A077953 -id:A077953 - cp's OEIS frontend

A077925 Expansion of 1/((1-x)(1+2x)).

1, -1, 3, -5, 11, -21, 43, -85, 171, -341, 683, -1365, 2731, -5461, 10923, -21845, 43691, -87381, 174763, -349525, 699051, -1398101, 2796203, -5592405, 11184811, -22369621, 44739243, -89478485, 178956971, -357913941, 715827883, -1431655765, 2863311531, -5726623061

Offset: 0

N. J. A. Sloane, Nov 17 2002

a(n+1) is the reflection of a(n) through a(n-1) on the numberline. - Floor van Lamoen, Aug 31 2004
If a zero is added as the (new) a(0) in front, the sequence represents the inverse binomial transform of A001045. Partial sums are in A077898. - R. J. Mathar, Aug 30 2008
a(n) = A077953(2*n+3). - Reinhard Zumkeller, Oct 07 2008
Related to the Fibonacci sequence by an INVERT transform: if A(x) = 1+x^2*g(x) is the generating function of the a(n) prefixed with 1, 0, then 1/A(x) = 2+(x+1)/(x^2-x+1) is the generating function of 1, 0, -1, 1, -2, 3, ..., the signed Fibonacci sequence A000045 prefixed with 1. - Gary W. Adamson, Jan 07 2011
Also: Gaussian binomial coefficients [n+1,1], or q-integers, for q=-2, diagonal k=1 in the triangular (or column r=1 in the square) array A015109. - M. F. Hasler, Nov 04 2012
With a leading zero, 0, 1, -1, 3, -5, 11, -21, 43, -85, 171, -341, 683, ... we obtain the Lucas U(-1,-2) sequence. - R. J. Mathar, Jan 08 2013
Let m = a(n). Then 18*m^2 - 12*m + 1 = A000225(2n+3). - Roderick MacPhee, Jan 17 2013

			G.f. = 1 - x + 3*x^2 - 5*x^3 + 11*x^4 - 21*x^5 + 43*x^6 - 85*x^7 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Lucas sequence
Index entries for linear recurrences with constant coefficients, signature (-1,2)
Index entries related to Gaussian binomial coefficients.
Index entries for Lucas sequences

Cf. A001045 (unsigned version).
Cf. A014983, A014985, A014986. - Zerinvary Lajos, Dec 16 2008
Cf. A232600, A232601, A232602.

[(1-(-2)^(n+1))/3: n in [0..40]]; // Vincenzo Librandi, Jun 21 2011

a:=n->sum ((-2)^j, j=0..n): seq(a(n), n=0..35); # Zerinvary Lajos, Dec 16 2008

CoefficientList[Series[(1 - x)^(-1)/(1 + 2 x), {x, 0, 50}], x] (* Vladimir Joseph Stephan Orlovsky, Jun 20 2011 *)

a(n)=(1+(-2)^n*2)/3 \\ Charles R Greathouse IV, Jun 21 2011

[gaussian_binomial(n,1,-2) for n in range(1,35)] # Zerinvary Lajos, May 28 2009

G.f.: 1/(1+x-2*x^2).
a(n) = (1-(-2)^(n+1))/3. - Vladeta Jovovic, Apr 17 2003
a(n) = Sum_{k=0..n} (-2)^k. - Paul Barry, May 26 2003
a(n+1) - a(n) = A122803(n). - R. J. Mathar, Aug 30 2008
a(n) = Sum_{k=0..n} A112555(n,k)*(-2)^k. - Philippe Deléham, Sep 11 2009
a(n) = A082247(n+1) - 1. - Philippe Deléham, Oct 07 2009
G.f.: Q(0)/(3*x), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k + 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k - 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 22 2013
G.f.: Q(0)/2 , where Q(k) = 1 + 1/(1 - x*(4*k-1 + 2*x)/( x*(4*k+1 + 2*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 08 2013
E.g.f.: (2*exp(-2*x) + exp(x))/3. - Ilya Gutkovskiy, Nov 12 2016
a(n) = A086893(n+2) - A061547(n+3), n >= 0. - Yosu Yurramendi, Jan 16 2017
a(n) = (-1)^n*A001045(n+1). - M. F. Hasler, Feb 13 2020
a(n) - a(n-1) = a(n-1) - a(n+1) = (-2)^n, a(n+1) = - a(n) + 2*a(n-1) = 1 - 2*a(n). - Michael Somos, Feb 22 2023

A052551 Expansion of 1/((1 - x)(1 - 2x^2)).

