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.

Showing 1-6 of 6 results.

A198258 Row sums of A197655.

Original entry on oeis.org

1, 7, 190, 5831, 219626, 8976562, 394800204, 18211045575, 873216720082, 43136486269382, 2183722698469676, 112795257850321202, 5925951358743662260, 315869014732813229716, 17048301919464100932440, 930217336628892162216135, 51244644190683748419791970
Offset: 0

Views

Author

Susanne Wienand, Oct 24 2011

Keywords

Comments

Number of meanders of length (n+1)*6 which are composed by arcs of equal length and a central angle of 60 degrees.
Definition of a meander:
A binary curve C is a triple (m, S, dir) such that
(a) S is a list with values in {L,R} which starts with an L,
(b) dir is a list of m different values, each value of S being allocated a value of dir,
(c) consecutive Ls increment the index of dir,
(d) consecutive Rs decrement the index of dir,
(e) the integer m>0 divides the length of S and
(f) C is a meander if each value of dir occurs length(S)/m times.
For this sequence, m = 6.
The terms are proved by brute force for 0 <= n <= 5, but not yet in general. - Susanne Wienand, Oct 29 2011

Examples

			Some examples of list S and allocated values of dir if n = 4:
Length(S) = (4+1)*6 = 30.
  S: L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L
dir: 1,2,3,4,5,0,1,2,3,4,5,0,1,2,3,4,5,0,1,2,3,4,5,0,1,2,3,4,5,0
  S: L,L,L,L,L,L,L,R,R,R,R,L,R,R,R,R,R,L,L,L,L,R,R,L,L,L,L,R,L,L
dir: 1,2,3,4,5,0,1,1,0,5,4,4,4,3,2,1,0,0,1,2,3,3,2,2,3,4,5,5,5,0
  S: L,L,L,L,L,R,R,R,R,R,R,R,L,L,R,L,L,R,L,L,R,L,L,L,R,R,L,L,L,L
dir: 1,2,3,4,5,5,4,3,2,1,0,5,5,0,0,0,1,1,1,2,2,2,3,4,4,3,3,4,5,0
Each value of dir occurs 30/6 = 5 times.

Links

Crossrefs

Cf. A198060, A198256, A198257.

Programs

  • Maple
    A198258 := proc(n) local i, j, k, pow;
pow := (a, b) -> if a=0 and b=0 then 1 else a^b fi;
add(add(add((-1)^(j+i)*binomial(i,j)*binomial(n,k)^6*pow(n+1,j)*pow(k+1,5-j)/(k+1)^5, i=0..5),j=0..5),k=0..n) end:
seq(A198258(n), n=0..16); # Peter Luschny, Nov 02 2011
  • Mathematica
    T[n_, k_] := (1 + n)(1 + 3 k + 3 k^2 - n - 3 k*n + n^2)(1 + k + k^2 + n - k*n + n^2) Binomial[n, k]^6/(1 + k)^5;
a[n_] := Sum[T[n, k], {k, 0, n}];
Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Jun 29 2019 *)
  • PARI
    A198258(n) = {sum(k=0,n,if(n == 1+2*k,6,(1+k)*(1-((n-k)/(1+k))^6)/(1+2*k-n))*binomial(n,k)^6)} \\ Peter Luschny, Nov 24 2011

Formula

a(n) = Sum_{k=0..n} Sum_{j=0..5} Sum_{i=0..5} (-1)^(j+i)*C(i,j)*C(n,k)^6*(n+1)^j*(k+1)^(5-j)/(k+1)^5. - Peter Luschny, Nov 02 2011
a(n) = Sum_{k=0..n} h(n,k)*binomial(n,k)^6, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^6)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 6. - Peter Luschny, Nov 24 2011
Conjecture: working with offset 1, that is, a(1) = 1, a(2) = 7, ..., then the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for positive integers n and r and all primes p >= 3. - Peter Bala, Mar 21 2023
a(n) ~ sqrt(3) * 2^(6*n+3) / (Pi*n)^(5/2). - Vaclav Kotesovec, Apr 17 2023

A194595 Triangle by rows T(n,k), showing the number of meanders with length (n+1)*3 and containing (k+1)*3 L's and (n-k)*3 R's, where L's and R's denote arcs of equal length and a central angle of 120 degrees which are positively or negatively oriented.

