A000749 -id:A000749 - cp's OEIS frontend

A038505 Sum of every 4th entry of row n in Pascal's triangle, starting at binomial(n,2).

0, 0, 1, 3, 6, 10, 16, 28, 56, 120, 256, 528, 1056, 2080, 4096, 8128, 16256, 32640, 65536, 131328, 262656, 524800, 1048576, 2096128, 4192256, 8386560, 16777216, 33558528, 67117056, 134225920, 268435456, 536854528, 1073709056

Offset: 0

Frank Ruskey

Number of strings over Z_2 of length n with trace 0 and subtrace 1.
Same as number of strings over GF(2) of length n with trace 0 and subtrace 1.
Binomial transform of (0,1,1,0,0,1,1,0,...) gives a(n) for n >= 1. - Paul Barry, Jul 07 2003
From Gary W. Adamson, Mar 13 2009: (Start)
M^n * [1,0,0,0] = [A038503(n), A000749(n), a(n), A038504(n)]; where M = a 4 X 4 matrix [1,1,0,0; 0,1,1,0; 0,0,1,1; 1,0,0,1]. Sum of terms = 2^n.
Example: M^6 * [1,0,0,0] [16, 20, 16, 12]; sum = 2^6 = 64. (End)
{A038503, A038504, A038505, A000749} is the difference analog of the hyperbolic functions of order 4, {h_1(x), h_2(x), h_3(x), h_4(x)}. For a definition of {h_i(x)} and the difference analog {H_i (n)} see [Erdelyi] and the Shevelev link respectively. - Vladimir Shevelev, Jun 14 2017

			a(3; 0, 1) = 3 since the three binary strings of trace 0, subtrace 1 and length 3 are { 011, 101, 110 }.

A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.

Peter Luschny, Table of n, a(n) for n = 0..1000
F. Ruskey, Strings over Z_2 with given trace and subtrace
F. Ruskey, Strings over GF(2) with given trace and subtrace
Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4).

Cf. A000749, A009116, A009545, A038503, A038504.

List([0..35],n->Sum([0..n],k->Binomial(n,2+4*k))); # Muniru A Asiru, Feb 21 2019

a038505 n = a038505_list !! n
a038505_list = tail $ zipWith (-) (tail a000749_list) a000749_list
-- Reinhard Zumkeller, Jul 15 2013

I:=[0, 0, 1, 3]; [n le 3 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jun 22 2012

# From Peter Luschny, Jun 15 2017: (Start)
s := sqrt(2): h := n -> [-2, -s, 0, s, 2, s, 0, -s][1 + (n mod 8)]:
a := n -> `if`(n=0, 0, (2^n + 2^(n/2)*h(n))/4): seq(a(n), n=0..32);
# Alternatively:
egf := (1 + exp(2*x) - 2*exp(x)*cos(x))/4:
series(egf, x, 33): seq(n!*coeff(%,x,n), n=0..32); # (End)

LinearRecurrence[{4, -6, 4}, {0, 0, 1, 3}, 40] (* Vincenzo Librandi, Jun 22 2012 *)
Table[If[n==0, 0, 2^(n-2) - 2^(n/2-1) Cos[Pi*n/4]], {n, 0, 32}] (* Vladimir Reshetnikov, Sep 16 2016 *)

A = lambda n: (2^n - (1-I)^n - (1+I)^n) / 4 if n != 0 else 0
print([A(n) for n in (0..32)]) # Peter Luschny, Jun 16 2017

