A126443 -id:A126443 - cp's OEIS frontend

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.)

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

A352860 a(0) = 1; a(n) = Sum_{k=0..n-1} binomial(n,k) * 2^k * a(k).

Original entry on oeis.org

1, 1, 5, 67, 2273, 187411, 36539465, 16496912587, 16958655627233, 39148957534778851, 200638280176080172025, 2261092739579072893806907, 55582179517311967755693514193, 2960001149710485505367113202321491, 339497331023047752386812273780566932585

Offset: 0

Ilya Gutkovskiy, Apr 06 2022

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..77

Cf. A000670, A122704, A126443, A352859.

a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k] 2^k a[k], {k, 0, n - 1}]; Table[a[n], {n, 0, 14}]
nmax = 14; A[] = 0; Do[A[x] = 1 + (Exp[x] - 1) A[2 x] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x] Range[0, nmax]!

a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=0, i-1, 2^j*binomial(i, j)*v[j+1])); v; \\ Seiichi Manyama, Jun 18 2022

E.g.f. A(x) satisfies: A(x) = 1 + (exp(x) - 1) * A(2*x).

a(n) ~ c * n! * 2^(n*(n-1)/2), where c = 1.572986203588985421674040830458773854660492965929302012... - Vaclav Kotesovec, Apr 07 2022

A306245 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where A(0,k) = 1 and A(n,k) = Sum_{j=0..n-1} k^j * binomial(n-1,j) * A(j,k) for n > 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 5, 1, 1, 1, 4, 17, 15, 1, 1, 1, 5, 43, 179, 52, 1, 1, 1, 6, 89, 1279, 3489, 203, 1, 1, 1, 7, 161, 5949, 108472, 127459, 877, 1, 1, 1, 8, 265, 20591, 1546225, 26888677, 8873137, 4140, 1

Offset: 0

Seiichi Manyama, Jul 28 2019

			Square array begins:
   1,  1,    1,      1,       1,        1, ...
   1,  1,    1,      1,       1,        1, ...
   1,  2,    3,      4,       5,        6, ...
   1,  5,   17,     43,      89,      161, ...
   1, 15,  179,   1279,    5949,    20591, ...
   1, 52, 3489, 108472, 1546225, 12950796, ...

Seiichi Manyama, Antidiagonals n = 0..55, flattened

Columns k=0..4 give A000012, A000110, A126443, A355081, A355082.
Rows n=0+1, 2 give A000012, A000027(n+1).
Main diagonal gives A309401.
Cf. A309386.

A:= proc(n, k) option remember; `if`(n=0, 1,
      add(k^j*binomial(n-1, j)*A(j, k), j=0..n-1))
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);  # Alois P. Heinz, Jul 28 2019

A[0, _] = 1;
A[n_, k_] := A[n, k] = Sum[k^j Binomial[n-1, j] A[j, k], {j, 0, n-1}];
Table[A[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, May 29 2020 *)

G.f. A_k(x) of column k satisfies A_k(x) = 1 + x * A_k(k * x / (1 - x)) / (1 - x). - Seiichi Manyama, Jun 18 2022

A355081 G.f. A(x) satisfies A(x) = 1 + x * A(3 * x / (1 - x)) / (1 - x).

Original entry on oeis.org

1, 1, 4, 43, 1279, 108472, 26888677, 19761575473, 43356335678176, 284807217244068223, 5608422162798704960959, 331227791701602557410058404, 58679652813856265804094312228601, 31185477505022553490008128886444268657

Offset: 0

Seiichi Manyama, Jun 18 2022

nonn

Column k=3 of A306245.

Cf. A000110, A126443, A355082.

a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=0, i-1, 3^j*binomial(i-1, j)*v[j+1])); v;

a(0) = 1; a(n) = Sum_{k=0..n-1} 3^k * binomial(n-1,k) * a(k).

A355082 G.f. A(x) satisfies A(x) = 1 + x * A(4 * x / (1 - x)) / (1 - x).

Original entry on oeis.org

1, 1, 5, 89, 5949, 1546225, 1591006901, 6526287232201, 106972340665773165, 7011394913950382306529, 1838058207026378316690626149, 1927362102757461997768349891040825, 8083963777926072174628168609626454270621

Offset: 0

Seiichi Manyama, Jun 18 2022

nonn

Column k=4 of A306245.

Cf. A000110, A126443, A355081.

a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=0, i-1, 4^j*binomial(i-1, j)*v[j+1])); v;

a(0) = 1; a(n) = Sum_{k=0..n-1} 4^k * binomial(n-1,k) * a(k).

A309401 a(n) = A306245(n,n).