Original entry on oeis.org

1, 1, 3, 3, 7, 7, 15, 15, 31, 31, 63, 63, 127, 127, 255, 255, 511, 511, 1023, 1023, 2047, 2047, 4095, 4095, 8191, 8191, 16383, 16383, 32767, 32767, 65535, 65535, 131071, 131071, 262143, 262143, 524287, 524287, 1048575, 1048575, 2097151, 2097151

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Equals row sums of triangle A137865. - Gary W. Adamson, Feb 18 2008
Also, the decimal representation of the diagonal from the corner to the origin of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 566", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. - Robert Price, Jul 05 2017
Number of nonempty subsets of {1,2,...,n+1} that contain only odd numbers. a(0) = a(1) = 1: {1}; a(6) = a(7) = 15: {1}, {3}, {5}, {7}, {1,3}, {1,5}, {1,7}, {3,5}, {3,7}, {5,7}, {1,3,5}, {1,3,7}, {1,5,7}, {3,5,7}, {1,3,5,7}. - Enrique Navarrete, Mar 16 2018
Number of nonempty subsets of {1,2,...,n+2} that contain only even numbers. a(0) = a(1) = 1: {2}; a(4) = a(5) = 7: {2}, {4}, {6}, {2,4}, {2,6}, {4,6}, {2,4,6}. - Enrique Navarrete, Mar 26 2018
Doubling of A000225(n+1), n >= 0 entries. First differences give A077957. - Wolfdieter Lang, Apr 08 2018
a(n-2) is the number of achiral rows or cycles of length n partitioned into two sets or the number of color patterns using exactly 2 colors. An achiral row or cycle is equivalent to its reverse. Two color patterns are equivalent if the colors are permuted. For n = 4, the a(n-2) = 3 row patterns are AABB, ABAB, and ABBA; the cycle patterns are AAAB, AABB, and ABAB. For n = 5, the a(n-2) = 3 patterns for both rows and cycles are AABAA, ABABA, and ABBBA. For n = 6, the a(n-2) = 7 patterns for rows are AAABBB, AABABB, AABBAA, ABAABA, ABABAB, ABBAAB, and ABBBBA; the cycle patterns are AAAAAB, AAAABB, AAABAB, AAABBB, AABAAB, AABABB, and ABABAB. - Robert A. Russell, Oct 15 2018
For integers m > 1, the expansion of 1/((1 - x)*(1 - m*x^2)) generates a(n) = (sqrt(m)^(n + 1)*((-1)^n*(sqrt(m) - 1) + sqrt(m) + 1) - 2)/(2*(m - 1)). It appears, for integer values of n >= 0 and m > 1, that it could be simplified in the integral domain a(n) = (m^(1 + floor(n/2)) - 1)/(m - 1). - Federico Provvedi, Nov 23 2018
From Werner Schulte, Mar 04 2019: (Start)
More generally: For some fixed integers q and r > 0 the expansion of A(q,r; x) = 1/((1-x)*(1-q*x^r)) generates coefficients a(q,r; n) = (q^(1+floor(n/r))-1)/(q-1) for n >= 0; the special case q = 1 leads to a(1,r; n) = 1 + floor(n/r).
The a(q,r; n) satisfy for n > r a linear recurrence equation with constant coefficients. The signature vector is given by the sum of two vectors v and w where v has terms 1 followed by r zeros, i.e., (1,0,0,...,0), and w has r-1 leading zeros followed by q and -q, i.e., (0,0,...,0,q,-q).
Let a_i(q,r; n) be the convolution inverse of a(q,r; n). The terms are given by the sum a_i(q,r; n) = b(n) + c(n) for n >= 0 where b(n) has terms 1 and -1 followed by infinitely zeros, i.e., (1,-1,0,0,0,...), and c(n) has r leading zeros followed by -q, q and infinitely zeros, i.e., (0,0,...,0,-q,q,0,0,0,...).
Here is an example for q = 3 and r = 5: The expansion of A(3,5; x) = 1/((1-x)*(1-3*x^5)) = Sum_{n>=0} a(3,5; n)*x^n generates the sequence of coefficients (a(3,5; n)) = (1,1,1,1,1,4,4,4,4,4,13,13,13,13,13,40,...) where r = 5 controls the repetition and q = 3 the different values.
The a(3,5; n) satisfy for n > 5 the linear recurrence equation with constant coefficients and signature (1,0,0,0,0,0) + (0,0,0,0,3,-3) = (1,0,0,0,3,-3).
The convolution inverse a_i(3,5; n) has terms (1,-1,0,0,0,0,0,0,0,...) + (0,0,0,0,0,-3,3,0,0,...) = (1,-1,0,0,0,-3,3,0,0,...).
For further examples and informations see A014983 (q,r = -3,1), A077925 (q,r = -2,1), A000035 (q,r = -1,1), A000012 (q,r = 0,1), A000027 (q,r = 1,1), A000225 (q,r = 2,1), A003462 (q,r = 3,1), A077953 (q,r = -2,2), A133872 (q,r = -1,2), A004526 (q,r = 1,2), A052551 (this sequence with q,r = 2,2), A077886 (q,r = -2,3), A088911 (q,r = -1,3), A002264 (q,r = 1,3) and A077885 (q,r = 2,3). The offsets might be different.
(End)
a(n) is the number of palindromes of length n over the alphabet {1,2} containing the letter 1. More generally, the number of palindromes of length n over the alphabet {1,2,...,k} containing the letter 1 is given by k^ceiling(n/2)-(k-1)^ceiling(n/2). - Sela Fried, Dec 10 2024

