A000078 -id:A000078 - cp's OEIS frontend

A074193 Sum of determinants of 2nd order principal minors of powers of the matrix ((1,1,0,0),(1,0,1,0),(1,0,0,1),(1,0,0,0)).

6, -1, -3, -1, 17, -16, -15, 13, 81, -127, -58, 175, 329, -885, -31, 1424, 833, -5543, 2181, 9233, -2298, -31025, 27893, 49495, -54879, -150416, 245697, 204965, -526887, -570895, 1801670, 407711, -3882303, -946397, 11542929, -3442672, -24121039, 10317745, 64959629, -56727711, -127083514

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 20 2002

From Kai Wang, Oct 21 2020: (Start)
Let f(x) = x^4 - x^3 - x^2 - x - 1 be the characteristic polynomial for Tetranacci numbers (A000078). Let {x1,x2,x3,x4} be the roots of f(x). Then a(n) = (x1*x2)^n + (x1*x3)^n + (x1*x4)^n + (x2*x3)^n + (x2*x4)^n + (x3*x4)^n.
Let g(y) = y^6 + y^5 + 2*y^4 + 2*y^3 - 2*y^2 + y - 1 be the characteristic polynomial for a(n). Let {y1,y2,y3,y4,y5,y6} be the roots of g(y). Then a(n) = y1^n + y2^n + y3^n + y4^n + y5^n + y6^n. (End)

Michael De Vlieger, Table of n, a(n) for n = 0..5052
Kai Wang, Identities, generating functions and Binet formula for generalized k-nacci sequences, 2020.
Index entries for linear recurrences with constant coefficients, signature (-1,-2,-2,2,-1,1).

Cf. A073817, A073937, A000078, A074081.

CoefficientList[Series[(6+5*x+8*x^2+6*x^3-4*x^4+x^5)/(1+x+2*x^2+2*x^3-2*x^4+x^5-x^6), {x, 0, 50}], x]
LinearRecurrence[{-1,-2,-2,2,-1,1},{6,-1,-3,-1,17,-16},50] (* Harvey P. Dale, Sep 06 2025 *)

polsym(x^6 + x^5 + 2*x^4 + 2*x^3 - 2*x^2 + x - 1,44) \\ Joerg Arndt, Oct 22 2020

a(n) = -a(n-1)-2a(n-2)-2a(n-3)+2a(n-4)-a(n-5)+a(n-6).
G.f.: (6+5x+8x^2+6x^3-4x^4+x^5)/(1+x+2x^2+2x^3-2x^4+x^5-x^6).
abs(a(n)) = abs(A074453(n)). - Joerg Arndt, Oct 22 2020

A102036 Triangle, read by rows, where the terms are generated by the rule: T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k-1) + T(n-3,k-1), with T(0,0)=1.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 9, 15, 7, 1, 1, 12, 33, 28, 9, 1, 1, 15, 60, 81, 45, 11, 1, 1, 18, 96, 189, 161, 66, 13, 1, 1, 21, 141, 378, 459, 281, 91, 15, 1, 1, 24, 195, 675, 1107, 946, 449, 120, 17, 1, 1, 27, 258, 1107, 2349, 2673, 1742, 673, 153, 19, 1

Offset: 0

Paul D. Hanna, Dec 30 2004

Row sums form A077939. This sequence was inspired by Luke Hanna.
Diagonal sums are A000078(n+3). - Philippe Deléham, Feb 16 2014
Riordan array (1/(1-x), x*(1+x+x^2)/(1-x)). - Philippe Deléham, Feb 16 2014

			Generated by adding preceding terms in the triangle at positions that form the letter 'L':
T(n,k) =
T(n-3,k-1) +
T(n-2,k-1) +
T(n-1,k-1) + T(n-1,k).
Rows begin:
  [1],
  [1,  1],
  [1,  3,   1],
  [1,  6,   5,   1],
  [1,  9,  15,   7,   1],
  [1, 12,  33,  28,   9,   1],
  [1, 15,  60,  81,  45,  11,  1],
  [1, 18,  96, 189, 161,  66, 13,  1],
  [1, 21, 141, 378, 459, 281, 91, 15, 1], ...

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Kuhapatanakul, Kantaphon; Anantakitpaisal, Pornpawee The k-nacci triangle and applications. Cogent Math. 4, Article ID 1333293, 13 p. (2017).
J. L. Ramírez, V. F. Sirvent, A Generalization of the k-Bonacci Sequence from Riordan Arrays, The Electronic Journal of Combinatorics, 22(1) (2015), #P1.38.

