cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A097353 Number of digits of the (10^n)-th tetranacci number (A000078(10^n)).

Original entry on oeis.org

1, 2, 28, 284, 2849, 28500, 285008, 2850083, 28500834, 285008350, 2850083504, 28500835049, 285008350498, 2850083504986, 28500835049863, 285008350498633, 2850083504986335, 28500835049863359, 285008350498633597, 2850083504986335973
Offset: 0

Views

Author

Michael Taktikos, Sep 17 2004

Keywords

Comments

a(n)/10^n converges to 0.28500835...

Examples

			Let t(n) = A000078(n). Then we have t(1) = 0, t(10) = 56, t(100) = 2505471397838180985096739296, with respectively 1, 2, 28 and 284 digits.
		

Programs

  • Maple
    # This Maple code will at least get the first few terms correctly!
    f:=proc(n) option remember; if n <= 2 then RETURN(0); fi; if n = 3 then RETURN(1); fi; f(n-1) + f(n-2) + f(n-3) +f(n-4); end; for n from 0 to 4 do lprint(f(10^n), length(f(10^n))); od;
  • Mathematica
    a = b = c = 0; d = i = 1; Do[e = a + b + c + d; a = b; b = c; c = d; d = e; If[n == 10^i, Print[Length[IntegerDigits[e]]]; i++ ], {n, 4, 10^6}] (* Ryan Propper, Jul 22 2005 *)
  • PARI
    \p 100 x=solve(x=1.9274,1.9276,x^4-x^3-x^2-x-1); r=solve(x=0.2937,0.2939,563*x^4-20*x^2-5*x-1); for(k=1,25,n=10^k;print1(floor( (log(r)+(n-2)*log(x))/log(10) )+1",")) \\ Herman Jamke (hermanjamke(AT)fastmail.fm), May 01 2007

Formula

a(n) = floor(log_10(r) + (10^n-2)*log_10(x)) + 1 for n >= 1, where x is the positive real root of the tetranacci limit equation x^4 - x^3 - x^2 - x - 1 = 0, x = 1.92756... and r is the positive real root of the tetranacci auxiliary equation 563r^4 - 20r^2 - 5r - 1 = 0, r = 0.293813... - Herman Jamke (hermanjamke(AT)fastmail.fm), May 01 2007

Extensions

2 more terms from Ryan Propper, Jul 22 2005
More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), May 01 2007

A116521 Binomial transform of tetranacci sequence A000078.

Original entry on oeis.org

0, 0, 0, 1, 5, 17, 51, 148, 429, 1250, 3655, 10701, 31336, 91752, 268623, 786414, 2302262, 6739984, 19731685, 57765711, 169112717, 495088023, 1449400960, 4243211207, 12422263776, 36366946961, 106466490879, 311687250156
Offset: 0

Views

Author

Jonathan Vos Post, Mar 10 2006

Keywords

Comments

See also A115390, the binomial transform of tribonacci sequence A000073. Tetranacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4), with a(0) = a(1) = a(2) = 0, a(3) = 1.

Examples

			Table shows the tetranacci numbers multiplied into rows of Pascal's triangle.
1*0 = 0.
1*0 + 1*0 = 0.
1*0 + 2*0 + 1*0 = 0.
1*0 + 3*0 + 3*0 + 1* 1 = 1.
1*0 + 4*0 + 6*0 + 4*1 + 1*1 = 5.
1*0 + 5*0 + 10*0 + 10*1 + 5*1 + 1*2 = 17.
		

