cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A015585 a(n) = 9*a(n-1) + 10*a(n-2).

Original entry on oeis.org

0, 1, 9, 91, 909, 9091, 90909, 909091, 9090909, 90909091, 909090909, 9090909091, 90909090909, 909090909091, 9090909090909, 90909090909091, 909090909090909, 9090909090909091, 90909090909090909, 909090909090909091, 9090909090909090909, 90909090909090909091
Offset: 0

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Author

Keywords

Comments

Number of walks of length n between any two distinct nodes of the complete graph K_11. Example: a(2)=9 because the walks of length 2 between the nodes A and B of the complete graph ABCDEFGHIJK are: ACB, ADB, AEB, AFB, AGB, AHB, AIB, AJB and AKB. - Emeric Deutsch, Apr 01 2004
Beginning with n=1 and a(1)=1, these are the positive integers whose balanced base-10 representations (A097150) are the first n digits of 1,-1,1,-1,.... Also, a(n) = (-1)^(n-1)*A014992(n) = |A014992(n)| for n >= 1. - Rick L. Shepherd, Jul 30 2004

Crossrefs

Programs

Formula

a(n) = 9*a(n-1) + 10*a(n-2).
From Emeric Deutsch, Apr 01 2004: (Start)
a(n) = 10^(n-1) - a(n-1).
G.f.: x/(1 - 9x - 10x^2). (End)
From Henry Bottomley, Sep 17 2004: (Start)
a(n) = round(10^n/11).
a(n) = (10^n - (-1)^n)/11.
a(n) = A098611(n)/11 = 9*A094028(n+1)/A098610(n). (End)
E.g.f.: exp(-x)*(exp(11*x) - 1)/11. - Elmo R. Oliveira, Aug 17 2024

Extensions

Extended by T. D. Noe, May 23 2011

A109501 Number of closed walks of length n on the complete graph on 7 nodes from a given node.

Original entry on oeis.org

1, 0, 6, 30, 186, 1110, 6666, 39990, 239946, 1439670, 8638026, 51828150, 310968906, 1865813430, 11194880586, 67169283510, 403015701066, 2418094206390, 14508565238346, 87051391430070, 522308348580426, 3133850091482550, 18803100548895306, 112818603293371830
Offset: 0

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Author

Mitch Harris, Jun 30 2005

Keywords

Crossrefs

Cf. A109502.
Cf. sequences with the same recurrence form: A001045, A078008, A097073, A115341, A015518, A054878, A015521, A109499, A015531, A109500, A015540. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008

Programs

  • Magma
    [(6^n + 6*(-1)^n)/7: n in [0..30]]; // G. C. Greubel, Dec 30 2017
  • Mathematica
    k=0;lst={k};Do[k=6^n-k;AppendTo[lst, k], {n, 1, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 11 2008 *)
    CoefficientList[Series[(1 - 5*x)/(1 - 5*x - 6*x^2), {x, 0, 50}], x] (* or *) LinearRecurrence[{5,6}, {1,0}, 30] (* G. C. Greubel, Dec 30 2017 *)
  • PARI
    for(n=0,30, print1((6^n + 6*(-1)^n)/7, ", ")) \\ G. C. Greubel, Dec 30 2017
    

Formula

G.f.: (1 - 5*x)/(1 - 5*x - 6*x^2).
a(n) = (6^n + 6*(-1)^n)/7.
a(n) = 6^(n-1) - a(n-1), a(0) = 1. - Jon E. Schoenfield, Feb 09 2015
a(n) = 5*a(n-1) + 6*a(n-2). - G. C. Greubel, Dec 30 2017
E.g.f.: exp(-x)*(exp(7*x) + 6)/7. - Elmo R. Oliveira, Aug 17 2024

Extensions

Corrected by Franklin T. Adams-Watters, Sep 18 2006
Edited by Jon E. Schoenfield, Feb 09 2015

A299960 a(n) = (4^(2*n+1) + 1) / 5.