Cf. A077939, A103141.

[[(&+[Binomial(n-m,k)*(&+[Binomial(j,m-j)*Binomial(k,j):j in [0..k]]): m in [0..n-k]]): k in [0..n]]: n in [0..15]]; // G. C. Greubel, Dec 11 2018

T:=(n,k)->add(add((binomial(j,m-j)*binomial(k,j))*binomial(n-m,k),j=0..k),m=0..n-k): seq(seq(T(n,k),k=0..n),n=0..10); # Muniru A Asiru, Dec 11 2018

T[n_, k_] := If[n < k || k < 0, 0, If[n == 0, 1, T[n - 1, k] + T[n - 1, k - 1] + T[n - 2, k - 1] + T[n - 3, k - 1]]]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Dec 07 2018 *)
Table[Sum[Binomial[n-m, k]*Sum[Binomial[j, m-j]*Binomial[k, j], {j, 0, k}], {m, 0, n-k}], {n, 0, 15}, {k, 0, n}]//Flatten (* G. C. Greubel, Dec 11 2018 *)

T(n,k):=sum((sum(binomial(j,m-j)*binomial(k,j),j,0,k))*binomial(n-m,k),m,0,n-k); /* Vladimir Kruchinin, Apr 21 2015 */

PARI
```
{T(n,k)=if(n
				
```

[[sum(binomial(n-m,k)*sum(binomial(j,m-j)*binomial(k,j) for j in (0..k)) for m in (0..n-k)) for k in (0..n)] for n in range(15)] # G. C. Greubel, Dec 11 2018

G.f.: 1/(1-y-x*(1+y+y^2)). - Vladimir Kruchinin, Apr 21 2015
T(n,k) = Sum_{m=0..(n-k)} (Sum_{j=0..k} C(j,m-j)*C(k,j))*C(n-m,k). - Vladimir Kruchinin, Apr 21 2015
From Werner Schulte, Dec 07 2018: (Start)
G.f. of column k: Sum_{n>=0} T(n+k,k) * x^n = (1+x+x^2)^k / (1-x)^(k+1) = (1-x^3)^k / (1-x)^(2*k+1).
Let k >= 0 be some fixed integer and a_k(n) be multiplicative with a_k(p^e) = T(e+k,k) for prime p and e >= 0. Then we have the Dirichlet g.f.: Sum{n>0} a_k(n) / n^s = (zeta(s))^(2*k+1) / (zeta(3*s))^k. (End)

A113243 Differences of nonzero tetranacci numbers.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 11, 13, 14, 21, 25, 27, 28, 41, 48, 52, 54, 55, 79, 93, 100, 104, 106, 107, 152, 179, 193, 200, 204, 206, 207, 293, 345, 372, 386, 393, 397, 399, 400, 565, 665, 717, 744, 758, 765, 769, 771, 772, 1089, 1282, 1382, 1434, 1461, 1475, 1482

Offset: 1

Jonathan Vos Post, Oct 19 2005; corrected Oct 20 2005

R. J. Mathar, Table of n, a(n) for n = 1..197

Cf. A000078, A113244.

isA113243 := proc(n)
    local i,j ;
    for j from 3 do
        for i from 3 to j do
            if A000078(j) - A000078(i) = n then
                return true;
            elif A000078(j) - A000078(i) < n then
                break ;
            end if;
        end do:
        if A000078(j) - A000078(j-1) > n then
            return false;
        end if;
    end do:
end proc:
for n from 0 to 10000 do
    if isA113243(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Oct 04 2014

Union[Flatten[Differences/@Subsets[Drop[LinearRecurrence[{1, 1, 1,1}, {0,0, 0, 1}, 16],3],{2}]]] (* James C. McMahon, Jun 23 2024 *)

{a(n)} = { A000078(i) - A000078(j) such that i>=j>=3 }.

A118897 Triangle read by rows: T(n,k) is the number of binary sequences of length n containing k subsequences 0000 (n,k>=0).

