A074048 -id:A074048 - cp's OEIS frontend

A123126 Absolute value of coefficient of X^2 in the characteristic polynomial of the n-th power of the matrix M = {{1,1,1,1,1}, {1,0,0,0,0}, {0,1,0,0,0}, {0,0,1,0,0}, {0,0,0,1,0}}.

1, 1, 4, 1, 31, 22, 1, 33, 4, 141, 199, 10, 209, 113, 604, 1473, 375, 1174, 1521, 2721, 9580, 5501, 6671, 14346, 15681, 57409, 56596, 44577, 112463, 119382, 333313, 480641, 360628, 800973, 1007191, 1988362, 3628369, 3160689, 5525420, 8309793

Offset: 1

Artur Jasinski, Sep 30 2006

nonn

Let P(x) = X^5 - X^4 - X^3 - X^2 - X - 1 and X1, X2, X3, X4, X5 its roots. Then a(n) = (X1*X2*X3)^n + (X1*X2*X4)^n + (X1*X2*X5)^n + ... + (X3*X4*X5)^n.

			a(5) = 31 because the characteristic polynomial of M^5 is X^5 - 31*X^4 + 49*X^3 - 31*X^2 + 9*X - 1.

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A074048, A123127.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( x*(1 +3*x^2 -4*x^3 +30*x^4 -18*x^5 -21*x^6 -16*x^7 -9*x^8 -10*x^9)/(1 -x -x^3 +x^4 -6*x^5 +3*x^6 +3*x^7 +2*x^8 +x^9 +x^10) )); // G. C. Greubel, Aug 03 2021

with(linalg): M[1]:=matrix(5,5,[1,1,1,1,1,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0]): for n from 2 to 45 do M[n]:=multiply(M[n-1],M[1]) od: seq(-coeff(charpoly(M[n],x),x,2),n=1..45); # Emeric Deutsch

f[n_]:= CoefficientList[CharacteristicPolynomial[MatrixPower[{{1,1,1,1,1}, {1,0,0, 0,0}, {0,1,0,0,0}, {0,0,1,0,0}, {0,0,0,1,0}}, n], x], x][[3]]; Array[f, 40] (* Robert G. Wilson v *)

def A123126_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(1 +3*x^2 -4*x^3 +30*x^4 -18*x^5 -21*x^6 -16*x^7 -9*x^8 -10*x^9)/(1 -x -x^3 +x^4 -6*x^5 +3*x^6 +3*x^7 +2*x^8 +x^9 +x^10) ).list()
a=A123126_list(40); a[1:] # G. C. Greubel, Aug 03 2021

G.f.: x*(1 +3*x^2 -4*x^3 +30*x^4 -18*x^5 -21*x^6 -16*x^7 -9*x^8 -10*x^9)/(1 -x -x^3 +x^4 -6*x^5 +3*x^6 +3*x^7 +2*x^8 +x^9 +x^10). - Colin Barker, May 16 2013

Edited by N. J. A. Sloane, Oct 24 2006

More terms from Emeric Deutsch and Robert G. Wilson v, Oct 24 2006

A123127 Coefficient of X^3 in the characteristic polynomial of the n-th power of the matrix M = {{1,1,1,1,1}, {1,0,0,0,0}, {0,1,0,0,0}, {0,0,1,0,0}, {0,0,0,1,0}}.

Original entry on oeis.org

-1, -3, -4, 1, 49, -42, -57, -31, 140, 497, -815, -758, 311, 3021, 3796, -13759, -7039, 16086, 45295, 3681, -204684, -10431, 365377, 507914, -618001, -2642435, 1427468, 6214881, 3341553, -16185322, -27959273, 42625665, 85186108, -23867663, -286766767, -193092086, 854985639, 900760205

Offset: 1

Artur Jasinski, Sep 30 2006

Also sum of the successive powers of all combinations of products of two different roots of the quintic pentanacci polynomial X^5 -X^4 -X^3 -X^2 -X -1; namely (X1*X2)^n + (X1*X3)^n + (X1*X4)^n + (X1*X5)^n + (X2*X3)^n + (X2*X4)^n + (X2*X5)^n + (X3*X4)^n + (X3*X5)^n + (X4*X5)^n, where X1, X2, X3, X4, X5 are the roots. A074048 are the coefficients, with changed signs, of X^4 in the characteristic polynomials of the successive powers of the pentanacci matrix or (X1)^n + (X2)^n + (X3)^n + (X4)^n + (X5)^n.