Crossrefs

Programs

  • Maple
    t[0]:=0: t[1]:=0: t[2]:=0: t[3]:=1: for n from 4 to 35 do t[n]:=t[n-1]+t[n-2]+t[n-3]+t[n-4] od: seq(add(binomial(n,k)*t[k],k=0..n),n=0..30); # end of first Maple program
    G:=x^3/(1-5*x+8*x^2-6*x^3+x^4): Gser:=series(G,x=0,33): seq(coeff(Gser,x,n),n=0..30); # Emeric Deutsch, Apr 09 2006
  • Mathematica
    LinearRecurrence[{5,-8,6,-1}, {0,0,0,1}, 25] (* G. C. Greubel, Nov 03 2016 *)
  • PARI
    a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,6,-8,5]^n*[0;0;0;1])[1,1] \\ Charles R Greathouse IV, Jun 28 2017

Formula

a(n) = Sum_{k=0..n} C(n,k) * A000078(k).
G.f.: x^3/(1-5*x+8*x^2-6*x^3+x^4). - Emeric Deutsch, Apr 09 2006
a(n) = 5*a(n-1) - 8*a(n-2) + 6*a(n-3) - a(n-4). - G. C. Greubel, Nov 03 2016

Extensions

Definition corrected by Franklin T. Adams-Watters, Mar 13 2006
More terms from Emeric Deutsch, Apr 09 2006

A140285 Numbers which are both lucky (A000959) and tetranacci (A000078).

Original entry on oeis.org

1, 15, 10671
Offset: 1

Views

Author

Jonathan Vos Post, May 24 2008

Keywords

Comments

Lucky (A000959) and Fibonacci (A000045) = A057589; lucky and tribonacci (A000073) = A130594. No more through 10^6.
a(4) > 4*10^9, if it exists. - Giovanni Resta, May 10 2020
a(4) >= 10312882481, if it exists. - Kevin P. Thompson, Nov 24 2021

Examples

			a(3) = 10671 = A000078(18) = A000959(1187).
		

Crossrefs

A193993 Number of zeros in the period of Fibonacci 4-step sequence A000078 mod n.

Original entry on oeis.org

1, 3, 9, 5, 72, 27, 49, 8, 12, 216, 11, 24, 7, 147, 36, 10, 336, 36, 361, 188, 231, 8, 529, 39, 80, 21, 21, 126, 11, 108, 1986, 14, 51, 1008, 636, 31, 34, 1083, 36, 152, 11, 693, 3786, 8, 24, 1587, 2209, 51, 56, 56, 1440, 19, 5832, 63, 33, 203, 1653, 9, 3481
Offset: 1

Views

Author

T. D. Noe, Aug 18 2011

Keywords

Crossrefs

Cf. A106295 (period of Lucas 4-step sequence).

Programs

  • Mathematica
    n = 4; Table[a = Join[{1}, Table[0, {n - 1}]]; a = Mod[a, i]; a0 = a; k = 0; zeros = 0; While[k++; s = Mod[Plus @@ a, i]; a = RotateLeft[a]; If[s == 0, zeros++]; a[[n]] = s; a != a0]; zeros, {i, 100}]

A353593 Triangle read by rows. The Riordan square of the tetranacci numbers (A000078).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 4, 8, 5, 1, 8, 20, 18, 7, 1, 15, 47, 56, 32, 9, 1, 29, 107, 158, 120, 50, 11, 1, 56, 238, 419, 397, 220, 72, 13, 1, 108, 520, 1063, 1207, 836, 364, 98, 15, 1, 208, 1120, 2608, 3452, 2871, 1563, 560, 128, 17, 1, 401, 2386, 6233, 9424, 9153, 5987, 2682, 816, 162, 19, 1
Offset: 0

Views

Author

Peter Luschny, Apr 29 2022

Keywords

Comments

See A321620 for the definition of the Riordan square.

Examples

			Triangle starts:
[0]   1;
[1]   1,    1;
[2]   2,    3,    1;
[3]   4,    8,    5,    1;
[4]   8,   20,   18,    7,    1;
[5]  15,   47,   56,   32,    9,    1;
[6]  29,  107,  158,  120,   50,   11,   1;
[7]  56,  238,  419,  397,  220,   72,  13,   1;
[8] 108,  520, 1063, 1207,  836,  364,  98,  15,  1;
[9] 208, 1120, 2608, 3452, 2871, 1563, 560, 128, 17, 1.
		

Crossrefs

Cf. A000078 (tetranacci), A160175 (row sums), A321620 (Riordan square).