Original entry on oeis.org

1, 2, 4, 8, 15, 1, 29, 2, 1, 56, 5, 2, 1, 108, 12, 5, 2, 1, 208, 28, 12, 5, 2, 1, 401, 62, 29, 12, 5, 2, 1, 773, 136, 65, 30, 12, 5, 2, 1, 1490, 294, 145, 68, 31, 12, 5, 2, 1, 2872, 628, 319, 154, 71, 32, 12, 5, 2, 1, 5536, 1328, 694, 344, 163, 74, 33, 12, 5, 2, 1, 10671, 2787

Offset: 0

Emeric Deutsch, May 04 2006

Row n has n-2 terms (n>=3). Sum of entries in row n is 2^n (A000079). T(n,0) = A000078(n+4) (tetranacci numbers). T(n,1) = A118898(n). Sum(k*T(n,k),n>=0) = (n-3)*2^(n-4) (A001787).

			T(7,2) = 5 because we have 0000010, 0000011, 0100000, 1100000 and 1000001.
Triangle starts:
    1;
    2;
    4;
    8;
   15,  1;
   29,  2, 1;
   56,  5, 2, 1;
  108, 12, 5, 2, 1;
  ...

Alois P. Heinz, Rows n = 0..143, flattened

Cf. A000079, A000078, A118898, A001787, A076791, A118390.

G:=(1+(1-t)*(z+z^2+z^3))/(1-(1+t)*z-(1-t)*(z^2+z^3+z^4)): Gser:=simplify(series(G,z=0,17)): P[0]:=1: for n from 1 to 14 do P[n]:=sort(coeff(Gser,z^n)) od: 1;2;4;8; for n from 4 to 14 do seq(coeff(P[n],t,j),j=0..n-3) od; # yields sequence in triangular form
# second Maple program:
b:= proc(n, t) option remember; `if`(n=0, 1,
      expand(b(n-1, min(3, t+1))*`if`(t>2, x, 1))+b(n-1, 0))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0)):
seq(T(n), n=0..14);  # Alois P. Heinz, Sep 17 2019

nn=15;a=x^3/(1-y x)+x+x^2;b=1/(1-x);f[list_]:=Select[list,#>0&];Map[f,CoefficientList[Series[b (1+a)/(1-a x/(1-x)) ,{x,0,nn}],{x,y}]]//Grid  (* Geoffrey Critzer, Nov 18 2012 *)

G.f.: G(t,z) = [1+(1-t)(z+z^2+z^3)]/[1-(1+t)z-(1-t)(z^2+z^3+z^4)].

A144406 Rectangular array A read by upward antidiagonals: entry A(n,k) in row n and column k gives the number of compositions of k in which no part exceeds n, n>=1, k>=0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 2, 4, 5, 1, 1, 1, 2, 4, 7, 8, 1, 1, 1, 2, 4, 8, 13, 13, 1, 1, 1, 2, 4, 8, 15, 24, 21, 1, 1, 1, 2, 4, 8, 16, 29, 44, 34, 1, 1, 1, 2, 4, 8, 16, 31, 56, 81, 55, 1, 1, 1, 2, 4, 8, 16, 32, 61, 108, 149, 89, 1, 1, 1, 2, 4, 8, 16, 32, 63, 120, 208, 274, 144, 1

Offset: 1

Roger L. Bagula and Gary W. Adamson, Sep 29 2008

Polynomial expansion as antidiagonal of p(x,n) = (x-1)/(x^n*(-x+(2*x-1)/x^n)). Based on the Pisot general polynomial type q(x,n) = x^n - (x^n-1)/(x-1) (the original name of the sequence).
Row sums are 1, 2, 3, 5, 8, 14, ... (A079500).
Conjecture: Since the array row sequences successively tend to A000079, the absolute values of nonzero differences between two successive row sequences tend to A045623 = {1,2,5,12,28,64,144,320,704,1536,...}, as k -> infinity. - L. Edson Jeffery, Dec 26 2013