Original entry on oeis.org

1, 13, 205, 3277, 52429, 838861, 13421773, 214748365, 3435973837, 54975581389, 879609302221, 14073748835533, 225179981368525, 3602879701896397, 57646075230342349, 922337203685477581, 14757395258967641293, 236118324143482260685, 3777893186295716170957
Offset: 0

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Author

M. F. Hasler, Feb 22 2018

Keywords

Comments

It is easily seen that 4^(2n+1)+1 is divisible by 5 for all n, since 4 = -1 (mod 5). For even powers this does not hold.
The aerated sequence 1, 0, 13, 0, 205, 0, 3277, ... is a linear divisibility sequence of order 4. It is the case P1 = 0, P2 = -5^2, Q = 4 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. Cf. A007583, A095372 and A100706. - Peter Bala, Aug 28 2019
Let G be a sequence satisfying G(i) = 2*G(i-1) + G(i-2) - 2*G(i-3) for arbitrary integers i and without regard to the initial values of G. Then a(n) = (G(i)*2^(4*n+2) + G(i+8*n+4))/(5*G(i+4*n+2)) as long as G(i+4*n+2) != 0. - Klaus Purath, Feb 02 2021
Ch. Gerbr asks (personal comm.) whether we can prove that 13 is the only prime in this sequence. We can prove divisibility conditions for many residue classes of the index n (cf. formulas), but have not yet found a complete covering set. - M. F. Hasler, Jan 07 2025

Examples

			For n = 0, a(0) = (4^1+1)/5 = 5/5 = 1.
For n = 1, a(1) = (4^3+1)/5 = 65/5 = 13.
		

Crossrefs

Cf. A299959 for the smallest prime factor.

Programs

Formula

a(n) = A052539(2*n+1)/5 = A015521(2*n+1) = A014985(2*n+1) = A007910(4*n+1) = A007909(4*n+1) = A207262(n+1)/5.
O.g.f.: (1 - 4*x)/(1 - 17*x + 16*x^2). - Peter Bala, Aug 28 2019
a(n) = 17*a(n-1) - 16*a(n-2). - Wesley Ivan Hurt, Oct 02 2020
From Klaus Purath, Feb 02 2021: (Start)
a(n) = (2^(4*n+2)+1)/5.
a(n) = (A061654(n) + A001025(n))/2.
a(n) = A091881(n+1) + 7*A131865(n-1) for n > 0.
(End)
E.g.f.: (exp(x) + 4*exp(16*x))/5. - Stefano Spezia, Feb 02 2021
We have d | a(n) for all n in R, for the following pairs (d, R) of divisors d and residue classes R: (13, 1 + 3Z), (5, 2 + 5Z), (29, 3 + 7Z), (397, 5 + 11Z),
(53, 6 + 13Z), (137, 8 + 17Z), (229, 9 + 19Z), (277, 11 + 23Z),
(107367629, 14 + 29Z), (5581, 15 + 31Z), (149, 18 + 27Z), (10169, 20 + 41Z),
(173, 21 + 43Z), (3761, 23 + 47Z), (15358129, 26 + 53Z), (1181, 29 + 59Z),
(733, 30 + 61Z), (269, 33 + 67Z), (569, 35 + 71Z),(293, 36 + 73Z), (317, 39 + 79Z),
(997, 41 + 83Z), (1069, 44 + 89Z), (389, 48 + 97Z), (809, 50 + 101Z),
(41201, 51 + 103Z), (857, 53 + 107Z), (5669, 54 + 109Z), (58309, 56 + 113Z),
(509, 63 + 127Z), (269665073, 65 + 131Z), (189061, 68 + 137Z), (557, 69 + 139Z),
(1789, 74 + 149Z), (653, 81 + 163Z), (9413, 90 + 181Z), (3821, 95 + 191Z),
(773, 96 + 193Z), (4729, 98 + 197Z), (797, 99 + 199Z), ... - M. F. Hasler, Jan 07 2025

A015577 a(n+1) = 8*a(n) + 9*a(n-1), a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 8, 73, 656, 5905, 53144, 478297, 4304672, 38742049, 348678440, 3138105961, 28242953648, 254186582833, 2287679245496, 20589113209465, 185302018885184, 1667718169966657, 15009463529699912, 135085171767299209, 1215766545905692880, 10941898913151235921
Offset: 0

