A293411 -id:A293411 - cp's OEIS frontend

A126116 a(n) = a(n-1) + a(n-3) + a(n-4), with a(0)=a(1)=a(2)=a(3)=1.

1, 1, 1, 1, 3, 5, 7, 11, 19, 31, 49, 79, 129, 209, 337, 545, 883, 1429, 2311, 3739, 6051, 9791, 15841, 25631, 41473, 67105, 108577, 175681, 284259, 459941, 744199, 1204139, 1948339, 3152479, 5100817, 8253295, 13354113, 21607409, 34961521

Offset: 0

Luis A Restrepo (luisiii(AT)mac.com), Mar 05 2007

nonn

This sequence has the same growth rate as the Fibonacci sequence, since x^4 - x^3 - x - 1 has the real roots phi and -1/phi.

The Ca1 sums, see A180662 for the definition of these sums, of triangle A035607 equal the terms of this sequence without the first term. - Johannes W. Meijer, Aug 05 2011

			G.f. = 1 + x + x^2 + x^3 + 3*x^4 + 5*x^5 + 7*x^6 + 11*x^7 + 19*x^8 + 31*x^9 + ...

S. Wolfram, A New Kind of Science. Champaign, IL: Wolfram Media, pp. 82-92, 2002

Seiichi Manyama, Table of n, a(n) for n = 0..4786
K. T. Atanassov, D. R. Deford, A. G. Shannon, Pulsated Fibonacci recurrences, Fibonacci Quarterly, Vol. 52, No. 5, Dec. 2014, pp. 22-27.
Kelley L. Ross, The Golden Ratio and The Fibonacci Numbers
Eric Weisstein's World of Mathematics, Golden Ratio
Wikipedia, Golden Ratio
Index entries for linear recurrences with constant coefficients, signature (1,0,1,1).

Cf. Fibonacci numbers A000045; Lucas numbers A000032; tribonacci numbers A000213; tetranacci numbers A000288; pentanacci numbers A000322; hexanacci numbers A000383; 7th-order Fibonacci numbers A060455; octanacci numbers A079262; 9th-order Fibonacci sequence A127193; 10th-order Fibonacci sequence A127194; 11th-order Fibonacci sequence A127624, A128429.

a:=[1,1,1,1];; for n in [5..50] do a[n]:=a[n-1]+a[n-3]+a[n-4]; od; a; # G. C. Greubel, Jul 15 2019

[n le 4 select 1 else Self(n-1) + Self(n-3) + Self(n-4): n in [1..50]]; // Vincenzo Librandi, Dec 25 2015

# From R. J. Mathar, Jul 22 2010: (Start)
A010684 := proc(n) 1+2*(n mod 2) ; end proc:
A000032 := proc(n) coeftayl((2-x)/(1-x-x^2),x=0,n) ; end proc:
A126116 := proc(n) ((-1)^floor(n/2)*A010684(n)+2*A000032(n))/5 ; end proc: seq(A126116(n),n=0..80) ; # (End)
with(combinat): A126116 := proc(n): fibonacci(n-1) + fibonacci(floor((n-4)/2)+1)* fibonacci(ceil((n-4)/2)+2) end: seq(A126116(n), n=0..38); # Johannes W. Meijer, Aug 05 2011

LinearRecurrence[{1,0,1,1},{1,1,1,1},50] (* Harvey P. Dale, Nov 08 2011 *)

Vec((x-1)*(1+x+x^2)/((x^2+x-1)*(x^2+1)) + O(x^50)) \\ Altug Alkan, Dec 25 2015

((1-x)*(1+x+x^2)/((1-x-x^2)*(1+x^2))).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Jul 15 2019

From R. J. Mathar, Jul 22 2010: (Start)
G.f.: (1-x)*(1+x+x^2)/((1-x-x^2)*(1+x^2)).
a(n) = ( (-1)^floor(n/2) * A010684(n) + 2*A000032(n))/5.
a(2*n) = A061646(n). (End)
From Johannes W. Meijer, Aug 05 2011: (Start)
a(n) = F(n-1) + A070550(n-4) with F(n) = A000045(n).
a(n) = F(n-1) + F(floor((n-4)/2) + 1)*F(ceiling((n-4)/2) + 2). (End)
a(n) = (1/5)*((sqrt(5)-1)*(1/2*(1+sqrt(5)))^n - (1+sqrt(5))*(1/2*(1-sqrt(5)))^n + sin((Pi*n)/2) - 3*cos((Pi*n)/2)). - Harvey P. Dale, Nov 08 2011
(-1)^n * a(-n) = a(n) = F(n) - A070550(n - 6). - Michael Somos, Feb 05 2012
a(n)^2 + 3*a(n-2)^2 + 6*a(n-5)^2 + 3*a(n-7)^2 = a(n-8)^2 + 3*a(n-6)^2 + 6*a(n-3)^2 + 3*a(n-1)^2. - Greg Dresden, Jul 07 2021
a(n) = A293411(n)-A293411(n-1). - R. J. Mathar, Jul 20 2025