			Array A begins:
  {1, 1, 1, 1, 1,  1,  1,  1,   1,   1,   1, ...}
  {1, 1, 2, 3, 5,  8, 13, 21,  34,  55,  89, ...}
  {1, 1, 2, 4, 7, 13, 24, 44,  81, 149, 274, ...}
  {1, 1, 2, 4, 8, 15, 29, 56, 108, 208, 401, ...}
  {1, 1, 2, 4, 8, 16, 31, 61, 120, 236, 464, ...}
  {1, 1, 2, 4, 8, 16, 32, 63, 125, 248, 492, ...}
  {1, 1, 2, 4, 8, 16, 32, 64, 127, 253, 504, ...}
  {1, 1, 2, 4, 8, 16, 32, 64, 128, 255, 509, ...}
  {1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 511, ...}
  ... - _L. Edson Jeffery_, Dec 26 2013
As a triangle:
  {1},
  {1, 1},
  {1, 1, 1},
  {1, 1, 2, 1},
  {1, 1, 2, 3, 1},
  {1, 1, 2, 4, 5, 1},
  {1, 1, 2, 4, 7, 8, 1},
  {1, 1, 2, 4, 8, 13, 13, 1},
  {1, 1, 2, 4, 8, 15, 24, 21, 1},
  {1, 1, 2, 4, 8, 16, 29, 44, 34, 1},
  {1, 1, 2, 4, 8, 16, 31, 56, 81, 55, 1},
  {1, 1, 2, 4, 8, 16, 32, 61, 108, 149, 89, 1},
  {1, 1, 2, 4, 8, 16, 32, 63, 120, 208, 274, 144, 1},
  {1, 1, 2, 4, 8, 16, 32, 64, 125, 236, 401, 504, 233, 1},
  {1, 1, 2, 4, 8, 16, 32, 64, 127, 248, 464, 773, 927, 377, 1}

Cf. A000079, A045623, A092921, A175331.
Same as A048887 but with a column of 1's added on the left (the number of compositions of 0 is defined to be equal to 1).
Array rows (with appropriate offsets) are A000012, A000045, A000073, A000078, A001591, A001592, etc.

g[x_, n_] = x^(n) - (x^n - 1)/(x - 1);
h[x_, n_] = FullSimplify[ExpandAll[x^(n)*g[1/x, n]]];
f[t_, n_] := 1/h[t, n];
a = Table[CoefficientList[Series[f[t, m], {t, 0, 30}], t], {m, 1, 31}];
b = Table[Table[a[[n - m + 1]][[m]], {m, 1, n }], {n, 1, 15}];
Flatten[b] (* Triangle version *)
Grid[Table[CoefficientList[Series[(1 - x)/(1 - 2 x + x^(n + 1)), {x, 0, 10}], x], {n, 1, 10}]] (* Array version - L. Edson Jeffery, Jul 18 2014 *)

t(n,m) = antidiagonal_expansion of p(x,n) where p(x,n) = (x-1)/(x^n*(-x+(2*x-1)/x^n)).

G.f. for array A: (1-x)/(1 - 2*x + x^(n+1)), n>=1. - L. Edson Jeffery, Dec 26 2013

Definition changed by L. Edson Jeffery, Jul 18 2014

A168088 a(n) = 2^tetranacci(n).

Original entry on oeis.org

1, 1, 1, 2, 2, 4, 16, 256, 32768, 536870912, 72057594037927936, 324518553658426726783156020576256, 411376139330301510538742295639337626245683966408394965837152256

Offset: 0

Vladimir Joseph Stephan Orlovsky, Nov 18 2009

Harvey P. Dale, Table of n, a(n) for n = 0..16
Index to divisibility sequences

Cf. A000078.

Subsequence of A000079.

a={1,0,0,0};Flatten[Prepend[Table[s=Plus@@a;a=RotateLeft[a];z=a[[ -1]]=s;z=2^z,{n,12}],Table[0,{m,Length[a]-1}]]]
2^LinearRecurrence[{1,1,1,1},{0,0,0,1},15] (* Harvey P. Dale, Dec 12 2017 *)

a(n) = 2^A000078(n).

a(0)-a(2) and offset corrected by Charles R Greathouse IV, Jul 19 2012

A220493 Fibonacci 15-step numbers, a(n) = a(n-1) + a(n-2) + ... + a(n-15).

Original entry on oeis.org

1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32767, 65533, 131064, 262124, 524240, 1048464, 2096896, 4193728, 8387328, 16774400, 33548288, 67095552, 134189056, 268374016, 536739840, 1073463296, 2146893825, 4293722117, 8587313170

Offset: 1

Ruskin Harding, Feb 20 2013

Also called Pentadecanacci numbers. In previous similar sequences, a(1), ..., a(n-1) have been set equal to zero and a(n)=1. For example, A168084 (Fibonacci 13-step numbers) has 12 0's as the first 12 terms and a(13)=1.

Robert Israel, Table of n, a(n) for n = 1..3320
M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, J. Int. Seq. 18 (2015) # 15.4.7.
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.
Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1,1,1,1,1,1,1,1,1,1,1).