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 488
Donatella Merlini and Massimo Nocentini, Algebraic Generating Functions for Languages Avoiding Riordan Patterns, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.3.
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
S. Wolfram, A New Kind of Science
Wolfram Research, Wolfram Atlas of Simple Programs
Index entries for sequences related to cellular automata
Index to 2D 5-Neighbor Cellular Automata
Index to Elementary Cellular Automata
Index entries for linear recurrences with constant coefficients, signature (1,2,-2).

Cf. A000225, A077957, A136865, A289404, A289405, A032085, A050686.
Column 2 (offset by two) of A304972.
Cf. A000225 (oriented), A056326 (unoriented), and A122746(n-2) (chiral) for rows.
Cf. A056295 (oriented), A056357 (unoriented), and A059053 (chiral) for cycles.

Flat(List([1..21],n->[2^n-1,2^n-1])); # Muniru A Asiru, Oct 16 2018

[2^Floor(n/2)-1: n in [2..50]]; // Vincenzo Librandi, Aug 16 2011

spec := [S,{S=Prod(Sequence(Prod(Z,Union(Z,Z))),Sequence(Z))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);

Table[StirlingS2[Floor[n/2] + 2, 2], {n, 0, 50}] (* Robert A. Russell, Dec 20 2017 *)
Drop[LinearRecurrence[{1, 2, -2}, {0, 1, 1}, 50], 1] (* Robert A. Russell, Oct 14 2018 *)
CoefficientList[Series[1/((1-x)*(1-2*x^2)), {x, 0, 50}], x] (* Stefano Spezia, Oct 16 2018 *)
2^(1+Floor[(Range[0,50])/2])-1 (* Federico Provvedi, Nov 22 2018 *)
((-1)^#(Sqrt[2]-1)+Sqrt[2]+1)2^((#-1)/2)-1&@Range[0, 50] (* Federico Provvedi, Nov 23 2018 *)