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Author

Keywords

Comments

Binomial transform is A011557, with a leading zero. - Paul Barry, Jul 09 2003
Number of walks of length n between any two distinct nodes of the complete graph K_10. Example: a(2) = 8 because the walks of length 2 between the nodes A and B of the complete graph ABCDEFGHIJ are: ACB, ADB, AEB, AFB, AGB, AHB, AIB and AJB. - Emeric Deutsch, Apr 01 2004
The ratio a(n+1)/a(n) converges to 9 as n approaches infinity. - Felix P. Muga II, Mar 09 2014

Crossrefs

Programs

Formula

From Paul Barry, Jul 09 2003: (Start)
G.f.: x/((1+x)*(1-9*x)).
E.g.f. exp(4*x)*sinh(5*x)/5.
a(n) = (9^n - (-1)^n)/10. (End)
a(n) = 9^(n-1)-a(n-1). - Emeric Deutsch, Apr 01 2004
a(n) = round(9^n/10). - Mircea Merca, Dec 28 2010

Extensions

Extended by T. D. Noe, May 23 2011

A135030 Generalized Fibonacci numbers: a(n) = 6*a(n-1) + 2*a(n-2).

Original entry on oeis.org

0, 1, 6, 38, 240, 1516, 9576, 60488, 382080, 2413456, 15244896, 96296288, 608267520, 3842197696, 24269721216, 153302722688, 968355778560, 6116740116736, 38637152257536, 244056393778688, 1541612667187200
Offset: 0

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Author

Rolf Pleisch, Feb 10 2008, Feb 14 2008

Keywords

Comments

For n>0, a(n) equals the number of words of length n-1 over {0,1,...,7} in which 0 and 1 avoid runs of odd lengths. - Milan Janjic, Jan 08 2017

Crossrefs

Programs

  • Magma
    [n le 2 select n-1 else 6*Self(n-1) + 2*Self(n-2): n in [1..35]]; // Vincenzo Librandi, Sep 18 2016
    
  • Maple
    A:= gfun:-rectoproc({a(0) = 0, a(1) = 1, a(n) = 2*(3*a(n-1) + a(n-2))},a(n),remember):
    seq(A(n),n=1..30); # Robert Israel, Sep 16 2014
  • Mathematica
    Join[{a=0,b=1},Table[c=6*b+2*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 16 2011 *)
    LinearRecurrence[{6,2},{0,1},30] (* or *) CoefficientList[Series[ -(x/(2x^2+6x-1)),{x,0,30}],x] (* Harvey P. Dale, Jun 20 2011 *)
  • PARI
    a(n)=([0,1; 2,6]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Oct 03 2016
  • Sage
    [lucas_number1(n,6,-2) for n in range(0, 21)] # Zerinvary Lajos, Apr 24 2009
    

Formula

a(0) = 0; a(1) = 1; a(n) = 2*(3*a(n-1) + a(n-2)).
a(n) = 1/(2*sqrt(11))*( (3 + sqrt(11))^n - (3 - sqrt(11))^n ).
G.f.: x/(1 - 6*x - 2*x^2). - Harvey P. Dale, Jun 20 2011
a(n+1) = Sum_{k=0..n} A099097(n,k)*2^k. - Philippe Deléham, Sep 16 2014
E.g.f.: (1/sqrt(11))*exp(3*x)*sinh(sqrt(11)*x). - G. C. Greubel, Sep 17 2016

Extensions

More terms from Joshua Zucker, Feb 23 2008

A239473 Triangle read by rows: signed version of A059260: coefficients for expansion of partial sums of sequences a(n,x) in terms of their binomial transforms (1+a(.,x))^n ; Laguerre polynomial expansion of the truncated exponential.

Original entry on oeis.org

1, 0, 1, 1, -1, 1, 0, 2, -2, 1, 1, -2, 4, -3, 1, 0, 3, -6, 7, -4, 1, 1, -3, 9, -13, 11, -5, 1, 0, 4, -12, 22, -24, 16, -6, 1, 1, -4, 16, -34, 46, -40, 22, -7, 1, 0, 5, -20, 50, -80, 86, -62, 29, -8, 1, 1, -5, 25, -70, 130, -166, 148, -91, 37, -9, 1, 0, 6, -30, 95, -200, 296, -314, 239, -128, 46, -10, 1
Offset: 0