Programs

  • Maple
    # The function RiordanSquare is defined in A321620.
    RiordanSquare(1/(1 - x - x^2 - x^3 - x^4), 10);

A357239 Inverse Moebius transform of tetranacci number (A000078).

Original entry on oeis.org

0, 0, 1, 1, 2, 5, 8, 16, 30, 58, 108, 214, 401, 781, 1493, 2888, 5536, 10705, 20569, 39707, 76433, 147420, 283953, 547566, 1055028, 2034029, 3919974, 7556717, 14564533, 28075593, 54114452, 104311848, 201062094, 387564973, 747044844, 1439986130, 2775641472, 5350241528, 10312882883
Offset: 1

Views

Author

Ilya Gutkovskiy, Sep 19 2022

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 39; CoefficientList[Series[Sum[x^(3 k)/(1 - x^k - x^(2 k) - x^(3 k) - x^(4 k)), {k, 1, nmax}], {x, 0, nmax}], x] // Rest

Formula

G.f.: Sum_{k>=1} x^(3*k) / (1 - x^k - x^(2*k) - x^(3*k) - x^(4*k)).
G.f.: Sum_{k>=1} A000078(k) * x^k / (1 - x^k).
a(n) = Sum_{d|n} A000078(d).

A373054 Numbers k that divide the k-th tetranacci number (A000078).

Original entry on oeis.org

1, 2, 22, 32, 80, 137, 179, 272, 320, 352, 600, 653, 859, 936, 991, 1279, 1280, 1306, 1601, 1609, 1632, 1672, 1982, 2089, 2152, 2437, 2560, 2591, 2693, 2789, 2897, 3120, 3202, 3701, 3823, 3847, 4110, 4212, 4451, 4691, 4751, 4919, 5120, 5182, 5280, 5386, 5431, 5479
Offset: 1

Views

Author

Amiram Eldar, May 20 2024

Keywords

Comments

Numbers k such that k | A000078(k).

Examples

			22 is a term since A000078(22) = 147312 = 22 * 6696 is divisible by 22.
		

Crossrefs

Cf. A000078.
Similar sequences: A014847 (Catalan), A016089 (Lucas), A023172 (Fibonacci), A051177 (partition), A232570 (tribonacci), A246692 (Pell), A266969 (Motzkin).

Programs

  • Mathematica
    With[{m = 10000}, Position[LinearRecurrence[{1, 1, 1, 1}, {0, 0, 1, 1}, m]/Range[m], _?IntegerQ] // Flatten]
  • PARI
    lista(kmax) = {my(t0 = 0, t1 = 0, t2 = 0, t3 = 1, t4 = 0); print1(1, ", ",  2, ", "); for(k = 4, kmax, t4 = t0 + t1 + t2 + t3; if(!(t4%k), print1(k, ", ")); t0 = t1; t1 = t2; t2 = t3; t3 = t4);}

A112374 Let T(n) = A000078(n+2), n >= 1; a(n) = smallest k such that n divides T(k).

Original entry on oeis.org

1, 3, 6, 4, 6, 9, 8, 5, 9, 13, 20, 9, 10, 8, 6, 10, 53, 9, 48, 28, 18, 20, 35, 18, 76, 10, 9, 8, 7, 68, 20, 15, 20, 53, 30, 9, 58, 48, 78, 28, 19, 18, 63, 20, 68, 35, 28, 18, 46, 108, 76, 10, 158, 9, 52, 8, 87, 133, 18, 68, 51, 20, 46, 35, 78, 20, 17, 138, 35, 30, 230, 20, 72, 58, 76
Offset: 1

Views

Author

Jonathan Vos Post, Dec 02 2005

Keywords

Comments

Rank of apparition of n in the tetranacci numbers. - T. D. Noe, Dec 05 2005
This sequence is well-defined. Proof by T. D. Noe: for every prime p, Brenner proves we can find k(p) such that p divides the k(p)-th term of n-step Fibonacci. Using Brenner's methods, we know that p will also divide every j*k(p)-th term of the sequence for any j>0. We use this last fact to go to the general case: For integer m, we can find a term that m divides as follows: (1) factor m into primes: m = p1^e1 p2^e2...pr^er, (2) let K = m k(p1) k(p2)...k(pr) / (p1 p2 ... pr) (3) then m will divide the K-th term of the sequence. In general, K is much too large. However, it does show that every prime divides a term of every n-step Fibonacci sequence for n>1. - T. D. Noe, Dec 05 2005