a(n; t, s) = a(n-1; t, s) + a(n-1; t+1, s+t+1) where t is the trace and s is the subtrace.
a(n) = Sum_{k=0..n} binomial(n, 2 + 4*k), n >= 0.
a(n) = Sum_{k=0..n} (1/2)*C(n, k)*(-1)^C(k+3, 3) for n >= 1. - Paul Barry, Jul 07 2003
From Paul Barry, Nov 29 2004: (Start)
G.f.: x^2*(1-x)/((1-x)^4-x^4) = x^2*(1-x)/((1-2*x)*(1-2*x+2*x^2));
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*(1-(-1)^k)/2. (End)
Conjecture: 2*a(n+2) = A038504(n+2) + A000749(n+2) + 2*A009545(n). - Creighton Dement, May 22 2005
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3), n > 3; sequence is identical to its fourth differences. - Paul Curtz, Dec 21 2007
a(n) = A000749(n+1) - A000749(n). - Reinhard Zumkeller, Jul 15 2013
a(n+m) = a(n)*H_1(m) + H_2(n)*H_2(m) + H_1(n)*a(m) + H_4(n)*H_4(m),
where H_1=A038503, H_2=A038504, H_4=A000749. - Vladimir Shevelev, Jun 14 2017
From Peter Luschny, Jun 15 2017: (Start)
a(n) = n! [x^n] (1 + exp(2*x) - 2*exp(x)*cos(x))/4.
a(n) = A038503(n+2) - 2*A038503(n+1) + A038503(n).
a(n) = 2^(n-2) - A046980(n)*2^(A004525(n-3)) for n >= 1.
a(n) = (2^n - (1-i)^n - (1+i)^n) / 4 for n >= 1. Compare V. Shevelevs' formula (1) in A000749. (End)
From Vladimir Shevelev, Jun 16 2017: (Start)
Proof of the conjecture by Creighton Dement (May 22 2005): using the first formula of Theorem 1 in [Shevelev link] for n=4, omega=i=sqrt(-1), i:=1,2,3,4, m:=n>=1, we have
a(n) = (1/2)*(2^(n-1)-2^(n/2)*cos(Pi*n/4)), A038504(n) = (1/2)*(2^(n-1)+2^(n/2)* sin(Pi*n/4)), A000749(n) = (1/2)*(2^(n-1)-2^(n/2)*sin(Pi*n/4)). Finally we use the formula by Paul Barry: A009545(n) = 2^(n/2)*sin(Pi*n/4) = 2^(n/2)*(-cos(Pi*(n+2)/4)). Now it is easy to obtain the hypothetical formula. (End)

Missing 0 prepended by Vladimir Shevelev, Jun 14 2017

Edited by Peter Luschny, Jun 16 2017

A139398 a(n) = Sum_{k >= 0} binomial(n,5*k).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 7, 22, 57, 127, 254, 474, 859, 1574, 3004, 6008, 12393, 25773, 53143, 107883, 215766, 427351, 843756, 1669801, 3321891, 6643782, 13333932, 26789257, 53774932, 107746282, 215492564, 430470899, 859595529, 1717012749, 3431847189, 6863694378

Offset: 0

N. J. A. Sloane, Jun 13 2008

From Gary W. Adamson, Mar 13 2009: (Start)
M^n * [1,0,0,0,0] = [a(n), A139761(n), A139748(n), A139714(n), A133476(n)]
where M = the 5 X 5 matrix [1,1,0,0,0; 0,1,1,0,0; 0,0,1,1,0; 0,0,0,1,1; 1,0,0,0,1]
Sum of terms = 2^n. Example: M^6 * [1,0,0,0,0] = [7, 15, 20, 15, 7]; sum = 2^6 = 64. (End)
{A139398, A133476, A139714, A139748, A139761} is the difference analog of the hyperbolic functions of order 5, {h_1(x), h_2(x), h_3(x), h_4(x), h_5 (x)}. For a definition see the reference "Higher Transcendental Functions" and the Shevelev link. - Vladimir Shevelev, Jun 14 2017

A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Ch. 18.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
John B. Dobson, A matrix variation on Ramus's identity for lacunary sums of binomial coefficients, arXiv preprint arXiv:1610.09361 [math.NT], 2016.
Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,2).

Cf. A000749, A024493, A024494, A024495, A038503, A038504, A038505, A133476, A139714, A139748, A139761.