Views

Author

Tom Copeland, Mar 19 2014

Keywords

Comments

With T the lower triangular array above and the Laguerre polynomials L(k,x) = Sum_{j=0..k} (-1)^j binomial(k, j) x^j/j!, the following identities hold:
(A) Sum_{k=0..n} (-1)^k L(k,x) = Sum_{k=0..n} T(n,k) x^k/k!;
(B) Sum_{k=0..n} x^k/k! = Sum_{k=0..n} T(n,k) L(k,-x);
(C) Sum_{k=0..n} x^k = Sum_{k=0..n} T(n,k) (1+x)^k = (1-x^(n+1))/(1-x).
More generally, for polynomial sequences,
(D) Sum_{k=0..n} P(k,x) = Sum_{k=0..n} T(n,k) (1+P(.,x))^k,
where, e.g., for an Appell sequence, such as the Bernoulli polynomials, umbrally, (1+ Ber(.,x))^k = Ber(k,x+1).
Identity B follows from A through umbral substitution of j!L(j,-x) for x^j in A. Identity C, related to the cyclotomic polynomials for prime index, follows from B through the Laplace transform.
Integrating C gives Sum_{k=0..n} T(n,k) (2^(k+1)-1)/(k+1) = H(n+1), the harmonic numbers.
Identity A >= 0 for x >= 0 (see MathOverflow link for evaluation in terms of Hermite polynomials).
From identity C, W(m,n) = (-1)^n Sum_{k=0..n} T(n,k) (2-m)^k = number of walks of length n+1 between any two distinct vertices of the complete graph K_m for m > 2.
Equals A112468 with the first column of ones removed. - Georg Fischer, Jul 26 2023

Examples

			Triangle begins:
   1
   0    1
   1   -1    1
   0    2   -2    1
   1   -2    4   -3    1
   0    3   -6    7   -4    1
   1   -3    9  -13   11   -5    1
   0    4  -12   22  -24   16   -6    1
   1   -4   16  -34   46  -40   22   -7    1
   0    5  -20   50  -80   86  -62   29   -8    1
   1   -5   25  -70  130 -166  148  -91   37   -9    1
		

Crossrefs

For column 2: A001057, A004526, A008619, A140106.
Column 3: A002620, A087811.
Column 4: A002623, A173196.
Column 5: A001752.
Column 6: A001753.
Cf. Bottomley's cross-references in A059260.
Embedded in alternating antidiagonals of T are the reversals of arrays A071921 (A225010) and A210220.

Programs

  • Magma
    [[(&+[(-1)^(j+k)*Binomial(j,k): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Feb 06 2018
    
  • Maple
    A239473 := proc(n,k)
        add(binomial(j,k)*(-1)^(j+k),j=k..n) ;
    end proc; # R. J. Mathar, Jul 21 2016
  • Mathematica
    Table[Sum[(-1)^(j+k)*Binomial[j,k], {j,0,n}], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 06 2018 *)
  • PARI
    for(n=0,10, for(k=0,n, print1(sum(j=0,n, (-1)^(j+k)*binomial(j, k)), ", "))) \\ G. C. Greubel, Feb 06 2018
    
  • Sage
    Trow = lambda n: sum((x-1)^j for j in (0..n)).list()
    for n in (0..10): print(Trow(n)) # Peter Luschny, Jul 09 2019

