A005708 -id:A005708 - cp's OEIS frontend

A193941 G.f.: (1+x^3)/(1-x-x^6).

1, 1, 1, 2, 2, 2, 3, 4, 5, 7, 9, 11, 14, 18, 23, 30, 39, 50, 64, 82, 105, 135, 174, 224, 288, 370, 475, 610, 784, 1008, 1296, 1666, 2141, 2751, 3535, 4543, 5839, 7505, 9646, 12397, 15932, 20475, 26314, 33819, 43465, 55862, 71794, 92269, 118583, 152402

Offset: 0

Johannes W. Meijer, Aug 11 2011

The Ca1 sums, see A180662, of triangle A065941 equal the terms of this sequence.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,1).

Cf. A005708.

A193941 := proc(n): coeftayl((1+x^3)/(1-x-x^6),x=0,n) end: seq(A193941(n),n=0..49);

CoefficientList[Series[(1+x^3)/(1-x-x^6),{x,0,50}],x] (* or *) LinearRecurrence[{1,0,0,0,0,1},{1,1,1,2,2,2},50] (* Harvey P. Dale, Apr 25 2014 *)

Vec((1+x^3)/(1-x-x^6) + O(x^50)) \\ Jinyuan Wang, Apr 01 2020

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

a(n) = A005708(n) + A005708(n-3).

A242763 a(n) = 1 for n <= 7; a(n) = a(n-5) + a(n-7) for n>7.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 7, 7, 8, 9, 9, 12, 12, 15, 16, 17, 21, 21, 27, 28, 32, 37, 38, 48, 49, 59, 65, 70, 85, 87, 107, 114, 129, 150, 157, 192, 201, 236, 264, 286, 342, 358, 428, 465, 522, 606, 644, 770, 823, 950, 1071, 1166, 1376

Offset: 1

Vicente Jesús Maniega Granado, Oct 03 2016

Generalized Fibonacci growth sequence using i = 2 as maturity period, j = 5 as conception period, and k = 2 as growth factor.

Maturity period is the number of periods that a Fibonacci tree node needs for being able to start developing branches. Conception period is the number of periods in a Fibonacci tree node needed to develop new branches since its maturity. Growth factor is the number of additional branches developed by a Fibonacci tree node, plus 1, and equals the base of the exponential series related to the given tree if maturity factor would be zero. Standard Fibonacci would use 1 as maturity period, 1 as conception period, and 2 as growth factor as the series becomes equal to 2^n with a maturity period of 0. Related to Lucas sequences.

			For n = 13 the a(13) = a(8) + a(6) = 2 + 1 = 3.

Colin Barker, Table of n, a(n) for n = 1..1000
Julia Collins, Fibonacci Tree
Fractal Foundation, Fibonacci Fractals
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1,0,1).