Crossrefs

Programs

  • Mathematica
    n=4; Table[a=Join[{1}, Table[0, {n-1}]]; k=0; While[k++; s=Mod[Plus@@a, i]; a=RotateLeft[a]; a[[n]]=s; s!=0]; k, {i, 100}] (* T. D. Noe, Dec 05 2005 *)

Formula

a(n) = Min{k: n | A000078(k)}.

Extensions

Corrected by T. D. Noe, Dec 05 2005

A280426 Digital roots of tetranacci numbers A000078.

Original entry on oeis.org

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

Views

Author

Peter M. Chema, Jan 02 2017

Keywords

Comments

From a(3) onwards, periodic with period length 78, twice the periodicity of the digital roots of tribonacci numbers A222407 (also the 12th triangular number).

Crossrefs

Formula

a(n) = A010888(A000078(n)).

A000073 Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) for n >= 3 with a(0) = a(1) = 0 and a(2) = 1.

Original entry on oeis.org

0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012, 121415, 223317, 410744, 755476, 1389537, 2555757, 4700770, 8646064, 15902591, 29249425, 53798080, 98950096, 181997601, 334745777, 615693474, 1132436852
Offset: 0

Views

Author

Keywords

Comments

The name "tribonacci number" is less well-defined than "Fibonacci number". The sequence A000073 (which begins 0, 0, 1) is probably the most important version, but the name has also been applied to A000213, A001590, and A081172. - N. J. A. Sloane, Jul 25 2024
Also (for n > 1) number of ordered trees with n+1 edges and having all leaves at level three. Example: a(4)=2 because we have two ordered trees with 5 edges and having all leaves at level three: (i) one edge emanating from the root, at the end of which two paths of length two are hanging and (ii) one path of length two emanating from the root, at the end of which three edges are hanging. - Emeric Deutsch, Jan 03 2004
a(n) is the number of compositions of n-2 with no part greater than 3. Example: a(5)=4 because we have 1+1+1 = 1+2 = 2+1 = 3. - Emeric Deutsch, Mar 10 2004
Let A denote the 3 X 3 matrix [0,0,1;1,1,1;0,1,0]. a(n) corresponds to both the (1,2) and (3,1) entries in A^n. - Paul Barry, Oct 15 2004
Number of permutations satisfying -k <= p(i)-i <= r, i=1..n-2, with k=1, r=2. - Vladimir Baltic, Jan 17 2005
Number of binary sequences of length n-3 that have no three consecutive 0's. Example: a(7)=13 because among the 16 binary sequences of length 4 only 0000, 0001 and 1000 have 3 consecutive 0's. - Emeric Deutsch, Apr 27 2006
Therefore, the complementary sequence to A050231 (n coin tosses with a run of three heads). a(n) = 2^(n-3) - A050231(n-3) - Toby Gottfried, Nov 21 2010
Convolved with the Padovan sequence = row sums of triangle A153462. - Gary W. Adamson, Dec 27 2008
For n > 1: row sums of the triangle in A157897. - Reinhard Zumkeller, Jun 25 2009
a(n+2) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 1; 0, 0, 1; 1, 0, 0] or [1, 1, 0; 1, 0, 1; 1, 0, 0] or [1, 1, 1; 1, 0, 0; 0, 1, 0] or [1, 0, 1; 1, 0, 0; 1, 1, 0]. - R. J. Mathar, Feb 03 2014
a(n-1) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 0, 1; 1, 1, 1; 0, 1, 0], [0, 1, 0; 0, 1, 1; 1, 1, 0], [0, 0, 1; 1, 0, 1; 0, 1, 1] or [0, 1, 0; 0, 0, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
Also row sums of A082601 and of A082870. - Reinhard Zumkeller, Apr 13 2014
Least significant bits are given in A021913 (a(n) mod 2 = A021913(n)). - Andres Cicuttin, Apr 04 2016
The nonnegative powers of the tribonacci constant t = A058265 are t^n = a(n)*t^2 + (a(n-1) + a(n-2))*t + a(n-1)*1, for n >= 0, with a(-1) = 1 and a(-2) = -1. This follows from the recurrences derived from t^3 = t^2 + t + 1. See the example in A058265 for the first nonnegative powers. For the negative powers see A319200. - Wolfdieter Lang, Oct 23 2018
The term "tribonacci number" was coined by Mark Feinberg (1963), a 14-year-old student in the 9th grade of the Susquehanna Township Junior High School in Pennsylvania. He died in 1967 in a motorcycle accident. - Amiram Eldar, Apr 16 2021
Andrews, Just, and Simay (2021, 2022) remark that it has been suggested that this sequence is mentioned in Charles Darwin's Origin of Species as bearing the same relation to elephant populations as the Fibonacci numbers do to rabbit populations. - N. J. A. Sloane, Jul 12 2022