Formula

T(n, k) = Sum_{j=0..n} (-1)^(j+k) * binomial(j, k).
E.g.f: (exp(t) - (x-1)*exp((x-1)*t))/(2-x).
O.g.f. (n-th row): (1-(x-1)^(n+1))/(2-x).
Associated operator identities:
With D=d/dx, :xD:^n=x^n*D^n, and :Dx:^n=D^n*x^n, then bin(xD,n)= binomial(xD,n)=:xD:^n/n! and L(n,-:xD:)=:Dx:^n/n!=bin(xD+n,n)=(-1)^n bin(-xD-1,n),
A-o) Sum_{k=0..n} (-1)^k L(k,-:xD:) = Sum_{k=0..n} :-Dx:^k/k!
= Sum_{k=0..n} T(n,k) :-xD:^k/k! = Sum_{k=0..n} (-1)^k T(n,k)bin(xD,k)
B-o) Sum_{k=0..n} :xD:^k/k! = Sum_{k=0..n}, T(n,k) L(k,-:xD:)
= Sum_{k=0..n} T(n,k) :Dx:^k/k! = Sum_{k=0..n}, bin(xD,k).
Associated binomial identities:
A-b) Sum_{k=0..n} (-1)^k bin(s+k,k) = Sum_{k=0..n} (-1)^k T(n,k) bin(s,k)
= Sum_{k=0..n} bin(-s-1,k) = Sum{k=0..n} T(n,k) bin(-s-1+k,k)
B-b) Sum_{k=0..n} bin(s,k) = Sum_{k=0..n} T(n,k) bin(s+k,k)
= Sum_{k=0..n} (-1)^k bin(-s-1+k,k)
= Sum_{k=0..n} (-1)^k T(n,k) bin(-s-1,k).
In particular, from B-b with s=n, Sum_{k=0..n} T(n,k) bin(n+k,k) = 2^n. From B-b with s=0, row sums are all 1.
From identity C with x=-2, the unsigned row sums are the Jacobsthal sequence, i.e., Sum_{k=0..n} T(n,k) (1+(-2))^k = (-1)^n A001045(n+1); for x=2, the Mersenne numbers A000225; for x=-3, A014983 or signed A015518; for x=3, A003462; for x=-4, A014985 or signed A015521; for x=4, A002450; for x=-5, A014986 or signed A015531; and for x=5, A003463; for x=-6, A014987 or signed A015540; and for x=6, A003464.
With -s-1 = m = 0,1,2,..., B-b gives finite differences (recursions):
Sum_{k=0..n} (-1)^k T(n,k) bin(m,k) = Sum_{k=0..n} (-1)^k bin(m+k,k) = T(n+m,m), i.e., finite differences of the columns of T generate shifted columns of T. The columns of T are signed, shifted versions of sequences listed in the cross-references. Since the finite difference is an involution, T(n,k) = Sum_{j=0..k} (-1)^j T(n+j,j) bin(k,j)}. Gauss-Newton interpolation can be applied to give a generalized T(n,s) for s noninteger.
From identity C, S(n,m) = Sum_{k=0..n} T(n,k) bin(k,m) = 1 for m < n+1 and 0 otherwise, i.e., S = T*P, where S = A000012, as a lower triangular matrix and P = Pascal = A007318, so T = S*P^(-1), where P^(-1) = A130595, the signed Pascal array (see A132440), the inverse of P, and T^(-1) = P*S^(-1) = P*A167374 = A156644.
U(n,cos(x)) = e^(-n*i*x)*Sum_{k=0..n} T(n,k)*(1+e^(2*i*x))^k = sin((n+1)x)/sin(x), where U is the Chebyschev polynomial of the second kind A053117 and i^2 = -1. - Tom Copeland, Oct 18 2014
From Tom Copeland, Dec 26 2015: (Start)
With a(n,x) = e^(nx), the partial sums are 1+e^x+...+e^(nx) = Sum_{k=0..n} T(n,k) (1+e^x)^k = [ x / (e^x-1) ] [ e^((n+1)x) -1 ] / x = [ (x / (e^x-1)) e^((n+1)x) - (x / (e^x-1)) ] / x = Sum_{k>=0} [ (Ber(k+1,n+1) - Ber(k+1,0)) / (k+1) ] * x^k/k!, where Ber(n,x) are the Bernoulli polynomials (cf. Adams p. 140). Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of the m-th powers of the integers, their binomial transforms, and the Bernoulli polynomials.
With a(n,x) = (-1)^n e^(nx), the partial sums are 1-e^x+...+(-1)^n e^(nx) = Sum_{k=0..n} T(n,k) (1-e^x)^k = [ (-1)^n e^((n+1)x) + 1 ] / (e^x+1) = [ (-1)^n (2 / (e^x+1)) e^((n+1)x) + (2 / (e^x+1)) ] / 2 = (1/2) Sum_{k>=0} [ (-1)^n Eul(k,n+1) + Eul(k,0) ] * x^k/k!, where Eul(n,x) are the Euler polynomials. Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of signed m-th powers of the integers; their binomial transforms, related to the Stirling numbers of the second kind and face numbers of the permutahedra; and the Euler polynomials. (End)
As in A059260, a generator in terms of bivariate polynomials with the coefficients of this entry is given by (1/(1-y))*1/(1 + (y/(1-y))*x - (1/(1-y))*x^2) = 1 + y + (x^2 - x*y + y^2) + (2*x^2*y - 2*x*y^2 + y^3) + (x^4 - 2*x^3*y + 4*x^2*y^2 - 3*x*y^3 + y^4) + ... . This is of the form -h2 * 1 / (1 + h1*x + h2*x^2), related to the bivariate generator of A049310 with h1 = y/(1-y) and h2 = -1/(1-y) = -(1+h1). - Tom Copeland, Feb 16 2016
From Tom Copeland, Sep 05 2016: (Start)
Letting P(k,x) = x in D gives Sum_{k=0..n} T(n,k)*Sum_{j=0..k} binomial(k,j) = Sum_{k=0..n} T(n,k) 2^k = n + 1.
The quantum integers [n+1]q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)) = q^(-n)*(1 - q^(2*(n+1))) / (1 - q^2) = q^(-n)*Sum{k=0..n} q^(2k) = q^(-n)*Sum_{k=0..n} T(n,k)*(1 + q^2)^k. (End)
T(n, k) = [x^k] Sum_{j=0..n} (x-1)^j. - Peter Luschny, Jul 09 2019
a(n) = -n + Sum_{k=0..n} A341091(k). - Thomas Scheuerle, Jun 17 2022