Edited by Don Reble, Mar 09 2007

A295754 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 3, 4, 12, 23, 37, 61, 105, 175, 284, 463, 757, 1231, 1994, 3231, 5237, 8481, 13726, 22215, 35955, 58186, 94152, 152348, 246516, 398882, 645411, 1044305, 1689734, 2734059, 4423808, 7157881, 11581709, 18739612, 30321339, 49060968, 79382329, 128443321

Offset: 0

Clark Kimberling, Nov 30 2017

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).

			a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, so that
b(4) = 9 (least "new number")
a(4) = a(3) + a(1) + a(0) + b(0) = 12
Complement: (b(n)) = (5, 6, 7, 8, 9, 10, 11, 13, 14, 15, ...)

Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Cf. A001622, A000045, A293411, A295755.

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4;
b[0] = 5; b[1] = 6; b[2] = 7; b[3] = 8;
a[n_] := a[n] = a[n - 1] + a[n - 3] + a[n - 4] + b[n - 4];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
z = 36;  Table[a[n], {n, 0, z}]   (* A295754 *)
Table[b[n], {n, 0, 20}]  (*complement *)

A295755 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-3), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 3, 4, 13, 25, 40, 66, 114, 190, 308, 502, 821, 1335, 2162, 3503, 5678, 9195, 14881, 24084, 38980, 63080, 102071, 165162, 267250, 432430, 699693, 1132136, 1831848, 2964004, 4795867, 7759886, 12555774, 20315682, 32871473, 53187172, 86058669, 139245866

Offset: 0

Clark Kimberling, Nov 30 2017

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).

			a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, so that
b(4) = 9 (least "new number")
a(4) = a(3) + a(1) + a(0) + b(1) = 13
Complement: (b(n)) = (5, 6, 7, 8, 9, 10, 11, 12, 14, 15, ...)

Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Cf. A001622, A000045, A293411, A295754.

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4;
b[0] = 5; b[1] = 6; b[2] = 7; b[3] = 8;
a[n_] := a[n] = a[n - 1] + a[n - 3] + a[n - 4] + b[n - 3];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
z = 36;  Table[a[n], {n, 0, z}]   (* A295755 *)
Table[b[n], {n, 0, 20}]  (*complement *)

A295756 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-2), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 3, 4, 14, 27, 43, 71, 123, 205, 332, 541, 885, 1439, 2330, 3775, 6119, 9909, 16036, 25953, 42005, 67975, 109990, 177976, 287985, 465980, 753977, 1219970, 1973968, 3193959, 5167941, 8361915, 13529879, 21891817, 35421712, 57313546, 92735283, 150048854

Offset: 0

Clark Kimberling, Dec 01 2017

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).

			a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, so that
b(4) = 9 (least "new number")
a(4) = a(3) + a(1) + a(0) + b(2) = 14
Complement: (b(n)) = (5, 6, 7, 8, 9, 10, 11, 12, 13, 15, ...)

Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Cf. A001622, A000045, A293411, A295754.

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4;
b[0] = 5; b[1] = 6; b[2] = 7; b[3] = 8;
a[n_] := a[n] = a[n - 1] + a[n - 3] + a[n - 4] + b[n - 2];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
z = 36;  Table[a[n], {n, 0, z}]   (* A295756 *)
Table[b[n], {n, 0, 20}]  (*complement *)

A295757 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-1), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 3, 4, 15, 29, 46, 76, 132, 220, 356, 580, 949, 1543, 2498, 4047, 6560, 10623, 17191, 27822, 45030, 72870, 117910, 190790, 308720, 499531, 808263, 1307806, 2116091, 3423920, 5540025, 8963959, 14504008, 23467992, 37972016, 61440024, 99412066, 160852117

Offset: 0

Clark Kimberling, Dec 01 2017

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).

			a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, so that
b(4) = 9 (least "new number")
a(4) = a(3) + a(1) + a(0) + b(3) = 15
Complement: (b(n)) = (5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, ...)

Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Cf. A001622, A000045, A293411, A295754.

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4;
b[0] = 5; b[1] = 6; b[2] = 7; b[3] = 8;
a[n_] := a[n] = a[n - 1] + a[n - 3] + a[n - 4] + b[n - 1];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
z = 36;  Table[a[n], {n, 0, z}]   (* A295757 *)
Table[b[n], {n, 0, 20}]  (*complement *)

cp's OEIS Frontend

A126116 a(n) = a(n-1) + a(n-3) + a(n-4), with a(0)=a(1)=a(2)=a(3)=1.

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A295754 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

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A295755 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-3), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

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A295756 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-2), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

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A295757 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-1), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

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