Original entry on oeis.org

1, 3, 1, 7, 14, 1, 13, 81, 39, 1, 21, 304, 456, 84, 1, 31, 875, 3000, 1750, 155, 1, 43, 2106, 13875, 18500, 5265, 258, 1, 57, 4459, 50421, 128625, 84035, 13377, 399, 1, 73, 8576, 153664, 669536, 836920, 307328, 30016, 584, 1, 91, 15309, 409536, 2815344, 6001128, 4223016, 955584, 61236, 819, 1
Offset: 0

Views

Author

Susanne Wienand, Oct 10 2011

Keywords

Comments

Definition of a meander:
A binary curve C is a triple (m, S, dir) such that
(a) S is a list with values in {L,R} which starts with an L,
(b) dir is a list of m different values, each value of S being allocated
a value of dir,
(c) consecutive L's increment the index of dir,
(d) consecutive R's decrement the index of dir,
(e) the integer m > 0 divides the length of S and
(f) C is a meander if each value of dir occurs length(S)/m times.
For this sequence, m = 3.
The values in the triangle are proved by brute force for 0 <= n <= 11. The formulas are not yet proved in general. - Susanne Wienand
Let S(N,n,k) = C(n,k)^(N+1)*Sum_{j=0..N} Sum_{i=0..N} (-1)^(N-j+i)*C(N-i,j)*((n+1)/(k+1))^j. Then S(0,n,k) = A007318(n,k), S(1,n,k) = A103371(n,k), S(2,n,k) = T(n,k), S(3,n,k) = A197653(n,k), S(4,n,k) = A197654(n,k), S(5,n,k) = A197655(n,k). - Peter Luschny, Oct 21 2011
The number triangle can be calculated recursively by the number triangles A103371 and A007318. The first column of the triangle contains the central polygonal numbers A002061. The diagonal right hand is A000012. The diagonal with k = n-1 seems to be A027444. Row sums are in A197657. - Susanne Wienand, Nov 24 2011
The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 62. - Susanne Wienand, Jun 24 2015

Examples

			For n = 4 and k = 2, T(3,4,2) = 456.
Recursive example:
T(1,4,0) = 1
T(1,4,1) = 4
T(1,4,2) = 6
T(1,4,3) = 4
T(1,4,4) = 1
T(2,4,0) = 5
T(2,4,1) = 40
T(2,4,2) = 60
T(2,4,3) = 20
T(2,4,4) = 1
T(3,4,0) = T(1,4,0)^3 + T(1,4,0)*T(2,4,4-1-0) = 1^3 + 1*20 = 21
T(3,4,1) = T(1,4,1)^3 + T(1,4,1)*T(2,4,4-1-1) = 4^3 + 4*60 = 304
T(3,4,2) = T(1,4,2)^3 + T(1,4,2)*T(2,4,4-1-2) = 6^3 + 6*40 = 456
T(3,4,3) = T(1,4,3)^3 + T(1,4,3)*T(2,4,4-1-3) = 4^3 + 4*5  = 84
T(3,4,4) = 1.
Example for closed formula:
T(4,2) = (C(4,2))^3 + C(4,2) * C(4,3) * C(5,3) = 6^3 + 6 * 4 * 10 = 456.
Some examples of list S and allocated values of dir if n = 4 and k = 2:
Length(S) = (4+1)*3 = 15 and S contains (2+1)*3 = 9 L's.
  S: L,L,L,L,L,L,L,L,L,R,R,R,R,R,R
dir: 1,2,0,1,2,0,1,2,0,0,2,1,0,2,1
  S: L,L,R,L,L,L,L,R,R,L,R,R,L,R,L
dir: 1,2,2,2,0,1,2,2,1,1,1,0,0,0,0
  S: L,R,R,R,L,L,L,L,R,R,L,L,L,R,L
dir: 1,1,0,2,2,0,1,2,2,1,1,2,0,0,0
Each value of dir occurs 15/3 = 5 times.

Links

Crossrefs

Cf. A103371, A197653, A197654, A197655, A197657, A007318, A002061, A000012, A027444, A073254.

Programs

  • Maple
    A194595 := (n,k)->binomial(n,k)^3*(k^2+k+1+n^2+n-k*n)/((k+1)^2);
seq(print(seq(A194595(n,k),k=0..n)),n=0..7); # Peter Luschny, Oct 14 2011
  • Mathematica
    T[n_, k_] := Binomial[n, k]^3*(k^2 + k + 1 + n^2 + n - k*n)/((k + 1)^2);
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2018, after Peter Luschny *)
  • PARI
    A194595(n,k) = {if(n == 1+2*k,3,(1+k)*(1-((n-k)/(1+k))^3)/(1+2*k-n))*binomial(n,k)^3} \\ Peter Luschny, Nov 24 2011

Formula

Recursive formula (conjectured):
T(n,k) = T(3,n,k) = T(1,n,k)^3 + T(1,n,k)*T(2,n,n-1-k), 0 <= k < n
T(3,n,n) = 1, k = n
T(2,n,k) = T(1,n,k)^2 + T(1,n,k) * T(1,n,n-1-k), 0 <= k < n
T(2,n,n) = 1, k = n
T(2,n,k) = A103371,
T(1,n,k) = A007318 (Pascal's Triangle).
Closed formula (conjectured): T(n,k) = (C(n,k))^3 + C(n,k) * C(n,k+1) * C(n+1,k+1). - Susanne Wienand
Let S(n,k) = binomial(2*n,n)^(k+1)*((n+1)^(k+1)-n^(k+1))/(n+1)^k. Then T(2*n,n) = S(n,2). - Peter Luschny, Oct 20 2011
T(n,k) = A073254(n+1,k+1)C(n,k)^3/(k+1)^2. - Peter Luschny, Oct 29 2011
T(n,k) = h(n,k)*binomial(n,k)^3, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^3)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 3. - Peter Luschny, Nov 24 2011

A197653 Triangle by rows T(n,k), showing the number of meanders with length (n+1)*4 and containing (k+1)*4 Ls and (n-k)*4 Rs, where Ls and Rs denote arcs of equal length and a central angle of 90 degrees which are positively or negatively oriented.