Cf. A000079 (i = 0, j = 1, k = 2), A000244 (i = 0, j = 1, k = 3), A000302 (i = 0, j = 1, k = 4), A000351 (i = 0, j = 1, k = 5), A000400 (i = 0, j = 1, k = 6), A000420 (i = 0, j = 1, k = 7), A001018 (i = 0, j = 1, k = 8), A001019 (i = 0, j = 1, k = 9), A011557 (i = 0, j = 1, k = 10), A001020 (i = 0, j = 1, k = 11), A001021 (i = 0, j = 1, k = 12), A016116 (i = 0, j = 2, k = 2), A108411 (i = 0, j = 2, k = 3), A213173 (i = 0, j = 2, k = 4), A074872 (i = 0, j = 2, k = 5), A173862 (i = 0, j = 3, k = 2), A127975 (i = 0, j = 3, k = 3), A200675 (i = 0, j = 4, k = 2), A111575 (i = 0, j = 4, k = 3), A000045 (i = 1, j = 1, k = 2), A001045 (i = 1, j = 1, k = 3), A006130 (i = 1, j = 1, k = 4), A006131 (i = 1, j = 1, k = 5), A015440 (i = 1, j = 1, k = 6), A015441 (i = 1, j = 1, k = 7), A015442 (i = 1, j = 1, k = 8), A015443 (i = 1, j = 1, k = 9), A015445 (i = 1, j = 1, k = 10), A015446 (i = 1, j = 1, k = 11), A015447 (i = 1, j = 1, k = 12), A000931 (i = 1, j = 2, k = 2), A159284 (i = 1, j = 2, k = 3), A238389 (i = 1, j = 2, k = 4), A097041 (i = 1, j = 2, k = 10), A079398 (i = 1, j = 3, k = 2), A103372 (i = 1, j = 4, k = 2), A103373 (i = 1, j = 5, k = 2), A103374 (i = 1, j = 6, k = 2), A000930 (i = 2, j = 1, k = 2), A077949 (i = 2, j = 1, k = 3), A084386 (i = 2, j = 1, k = 4), A089977 (i = 2, j = 1, k = 5), A178205 (i = 2, j = 1, k = 11), A103609 (i = 2, j = 2, k = 2), A077953 (i = 2, j = 2, k = 3), A226503 (i = 2, j = 3, k = 2), A122521 (i = 2, j = 6, k = 2), A003269 (i = 3, j = 1, k = 2), A052942 (i = 3, j = 1, k = 3), A005686 (i = 3, j = 2, k = 2), A237714 (i = 3, j = 2, k = 3), A238391 (i = 3, j = 2, k = 4), A247049 (i = 3, j = 3, k = 2), A077886 (i = 3, j = 3, k = 3), A003520 (i = 4, j = 1, k = 2), A108104 (i = 4, j = 2, k = 2), A005708 (i = 5, j = 1, k = 2), A237716 (i = 5, j = 2, k = 3), A005709 (i = 6, j = 1, k = 2), A122522 (i = 6, j = 2, k = 2), A005710 (i = 7, j = 1, k = 2), A237718 (i = 7, j = 2, k = 3), A017903 (i = 8, j = 1, k = 2).

[n le 7 select 1 else Self(n-5)+Self(n-7): n in [1..70]]; // Vincenzo Librandi, Nov 30 2016

LinearRecurrence[{0, 0, 0, 0, 1, 0, 1}, {1, 1, 1, 1, 1, 1, 1}, 70] (*  or *)
CoefficientList[ Series[(1+x+x^2+x^3+x^4)/(1-x^5-x^7), {x, 0, 70}], x] (* Robert G. Wilson v, Nov 25 2016 *)
nxt[{a_,b_,c_,d_,e_,f_,g_}]:={b,c,d,e,f,g,a+c}; NestList[nxt,{1,1,1,1,1,1,1},70][[;;,1]] (* Harvey P. Dale, Oct 22 2024 *)

Vec(x*(1+x+x^2+x^3+x^4)/((1-x+x^2)*(1+x-x^3-x^4-x^5)) + O(x^100)) \\ Colin Barker, Oct 27 2016

@CachedFunction # a = A242763
def a(n): return 1 if n<8 else a(n-5) +a(n-7)
[a(n) for n in range(1,76)] # G. C. Greubel, Oct 23 2024

Generic a(n) = 1 for n <= i+j; a(n) = a(n-j) + (k-1)*a(n-(i+j)) for n>i+j where i = maturity period, j = conception period, k = growth factor.
G.f.: x*(1+x+x^2+x^3+x^4) / ((1-x+x^2)*(1+x-x^3-x^4-x^5)). - Colin Barker, Oct 09 2016
Generic g.f.: x*(Sum_{l=0..j-1} x^l) / (1-x^j-(k-1)*x^(i+j)), with i > 0, j > 0 and k > 1.

A247489 Square array read by antidiagonals: A(k, n) = hypergeometric(P, Q, -k^k/(k-1)^(k-1)) rounded to the nearest integer, P = [(j-n)/k, j=0..k-1] and Q = [(j-n)/(k-1), j=0..k-2], k>=1, n>=0.

Original entry on oeis.org

1, 0, 2, 0, 1, 4, 0, 1, 2, 8, 0, 1, 1, 3, 16, 0, 1, 1, 2, 5, 32, 0, 1, 1, 1, 3, 8, 64, 0, 0, 1, 1, 2, 4, 13, 128, 0, 0, 1, 1, 1, 3, 6, 21, 256, 0, 0, 1, 1, 1, 2, 4, 9, 34, 512, 0, 0, 1, 1, 1, 1, 3, 5, 13, 55, 1024, 0, 0, 1, 1, 1, 1, 2, 4, 7, 19, 89, 2048

Offset: 0

Peter Luschny, Sep 19 2014

Conjecture: hypergeometric(P, Q, -k^k/(k-1)^(k-1)) = sum_{j=0.. floor(n/k)} binomial(n-(k-1)*j, j) for n>=(k-1)^2, P and Q as above. (This means for n>=(k-1)^2 the representation is exact without rounding.)