Examples

			G.f. = x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 13*x^7 + 24*x^8 + 44*x^9 + 81*x^10 + ...
		

References

  • M. Agronomof, Sur une suite récurrente, Mathesis (Series 4), Vol. 4 (1914), pp. 125-126.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 47, ex. 4.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.2.2.
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • J. Riordan, An Introduction to Combinatorial Analysis, Princeton University Press, Princeton, NJ, 1978.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000045, A000078, A000213, A000931, A001590 (first differences, also a(n)+a(n+1)), A001644, A008288 (tribonacci triangle), A008937 (partial sums), A021913, A027024, A027083, A027084, A046738 (Pisano periods), A050231, A054668, A062544, A063401, A077902, A081172, A089068, A118390, A145027, A153462, A230216.
A057597 is this sequence run backwards: A057597(n) = a(1-n).
Row 3 of arrays A048887 and A092921 (k-generalized Fibonacci numbers).
Partitions: A240844 and A117546.
Cf. also A092836 (subsequence of primes), A299399 = A092835 + 1 (indices of primes).

Programs

  • GAP
    a:=[0,0,1];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # Muniru A Asiru, Oct 24 2018
  • Haskell
    a000073 n = a000073_list !! n
    a000073_list = 0 : 0 : 1 : zipWith (+) a000073_list (tail
                              (zipWith (+) a000073_list $ tail a000073_list))
    -- Reinhard Zumkeller, Dec 12 2011
    
  • Magma
    [n le 3 select Floor(n/3) else Self(n-1)+Self(n-2)+Self(n-3): n in [1..70]]; // Vincenzo Librandi, Jan 29 2016
    
  • Maple
    a:= n-> (<<0|1|0>, <0|0|1>, <1|1|1>>^n)[1,3]:
    seq(a(n), n=0..40);  # Alois P. Heinz, Dec 19 2016
    # second Maple program:
    A000073:=proc(n) option remember; if n <= 1 then 0 elif n=2 then 1 else procname(n-1)+procname(n-2)+procname(n-3); fi; end; # N. J. A. Sloane, Aug 06 2018
  • Mathematica
    CoefficientList[Series[x^2/(1 - x - x^2 - x^3), {x, 0, 50}], x]
    a[0] = a[1] = 0; a[2] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] + a[n - 3]; Array[a, 36, 0] (* Robert G. Wilson v, Nov 07 2010 *)
    LinearRecurrence[{1, 1, 1}, {0, 0, 1}, 60] (* Vladimir Joseph Stephan Orlovsky, May 24 2011 *)
    a[n_] := SeriesCoefficient[If[ n < 0, x/(1 + x + x^2 - x^3), x^2/(1 - x - x^2 - x^3)], {x, 0, Abs @ n}] (* Michael Somos, Jun 01 2013 *)
    Table[-RootSum[-1 - # - #^2 + #^3 &, -#^n - 9 #^(n + 1) + 4 #^(n + 2) &]/22, {n, 0, 20}] (* Eric W. Weisstein, Nov 09 2017 *)
  • Maxima
    A000073[0]:0$
    A000073[1]:0$
    A000073[2]:1$
    A000073[n]:=A000073[n-1]+A000073[n-2]+A000073[n-3]$
      makelist(A000073[n], n, 0, 40);  /* Emanuele Munarini, Mar 01 2011 */
    