Let g(y) = y^10 + y^9 + 2*y^8 + 3*y^7 + 3*y^6 - 6*y^5 + y^4 - y^3 - y + 1 and {y1,...,y10} be the roots of g(y). Then a(n) = y1^n + ... + y10^n. - Kai Wang, Nov 01 2020

			a(5) = 49 because the characteristic polynomial of fifth power of pentanacci matrix M^5 is X^5 -31*X^4 +49*X^3 -31*X^2 +9*X -1 in which coefficient of X^3 is 49.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (-1,-2,-3,-3,6,-1,1,0,1,-1).

Cf. A074048, A123126.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( -x*(1+4*x+9*x^2 +12*x^3-30*x^4+6*x^5-7*x^6-9*x^8+10*x^9)/(1+x+2*x^2+3*x^3+3*x^4-6*x^5 +x^6 -x^7 -x^9+x^10) )); // G. C. Greubel, Aug 03 2021

with(linalg): M[1]:=matrix(5,5,[1,1,1,1,1,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0]): for n from 2 to 40 do M[n]:=multiply(M[n-1],M[1]) od: seq(coeff(charpoly(M[n],x),x,3),n=1..40); # Emeric Deutsch, Oct 24 2006

f[n_]:= CoefficientList[CharacteristicPolynomial[MatrixPower[{{1,1,1,1,1}, {1,0,0, 0,0}, {0,1,0,0,0}, {0,0,1,0,0}, {0,0,0,1,0}}, n], x], x][[4]]; Array[f, 36] (* Robert G. Wilson v, Oct 24 2006 *)
LinearRecurrence[{-1,-2,-3,-3,6,-1,1,0,1,-1},{-1,-3,-4,1,49,-42,-57,-31,140,497},40] (* Harvey P. Dale, Apr 10 2023 *)

g(y) = y^10 + y^9 + 2*y^8 + 3*y^7 + 3*y^6 - 6*y^5 + y^4 - y^3 - y + 1;
my(v=polsym(g(y),33)); vector(#v-1,n,v[n+1]) \\ Joerg Arndt, Nov 02 2020

def A123127_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( -x*(1+4*x+9*x^2+12*x^3-30*x^4+6*x^5-7*x^6-9*x^8+10*x^9)/(1+x+2*x^2 +3*x^3+3*x^4-6*x^5+x^6-x^7-x^9+x^10) ).list()
a=A123127_list(40); a[1:] # G. C. Greubel, Aug 03 2021

G.f.: -x*(1 +4*x +9*x^2 +12*x^3 -30*x^4 +6*x^5 -7*x^6 -9*x^8 +10*x^9)/(1 +x +2*x^2 +3*x^3 +3*x^4 -6*x^5 +x^6 -x^7 -x^9 +x^10). - Colin Barker, May 16 2013

Edited by N. J. A. Sloane, Oct 24 2006

More terms from Emeric Deutsch and Robert G. Wilson v, Oct 24 2006

A280303 Number of binary necklaces of length n with no subsequence 00000.

