A048495 -id:A048495 - cp's OEIS frontend

A060747 a(n) = 2*n - 1.

-1, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131, 133, 135, 137, 139, 141, 143, 145, 147, 149, 151

Offset: 0

Henry Bottomley, Apr 26 2001

If you put n red balls and n blue balls in a bag and draw them one by one without replacement, the probability of never having drawn equal numbers of the two colors before the final ball is drawn is 1/a(n) unsigned.
abs(a(n)) = 2n - 1 + 2*0^n. It has A048495 as binomial transform. - Paul Barry, Jun 09 2003
For n >= 1, a(n) = numbers k such that arithmetic mean of the first k positive integers is an integer. A040001(a(n)) = 1. See A145051 and A040001. -  Jaroslav Krizek, May 28 2010
From Jaroslav Krizek, May 28 2010: (Start)
For n >= 1, a(n) = corresponding values of antiharmonic means to numbers from A016777 (numbers k such that antiharmonic mean of the first k positive integers is an integer).
a(n) = A000330(A016777(n)) / A000217(A016777(n)) = A146535(A016777(n)+1). (End)

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A000217, A000330, A000984, A001477, A002420, A005408, A005843, A016777, A023443, A040001, A048495, A078008, A145051, A146535, A161680.

a060747 = subtract 1 . (* 2)
a060747_list = [-1, 1 ..] -- Reinhard Zumkeller, Jul 05 2015
-- Reinhard Zumkeller, Jul 05 2015

Table[2*n - 1, {n, 0, 200}] (* Vladimir Joseph Stephan Orlovsky, Feb 16 2012 *)
LinearRecurrence[{2,-1},{-1,1},80] (* Harvey P. Dale, Mar 27 2020 *)

a(n)=2*n-1 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = A005408(n)-2 = A005843(n)-1 = -A000984(n)/A002420(n) = A001477(n)+A023443(n).
G.f.: (3*x - 1)/(1 - x)^2.
Abs(a(n)) = Sum_{k=0..n} (A078008(k) mod 4). - Paul Barry, Mar 12 2004
E.g.f.: exp(x)*(2*x-1). - Paul Barry, Mar 31 2007
a(n) = 2*a(n-1) - a(n-2); a(0)=-1, a(1)=1. - Philippe Deléham, Nov 03 2008
a(n) = 4*n - a(n-1) - 4 for n>0, with a(0)=-1. - Vincenzo Librandi, Aug 07 2010
a(n) = A161680(A005843(n))/n for n > 0. - Stefano Spezia, Feb 14 2025

A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

Original entry on oeis.org

1, 1, 3, 5, 15, 33, 91, 221, 583, 1465, 3795, 9653, 24831, 63441, 162763, 416525, 1067575, 2733673, 7003971, 17938661, 45954543, 117709185, 301527355, 772364093, 1978473511

Offset: 0

Clark Kimberling, Aug 02 2011

nonn

Suppose that p=p(0)*x^n+p(1)*x^(n-1)+...+p(n-1)*x+p(n) is a polynomial of positive degree and that Q is a sequence of polynomials: q(k,x)=t(k,0)*x^k+t(k,1)*x^(k-1)+...+t(k,k-1)*x+t(k,k), for k=0,1,2,... The Q-downstep of p is the polynomial given by D(p)=p(0)*q(n-1,x)+p(1)*q(n-2,x)+...+p(n-1)*q(0,x)+p(n).

Since degree(D(p))

Example: let p(x)=2*x^3+3*x^2+4*x+5 and q(k,x)=(x+1)^k.

D(p)=2(x+1)^2+3(x+1)+4(1)+5=2x^2+7x+14

D(D(p))=2(x+1)+7(1)+14=2x+23

D(D(D(p)))=2(1)+23=25;

the Q-residue of p is 25.

We may regard the sequence Q of polynomials as the triangular array formed by coefficients:

t(0,0)

t(1,0)....t(1,1)

t(2,0)....t(2,1)....t(2,2)

t(3,0)....t(3,1)....t(3,2)....t(3,3)

and regard p as the vector (p(0),p(1),...,p(n)). If P is a sequence of polynomials [or triangular array having (row n)=(p(0),p(1),...,p(n))], then the Q-residues of the polynomials form a numerical sequence.