[n le 5 select 1 else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+2*Self(n-5): n in [1..40]]; // Vincenzo Librandi, Jun 27 2017

f:=(n,r,a) -> add(binomial(n,r*k+a),k=0..n); fs:=(r,a)->[seq(f(n,r,a),n=0..40)];
A139398_list := proc(n) local i; (exp(z)^2+2*exp(3/4*z+1/4*z*sqrt(5))* cos(1/4*z*sqrt(2)*sqrt(5+sqrt(5)))+2*exp(3/4*z-1/4*z*sqrt(5))* cos(1/4*z*sqrt(2)*sqrt(5-sqrt(5))))/5; series(%,z,n+2): seq(simplify(i!*coeff(%,z,i)), i=0..n) end: A139398_list(35); # Peter Luschny, Jul 10 2012

LinearRecurrence[{5,-10,10,-5,2},{1,1,1,1,1},40] (* Harvey P. Dale, Jun 11 2015 *)
Expand@Table[(2^n + Sqrt[5] (Cos[Pi n/5] - (-1)^n Cos[2 Pi n/5]) Fibonacci[n] + (Cos[Pi n/5] + (-1)^n Cos[2 Pi n/5]) LucasL[n])/5, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 04 2016 *)

G.f.: -(x-1)^4/((2*x-1)*(x^4-2*x^3+4*x^2-3*x+1)). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 12 2009
E.g.f.: (exp(z)^2+2*exp(3/4*z+1/4*z*sqrt(5))*cos(1/4*z*sqrt(2)*sqrt(5+sqrt(5)))+ 2*exp(3/4*z-1/4*z*sqrt(5))*cos(1/4*z*sqrt(2)*sqrt(5-sqrt(5))))/5. - Peter Luschny, Jul 10 2012
a(n) = (2^n + sqrt(5)*(cos(Pi*n/5) - (-1)^n*cos(2*Pi*n/5))*A000045(n) + (cos(Pi*n/5) + (-1)^n*cos(2*Pi*n/5))*A000032(n))/5. - Vladimir Reshetnikov, Oct 04 2016
From Vladimir Shevelev, Jun 17 2017: (Start)
a(n) = round((2/5)*(2^(n-1) + phi^n*cos(Pi*n/5))), where phi is the golden ratio and round(x) is the integer nearest to x.
The formula follows from the identity a(n)=1/5*Sum_{j=1..5}((omega_5)^j + 1)^n, where omega_5=exp(2*Pi*i)/5 (cf. Theorem 1 of [Shevelev] link for i=1, n=5, m:=n). Further note that for a=cos(x)+i*sin(x), a+1 = 2*cos ^2 (x/2) + i*sin(x), and for the argument y of a+1 we have tan(y)=tan(x/2) and r^2 = 4*cos^4(x/2) + sin^2(x) = 4*cos^2(x/2). So (a+1)^n = (2*cos(x /2))^n*(cos(n*x/2) + i*sin(n*x/2)). Using this, for x=2*Pi/5, we have (omega_5+1)^n = phi^n(cos(Pi*n/5) + i*sin(Pi*n/5)). Since (omega_5)^4+1=(1+omega_5)/omega_5, we easily find that ((omega_5)^4+1)^n is conjugate to (omega_5+1)^n. So (omega_5+1)^n+((omega_5)^4+1)^n = phi^n*cos(Pi*n/5). Further, we similarly obtain that (omega_5)^2+1 is conjugate to (omega_5) ^3+1=(1+(omega_5)^2)/(omega_5)^2 and ((omega_5)^2+1)^n +((omega_5)^3+1)^n = 2*(sqrt(2-phi))^n*cos(2*Pi*n/5). The absolute value of the latter <= 2*(2-phi)^(n/2) and quickly tends to 0. Finally, ((omega_5)^5+1)^n=2^n, and the formula follows. (End)
a(n+m) = a(n)*a(m) + H_2(n)*H_5(m) + H_3(n)*H_4(m) + H_4(n)*H_3(m) + H_5(n)*H_2(m), where H_2=A133476, H_3=A139714, H_4=A139748, H_5=A139761. - Vladimir Shevelev, Jun 17 2017

A049016 Expansion of 1/((1-x)^5 - x^5).