Original entry on oeis.org

1, 2, 3, 5, 7, 12, 17, 31, 51, 91, 155, 287, 505, 930, 1695, 3129, 5759, 10724, 19913, 37239, 69643, 130745, 245715, 463099, 873705, 1651838, 3126707, 5927817, 11251031, 21382558, 40679233, 77475673, 147694719, 281822847, 538213671, 1028714071, 1967728553

Offset: 1

Petros Hadjicostas and Lingyun Zhang, Dec 31 2016

nonn

a(n) is the number of cyclic sequences of length n consisting of zeros and ones that do not contain five consecutive zeros provided we consider as equivalent those sequences that are cyclic shifts of each other.

			a(5)=7 because we have seven binary cyclic sequences (necklaces) of length 5 that avoid five consecutive zeros: 00001, 00011, 00101, 00111, 01101, 01111, 11111.

Andrew Howroyd, Table of n, a(n) for n = 1..1000
P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 58-60.
Petros Hadjicostas, Proof of the formula for the generating function from the formula for a(n)
F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc.
F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc. [Cached copy, with permission, pdf format only]
L. Zhang and P. Hadjicostas, On sequences of independent Bernoulli trials avoiding the pattern '11..1', Math. Scientist, 40 (2015), 89-96.

Row 5 of A322057.

Cf. A000358, A093305, A280218, A074048.

a(n) = (1/n) * Sum_{d divides n} totient(n/d) * A074048(d).

G.f.: Sum_{k>=1} (phi(k)/k) * log(1/(1-B(x^k))) where B(x) = x*(1+x+x^2+x^3+x^4).

a(34) onwards from Andrew Howroyd, Jan 25 2024

A127208 Union of all n-step Lucas sequences, that is, all sequences s(1-n) = s(2-n) = ... = s(-1) = -1, s(0) = n and for k > 0, s(k) = s(k-1) + ... + s(k-n).

Original entry on oeis.org

1, 3, 4, 7, 11, 15, 18, 21, 26, 29, 31, 39, 47, 51, 57, 63, 71, 76, 99, 113, 120, 123, 127, 131, 191, 199, 223, 239, 241, 247, 255, 322, 367, 439, 443, 475, 493, 502, 511, 521, 708, 815, 843, 863, 943, 983, 1003, 1013, 1023, 1364, 1365, 1499, 1695, 1871, 1959

Offset: 1

T. D. Noe, Jan 09 2007

nonn

Noe and Post conjectured that the only positive terms that are common to any two distinct n-step Lucas sequences are the Mersenne numbers (A001348) that begin each sequence and 7 and 11 (in 2- and 3-step) and 5071 (in 3- and 4-step). The intersection of this sequence with the union of all the n-step Fibonacci sequences (A124168) appears to consist of 4, 21, 29, the Mersenne numbers 2^n-1 for all n and the infinite set of Eulerian numbers in A127232.

T. D. Noe, Table of n, a(n) for n=1..1000
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. A227885.

LucasSequence[n_,kMax_] := Module[{a,s,lst={}}, a=Join[Table[ -1,{n-1}],{n}]; While[s=Plus@@a; a=RotateLeft[a]; a[[n]]=s; s<=kMax, AppendTo[lst,s]]; lst]; nn=10; t={}; Do[t=Union[t,LucasSequence[n,2^(nn+1)]], {n,2,nn}]; t

Union(A000032, A001644, A073817, A074048, A074584, A104621, A105754, A105755,...)

A127221 a(n) = 2^npentanacci(n) or (2^n)A023424(n-1).

Original entry on oeis.org

2, 12, 56, 240, 992, 3648, 14464, 57088, 224768, 883712, 3471360, 13651968, 53682176, 211075072, 829915136, 3263102976, 12830244864, 50447253504, 198353354752, 779904614400, 3066503888896, 12057176965120, 47407572189184, 186401664532480, 732912043425792

Offset: 1

Artur Jasinski, Jan 09 2007

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,4,8,16,32).

Cf. A087131, A127210, A127211, A127212, A127213, A127214, A127216, A001648, A127220, A127222.

Cf. A023424, A074048.

