^4-5*x^5)/(1-2*x-3*x^2-6*x^3+x^4+x^6) )); // G. C. Greubel, Apr 21 2019

CoefficientList[Series[2*(5-5*x+4*x^2-18*x^3-x^4-5*x^5)/(1-2*x-3*x^2 -6*x^3+x^4+x^6), {x,0,40}], x] (* G. C. Greubel, Apr 21 2019 *)
Total/@Partition[LinearRecurrence[{1,1,1},{3,1,3},40]^2,2,1] (* Harvey P. Dale, Apr 03 2022 *)

my(x='x+O('x^40)); Vec(2*(5-5*x+4*x^2-18*x^3-x^4-5*x^5)/(1-2*x-3*x^2-6*x^3+x^4+x^6)) \\ G. C. Greubel, Apr 21 2019

(2*(5-5*x+4*x^2-18*x^3-x^4-5*x^5)/(1-2*x-3*x^2-6*x^3+x^4+x^6)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 21 2019

a(n) = A001644(n)^2 + A001644(n+1)^2.

G.f.: 2*(5 - 5*x + 4*x^2 - 18*x^3 - x^4 - 5*x^5)/((1+x+x^2-x^3)*(1-3*x-x^2-x^3)).

A073748 a(n) = S(n)*S(n-1), where S(n) are the generalized tribonacci numbers A001644.

Original entry on oeis.org

-3, 3, 3, 21, 77, 231, 819, 2769, 9301, 31571, 106763, 361045, 1221685, 4132743, 13980747, 47297217, 160004685, 541291715, 1831178355, 6194830005, 20956959933, 70896891079, 239842458947, 811381229009, 2744883043045, 9285872805715, 31413882695739, 106272403946805

Offset: 0

Maio Catalani (mario.catalani(AT)unito.it), Aug 08 2002

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2, 3, 6, -1, 0, -1).

Cf. A001644, A073145.

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (-3+9*x+6*x^2+24*x^3+5*x^4-x^5)/(1-2*x-3*x^2-6*x^3 +x^4+x^6) )); // G. C. Greubel, Apr 21 2019

CoefficientList[Series[(-3+9*x+6*x^2+24*x^3+5*x^4-x^5)/(1-2*x-3*x^2-6*x^3 +x^4+x^6), {x, 0, 30}], x]
Join[{-3},Times@@@Partition[LinearRecurrence[{1,1,1},{3,1,3},30],2,1]] (* or *) LinearRecurrence[{2,3,6,-1,0,-1},{-3,3,3,21,77,231},30] (* Harvey P. Dale, Nov 18 2013 *)

my(x='x+O('x^30)); Vec((-3+9*x+6*x^2+24*x^3+5*x^4-x^5)/(1-2*x-3*x^2-6*x^3 +x^4+x^6)) \\ G. C. Greubel, Apr 21 2019

((-3+9*x+6*x^2+24*x^3+5*x^4-x^5)/(1-2*x-3*x^2-6*x^3 +x^4+x^6)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Apr 21 2019

a(n) = S(2*n-1) + C(n-1) - C(n-2), where S(n) is A001644, C(n) is A073145.
G.f.: (-3 + 9*x + 6*x^2 + 24*x^3 + 5*x^4 - x^5)/(1 - 2*x - 3*x^2 - 6*x^3 + x^4 + x^6).
a(0)=-3, a(1)=3, a(2)=3, a(3)=21, a(4)=77, a(5)=231, a(n) = 2*a(n-1) + 3*a(n-2) + 6*a(n-3) - a(n-4) - a(n-6). - Harvey P. Dale, Nov 18 2013

A073782 a(n) = Sum_{k=0..n} S(k)*S(n-k), convolution of S=A001644 with itself.

Original entry on oeis.org

9, 6, 19, 48, 89, 190, 391, 784, 1577, 3142, 6219, 12256, 24041, 46974, 91471, 177568, 343753, 663814, 1278979, 2459152, 4719417, 9041470, 17294039, 33030320, 62999145, 120006214, 228327099, 433939904, 823854793, 1562602238

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 11 2002

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,1,0,-3,-2,-1).

Cf. A001644.

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (3-2*x-x^2)^2/(1-x-x^2-x^3)^2 )); // G. C. Greubel, Apr 12 2019

CoefficientList[Series[(3-2x-x^2)^2/(1-x-x^2-x^3)^2, {x, 0, 30}], x]

my(x='x+O('x^30)); Vec((3-2*x-x^2)^2/(1-x-x^2-x^3)^2) \\ G. C. Greubel, Apr 12 2019

((3-2*x-x^2)^2/(1-x-x^2-x^3)^2).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Apr 12 2019

G.f.: (3-2*x-x^2)^2/(1-x-x^2-x^3)^2.