x='x+O('x^50); Vec(1/((1-x)*(1-2*x^2))) \\ Altug Alkan, Mar 19 2018

[2^(floor(n/2)) -1 for n in (2..50)] # G. C. Greubel, Mar 04 2019

G.f.: 1/((1 - x)*(1 - 2*x^2)).
Recurrence: a(1) = 1, a(0) = 1, -2*a(n) - 1 + a(n+2) = 0.
a(n) = -1 + Sum((1/2)*(1 + 2*alpha)*alpha^(-1 - n)) where the sum is over alpha = the two roots of -1 + 2*x^2.
a(n) = A016116(n+2) - 1. - R. J. Mathar, Jun 15 2009
a(n) = A060546(n+1) - 1. - Filip Zaludek, Dec 10 2016
From Robert A. Russell, Oct 15 2018: (Start)
a(n-2) = S2(floor(n/2)+1,2), where S2 is the Stirling subset number A008277.
a(n-2) = 2*A056326(n) - A000225(n) = A000225(n) - 2*A122746(n-2) = A056326(n) - A122746(n-2).
a(n-2) = 2*A056357(n) - A056295(n) = A056295(n) - 2*A059053(n) = A056357(n) - A059053(n). (End)
From Federico Provvedi, Nov 22 2018: (Start)
a(n) = 2^( 1 + floor(n/2) ) - 1.
a(n) = ( (-1)^n*(sqrt(2)-1) + sqrt(2) + 1 ) * 2^( (n - 1)/2 ) - 1.  (End)
E.g.f.: 2*cosh(sqrt(2)*x) + sqrt(2)*sinh(sqrt(2)*x) - cosh(x) - sinh(x). - Franck Maminirina Ramaharo, Nov 23 2018

More terms from James Sellers, Jun 06 2000

A007420 Berstel sequence: a(n+1) = 2a(n) - 4a(n-1) + 4*a(n-2).

Original entry on oeis.org

0, 0, 1, 2, 0, -4, 0, 16, 16, -32, -64, 64, 256, 0, -768, -512, 2048, 3072, -4096, -12288, 4096, 40960, 16384, -114688, -131072, 262144, 589824, -393216, -2097152, -262144, 6291456, 5242880, -15728640, -27262976, 29360128, 104857600, -16777216

Offset: 0

N. J. A. Sloane

a(n) = 0 only for n = 0,1,4,6,13 and 52. [Cassels, following Mignotte. See also Beukers] - N. J. A. Sloane, Aug 29 2010

J. W. S. Cassels, Local Fields, Cambridge, 1986, see p. 67.
G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; p. 28.
J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 193.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..500
F. Beukers, The zero-multiplicity of ternary recurrences, Compositio Math. 77 (1991), 165-177.
Daniel Birmajer, Juan B. Gil, and Michael D. Weiner, Linear recurrence sequences with indices in arithmetic progression and their sums, arXiv:1505.06339 [math.NT], 2015.
M. Mignotte, Suites récurrentes linéaires, Sém. Delange-Pisot-Poitou, 15th year (1973/1974), No. 14, 9 pages.
G. Myerson and A. J. van der Poorten, Some problems concerning recurrence sequences, Amer. Math. Monthly 102 (1995), no. 8, 698-705.
Index entries for linear recurrences with constant coefficients, signature (2,-4,4).

Cf. A035302, A077953.

a007420 n = a007420_list !! n
a007420_list = 0 : 0 : 1 : (map (* 2) $ zipWith (+) (drop 2 a007420_list)
   (map (* 2) $ zipWith (-) a007420_list (tail a007420_list)))
-- Reinhard Zumkeller, Oct 21 2011

I:=[0,0,1]; [n le 3 select I[n]  else 2*Self(n-1)-4*Self(n-2)+4*Self(n-3): n in [1..70]]; // Vincenzo Librandi, Oct 05 2015

A007420 := proc(n) options remember; if n <=1 then 0 elif n=2 then 1 else 2*A007420(n-1)-4*A007420(n-2)+4*A007420(n-3); fi; end;

a[0] = a[1] = 0; a[2] = 1; a[n_] := a[n] = 2*a[n - 1] - 4*a[n - 2] + 4*a[n - 3]; a /@ Range[0, 34] (* Jean-François Alcover, Apr 06 2011 *)
LinearRecurrence[{2, -4, 4}, {0, 0, 1}, 40] (* Harvey P. Dale, Oct 24 2011 *)
Table[RootSum[-4 + 4 # - 2 #^2 + #^3 &, 6 #^n - #^(n + 1) + 4 #^(n + 1) &]/44, {n, 0, 20}] (* Eric W. Weisstein, Nov 09 2017 *)

a(n)=([0,1,0; 0,0,1; 4,-4,2]^n*[0;0;1])[1,1] \\ Charles R Greathouse IV, Feb 19 2017