Original entry on oeis.org

1, 5, 15, 35, 70, 127, 220, 385, 715, 1430, 3004, 6385, 13380, 27370, 54740, 107883, 211585, 416405, 826045, 1652090, 3321891, 6690150, 13455325, 26985675, 53971350, 107746282, 214978335, 429124630, 857417220, 1714834440, 3431847189

Offset: 0

N. J. A. Sloane

Seiichi Manyama, Table of n, a(n) for n = 0..3000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,2).

Sequences of the form 1/((1-x)^m - x^m): A000079 (m=1,2), A024495 (m=3), A000749 (m=4), this sequence (m=5), A192080 (m=6), A049017 (m=7), A290995 (m=8), A306939 (m=9).

Cf. A000750, A078789.

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( 1/((1-x)^5-x^5) )); // G. C. Greubel, Apr 11 2023

CoefficientList[Series[1/((1-x)^5-x^5),{x,0,30}],x] (* or *) LinearRecurrence[ {5,-10,10,-5,2},{1,5,15,35,70},40] (* Harvey P. Dale, Jan 20 2014 *)

def A049016_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 1/((1-x)^5-x^5) ).list()
A049016_list(30) # G. C. Greubel, Apr 11 2023

G.f.: 1/((1-x)^5-x^5) = 1/( (1-2*x)*(1-3*x+4*x^2-2*x^3+x^4) ).
a(10*n+3) = A078789(5*n+3).
a(10*n+5) = A078789(5*n+4).
a(n) = (-1)^n * A000750(n).
Binomial transform of expansion of (1+x)^4/(1-x^5), or (1, 4, 6, 4, 1, 1, 4, 6, 4, 1, ...). - Paul Barry, Mar 19 2004
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + 2*a(n-5). - Paul Curtz, May 24 2008
G.f.: -1/( x^5 - 1 + 5*x/Q(0) ) where Q(k) = 1 + k*(x+1) + 5*x - x*(k+1)*(k+6)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Mar 15 2013

A135356 Triangle T(n,k) read by rows: coefficients in the recurrence of sequences which equal their n-th differences.

Original entry on oeis.org

2, 2, 0, 3, -3, 2, 4, -6, 4, 0, 5, -10, 10, -5, 2, 6, -15, 20, -15, 6, 0, 7, -21, 35, -35, 21, -7, 2, 8, -28, 56, -70, 56, -28, 8, 0, 9, -36, 84, -126, 126, -84, 36, -9, 2, 10, -45, 120, -210, 252, -210, 120, -45, 10, 0, 11, -55, 165, -330, 462, -462, 330, -165, 55, -11, 2

Offset: 1

Paul Curtz, Dec 08 2007, Mar 25 2008, Apr 28 2008

Sequences which equal their p-th differences obey recurrences a(n) = Sum_{s=1..p} T(p,s)*a(n-s).
This defines T(p,s) as essentially a signed version of a chopped Pascal triangle A014410, see A130785.
For cases like p=2, 4, 6, 8, 10, 12, 14, the denominator of the rational generating function of a(n) contains a factor 1-x; depending on the first terms in the sequences a(n), additional, simpler recurrences may exist if this cancels with a factor in the numerator. - R. J. Mathar, Jun 10 2008

			Triangle begins with row n=1:
  2;
  2,   0;
  3,  -3,  2;
  4,  -6,  4,    0;
  5, -10, 10,   -5,   2;
  6, -15, 20,  -15,   6,   0;
  7, -21, 35,  -35,  21,  -7,  2;
  8, -28, 56,  -70,  56, -28,  8,  0;
  9, -36, 84, -126, 126, -84, 36, -9, 2;

Alois P. Heinz, Rows n = 1..141, flattened

Related sequences: A000079 (n=1), A131577 (n=2), (A131708 , A130785, A131562, A057079) (n=3), (A000749, A038503, A009545, A038505) (n=4), A133476 (n=5), A140343 (n=6), A140342 (n=7).