I:=[2, 12, 56, 240, 992]; [n le 5 select I[n] else 2*Self(n-1) + 4*Self(n-2) + 8*Self(n-3) + 16*Self(n-4) + 32*Self(n-5): n in [1..30]]; // G. C. Greubel, Dec 19 2017

Table[Tr[MatrixPower[2*{{1, 1, 1, 1, 1}, {1, 0, 0, 0, 0}, {0, 1, 0, 0, 0}, {0, 0, 1, 0, 0}, {0, 0, 0, 1, 0}}, x]], {x, 1, 20}]
LinearRecurrence[{2, 4, 8, 16, 32}, {2, 12, 56, 240, 992}, 50] (* G. C. Greubel, Dec 19 2017 *)

x='x+O('x^30); Vec(-2*x*(1 +4*x +12*x^2 +32*x^3 +80*x^4)/(-1 +2*x +4*x^2 +8*x^3 +16*x^4 +32*x^5)) \\ G. C. Greubel, Dec 19 2017

a(n) = Trace of matrix [({2,2,2,2,2},{2,0,0,0,0},{0,2,0,0,0},{0,0,2,0,0},{0,0,0,2,0})^n].
a(n) = 2^n * Trace of matrix [({1,1,1,1,1},{1,0,0,0,0},{0,1,0,0,0},{0,0,1,0,0},{0,0,0,1,0})^n].
G.f.: -2*x*(1 +4*x +12*x^2 +32*x^3 +80*x^4)/(-1 +2*x +4*x^2 +8*x^3 +16*x^4 +32*x^5). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009; corrected by R. J. Mathar, Sep 16 2009
a(n) = 2*a(n-1)+4*a(n-2)+8*a(n-3)+16*a(n-4)+32*a(n-5). - Colin Barker, Sep 02 2013

Definition corrected by R. J. Mathar, Sep 17 2009

More terms from Colin Barker, Sep 02 2013

A127222 a(n) = 3^npentanacci(n) or (3^n)A023424(n-1).

Original entry on oeis.org

3, 27, 189, 1215, 7533, 41553, 247131, 1463103, 8640837, 50959287, 300264165, 1771292853, 10447598619, 61618989627, 363414767589, 2143339285311, 12641143135581, 74555586323649, 439717218548643, 2593383067853775, 15295369041550269, 90209719910309895

Offset: 1

Artur Jasinski, Jan 09 2007

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,9,27,81,243).

Cf. A087131, A127210, A127211, A127212, A127213, A127214, A127216, A001648, A127220, A127221.

Cf. A023424, A074048.

I:=[3, 27, 189, 1215, 7533]; [n le 5 select I[n] else 3*Self(n-1) + 9*Self(n-2) + 27*Self(n-3) + 81*Self(n-4) + 243*Self(n-5): n in [1..30]]; // G. C. Greubel, Dec 19 2017

Table[Tr[MatrixPower[3*{{1, 1, 1, 1, 1}, {1, 0, 0, 0, 0}, {0, 1, 0, 0, 0}, {0, 0, 1, 0, 0}, {0, 0, 0, 1, 0}}, x]], {x, 1, 20}]
LinearRecurrence[{3, 9, 27, 81, 243}, {3, 27, 189, 1215, 7533}, 50] (* G. C. Greubel, Dec 19 2017 *)

x='x+O('x^30); Vec(-3*x*(1 +6*x +27*x^2 +108*x^3 +405*x^4)/(-1 +3*x +9*x^2 +27*x^3 +81*x^4 +243*x^5)) \\ G. C. Greubel, Dec 19 2017

a(n) = Trace of matrix [({3,3,3,3,3},{3,0,0,0,0},{0,3,0,0,0},{0,0,3,0,0},{0,0,0,3,0})^n].
a(n) = 3^n * Trace of matrix [({1,1,1,1,1},{1,0,0,0,0},{0,1,0,0,0},{0,0,1,0,0},{0,0,0,1,0})^n].
G.f.: -3*x*(1 +6*x +27*x^2 +108*x^3 +405*x^4)/(-1 +3*x +9*x^2 +27*x^3 +81*x^4 +243*x^5). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 28 2009
a(n) = 3*a(n-1)+9*a(n-2)+27*a(n-3)+81*a(n-4)+243*a(n-5). - Colin Barker, Sep 02 2013