  • PARI
    {a(n) = polcoeff( if( n<0, x / ( 1 + x + x^2 - x^3), x^2 / ( 1 - x - x^2 - x^3) ) + x * O(x^abs(n)), abs(n))}; /* Michael Somos, Sep 03 2007 */
    
  • PARI
    my(x='x+O('x^99)); concat([0, 0], Vec(x^2/(1-x-x^2-x^3))) \\ Altug Alkan, Apr 04 2016
    
  • PARI
    a(n)=([0,1,0;0,0,1;1,1,1]^n)[1,3] \\ Charles R Greathouse IV, Apr 18 2016, simplified by M. F. Hasler, Apr 18 2018
    
  • Python
    def a(n, adict={0:0, 1:0, 2:1}):
        if n in adict:
            return adict[n]
        adict[n]=a(n-1)+a(n-2)+a(n-3)
        return adict[n] # David Nacin, Mar 07 2012
    from functools import cache
    @cache
    def A000073(n: int) -> int:
        if n <= 1: return 0
        if n == 2: return 1
        return A000073(n-1) + A000073(n-2) + A000073(n-3) # Peter Luschny, Nov 21 2022
    

Formula

G.f.: x^2/(1 - x - x^2 - x^3).
G.f.: x^2 / (1 - x / (1 - x / (1 + x^2 / (1 + x)))). - Michael Somos, May 12 2012
G.f.: Sum_{n >= 0} x^(n+2) *[ Product_{k = 1..n} (k + k*x + x^2)/(1 + k*x + k*x^2) ] = x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 13*x^7 + ... may be proved by the method of telescoping sums. - Peter Bala, Jan 04 2015
a(n+1)/a(n) -> A058265. a(n-1)/a(n) -> A192918.
a(n) = central term in M^n * [1 0 0] where M = the 3 X 3 matrix [0 1 0 / 0 0 1 / 1 1 1]. (M^n * [1 0 0] = [a(n-1) a(n) a(n+1)].) a(n)/a(n-1) tends to the tribonacci constant, 1.839286755... = A058265, an eigenvalue of M and a root of x^3 - x^2 - x - 1 = 0. - Gary W. Adamson, Dec 17 2004
a(n+2) = Sum_{k=0..n} T(n-k, k), where T(n, k) = trinomial coefficients (A027907). - Paul Barry, Feb 15 2005
A001590(n) = a(n+1) - a(n); A001590(n) = a(n-1) + a(n-2) for n > 1; a(n) = (A000213(n+1) - A000213(n))/2; A000213(n-1) = a(n+2) - a(n) for n > 0. - Reinhard Zumkeller, May 22 2006
Let C = the tribonacci constant, 1.83928675...; then C^n = a(n)*(1/C) + a(n+1)*(1/C + 1/C^2) + a(n+2)*(1/C + 1/C^2 + 1/C^3). Example: C^4 = 11.444...= 2*(1/C) + 4*(1/C + 1/C^2) + 7*(1/C + 1/C^2 + 1/C^3). - Gary W. Adamson, Nov 05 2006
a(n) = j*C^n + k*r1^n + L*r2^n where C is the tribonacci constant (C = 1.8392867552...), real root of x^3-x^2-x-1=0, and r1 and r2 are the two other roots (which are complex), r1 = m+p*i and r2 = m-p*i, where i = sqrt(-1), m = (1-C)/2 (m = -0.4196433776...) and p = ((3*C-5)*(C+1)/4)^(1/2) = 0.6062907292..., and where j = 1/((C-m)^2 + p^2) = 0.1828035330..., k = a+b*i, and L = a-b*i, where a = -j/2 = -0.0914017665... and b = (C-m)/(2*p*((C-m)^2 + p^2)) = 0.3405465308... . - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007