Cf. A000225, A000984, A051049, A130785.

A135356:= func< n,k | k eq n select 1-(-1)^n else (-1)^(k+1)*Binomial(n,k) >;
[A135356(n,k): k in [1..n], n in [1..12]]; // G. C. Greubel, Apr 09 2023

T:= (p, s)->  `if`(p=s, 2*irem(p, 2), (-1)^(s+1) *binomial(p, s)):
seq(seq(T(p, s), s=1..p), p=1..11);  # Alois P. Heinz, Aug 26 2011

T[p_, s_]:= If[p==s, 2*Mod[s, 2], (-1)^(s+1)*Binomial[p, s]];
Table[T[p, s], {p, 12}, {s, p}]//Flatten (* Jean-François Alcover, Feb 19 2015, after Alois P. Heinz *)

def A135356(n,k):
    if (k==n): return 2*(n%2)
    else: return (-1)^(k+1)*binomial(n,k)
flatten([[A135356(n,k) for k in range(1,n+1)] for n in range(1,13)]) # G. C. Greubel, Apr 09 2023

T(n,k) = (-1)^(k+1)*A007318(n, k). T(n,n) = 1 - (-1)^n.
Sum_{k=1..n} T(n, k) = 2.
From G. C. Greubel, Apr 09 2023: (Start)
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = 2*A051049(n-1).
Sum_{k=1..n-1} T(n, k) = (1 + (-1)^n).
Sum_{k=1..n-1} (-1)^(k-1)*T(n, k) = A000225(n-1).
T(2*n, n) = (-1)^(n-1)*A000984(n), n >= 1. (End)

Edited by R. J. Mathar, Jun 10 2008

A049017 Expansion of 1/((1-x)^7 - x^7).

Original entry on oeis.org

1, 7, 28, 84, 210, 462, 924, 1717, 3017, 5110, 8568, 14756, 27132, 54264, 116281, 257775, 572264, 1246784, 2641366, 5430530, 10861060, 21242341, 40927033, 78354346, 150402700, 291693136, 574274008, 1148548016, 2326683921, 4749439975, 9714753412, 19818498700, 40199107690

Offset: 0

N. J. A. Sloane

Differs for n >= 7 (1717 vs. 1716) from A000579(n+6) = binomial(n+6,6); see also row 6 of A027555, A059481 and A213808. - M. F. Hasler, Mar 05 2017

Seiichi Manyama, Table of n, a(n) for n = 0..3000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,2).

Cf. A000579, A000750, A027555, A049018, A059481, A213808.

Sequences of the form 1/((1-x)^m - x^m): A000079 (m=1,2), A024495 (m=3), A000749 (m=4), A049016 (m=5), A192080 (m=6), this sequence (m=7), A290995 (m=8), A306939 (m=9).

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 1/((1-x)^7 - x^7) )); // G. C. Greubel, Apr 11 2023

CoefficientList[Series[1/((1-x)^7-x^7),{x,0,30}],x]  (* Harvey P. Dale, Feb 18 2011 *)

Vec(1/((1-x)^7-x^7)+O(x^99)) \\ M. F. Hasler, Mar 05 2017

def A049017_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 1/((1-x)^7 - x^7) ).list()
A049017_list(40) # G. C. Greubel, Apr 11 2023

G.f.: 1/((1-x)^7 - x^7) = 1/((1-2*x)*(1-5*x+11*x^2-13*x^3+9*x^4-3*x^5+x^6)).

A084101 Expansion of (1+x)^2/((1-x)*(1+x^2)).