G.f. proposed by Maksym Voznyy checked and corrected by R. J. Mathar, Sep 16 2009
Definition corrected by R. J. Mathar, Sep 17 2009
More terms from Colin Barker, Sep 02 2013

A247505 Generalized Lucas numbers: square array A(n,k) read by antidiagonals, A(n,k)=(-1)^(k+1)k[x^k](-log((1+sum_{j=1..n}(-1)^(j+1)*x^j)^(-1))), (n>=0, k>=0).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 3, 1, 0, 0, 1, 3, 4, 1, 0, 0, 1, 3, 7, 7, 1, 0, 0, 1, 3, 7, 11, 11, 1, 0, 0, 1, 3, 7, 15, 21, 18, 1, 0, 0, 1, 3, 7, 15, 26, 39, 29, 1, 0, 0, 1, 3, 7, 15, 31, 51, 71, 47, 1, 0, 0, 1, 3, 7, 15, 31, 57, 99, 131, 76, 1, 0

Offset: 0

Peter Luschny, Nov 02 2014

			n\k[0][1][2][3] [4] [5] [6]  [7]  [8]  [9]  [10]  [11]  [12]
[0] 0, 0, 0, 0,  0,  0,  0,   0,   0,   0,    0,    0,    0
[1] 0, 1, 1, 1,  1,  1,  1,   1,   1,   1,    1,    1,    1
[2] 0, 1, 3, 4,  7, 11, 18,  29,  47,  76,  123,  199,  322 [A000032]
[3] 0, 1, 3, 7, 11, 21, 39,  71, 131, 241,  443,  815, 1499 [A001644]
[4] 0, 1, 3, 7, 15, 26, 51,  99, 191, 367,  708, 1365, 2631 [A073817]
[5] 0, 1, 3, 7, 15, 31, 57, 113, 223, 439,  863, 1695, 3333 [A074048]
[6] 0, 1, 3, 7, 15, 31, 63, 120, 239, 475,  943, 1871, 3711 [A074584]
[7] 0, 1, 3, 7, 15, 31, 63, 127, 247, 493,  983, 1959, 3903 [A104621]
[8] 0, 1, 3, 7, 15, 31, 63, 127, 255, 502, 1003, 2003, 3999 [A105754]
[.] .  .  .  .   .   .   .    .    .    .     .     .     .
oo] 0, 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 4095 [A000225]
'
As a triangular array, starts:
0,
0, 0,
0, 1, 0,
0, 1, 1, 0,
0, 1, 3, 1, 0,
0, 1, 3, 4, 1, 0,
0, 1, 3, 7, 7, 1,  0,
0, 1, 3, 7, 11, 11, 1, 0,
0, 1, 3, 7, 15, 21, 18, 1, 0,
0, 1, 3, 7, 15, 26, 39, 29, 1, 0,

Cf. A247506, A000225, A000032, A001644, A073817, A074048, A074584, A104621, A105754.

Cf. A125127.

A := proc(n, k) f := -log((1+add((-1)^(j+1)*x^j, j=1..n))^(-1));
(-1)^(k+1)*k*coeff(series(f,x,k+2),x,k) end:
seq(print(seq(A(n,k), k=0..12)), n=0..8);

A[n_, k_] := Module[{f, x}, f = -Log[(1+Sum[(-1)^(j+1) x^j, {j, 1, n}] )^(-1)]; (-1)^(k+1) k SeriesCoefficient[f, {x, 0, k}]];
Table[A[n-k, k], {n, 0, 12}, {k, 0, n}] (* Jean-François Alcover, Jun 28 2019, from Maple *)