Following are examples in which Q is the triangle given by t(i,j)=1 for 0<=i<=j:

Q.....P...................Q-residue of P

1.....1...................A000079, 2^n

1....(x+1)^n..............A007051, (1+3^n)/2

1....(x+2)^n..............A034478, (1+5^n)/2

1....(x+3)^n..............A034494, (1+7^n)/2

1....(2x+1)^n.............A007582

1....(3x+1)^n.............A081186

1....(2x+3)^n.............A081342

1....(3x+2)^n.............A081336

1.....A040310.............A193649

1....(x+1)^n+(x-1)^n)/2...A122983

1....(x+2)(x+1)^(n-1).....A057198

1....(1,2,3,4,...,n)......A002064

1....(1,1,2,3,4,...,n)....A048495

1....(n,n+1,...,2n).......A087323

1....(n+1,n+2,...,2n+1)...A099035

1....p(n,k)=(2^(n-k))*3^k.A085350

1....p(n,k)=(3^(n-k))*2^k.A090040

1....A008288 (Delannoy)...A193653

1....A054142..............A101265

1....cyclotomic...........A193650

1....(x+1)(x+2)...(x+n)...A193651

1....A114525..............A193662

More examples:

Q...........P.............Q-residue of P

(x+1)^n...(x+1)^n.........A000110, Bell numbers

(x+1)^n...(x+2)^n.........A126390

(x+2)^n...(x+1)^n.........A028361

(x+2)^n...(x+2)^n.........A126443

(x+1)^n.....1.............A005001

(x+2)^n.....1.............A193660

A094727.....1.............A193657

(k+1).....(k+1)...........A001906 (even-ind. Fib. nos.)

(k+1).....(x+1)^n.........A112091

(x+1)^n...(k+1)...........A029761

(k+1)......A049310........A193663

(In these last four, (k+1) represents the triangle t(n,k)=k+1, 0<=k<=n.)

A051162...(x+1)^n.........A193658

A094727...(x+1)^n.........A193659

A049310...(x+1)^n.........A193664

A075362....A075362........A193665

Changing the notation slightly leads to the Mathematica program below and the following formulation for the Q-downstep of p: first, write t(n,k) as q(n,k). Define r(k)=Sum{q(k-1,i)*r(k-1-i) : i=0,1,...,k-1} Then row n of D(p) is given by v(n)=Sum{p(n,k)*r(n-k) : k=0,1,...,n}.

			First five rows of Q, coefficients of Fibonacci polynomials (A049310):
1
1...0
1...0...1
1...0...2...0
1...0...3...0...1
To obtain a(4)=15, downstep four times:
D(x^4+3*x^2+1)=(x^3+x^2+x+1)+3(x+1)+1: (1,1,4,5) [coefficients]
DD(x^4+3*x^2+1)=D(1,1,4,5)=(1,2,11)
DDD(x^4+3*x^2+1)=D(1,2,11)=(1,14)
DDDD(x^4+3*x^2+1)=D(1,14)=15.

Cf. A192872 (polynomial reduction), A193091 (polynomial augmentation), A193722 (the upstep operation and fusion of polynomial sequences or triangular arrays).

q[n_, k_] := 1;
r[0] = 1; r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}];
f[n_, x_] := Fibonacci[n + 1, x];
p[n_, k_] := Coefficient[f[n, x], x, k]; (* A049310 *)
v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
Table[v[n], {n, 0, 24}]    (* A193649 *)
TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
Table[r[k], {k, 0, 8}]  (* 2^k *)
TableForm[Table[p[n, k], {n, 0, 6}, {k, 0, n}]]

Conjecture: G.f.: -(1+x)*(2*x-1) / ( (x-1)*(4*x^2+x-1) ). - R. J. Mathar, Feb 19 2015

A097067 Expansion of g.f. (1-4x+5x^2)/(1-2*x)^2.

Original entry on oeis.org

1, 0, 1, 4, 12, 32, 80, 192, 448, 1024, 2304, 5120, 11264, 24576, 53248, 114688, 245760, 524288, 1114112, 2359296, 4980736, 10485760, 22020096, 46137344, 96468992, 201326592, 419430400, 872415232, 1811939328, 3758096384, 7784628224, 16106127360, 33285996544, 68719476736

Offset: 0

Paul Barry, Jul 22 2004

Binomial transform of A097065. Binomial transform is (n-2)*2^(n-1)+2, or A048495 with an extra leading 1.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (4,-4).

Essentially the same as A001787.

Cf. A048495, A097065.

[(n-1)*2^(n-2)+5*0^n/4 : n in [0..30]]; // Vincenzo Librandi, Sep 25 2011

a:=n->abs(floor(sum (2^(n-1),j=1..n))): seq(a(n),n=-1..28); # Zerinvary Lajos, Jun 27 2007

Vec((1-4*x+5*x^2)/(1-2*x)^2 + O(x^50)) \\ Altug Alkan, Nov 13 2015

a(n) = (n-1)*2^(n-2) + 5*0^n/4.
a(n) = 4*a(n-1) - 4*a(n-2), n > 1.
a(n+1) = A001787(n).
E.g.f.: (5 - exp(2*x)*(1 - 2*x))/4. - Stefano Spezia, Jul 01 2023

A048481 a(n) = T(0,n) + T(1,n-1) + ... + T(n,0), array T given by A048472.