a(n+1) = 3*c*((1/3)*(a+b+1))^n/(c^2-2*c+4) where a=(19+3*sqrt(33))^(1/3), b=(19-3*sqrt(33))^(1/3), c=(586+102*sqrt(33))^(1/3). Round to the nearest integer. - Al Hakanson (hawkuu(AT)gmail.com), Feb 02 2009
a(n) = round(3*((a+b+1)/3)^n/(a^2+b^2+4)) where a=(19+3*sqrt(33))^(1/3), b=(19-3*sqrt(33))^(1/3).. - Anton Nikonov
Another form of the g.f.: f(z) = (z^2-z^3)/(1-2*z+z^4). Then we obtain a(n) as a sum: a(n) = Sum_{i=0..floor((n-2)/4)} ((-1)^i*binomial(n-2-3*i,i)*2^(n-2-4*i)) - Sum_{i=0..floor((n-3)/4)} ((-1)^i*binomial(n-3-3*i,i)*2^(n-3-4*i)) with natural convention: Sum_{i=m..n} alpha(i) = 0 for m > n. - Richard Choulet, Feb 22 2010
a(n+2) = Sum_{k=0..n} Sum_{i=k..n, mod(4*k-i,3)=0} binomial(k,(4*k-i)/3)*(-1)^((i-k)/3)*binomial(n-i+k-1,k-1). - Vladimir Kruchinin, Aug 18 2010
a(n) = 2*a(n-2) + 2*a(n-3) + a(n-4). - Gary Detlefs, Sep 13 2010
Sum_{k=0..2*n} a(k+b)*A027907(n,k) = a(3*n+b), b >= 0 (see A099464, A074581).
a(n) = 2*a(n-1) - a(n-4), with a(0)=a(1)=0, a(2)=a(3)=1. - Vincenzo Librandi, Dec 20 2010
Starting (1, 2, 4, 7, ...) is the INVERT transform of (1, 1, 1, 0, 0, 0, ...). - Gary W. Adamson, May 13 2013
G.f.: Q(0)*x^2/2, where Q(k) = 1 + 1/(1 - x*(4*k+1 + x + x^2)/( x*(4*k+3 + x + x^2) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 09 2013
a(n+2) = Sum_{j=0..floor(n/2)} Sum_{k=0..j} binomial(n-2*j,k)*binomial(j,k)*2^k. - Tony Foster III, Sep 08 2017
Sum_{k=0..n} (n-k)*a(k) = (a(n+2) + a(n+1) - n - 1)/2. See A062544. - Yichen Wang, Aug 20 2020
a(n) = A008937(n-1) - A008937(n-2) for n >= 2. - Peter Luschny, Aug 20 2020
From Yichen Wang, Aug 27 2020: (Start)
Sum_{k=0..n} a(k) = (a(n+2) + a(n) - 1)/2. See A008937.
Sum_{k=0..n} k*a(k) = ((n-1)*a(n+2) - a(n+1) + n*a(n) + 1)/2. See A337282. (End)
For n > 1, a(n) = b(n) where b(1) = 1 and then b(n) = Sum_{k=1..n-1} b(n-k)*A000931(k+2). - J. Conrad, Nov 24 2022
Conjecture: the congruence a(n*p^(k+1)) + a(n*p^k) + a(n*p^(k-1)) == 0 (mod p^k) holds for positive integers k and n and for all the primes p listed in A106282. - Peter Bala, Dec 28 2022
Sum_{k=0..n} k^2*a(k) = ((n^2-4*n+6)*a(n+1) - (2*n^2-2*n+5)*a(n) + (n^2-2*n+3)*a(n-1) - 3)/2. - Prabha Sivaramannair, Feb 10 2024
a(n) = Sum_{r root of x^3-x^2-x-1} r^n/(3*r^2-2*r-1). - Fabian Pereyra, Nov 23 2024

Extensions

Minor edits by M. F. Hasler, Apr 18 2018
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021
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