Original entry on oeis.org

1, 1, 3, 43, 5949, 12950796, 586826390263, 669793946192984257, 22558227235537152753501561, 25741074696455818592335996518315259, 1124843928218943684789052411802502269971863691, 2100464404490451025972467064515428575200326254804659324780

Offset: 0

Seiichi Manyama, Jul 28 2019

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..36

Main diagonal of A306245.

Cf. A126443, A301419.

b:= proc(n, k) option remember; `if`(n=0, 1,
      add(k^j*binomial(n-1, j)*b(j, k), j=0..n-1))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..12);  # Alois P. Heinz, Jul 28 2019

b[0, _] = 1;
b[n_, k_] := b[n, k] = Sum[k^j Binomial[n-1, j] b[j, k], {j, 0, n-1}];
a[n_] := b[n, n];
a /@ Range[0, 12] (* Jean-François Alcover, Nov 14 2020, after Alois P. Heinz *)

def ncr(n, r)
  return 1 if r == 0
  (n - r + 1..n).inject(:*) / (1..r).inject(:*)
end
def A(k, n)
  ary = [1]
  (1..n).each{|i| ary << (0..i - 1).inject(0){|s, j| s + k ** j * ncr(i - 1, j) * ary[j]}}
  ary
end
def A309401(n)
  (0..n).map{|i| A(i, i)}
end
p A309401(20)

A352859 a(0) = 1; a(n) = Sum_{k=0..n-1} binomial(n,k+1) * 2^k * a(k).

Original entry on oeis.org

1, 1, 4, 25, 280, 5665, 211516, 14907673, 2021820016, 535262714881, 279317901141172, 289064917007756761, 595455410823115765768, 2446703815513439818406305, 20077597428602000393057306476, 329252263598282049972950683567705, 10794203801863458962317873561872563680

Offset: 0

Ilya Gutkovskiy, Apr 06 2022

nonn

Cf. A040027, A126443, A351756, A352860.

a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k + 1] 2^k a[k], {k, 0, n - 1}]; Table[a[n], {n, 0, 16}]
nmax = 16; A[] = 0; Do[A[x] = 1 + x A[2 x/(1 - x)]/(1 - x)^2 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

G.f. A(x) satisfies: A(x) = 1 + x * A(2*x/(1 - x)) / (1 - x)^2.

a(n) ~ c * 2^(n*(n-1)/2), where c = 8.12511731924148105991770742530352144084320407825344... - Vaclav Kotesovec, Apr 07 2022

A193660 Q-residue of the triangle A038207 of coefficients of (x+2)^n, where Q is the triangle given by t(i,j)=1 for 0<=i<=j. (See Comments.)

Original entry on oeis.org

1, 2, 5, 22, 201, 3690, 131149, 9004286, 1204317329, 316525415890, 164556516205461, 169974659148800742, 349799994417738642265, 1436618749673583674658362, 11785996128174350460348176861, 193254862258295280115072223316430

Offset: 0

Clark Kimberling, Aug 02 2011

nonn

For the definition of Q-residue, see A193649.

Cf. A038207, A126443, A193649.

q[n_, k_] := Coefficient[(x + 2)^n, x, k]; (* A038207 *)
r[0] = 1; r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}]
p[n_, k_] := 1
v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
Table[v[n], {n, 0, 16}]    (* A038207 *)
TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
Table[r[k], {k, 0, 8}]  (* A126443 *)
TableForm[Table[p[n, k], {n, 0, 4}, {k, 0, n}]]

A376177 Triangle, read by rows, where T(n,k) = T(n-1,k-1) + 2*T(n,k-1) when k > 0, else T(n,0) = T(n-1,n-1) when n > 0, with T(0,0) = 1.

Original entry on oeis.org

1, 1, 3, 3, 7, 17, 17, 37, 81, 179, 179, 375, 787, 1655, 3489, 3489, 7157, 14689, 30165, 61985, 127459, 127459, 258407, 523971, 1062631, 2155427, 4372839, 8873137, 8873137, 17873733, 36005873, 72535717, 146134065, 294423557, 593219953, 1195313043, 1195313043, 2399499223, 4816872179, 9669750231, 19412036179, 38970206423, 78234836403, 157062892759, 315321098561, 315321098561

Offset: 0

George Plousos and Paul D. Hanna, Sep 22 2024

This triangle was found by George Plousos while exploring a variation of Aitken's array (A011971).