Cf. A000045 (Fibonacci), A000073 (tribonacci), A000078 (tetranacci), A001591 (pentanacci).

f:= gfun:-rectoproc({a(n) = add(a(n-i),i=1..15), seq(a(n)=0,n=-14..0),a(1)=1},a(n),remember):
map(f, [$1..100]); # Robert Israel, Feb 19 2019

FibonacciSequence[n_, kMax_] := Module[{a, s}, a = Join[{1}, Table[0, {n - 1}]]; lst = {}; Table[s = Plus @@ a; a = RotateLeft[a]; a[[n]] = s, {k, 1, kMax}]]; FibonacciSequence[15, 50] (* T. D. Noe, Feb 20 2013 *)

G.f.: x/(1-Sum_{k=1..15} x^k). - Robert Israel, Feb 19 2019

A227880 Primes in the union of all n-Fibonacci sequences.

Original entry on oeis.org

2, 3, 5, 7, 13, 29, 31, 61, 89, 127, 149, 233, 401, 509, 773, 1021, 1597, 4093, 8191, 16381, 28657, 31489, 128257, 131071, 514229, 524287, 1048573, 4194301, 5976577, 16777213, 433494437, 536870909, 2147483647, 2971215073, 4293722117, 5350220959, 13435170943

Offset: 1

Robert Price, Oct 25 2013

nonn

T. D. Noe, Table of n, a(n) for n = 1..127 (first 62 terms from Robert Price)
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. A124168, A000045, A000073, A000078, A001591, A001592, A124257.

plst = {};plimit=10^39; For[n = 2, n ≤ 1 + Log[2, plimit], n++,flst = {};For[i = 1, i < n, i++, AppendTo[flst, 0]];AppendTo[flst, 1];For[k = 2, k ≤ 1 + Log[GoldenRatio, plimit*Sqrt[5] + 0.5], k++,sum = 0;For[j = 0, j < n, j++, sum = sum + flst[[j + k - 1]]];AppendTo[flst, sum];If[sum ≤ plimit && PrimeQ[sum], AppendTo[plst, sum]]]];Union[plst]

Primes in A124168.

A299399 a(n) = a(n-1)a(n-2)a(n-3)*a(n-4); a(0..3) = (1, 1, 2, 3).

Original entry on oeis.org

1, 1, 2, 3, 6, 36, 1296, 839808, 235092492288, 9211413321697223245824, 2356948205087252000835395074931259831484416, 4286423488783965214900384842824017360544199884413056912194095171350270745233063936

Offset: 0

M. F. Hasler, Apr 22 2018

nonn

A variant of A000336 which uses initial values (1,2,3,4).

A multiplicative variant of the tetranacci sequences A000078, A001631 and other variants.

Seiichi Manyama, Table of n, a(n) for n = 0..14

Cf. A000336 (variant starting 1,2,3,4).
Cf. A000301 (order 2 variant), A000308 (order 3 variant).
Subsequence of A003586 (3-smooth numbers).
Cf. A000078, A001631 (additive variants).

nxt[{a_,b_,c_,d_}]:={b,c,d,a b c d}; NestList[nxt,{1,1,2,3},13][[All,1]] (* Harvey P. Dale, Jun 09 2022 *)

A299399(n,a=[1,1,2,3,6])={for(n=5,n,a[n%#a+1]=a[(n-1)%#a+1]^2\a[n%#a+1]);a[n%#a+1]}

a(n) = a(n-1)^2 / a(n-5) for n > 4.

a(n) = 2^A001631(n)*3^A000078(n).

A302990 a(n) = index of first odd prime number in the (n-th)-order Fibonacci sequence Fn, or 0 if no such index exists.

Original entry on oeis.org

0, 0, 4, 6, 9, 10, 40, 14, 17, 19, 361, 23, 90, 26, 373, 47, 288, 34, 75, 38, 251, 43, 67, 47, 74, 310, 511, 151534, 57, 20608, 1146, 62, 197, 94246, 9974, 287, 271172, 758