A251653 5-step Fibonacci sequence starting with 0,0,1,0,0.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 2, 4, 7, 14, 28, 55, 108, 212, 417, 820, 1612, 3169, 6230, 12248, 24079, 47338, 93064, 182959, 359688, 707128, 1390177, 2733016, 5372968, 10562977, 20766266, 40825404, 80260631, 157788246, 310203524, 609844071, 1198921876, 2357018348, 4633776065, 9109763884, 17909324244, 35208804417

Offset: 0

Arie Bos, Dec 06 2014

Doubling the entries > 1 as 1, 2, 2, 4, 4, 7, 7, 14, 14, 28, 28, 55, 55,... (offset 0) gives Nyblom's palindromic binary strings having no 5-runs of 1's. - R. J. Mathar, Mar 28 2025

G. C. Greubel, Table of n, a(n) for n = 0..1000
M. A. Nyblom, Counting Palindromic Binary Strings Without r-Runs of Ones, J. Int. Seq. 16 (2013) #13.8.7, P_5(n)
H. Prodinger, Counting Palindromes According to r-Runs of Ones Using Generating Functions, J. Int. Seq. 17 (2014) # 14.6.2, even length, r=4.
Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1).

Other 5-step sequences are A000322, A001591, A023424, A074048, A124312, A124313, A122997, A135055, A135056, A145029

(see www.jsoftware.com) First construct the generating matrix
1  1  1  1  1
1  2  2  2  2
2  3  4  4  4
4  6  7  8  8
8 12 14 15 16
Given that matrix one can produce the first 5*200 numbers by
, M(+/ . *)^:(i.250) 0 0 1 0 0x

LinearRecurrence[{1, 1, 1, 1, 1}, {0, 0, 1, 0, 0}, 100] (* G. C. Greubel, May 27 2016 *)

a(n+5) = a(n) + a(n+1) + a(n+2) + a(n+3) + a(n+4).

G.f.: x^2*(x^2 + x - 1)/(x^5 + x^4 + x^3 + x^2 + x - 1). - Chai Wah Wu, May 27 2016

A106284 Primes p such that the polynomial x^5-x^4-x^3-x^2-x-1 mod p has no zeros.

Original entry on oeis.org

3, 5, 7, 11, 13, 17, 31, 37, 41, 53, 71, 79, 83, 107, 151, 157, 199, 229, 233, 239, 241, 257, 263, 277, 281, 311, 317, 331, 337, 379, 389, 409, 431, 433, 463, 467, 521, 523, 541, 547, 557, 563, 571, 577, 607, 631, 659, 677, 727, 769, 787, 809, 827, 839, 853

Offset: 1

T. D. Noe, May 02 2005

nonn

This polynomial is the characteristic polynomial of the Fibonacci and Lucas 5-step sequences, A001591 and A074048.

Robert Israel, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number

Cf. A106278 (number of distinct zeros of x^5-x^4-x^3-x^2-x-1 mod prime(n)), A106298, A106304 (period of Lucas and Fibonacci 5-step sequence mod prime(n)), A003631 (primes p such that x^2-x-1 is irreducible mod p).

P:= x^5-x^4-x^3-x^2-x-1:
select(p -> [msolve(P,p)] = [], [seq(ithprime(i),i=1..10000)]); # Robert Israel, Mar 13 2024

t=Table[p=Prime[n]; cnt=0; Do[If[Mod[x^5-x^4-x^3-x^2-x-1, p]==0, cnt++ ], {x, 0, p-1}]; cnt, {n, 200}];Prime[Flatten[Position[t, 0]]]

from itertools import islice
from sympy import Poly, nextprime
from sympy.abc import x
def A106284_gen(): # generator of terms
    from sympy.abc import x
    p = 2
    while True:
        if len(Poly(x*(x*(x*(x*(x-1)-1)-1)-1)-1, x, modulus=p).ground_roots())==0:
            yield p
        p = nextprime(p)
A106284_list = list(islice(A106284_gen(),20)) # Chai Wah Wu, Mar 14 2024

Name corrected by Robert Israel, Mar 13 2024

A125129 Partial sums of diagonals of array of k-step Lucas numbers as in A125127, read by antidiagonals.