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

a(0)=0, a(1)=0, a(2)=1, a(n) = 2*a(n-1)-4*a(n-2)+4*a(n-3). - Harvey P. Dale, Jun 24 2015

A077980 Expansion of 1/(1 + x + 2x^2 + 2x^3).

Original entry on oeis.org

1, -1, -1, 1, 3, -3, -5, 5, 11, -11, -21, 21, 43, -43, -85, 85, 171, -171, -341, 341, 683, -683, -1365, 1365, 2731, -2731, -5461, 5461, 10923, -10923, -21845, 21845, 43691, -43691, -87381, 87381, 174763, -174763, -349525, 349525, 699051, -699051, -1398101, 1398101, 2796203, -2796203, -5592405

Offset: 0

N. J. A. Sloane, Nov 17 2002, Jun 17 2007

Essentially the same as A077953.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-1,-2,-2).

a:=[1,-1,-1];; for n in [4..50] do a[n]:=-a[n-1]-2*a[n-2]-2*a[n-3]; od; a; # G. C. Greubel, Jun 25 2019

I:=[1,-1,-1]; [n le 3 select I[n] else -Self(n-1)-2*Self(n-2) -2*Self(n-3): n in [1..50]]; // Vincenzo Librandi, Aug 17 2013

LinearRecurrence[{-1, -2, -2}, {1, -1, -1}, 50] (* Vincenzo Librandi, Aug 17 2013 *)

Vec(1/(1+x+2*x^2+2*x^3)+O(x^50)) \\ Charles R Greathouse IV, Sep 27 2012

a(n)=1/3*(-1)^floor((n+1)/2)*((2^floor(n/2+1)+(-1)^floor(n/2))) \\ Ralf Stephan, Aug 17 2013

(1/(1+x+2*x^2+2*x^3)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Jun 25 2019

a(n) = (1/3) * (-1)^floor((n+1)/2) * ((2^floor(n/2+1) + (-1)^floor(n/2))). - Ralf Stephan, Aug 17 2013

A373358 a(n) = 4a(n-1) -5a(n-2) +2a(n-3) +2a(n-4) for a(0) = a(1) = 0, a(2) = 1, a(3) = 4 for n >= 4.

Original entry on oeis.org

0, 0, 1, 4, 11, 26, 59, 136, 323, 782, 1903, 4620, 11175, 26970, 65051, 156944, 378811, 914566, 2208199, 5331476, 12871663, 31074802, 75020243, 181113240, 437244675, 1055602590, 2548453951, 6152518684, 14853499511, 35859517706, 86572518539, 209004522016, 504581529803, 1218167581622

Offset: 0

Paul Curtz, Jun 02 2024

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,2,2).

Cf. A000129, A009545, A077893, A077953, A077980, A114203, A146559, A373245.

LinearRecurrence[{4, -5, 2, 2}, {0, 0, 1, 4}, 50] (* Paolo Xausa, Jun 19 2024 *)
nxt[{a_,b_,c_,d_}]:={b,c,d,4d-5c+2b+2a}; NestList[nxt,{0,0,1,4},40][[;;,1]] (* Harvey P. Dale, Jan 11 2025 *)

a(n) = ((([2, 1; 1, 0]^(n+1))[2, 1]) - (1+I)^(n-1) - (1-I)^(n-1))/3 \\ Thomas Scheuerle, Jun 03 2024