Original entry on oeis.org

1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1

Offset: 0

Paul Barry, May 15 2003

Partial sums of A084099. Inverse binomial transform of A000749 (without leading zeros).
From Klaus Brockhaus, May 31 2010: (Start)
Periodic sequence: Repeat 1, 3, 3, 1.
Interleaving of A010684 and A176040.
Continued fraction expansion of (7 + 5*sqrt(29))/26.
Decimal expansion of 121/909.
a(n) = A143432(n+3) + 1 = 2*A021913(n+1) + 1 = 2*A133872(n+3) + 1.
a(n) = A165207(n+1) - 1.
First differences of A047538.
Binomial transform of A084102. (End)
From Wolfdieter Lang, Feb 09 2012: (Start)
a(n) = A045572(n+1) (Modd 5) := A203571(A045572(n+1)), n >= 0.
For general Modd n (not to be confused with mod n) see a comment on A203571. The nonnegative members of the five residue classes Modd 5, called [m] for m=0,1,...,4, are shown in the array A090298 if there the last row is taken as class [0] after inclusion of 0.
(End)

			From _Wolfdieter Lang_, Feb 09 2012: (Start)
Modd 5 of nonnegative odd numbers restricted mod 5:
A045572: 1, 3, 7, 9, 11, 13, 17, 19, 21, 23, ...
Modd 5:  1, 3, 3, 1,  1,  3,  3,  1,  1,  3, ...
(End)

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

Cf. A084102.

Cf. A010684 (repeat 1, 3), A176040 (repeat 3, 1), A178593 (decimal expansion of (7+5*sqrt(29))/26), A143432 (expansion of (1+x^4)/((1-x)*(1+x^2))), A021913 (repeat 0, 0, 1, 1), A133872 (repeat 1, 1, 0, 0), A165207 (repeat 2, 2, 4, 4), A047538 (congruent to 0, 1, 4 or 7 mod 8), A084099 (expansion of (1+x)^2/(1+x^2)), A000749 (expansion of x^3/((1-x)^4-x^4)). - Klaus Brockhaus, May 31 2010

R:=PowerSeriesRing(Integers(), 100); Coefficients(R!( (1+x)^2/((1-x)*(1+x^2)) )); // G. C. Greubel, Feb 28 2019

CoefficientList[Series[(1+x)^2/((1-x)(1+x^2)),{x,0,110}],x] (* or *) PadRight[{},110,{1,3,3,1}] (* Harvey P. Dale, Nov 21 2012 *)

x='x+O('x^100); Vec((1+x)^2/((1-x)*(1+x^2))) \\ Altug Alkan, Dec 24 2015

((1+x)^2/((1-x)*(1+x^2))).series(x, 100).coefficients(x, sparse=False) # G. C. Greubel, Feb 28 2019

a(n) = binomial(3, n mod 4). - Paul Barry, May 25 2003
From Klaus Brockhaus, May 31 2010: (Start)
a(n) = a(n-4) for n > 3; a(0) = a(3) = 1, a(1) = a(2) = 3.
a(n) = (4 - (1+i)*i^n - (1-i)*(-i)^n)/2 where i = sqrt(-1). (End)
E.g.f.: 2*exp(x) + sin(x) - cos(x). - Arkadiusz Wesolowski, Nov 04 2017
a(n) = 2 - (-1)^(n*(n+1)/2). - Guenther Schrack, Feb 26 2019

A290995 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^8.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 8, 36, 120, 330, 792, 1716, 3432, 6436, 11456, 19584, 32640, 54264, 93024, 170544, 341088, 735472, 1653632, 3749760, 8386560, 18289440, 38724480, 79594560, 159189120, 311058496, 597137408, 1133991936, 2147450880, 4089171840

Offset: 0

Clark Kimberling, Aug 22 2017

Suppose s = (c(0), c(1), c(2),...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A291000 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8).

Cf. A000012, A033453, A289780, A291000.

Sequences of the form 1/((1-x)^m - x^m): A000079 (m=1,2), A024495 (m=3), A000749 (m=4), A049016 (m=5), A192080 (m=6), A049017 (m=7), this sequence (m=8), A306939 (m=9).