			G.f.: A(x,y) = 1 + (3*y + 1)*x + (17*y^2 + 7*y + 3)*x^2 + (179*y^3 + 81*y^2 + 37*y + 17)*x^3 + (3489*y^4 + 1655*y^3 + 787*y^2 + 375*y + 179)*x^4 + (127459*y^5 + 61985*y^4 + 30165*y^3 + 14689*y^2 + 7157*y + 3489)*x^5 + (8873137*y^6 + 4372839*y^5 + 2155427*y^4 + 1062631*y^3 + 523971*y^2 + 258407*y + 127459)*x^6 + ...
which is defined by A(x,y) = (B(x) - 2*(B(x*y) - 1)/x) / (1 - (2+x)*y),
where B(x) = 1 + x*B( 2*x/(1-x) )/(1-x) is the g.f. B(x) for A126443,
B(x) = 1 + x + 3*x^2 + 17*x^3 + 179*x^4 + 3489*x^5 + 127459*x^6 + 8873137*x^7 + 1195313043*x^8 + 315321098561*x^9 + ... + A126443(n)*x^n + ...
This triangle begins
  1,
  1, 3,
  3, 7, 17,
  17, 37, 81, 179,
  179, 375, 787, 1655, 3489,
  3489, 7157, 14689, 30165, 61985, 127459,
  127459, 258407, 523971, 1062631, 2155427, 4372839, 8873137,
  8873137, 17873733, 36005873, 72535717, 146134065, 294423557, 593219953, 1195313043,
  ...

Paul D. Hanna, Table of n, a(n) for n = 0..1275

Cf. A011971, A126443.

{A126443(n) = if(n==0, 1, sum(k=0, n-1, binomial(n-1, k) * 2^k * A126443(k)))}
{T(n,k) = sum(j=0,k, binomial(k,j) * 2^j * A126443(n-k+j) )}
for(n=0,10, for(k=0,n, print1(T(n,k),", ")); print(""))

If k > 0, T(n,k) = T(n-1,k-1) + 2*T(n,k-1), else if n > 0, T(n,0) = T(n-1,n-1), with T(0,0) = 1.
T(n,k) = Sum_{j=0..k} binomial(k,j) * 2^j * A126443(n-k+j), where A126443(m) = Sum_{k=0..m-1} binomial(m-1, k) * 2^k * A126443(k) for m > 0 with A126443(0) = 1.
G.f. A(x,y) = (B(x) - 2*(B(x*y) - 1)/x) / (1 - (2+x)*y), where B(x) = 1 + x*B( 2*x/(1-x) )/(1-x) is the g.f. B(x) for A126443 given therein by Ilya Gutkovskiy.

A355109 a(n) = 1 + Sum_{k=1..n-1} binomial(n-1,k) * 2^(k-1) * a(k).

Original entry on oeis.org

1, 1, 2, 7, 44, 493, 9974, 372403, 26247008, 3559692121, 942403603562, 491777568765151, 508938530329020692, 1048381120745440503877, 4307758467916752367544414, 35349370769806113877653011083, 579693879415731511179957972407624

Offset: 0

Ilya Gutkovskiy, Jun 19 2022

nonn

Cf. A000110, A126443, A352859, A352860.

a:= proc(n) option remember; 1+add(a(k)*
      binomial(n-1, k)*2^(k-1), k=1..n-1)
    end:
seq(a(n), n=0..16);  # Alois P. Heinz, Jun 19 2022

a[n_] := a[n] = 1 + Sum[Binomial[n - 1, k] 2^(k - 1) a[k], {k, 1, n - 1}]; Table[a[n], {n, 0, 16}]
nmax = 16; A[] = 0; Do[A[x] = (2 - x + x A[2 x/(1 - x)])/(2 (1 - x)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

G.f. A(x) satisfies: A(x) = (2 - x + x * A(2*x/(1 - x))) / (2 * (1 - x)).

Showing 1-10 of 10 results.

cp's OEIS Frontend

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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A352860 a(0) = 1; a(n) = Sum_{k=0..n-1} binomial(n,k) * 2^k * a(k).

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A306245 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where A(0,k) = 1 and A(n,k) = Sum_{j=0..n-1} k^j * binomial(n-1,j) * A(j,k) for n > 0.

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A355081 G.f. A(x) satisfies A(x) = 1 + x * A(3 * x / (1 - x)) / (1 - x).

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A355082 G.f. A(x) satisfies A(x) = 1 + x * A(4 * x / (1 - x)) / (1 - x).

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A309401 a(n) = A306245(n,n).

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A352859 a(0) = 1; a(n) = Sum_{k=0..n-1} binomial(n,k+1) * 2^k * a(k).

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A193660 Q-residue of the triangle A038207 of coefficients of (x+2)^n, where Q is the triangle given by t(i,j)=1 for 0<=i<=j. (See Comments.)

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A376177 Triangle, read by rows, where T(n,k) = T(n-1,k-1) + 2*T(n,k-1) when k > 0, else T(n,0) = T(n-1,n-1) when n > 0, with T(0,0) = 1.

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