G.f.: x^2 / ( (1 - 2*x - x^2) * (1 - 2*x + 2*x^2) ).
E.g.f.: exp(x)*(2*cosh(sqrt(2)*x) - 2*(cos(x)+sin(x)) + sqrt(2)*sinh(sqrt(2)*x))/6.
a(n) = A373245(n+1) - A114203(n+1).
a(0) = 0, a(n) = A373245(n-1) + A146559(n-1).
Binomial transform of 0, 0, followed by A077893 = abs(A077953) = abs(A077980).
a(n) = 4*a(n-1) -5*a(n-2) +2*a(n-3) +2*a(n-4) for n >= 4.
From Thomas Scheuerle, Jun 03 2024: (Start)
a(n) = (A000129(n+1) - A009545(n+1))/3.
a(n) = (-i*sqrt(2)*(1-i)^(n+1) + i*sqrt(2)*(1+i)^(n+1) - (1-sqrt(2))^(n+1) + (1+sqrt(2))^(n+1))/(6*sqrt(2)).
a(n) = 2^n*(hypergeom([1/2 - n/2, -n/2], [-n], -1) - hypergeom([1/2 - n/2, -n/2], [-n], 2))/3. (End)

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.

A304223 Triangle read by rows: T(0,0)=1; T(n,k) = T(n-1,k)-2T(n-2,k-1)+2T(n-3,k-2) for k = 0..floor(2*n/3); T(n,k)=0 for n or k < 0.

Original entry on oeis.org

1, 1, 1, -2, 1, -4, 2, 1, -6, 8, 1, -8, 18, -8, 1, -10, 32, -32, 4, 1, -12, 50, -80, 36, 1, -14, 72, -160, 136, -24, 1, -16, 98, -280, 360, -160, 8, 1, -18, 128, -448, 780, -592, 128, 1, -20, 162, -672, 1484, -1632, 720, -64

Offset: 0

Shara Lalo, May 08 2018

The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle given in A304209.

The coefficients in the expansion of 1/(1-x+2*x^2-2*x^3) are given by the sequence generated by the row sums.

			Triangle begins:
  1;
  1;
  1,  -2;
  1,  -4,   2;
  1,  -6,   8;
  1,  -8,  18,   -8;
  1, -10,  32,  -32,    4;
  1, -12,  50,  -80,   36;
  1, -14,  72, -160,  136,   -24;
  1, -16,  98, -280,  360,  -160,    8;
  1, -18, 128, -448,  780,  -592,  128;
  1, -20, 162, -672, 1484, -1632,  720,  -64;
  1, -22, 200, -960, 2576, -3752, 2624, -640,  16;
  ...

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 204, 205.

Shara Lalo, Left-justified Triangle

Row sums is A077953.

Cf. A304209.

T(n,k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, T(n-1,k)-2*T(n-2,k-1)+2*T(n-3,k-2)));
tabf(nn) = for (n=0, nn, for (k=0, 2*n\3, print1(T(n,k), ", ")); print); \\ Michel Marcus, May 10 2018

cp's OEIS Frontend

A077893 Duplicate of A077953.

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A077925 Expansion of 1/((1-x)*(1+2*x)).

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A052551 Expansion of 1/((1 - x)*(1 - 2*x^2)).

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A007420 Berstel sequence: a(n+1) = 2*a(n) - 4*a(n-1) + 4*a(n-2).

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Haskell

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A077980 Expansion of 1/(1 + x + 2*x^2 + 2*x^3).

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A373358 a(n) = 4*a(n-1) -5*a(n-2) +2*a(n-3) +2*a(n-4) for a(0) = a(1) = 0, a(2) = 1, a(3) = 4 for n >= 4.

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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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A077925 Expansion of 1/((1-x)(1+2x)).

A052551 Expansion of 1/((1 - x)(1 - 2x^2)).

A007420 Berstel sequence: a(n+1) = 2a(n) - 4a(n-1) + 4*a(n-2).

A077980 Expansion of 1/(1 + x + 2x^2 + 2x^3).

A373358 a(n) = 4a(n-1) -5a(n-2) +2a(n-3) +2a(n-4) for a(0) = a(1) = 0, a(2) = 1, a(3) = 4 for n >= 4.

A304223 Triangle read by rows: T(0,0)=1; T(n,k) = T(n-1,k)-2T(n-2,k-1)+2T(n-3,k-2) for k = 0..floor(2*n/3); T(n,k)=0 for n or k < 0.