			First few rows of the square array:
[k\n]                                             if conjecture true
[1], 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, ...     A000079  n>=0
[2], 0, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...     'A000045' n>=1
[3], 0, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, ...    A000930  n>=4
[4], 0, 1, 1, 1, 2, 3, 4, 5, 7, 10, 14, 19, ...     A003269  n>=9
[5], 0, 1, 1, 1, 1, 2, 3, 4, 5, 6, 9, 11, 15, ...   A003520  n>=16
[6], 0, 1, 1, 1, 1, 1, 2, 3, 3, 4, 6, 7, 10, ...    A005708  n>=25
[7], 0, 0, 1, 1, 1, 1, 1, 2, 3, 3, 4, 5, 7, 8, ...  A005709  n>=36
[8], 0, 0, 1, 1, 1, 1, 2, 1, 2, 3, 3, 4, 5, 6, ...  A005710  n>=49
'A000045' means that the Fibonacci numbers as referenced here start 1, 1, 2, 3, ... for n>=0.

Cf. A000079, A000045, A000930, A003269, A003520, A005708, A005709, A005710.

A247489 := proc(k, n)
seq((j-n)/k, j=0..k-1); seq((j-n)/(k-1), j=0..k-2);
hypergeom([%%], [%], -k^k/(k-1)^(k-1));
round(evalf(%,100)) end: # Adjust precision if necessary!
for k from 1 to 9 do print(seq(A247489(k, n), n=0..16)) od;

def A247489(k, n):
    P = [(j-n)/k for j in range(k)]
    Q = [(j-n)/(k-1) for j in range(k-1)]
    H = hypergeometric(P, Q, -k^k/(k-1)^(k-1))
    return round(H.n(100)) # Adjust precision if necessary!

A292027 a(n) = a(n-7) + a(n-11), starting a(0)=a(1)=...= a(10) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5, 7, 7, 7, 8, 9, 9, 9, 12, 12, 12, 13, 16, 16, 16, 20, 21, 21, 22, 28, 28, 28, 33, 37, 37, 38, 48, 49, 49, 55, 65, 65, 66, 81, 86, 86, 93, 113, 114, 115, 136, 151, 151, 159, 194, 200, 201, 229, 264, 265, 274

Offset: 0

Jason Bruce, Sep 07 2017

Kenneth H. Rosen, Discrete Mathematics and its Applications, McGraw-Hill, 2012, 501-503.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,1,0,0,0,1).

Cf. A000930, A003269, A003520, A005708, A005709.

import java.util.Arrays;
public class IntegerSequences
{
    public static void main(String[] args)
    {
        int j = 7;
        int k = 11;
        // Set N to the number of terms you would like to generate.
        int N = 200;
        long[] G = new long[N];
        for(int i=0; i

LinearRecurrence[{0,0,0,0,0,0,1,0,0,0,1},{1,1,1,1,1,1,1,1,1,1,1},80] (* Harvey P. Dale, Oct 09 2018 *)

G.f.: (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)/(1 - x^7 - x^11). - R. J. Mathar and N. J. A. Sloane, Nov 10 2017

cp's OEIS Frontend

A193941 G.f.: (1+x^3)/(1-x-x^6).

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A242763 a(n) = 1 for n <= 7; a(n) = a(n-5) + a(n-7) for n>7.

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A247489 Square array read by antidiagonals: A(k, n) = hypergeometric(P, Q, -k^k/(k-1)^(k-1)) rounded to the nearest integer, P = [(j-n)/k, j=0..k-1] and Q = [(j-n)/(k-1), j=0..k-2], k>=1, n>=0.

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A292027 a(n) = a(n-7) + a(n-11), starting a(0)=a(1)=...= a(10) = 1.

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A365798 G.f. satisfies A(x) = 1 + x*A(x)*(1 + x^5*A(x)^4).

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A365798 G.f. satisfies A(x) = 1 + xA(x)(1 + x^5*A(x)^4).