Extensions

Inverse array added by Tom Copeland, Mar 26 2014
Formula re Euler polynomials corrected by Tom Copeland, Mar 08 2024

A033114 Base-4 digits are, in order, the first n terms of the periodic sequence with initial period 1,0.

Original entry on oeis.org

1, 4, 17, 68, 273, 1092, 4369, 17476, 69905, 279620, 1118481, 4473924, 17895697, 71582788, 286331153, 1145324612, 4581298449, 18325193796, 73300775185, 293203100740, 1172812402961, 4691249611844, 18764998447377, 75059993789508
Offset: 1

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Author

Keywords

Crossrefs

Programs

Formula

a(n) = floor(4^(n+1)/15) = 4^(n+1)/15 - 1/6 - (-1)^n/10. - Benoit Cloitre, Apr 18 2003
G.f.: 1/((1-x)*(1+x)*(1-4*x)); a(n) = 3*a(n-1) + 4*a(n-2)+1. Partial sum of A015521. - Paul Barry, Nov 12 2003
a(n) = Sum_{k=0..floor(n/2)} 4^(n-2*k); a(n) = Sum_{k=0..n} Sum_{j=0..k} (-1)^(j+k)*4^j. - Paul Barry, Nov 12 2003
Convolution of A000302 and A059841 (4^n and periodic{1, 0}). a(n) = Sum_{k=0..n} (1 + (-1)^(n-k))*4^k/2. - Paul Barry, Jul 19 2004
a(n) = Sum_{k=0..n} (-1)^(n-k)*(J(2*k+1)-1)/2, J(n)=A001045(n). - Paul Barry, Mar 06 2008
a(n) = round((8*4^n-5)/30) = ceiling((4*4^n-4)/15) = round((4*4^n-4)/15); a(n) = a(n-2) + 4^(n-1), n > 1. - Mircea Merca, Dec 28 2010
a(n) = A117616(n)/2. - J. M. Bergot, Apr 22 2015
a(n) = A043291(n)/3; a(n+1) = 4*a(n) + A000035(n). - Robert Israel, Apr 22 2015
a(n)+a(n+1) = A002450(n+1). - R. J. Mathar, Feb 27 2019

A062160 Square array T(n,k) = (n^k - (-1)^k)/(n+1), n >= 0, k >= 0, read by falling antidiagonals.