R:=PowerSeriesRing(Integers(), 60); [0,0,0,0,0,0,0] cat Coefficients(R!( x^7/((1-x)^8 - x^8) )); // G. C. Greubel, Apr 11 2023

z = 60; s = x/(1 - x); p = 1 - s^8;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290995 *)

concat(vector(7), Vec(x^7 / ((1 - 2*x)*(1 - 2*x + 2*x^2)*(1 - 4*x + 6*x^2 - 4*x^3 + 2*x^4)) + O(x^50))) \\ Colin Barker, Aug 22 2017

def A290995_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x^7/((1-x)^8 - x^8) ).list()
A290995_list(60) # G. C. Greubel, Apr 11 2023

a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) for n >= 9.
G.f.: x^7 / ((1 - 2*x)*(1 - 2*x + 2*x^2)*(1 - 4*x + 6*x^2 - 4*x^3 + 2*x^4)). - Colin Barker, Aug 22 2017
G.f.: x^7/((1-x)^8 - x^8). - G. C. Greubel, Apr 11 2023

A000750 Expansion of bracket function.

Original entry on oeis.org

1, -5, 15, -35, 70, -125, 200, -275, 275, 0, -1000, 3625, -9500, 21250, -42500, 76875, -124375, 171875, -171875, 0, 621875, -2250000, 5890625, -13171875, 26343750, -47656250, 77109375, -106562500, 106562500, 0

Offset: 0

N. J. A. Sloane

It appears that the (unsigned) sequence is identical to its 5th-order absolute difference. - John W. Layman, Sep 23 2003

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Seiichi Manyama, Table of n, a(n) for n = 0..3000
H. W. Gould, Binomial coefficients, the bracket function and compositions with relatively prime summands, Fib. Quart. 2(4) (1964), 241-260.
Index entries for linear recurrences with constant coefficients, signature (-5, -10, -10, -5).

Column 5 of A307047.

Cf. A000748, A000749, A001659, A006090, A049016.

LinearRecurrence[{-5, -10, -10, -5}, {1, -5, 15, -35}, 30] (* Jean-François Alcover, Feb 11 2016 *)

Vec(1/((1+x)^5-x^5) + O(x^40)) \\ Michel Marcus, Feb 11 2016

{a(n) = (-1)^n*sum(k=0, n\5, (-1)^k*binomial(n+4, 5*k+4))} \\ Seiichi Manyama, Mar 21 2019

G.f.: 1/((1+x)^5-x^5).

a(n) = (-1)^n * Sum_{k=0..floor(n/5)} (-1)^k * binomial(n+4,5*k+4). - Seiichi Manyama, Mar 21 2019

A038521 Number of elements of GF(2^n) with trace 1 and subtrace 1.

Original entry on oeis.org

0, 0, 2, 1, 4, 10, 12, 36, 64, 120, 272, 496, 1024, 2080, 4032, 8256, 16384, 32640, 65792, 130816, 262144, 524800, 1047552, 2098176, 4194304, 8386560, 16781312, 33550336, 67108864, 134225920, 268419072, 536887296, 1073741824, 2147450880, 4295032832, 8589869056

Offset: 0

Frank Ruskey

Colin Barker, Table of n, a(n) for n = 0..1001
K. Cattel, C. R. Miers, F. Ruskey, J. Sawada, M. Serra, The number of irreducible polynomials over Gf(2) with given trace and subtrace, J. Combin. Math. Combin. Comput. 47 (2003) 31-64.
Frank Ruskey, Number of irreducible polynomials over GF(2) with given trace and subtrace
Frank Ruskey, Number of elements of GF(2^n) with given trace and subtrace
Index entries for linear recurrences with constant coefficients, signature (0,2,4).

Cf. A038504, A000749, A108520.

Cf. A038518, A038519, A038520.