A074582 a(n) = S(3n), where S(n) is the generalized tribonacci sequence A001644.

Original entry on oeis.org

3, 7, 39, 241, 1499, 9327, 58035, 361109, 2246915, 13980895, 86992799, 541292033, 3368061131, 20956960551, 130399710235, 811381230021, 5048627019523, 31413882696791, 195465425009943, 1216237188605169, 7567747077883259

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 24 2002

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-5,1).

Cf. A001644, A074581.

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (3-14*x+5*x^2)/(1-7*x+5*x^2-x^3) )); // G. C. Greubel, Apr 13 2019

CoefficientList[Series[(3-14*x+5*x^2)/(1-7*x+5*x^2-x^3), {x, 0, 30}], x]
LinearRecurrence[{7,-5,1},{3,7,39},30] (* Harvey P. Dale, Mar 24 2022 *)

my(x='x+O('x^30)); Vec((3-14*x+5*x^2)/(1-7*x+5*x^2-x^3)) \\ G. C. Greubel, Apr 13 2019

((3-14*x+5*x^2)/(1-7*x+5*x^2-x^3)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Apr 13 2019

a(n) = 7*a(n-1) - 5*a(n-2) + a(n-3), a(0)=3, a(1)=7, a(2)=39.

G.f.: (3-14*x+5*x^2)/(1-7*x+5*x^2-x^3).

A075418 Sum of generalized tribonacci numbers A001644 and inverted tribonacci numbers A075298.

Original entry on oeis.org

4, 2, -2, 12, 12, 10, 54, 68, 108, 282, 422, 772, 1604, 2674, 5006, 9580, 16884, 31506, 58606, 105948, 196508, 362298, 662022, 1222772, 2249116, 4127210, 7605718, 13984148, 25701652, 47311458, 86994846, 159975004, 294336612, 541281698, 995529822, 1831291692, 3367998380, 6194717674

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Sep 14 2002

It seems that aside from a(2) the sequence is nonnegative.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0, 1, 4, 1, 0, -1).

Cf. A001644, A075298.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (4+2*x-6*x^2-6*x^3+2*x^4+4*x^5)/(1-x^2-4*x^3-x^4+x^6) )); // G. C. Greubel, Apr 21 2019

CoefficientList[Series[(4+2x-6x^2-6x^3+2x^4+4x^5)/(1-x^2-4x^3-x^4+x^6), {x, 0, 40}], x]

my(x='x+O('x^40)); Vec((4+2*x-6*x^2-6*x^3+2*x^4+4*x^5)/(1-x^2 -4*x^3-x^4+x^6)) \\ G. C. Greubel, Apr 21 2019

((4+2*x-6*x^2-6*x^3+2*x^4+4*x^5)/(1-x^2-4*x^3-x^4+x^6)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 21 2019

a(n) = a(n-2) + 4*a(n-3) + a(n-4) - a(n-6), a(0)=4, a(1)=2, a(2)=-2, a(3)=12, a(4)=12, a(5)=10.

O.g.f.: (4 + 2*x - 6*x^2 - 6*x^3 + 2*x^4 + 4*x^5)/(1 - x^2 - 4*x^3 - x^4 + x^6).

cp's OEIS Frontend

A074475 a(n) = Sum_{j=0..floor(n/2)} T(2*j + q), where T(n) are generalized tribonacci numbers (A001644) and q = n - 2*floor(n/2).

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A075092 Sum of generalized tribonacci numbers (A001644) and reflected generalized tribonacci numbers (A073145).

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A075536 a(n) = ((1+(-1)^n)*T(n+1) + (1-(-1)^n)*S(n))/2, where T(n) = tribonacci numbers A000073, S(n) = generalized tribonacci numbers A001644.

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A105285 Indices of Lucas 3-step numbers A001644 which have a nontrivial divisor in common with index.

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A105762 Primes in A001644 (the Lucas 3-step numbers).

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A106789 Sum of two consecutive squares of Lucas 3-step numbers (A001644).

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A073748 a(n) = S(n)*S(n-1), where S(n) are the generalized tribonacci numbers A001644.

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A074475 a(n) = Sum_{j=0..floor(n/2)} T(2j + q), where T(n) are generalized tribonacci numbers (A001644) and q = n - 2floor(n/2).

A075536 a(n) = ((1+(-1)^n)T(n+1) + (1-(-1)^n)S(n))/2, where T(n) = tribonacci numbers A000073, S(n) = generalized tribonacci numbers A001644.