Original entry on oeis.org

1, 3, 9, 27, 77, 207, 529, 1299, 3093, 7191, 16409, 36891, 81949, 180255, 393249, 852003, 1835045, 3932199, 8388649, 17825835, 37748781, 79691823, 167772209, 352321587, 738197557, 1543503927, 3221225529, 6710886459, 13958643773, 28991029311, 60129542209

Offset: 0

Clark Kimberling

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-13,12,-4).

Partial sums of A048495.

Cf. A134397, A048166.

LinearRecurrence[{6,-13,12,-4},{1,3,9,27},40] (* Harvey P. Dale, Aug 13 2015 *)

Vec((4*x^2-3*x+1)/((x-1)^2*(2*x-1)^2) + O(x^100)) \\ Colin Barker, Dec 04 2014

Row sums of triangle A134397. Also, binomial transform of A048166. - Gary W. Adamson, Oct 23 2007
a(n) = 6*a(n-1)-13*a(n-2)+12*a(n-3)-4*a(n-4). - Colin Barker, Dec 04 2014
G.f.: (4*x^2-3*x+1) / ((x-1)^2*(2*x-1)^2). - Colin Barker, Dec 04 2014
a(n) = 2^(n+1)*(n-2) + 2*n + 5. - Christian Krause, Oct 31 2023

Corrected by T. D. Noe, Nov 08 2006

A084643 a(n) = 3^(n-1)(2n-3) + 2^(n+1).

Original entry on oeis.org

1, 3, 11, 43, 167, 631, 2315, 8275, 28943, 99439, 336659, 1126027, 3728279, 12239527, 39890843, 129205699, 416249375, 1334710495, 4262149667, 13560765691, 43005771431, 135988785943, 428882869931, 1349402340403

Offset: 0

Paul Barry, Jun 09 2003

Binomial transform of A048495. Second binomial transform of 1, 1, 3, 5, 7, ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-21,18).

Cf. A048495, A060747.

[3^(n-1)*(2*n-3)+2^(n+1) : n in [0..30]]; // Vincenzo Librandi, Sep 25 2011

LinearRecurrence[{8,-21,18},{1,3,11},30] (* Harvey P. Dale, Dec 12 2015 *)

Vec((1-5*x+8*x^2)/(1-2*x)/(1-3*x)^2+O(x^99)) \\ Charles R Greathouse IV, Mar 22 2012

[2^(n+1) +3^(n-1)*(2*n-3) for n in range(41)] # G. C. Greubel, Mar 22 2023

G.f.: (1 - 5*x + 8*x^2)/((1-2*x)*(1-3*x)^2). - Colin Barker, Mar 22 2012

E.g.f.: 2*exp(2*x) + (2*x-1)*exp(3*x). - G. C. Greubel, Mar 22 2023

A343291 a(n) = (n-2)*2^(n-1) + n + 2.

Original entry on oeis.org

1, 2, 4, 9, 22, 55, 136, 329, 778, 1803, 4108, 9229, 20494, 45071, 98320, 213009, 458770, 983059, 2097172, 4456469, 9437206, 19922967, 41943064, 88080409, 184549402, 385875995, 805306396, 1677721629, 3489660958, 7247757343, 15032385568, 31138512929, 64424509474

Offset: 0

Alois P. Heinz, Apr 10 2021

a(n) is the cardinality of set s(n), where s(0) = {0} and s(n+1) = s(n) union {(i+j+1)/2 : i,j in s(n)}. s(4) = {0, 1/2, 3/4, 7/8, 15/16, 1, 17/16, 9/8, 19/16, 5/4, 21/16, 11/8, 23/16, 3/2, 25/16, 13/8, 27/16, 7/4, 29/16, 15/8, 31/16, 2} has cardinality a(4) = 22.

Total number of 0-bits in all numbers <= 2^n and for n >= 1 the total number of bits in all numbers <= 2^(n-1); similar to A048495. - Ruud H.G. van Tol, Apr 28 2025

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-13,12,-4).

Partial differences give A005183 (shifted).

Cf. A048495, A343264.

a:= n-> (n-2)*2^(n-1)+n+2:
seq(a(n), n=0..35);

G.f.: -(x^3-5*x^2+4*x-1)/((2*x-1)^2*(x-1)^2).

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A060747 a(n) = 2*n - 1.

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A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

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A097067 Expansion of g.f. (1-4*x+5*x^2)/(1-2*x)^2.

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A048481 a(n) = T(0,n) + T(1,n-1) + ... + T(n,0), array T given by A048472.

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A084643 a(n) = 3^(n-1)*(2*n-3) + 2^(n+1).

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A343291 a(n) = (n-2)*2^(n-1) + n + 2.

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A097067 Expansion of g.f. (1-4x+5x^2)/(1-2*x)^2.

A084643 a(n) = 3^(n-1)(2n-3) + 2^(n+1).