Original entry on oeis.org

0, 1, 0, -1, 1, 0, 1, 0, 1, 0, -1, 1, 1, 1, 0, 1, 0, 3, 2, 1, 0, -1, 1, 5, 7, 3, 1, 0, 1, 0, 11, 20, 13, 4, 1, 0, -1, 1, 21, 61, 51, 21, 5, 1, 0, 1, 0, 43, 182, 205, 104, 31, 6, 1, 0, -1, 1, 85, 547, 819, 521, 185, 43, 7, 1, 0, 1, 0, 171, 1640, 3277, 2604, 1111, 300, 57, 8, 1, 0, -1, 1, 341, 4921, 13107, 13021, 6665, 2101, 455, 73, 9, 1, 0
Offset: 0

Views

Author

Henry Bottomley, Jun 08 2001

Keywords

Comments

For n >= 1, T(n, k) equals the number of walks of length k between any two distinct vertices of the complete graph K_(n+1). - Peter Bala, May 30 2024

Examples

			From _Seiichi Manyama_, Apr 12 2019: (Start)
Square array begins:
   0, 1, -1,  1,  -1,    1,    -1,      1, ...
   0, 1,  0,  1,   0,    1,     0,      1, ...
   0, 1,  1,  3,   5,   11,    21,     43, ...
   0, 1,  2,  7,  20,   61,   182,    547, ...
   0, 1,  3, 13,  51,  205,   819,   3277, ...
   0, 1,  4, 21, 104,  521,  2604,  13021, ...
   0, 1,  5, 31, 185, 1111,  6665,  39991, ...
   0, 1,  6, 43, 300, 2101, 14706, 102943, ... (End)
		

Crossrefs

Related to repunits in negative bases (cf. A055129 for positive bases).
Main diagonal gives A081216.
Cf. A109502.

Programs

  • Maple
    seq(print(seq((n^k - (-1)^k)/(n+1), k = 0..10)), n = 0..10); # Peter Bala, May 31 2024
  • Mathematica
    T[n_,k_]:=(n^k - (-1)^k)/(n+1); Join[{0},Table[Reverse[Table[T[n-k,k],{k,0,n}]],{n,12}]]//Flatten (* Stefano Spezia, Feb 20 2024 *)

Formula

T(n, k) = n^(k-1) - n^(k-2) + n^(k-3) - ... + (-1)^(k-1) = n^(k-1) - T(n, k-1) = n*T(n, k-1) - (-1)^k = (n - 1)*T(n, k-1) + n*T(n, k-2) = round[n^k/(n+1)] for n > 1.
T(n, k) = (-1)^(k+1) * resultant( n*x + 1, (x^k-1)/(x-1) ). - Max Alekseyev, Sep 28 2021
G.f. of row n: x/((1+x) * (1-n*x)). - Seiichi Manyama, Apr 12 2019
E.g.f. of row n: (exp(n*x) - exp(-x))/(n+1). - Stefano Spezia, Feb 20 2024
From Peter Bala, May 31 2024: (Start)
Binomial transform of the m-th row: Sum_{k = 0..n} binomial(n, k)*T(m, k) = (m + 1)^(n-1) for n >= 1.
Let R(m, x) denote the g.f. of the m-th row of the square array. Then R(m_1, x) o R(m_2, x) = R(m_1 + m_2 + m_1*m_2, x), where o denotes the black diamond product of power series as defined by Dukes and White. Cf. A109502.
T(m_1 + m_2 + m_1*m_2, k) = Sum_{i = 0..k} Sum_{j = i..k} binomial(k, i)* binomial(k-i, j-i)*T(m_1, j)*T(m_2, k-i). (End)

A076026 Expansion of g.f.: (1-4*x*C)/(1-5*x*C) where C = (1/2-1/2*(1-4*x)^(1/2))/x = g.f. for Catalan numbers A000108.

Original entry on oeis.org

1, 1, 6, 37, 230, 1434, 8952, 55917, 349374, 2183230, 13643972, 85270626, 532926716, 3330739972, 20816939100, 130105200765, 813155081070, 5082210417270, 31763782696740, 198523522444950, 1240771573465140, 7754820693127020, 48467623215477120, 302922622226091090
Offset: 0

Views

Author

N. J. A. Sloane, Oct 29 2002

Keywords

Comments

a(n) is the number of Motzkin paths of length n-1 in which the (1,0)-steps at level 0 come in 6 colors and those at a higher level come in 2 colors. Example: a(4)=230 because, denoting U=(1,1), H=(1,0), and D=(1,-1), we have 6^3 = 216 paths of shape HHH, 6 paths of shape HUD, 6 paths of shape UDH, and 2 paths of shape UHD. - Emeric Deutsch, May 02 2011