Original entry on oeis.org

1, 1, 4, 1, 8, 11, 1, 12, 19, 26, 1, 19, 33, 45, 57, 1, 30, 58, 84, 102, 120, 1, 48, 101, 157, 197, 222, 247, 1, 77, 179, 292, 380, 436, 469, 502, 1, 124, 318, 546, 731, 855, 929, 971, 1013, 1, 200, 567, 1026, 1409, 1674, 1838, 1932, 1984, 2036

Offset: 1

Jonathan Vos Post, Nov 23 2006

Array of partial sums of diagonals of L(k,n) begins: 0.|.1...4..11...26...57..120..247..502.1013.2036.
|.1...8..19...45..102..222..469..971.1984.
|.1..12..33...84..197..436..929.1932.
|.1..19..58..157..380..855.1838.
|.1..30.101..292..731.1674.
|.1..48.179..546.1409.
|.1..77.318.1026.
|.1.124.567.
|.1.200.
|.1.

			Row 1 of the derived array is the partial sum of the diagonal above the main diagonal of array of k-step Lucas numbers as in A125127, hence the partial sums of: 1, 7, 11, 26, 57, 120, 247, 502, 103, ... are 1 = 1; 8 = 1 + 7; 19 = 1 + 7 + 11; 45 = 1 + 7 + 11 + 26; and so forth.

Cf. A000012, A000032, A000204, A001644, A001648, A048887, A048888, A074048, A074584, A092921, A104621, A105754, A105755, A125127, A000295.

Row 0 = SUM[i=1..n]L(i,i) = A127128 = partial sum of main diagonal of array of A125127. Row 1 = SUM[i=1..n]L(i,i+1) = partial sum of diagonal above main diagonal of array of A125127. Row 2 = SUM[i=1..n]L(i,i+2) = partial sum of diagonal 2 above main diagonal of array of A125127. .. Row m = SUM[i=1..n]L(i,i+m) = partial sum of diagonal 2 above main diagonal of array of A125127.

cp's OEIS Frontend

A123126 Absolute value of coefficient of X^2 in the characteristic polynomial of the n-th power of the matrix M = {{1,1,1,1,1}, {1,0,0,0,0}, {0,1,0,0,0}, {0,0,1,0,0}, {0,0,0,1,0}}.

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A123127 Coefficient of X^3 in the characteristic polynomial of the n-th power of the matrix M = {{1,1,1,1,1}, {1,0,0,0,0}, {0,1,0,0,0}, {0,0,1,0,0}, {0,0,0,1,0}}.

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A280303 Number of binary necklaces of length n with no subsequence 00000.

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A127208 Union of all n-step Lucas sequences, that is, all sequences s(1-n) = s(2-n) = ... = s(-1) = -1, s(0) = n and for k > 0, s(k) = s(k-1) + ... + s(k-n).

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A127221 a(n) = 2^n*pentanacci(n) or (2^n)*A023424(n-1).

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A127222 a(n) = 3^n*pentanacci(n) or (3^n)*A023424(n-1).

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A247505 Generalized Lucas numbers: square array A(n,k) read by antidiagonals, A(n,k)=(-1)^(k+1)*k*[x^k](-log((1+sum_{j=1..n}(-1)^(j+1)*x^j)^(-1))), (n>=0, k>=0).

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A127221 a(n) = 2^npentanacci(n) or (2^n)A023424(n-1).

A127222 a(n) = 3^npentanacci(n) or (3^n)A023424(n-1).

A247505 Generalized Lucas numbers: square array A(n,k) read by antidiagonals, A(n,k)=(-1)^(k+1)k[x^k](-log((1+sum_{j=1..n}(-1)^(j+1)*x^j)^(-1))), (n>=0, k>=0).