I:=[0,0,2,1]; [m le 4 select I[m] else  2*Self(m-2) + 4*Self(m-3): m in [1..33]]; // Marius A. Burtea, Aug 02 2019

A038521 := proc(n) local r,a,i ; if n mod 2 = 1 then r := 3 ; else r := 1 ; fi; a :=0 ; for i from r to n by 4 do a := a+binomial(n,i) ; od; a ; end: for n from 0 to 40 do printf("%d,",A038521(n)) ; od: # R. J. Mathar, Oct 20 2008

LinearRecurrence[{0, 2, 4}, {0, 0, 2, 1}, 33] (* Jean-François Alcover, May 08 2023 *)

concat([0, 0], Vec(x*(2 + x) / ((1 - 2*x)*(1 + 2*x + 2*x^2)) + O(x^35))) \\ Colin Barker, Aug 02 2019

a(n) = C(n, r+0) + C(n, r+4) + C(n, r+8) + ... where r = 3 if n odd, r = 1 if n even.
a(n) = (2^(n-1) - A108520(n-1))/2 if n > 0. - R. J. Mathar, Jan 29 2008
From Colin Barker, Aug 02 2019: (Start)
G.f.: x^2*(2 + x) / ((1 - 2*x)*(1 + 2*x + 2*x^2)).
a(n) = ((-1-i)^(n-2) + (-1+i)^(n-2) + 2^(n-1)) / 2 = 2*A176739(n-2) + A176739(n-3).
a(n) = 2*a(n-2) + 4*a(n-3) for n>3.
(End)

Values duplicated A038520 and were replaced by R. J. Mathar, Oct 20 2008

Missing a(0) = 0 inserted by Andrey Zabolotskiy, Nov 12 2024

A192080 Expansion of 1/((1-x)^6 - x^6).

Original entry on oeis.org

1, 6, 21, 56, 126, 252, 463, 804, 1365, 2366, 4368, 8736, 18565, 40410, 87381, 184604, 379050, 758100, 1486675, 2884776, 5592405, 10919090, 21572460, 43144920, 87087001, 176565486, 357913941, 723002336, 1453179126, 2906358252

Offset: 0

Bruno Berselli, Jun 23 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6).

Cf. A006090, A049016, A049017, A057079, A057083, A057681.

Sequences of the form 1/((1-x)^m - x^m): A000079 (m=1,2), A024495 (m=3), A000749 (m=4), A049016 (m=5), this sequence (m=6), A049017 (m=7), A290995 (m=8), A306939 (m=9).

m:=30; R:=PowerSeriesRing(Integers(),m); Coefficients(R!(1/((1-2*x)*(1-x+x^2)*(1-3*x+3*x^2))));

CoefficientList[Series[1/((1-2*x)*(1-x+x^2)*(1-3*x+3*x^2)), {x,0,50}], x] (* Vincenzo Librandi, Oct 15 2012 *)
LinearRecurrence[{6,-15,20,-15,6},{1,6,21,56,126},30] (* Harvey P. Dale, Feb 22 2017 *)

makelist(coeff(taylor(1/((1-2*x)*(1-x+x^2)*(1-3*x+3*x^2)), x, 0, n), x, n), n, 0, 29);

Vec(1/((1-2*x)*(1-x+x^2)*(1-3*x+3*x^2))+O(x^99)) \\ Charles R Greathouse IV, Jun 23 2011

def A192080_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 1/((1-x)^6-x^6) ).list()
A192080_list(51) # G. C. Greubel, Apr 11 2023

a(n) = abs(A006090(n)) = (-1)^n * A006090(n).
G.f.: 1/((1-2*x)*(1-x+x^2)*(1-3*x+3*x^2)).
From G. C. Greubel, Apr 11 2023: (Start)
a(n) = (2^(n+5) + A010892(n) - 2*A010892(n-1) - 27*(A057083(n) - 2*A057083(n-1)))/6.
a(n) = (2^(n+5) + A057079(n+2) - 27*A057681(n+1))/6. (End)

cp's OEIS Frontend

A038505 Sum of every 4th entry of row n in Pascal's triangle, starting at binomial(n,2).

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A139398 a(n) = Sum_{k >= 0} binomial(n,5*k).

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A049016 Expansion of 1/((1-x)^5 - x^5).

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A135356 Triangle T(n,k) read by rows: coefficients in the recurrence of sequences which equal their n-th differences.

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A049017 Expansion of 1/((1-x)^7 - x^7).

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A084101 Expansion of (1+x)^2/((1-x)*(1+x^2)).

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