References

  • L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( (2- 4*Sqrt(1-4*x))/(3-5*Sqrt(1-4*x)) )); // G. C. Greubel, May 04 2019
    
  • Mathematica
    CoefficientList[Series[(2-4*Sqrt[1-4*x])/(3-5*Sqrt[1-4*x]), {x, 0, 30}], x] (* Vaclav Kotesovec, Dec 09 2013 *)
    Flatten[{1,Table[FullSimplify[(2*n)!*Hypergeometric2F1Regularized[1, n+1/2, n+2, 16/25] / (25*n!) + 3*5^(2*n-1)/4^(n+1)], {n,1,30}]}] (* Vaclav Kotesovec, Dec 09 2013 *)
  • PARI
    my(x='x+O('x^30)); Vec((2-4*sqrt(1-4*x))/(3-5*sqrt(1-4*x))) \\ G. C. Greubel, May 04 2019
    
  • Sage
    ((2-4*sqrt(1-4*x))/(3-5*sqrt(1-4*x))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 04 2019

Formula

a(n+1) = Sum_{k=0..n} A039598(n,k)*4^k. - Philippe Deléham, Mar 21 2007
a(n) = Sum_{k=0..n} A039599(n,k)*A015521(k), for n >= 1. - Philippe Deléham, Nov 22 2007
Let A be the Toeplitz matrix of order n defined by: A[i,i-1]=-1, A[i,j]=Catalan(j-i), (i<=j), and A[i,j]=0, otherwise. Then, for n >= 1, a(n+1)=(-1)^n*charpoly(A,-5). - Milan Janjic, Jul 08 2010
From Gary W. Adamson, Jul 25 2011: (Start)
a(n) = upper left term in M^(n-1), M = an infinite square production matrix as follows:
6, 1, 0, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, ...
... (End)
D-finite with recurrence: 4*n*a(n) = (41*n-24)*a(n-1) - 50*(2*n-3)*a(n-2). - Vaclav Kotesovec, Dec 09 2013
a(n) ~ 3*5^(2*n-1)/4^(n+1). - Vaclav Kotesovec, Dec 09 2013
O.g.f. A(x) = (1 - *Sum_{n >= 1} binomial(2*n,n)*x^n)/(1 - (3/2)*Sum_{n >= 1} binomial(2*n,n)*x^n). - Peter Bala, Sep 01 2016

A093134 A Jacobsthal trisection.

Original entry on oeis.org

1, 0, 8, 56, 456, 3640, 29128, 233016, 1864136, 14913080, 119304648, 954437176, 7635497416, 61083979320, 488671834568, 3909374676536, 31274997412296, 250199979298360, 2001599834386888, 16012798675095096, 128102389400760776, 1024819115206086200, 8198552921648689608
Offset: 0

Views

Author

Paul Barry, Mar 23 2004

Keywords

Comments

Counts closed walks at a vertex of the complete graph on 9 nodes K_9.
Second binomial transform is A047855.

Crossrefs

Other sequences with a(n+1) = 8^n - a(n) are A001045, A078008, A097073, A115341, A015518, A054878, A015521, A109499, A015531, A109500, A109501, A015552, A015565. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008
Cf. A047855.

Programs

  • Magma
    [(8^n/9+8*(-1)^n/9): n in [0..20]]; // Vincenzo Librandi, Oct 11 2011
    
  • Mathematica
    k=0;lst={1, k};Do[k=8^n-k;AppendTo[lst, k], {n, 1, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 11 2008 *)
    Table[(8^n + 8*(-1)^n)/9, {n,0,30}] (* or *) LinearRecurrence[{7,8}, {1,0}, 30] (* G. C. Greubel, Jan 06 2018 *)
  • PARI
    for(n=0,30, print1((8^n + 8*(-1)^n)/9, ", ")) \\ G. C. Greubel, Jan 06 2018

Formula

G.f.: (1-7*x)/(1 - 7*x - 8*x^2).
a(n) = (8^n + 8*(-1)^n)/9.
a(n) = 8*A001045(3*n-3)/3.
From Elmo R. Oliveira, Aug 17 2024: (Start)
E.g.f.: exp(-x)*(exp(9*x) + 8)/9.
a(n) = 7*a(n-1) + 8*a(n-2) for n > 1. (End)
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