Offset: 0

Jacques Tramu, Apr 17 2018

Fn is defined by: Fn(0) = Fn(1) = ... = Fn(n-2) = 0, Fn(n-1) = 1, and Fn(k+1) = Fn(k) + Fn(k-1) + ... + Fn(k-n+1).
In general, Fn(k) is odd iff k == -1 or -2 (mod n+1), therefore a(n) = k*(n+1) - (1 or 2) for all n. Since Fn(n-1) = F(n) = 1, we must have a(n) >= 2n. Since Fn(k) = 2^(k-n) for n <= k < 2n, Fn(2n) = 2^n-1, so a(n) = 2n exactly for the Mersenne prime exponents A000043, while a(n) = 2n+1 when n is not in A000043 but n+1 is in A050414. - M. F. Hasler, Apr 18 2018
Further terms of the sequence: a(38) > 62000, a(39) > 72000, a(40) = 285, a(41) > 178000, a(42) = 558, a(44) = 19529, a(46) = 33369, a(47) = 239, a(48) = 6368, a(53) = 2860, a(54) = 2418, a(58) = 176, a(59) = 18418, a(60) = 1463, a(61) = 122, a(62) = 8755, a(63) = 5118,  a(64) = 25089, a(65) = 988, a(66) = 333, a(67) = 406, a(70) = 1632, a(74) = 374, a(76) = 13704, a(77) = 4991, a(86) = 347, a(89) = 178, a(92) = 1114, a(93) = 187, a(98) = 395, a(100) > 80000; a(n) > 10^4 for all other n up to 100. - Jacques Tramu and M. F. Hasler, Apr 18 2018

			a(2) = 4 because F2 (Fibonacci) = 0, 1, 1, 2, 3, 5, 8, ... and F2(4) = 3 is prime.
a(3) = 6 because F3 (tribonacci) = 0, 0, 1, 1, 2, 4, 7, 13, ... and F3(6) = 7 is prime.
a(4) = 9 because F4 (tetranacci) = 0, 0, 0, 1, 1, 2, 4, 8, 15, 29, 56, ...  and F4(9) = 29 is prime.
From _M. F. Hasler_, Apr 18 2018: (Start)
We see that Fn(k) = 2^(k-n) for n <= k < 2n and thus Fn(2n) = 2^n-1, so a(n) = 2n exactly for the Mersenne prime exponents A000043.
a(n) = 2n + 1 when 2^(n+1) - 3 is prime (n+1 in A050414) but 2^n-1 is not, i.e., n = 4, 8, 9, 11, 21, 23, 28, 93, 115, 121, 149, 173, 212, 220, 232, 265, 335, 451, 544, 688, 693, 849, 1735, ...
For other primes we have: a(29) = 687*30 - 2, a(37) = 20*38 - 2, a(41) > 10^4, a(43) > 10^4, a(47) = 5*48 - 1, a(53) = 53*54 - 2, a(59) = 307*60 - 2, a(67) = 6*67 - 1. (End)

Cf. A000045 (F2), A000073 (F3), A000078 (F4), A001591 (F5), A001592 (F6), A122189(F7), A079262 (F8), A104144 (F9), A122265 (F10).
(According to the definition, F0 = A000004 and F1 = A000012.)
Cf. A001605 (indices of prime numbers in F2).
Cf. A000043, A050414.

A302990(n,L=oo,a=vector(n+1,i,if(i1 && for(i=-2+2*n+=1,L, ispseudoprime(a[i%n+1]=2*a[(i-1)%n+1]-a[i%n+1]) && return(i))} \\ Testing primality only for i%n>n-3 is not faster, even for large n. - M. F. Hasler, Apr 17 2018; improved Apr 18 2018

a(n) == -1 or -2 (mod n+1). a(n) >= 2n, with equality iff n is in A000043. a(n) <= 2n+1 for n+1 in A050414. - M. F. Hasler, Apr 18 2018

a(29) from Jacques Tramu, Apr 19 2018
a(33) from Daniel Suteu, Apr 20 2018
a(36) from Jacques Tramu, Apr 25 2018

cp's OEIS Frontend

A074193 Sum of determinants of 2nd order principal minors of powers of the matrix ((1,1,0,0),(1,0,1,0),(1,0,0,1),(1,0,0,0)).

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A102036 Triangle, read by rows, where the terms are generated by the rule: T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k-1) + T(n-3,k-1), with T(0,0)=1.

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A113243 Differences of nonzero tetranacci numbers.

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A118897 Triangle read by rows: T(n,k) is the number of binary sequences of length n containing k subsequences 0000 (n,k>=0).

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A144406 Rectangular array A read by upward antidiagonals: entry A(n,k) in row n and column k gives the number of compositions of k in which no part exceeds n, n>=1, k>=0.

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A168088 a(n) = 2^tetranacci(n).

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A220493 Fibonacci 15-step numbers, a(n) = a(n-1) + a(n-2) + ... + a(n-15).

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A299399 a(n) = a(n-1)a(n-2)a(n-3)*a(n-4); a(0..3) = (1, 1, 2, 3).