Original entry on oeis.org

1, 4, 1, 15, 30, 1, 40, 324, 120, 1, 85, 2080, 3120, 340, 1, 156, 9375, 40000, 18750, 780, 1, 259, 32886, 328125, 437500, 82215, 1554, 1, 400, 96040, 1959216, 6002500, 3265360, 288120, 2800, 1
Offset: 0

Views

Author

Susanne Wienand, Oct 17 2011

Keywords

Comments

Definition of a meander:
A binary curve C is a triple (m, S, dir) such that
(a) S is a list with values in {L,R} which starts with an L,
(b) dir is a list of m different values, each value of S being allocated a value of dir,
(c) consecutive Ls increment the index of dir,
(d) consecutive Rs decrement the index of dir,
(e) the integer m > 0 divides the length of S and
(f) C is a meander if each value of dir occurs length(S)/m times.
For this sequence, m = 4.
The values in the triangle are proved by brute force for 0 <= n <= 8. The formulas are not yet proved in general.
The number triangle can be calculated recursively by the number triangles A007318, A103371 and A194595. The first column of the triangle seems to be A053698. The diagonal right hand is A000012. The diagonal with k = n-1 seems to be A027445. Row sums are in A198256.
The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 43. - Susanne Wienand, Jun 29 2015

Examples

			For n = 4 and k = 2, T(4,4,2) = 3120.
Recursive example:
T(1,4,0) = 1,
T(1,4,1) = 4,
T(1,4,2) = 6,
T(1,4,3) = 4,
T(1,4,4) = 1,
T(3,4,0) = 21,
T(3,4,1) = 304,
T(3,4,2) = 456,
T(3,4,3) = 84,
T(3,4,1) = 1,
T(4,4,2) = 6^4 + 6*304 = 3120.
Example for closed formula:
T(4,2) = 6^4 + 6^3 * 4 + 6^2 * 4^2 + 6 * 4^3 = 3120.
Some examples of list S and allocated values of dir if n = 4 and k = 2:
Length(S) = (4+1)*4 = 20 and S contains (2+1)*4 = 12 Ls.
  S: L,L,L,L,L,L,L,L,L,L,L,L,R,R,R,R,R,R,R,R
dir: 1,2,3,0,1,2,3,0,1,2,3,0,0,3,2,1,0,3,2,1
  S: L,L,L,R,L,L,R,L,L,R,R,L,L,L,R,L,L,R,R,R
dir: 1,2,3,3,3,0,0,0,1,1,0,0,1,2,2,2,3,3,2,1
  S: L,R,L,L,L,L,R,R,R,L,L,R,R,L,L,L,R,L,L,R
dir: 1,1,1,2,3,0,0,3,2,2,3,3,2,2,3,0,0,0,1,1
Each value of dir occurs 20/4 = 5 times.
Triangle begins:
   1;
   4,    1;
  15,   30,    1;
  40,  324,  120,   1;
  85, 2080, 3120, 340, 1;
  ...

Links

Crossrefs

Cf. A000012, A007318, A027445, A053698, A103371, A194595, A197654, A197655, A198063, A198256.

Programs

  • Maple
    A197653 := (n,k) -> binomial(n,k)^4*(n+1)*(n^2-2*n*k+1+2*k+2*k^2)/((1+k)^3);
seq(print(seq(A197653(n, k), k=0..n)), n=0..7); # Peter Luschny, Oct 19 2011
  • Mathematica
    T[n_, k_] := Binomial[n, k]^4 (n+1)(n^2 - 2n*k + 1 + 2k + 2k^2)/((1+k)^3);
Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2018, after Peter Luschny *)
  • PARI
    A197653(n,k) = {if(n==1+2*k,4,(1+k)*(1-((n-k)/(1+k))^4)/(1+2*k-n))*binomial(n,k)^4} \\ Peter Luschny, Nov 24 2011
  • Sage
    def S(N,n,k) : return binomial(n,k)^(N+1)*sum(sum((-1)^(N-j+i)*binomial(N-i,j)*((n+1)/(k+1))^j for i in (0..N) for j in (0..N)))
def A197653(n,k) : return S(3,n,k)
for n in (0..5) : print([A197653(n,k) for k in (0..n)])  ## Peter Luschny, Oct 24 2011

Formula

Recursive formula (conjectured):
T(n,k) = T(4,n,k)
T(4,n,k) = T(1,n,k)^4 + T(1,n,k)*T(3,n,n-1-k), 0 <= k < n
T(4,n,n) = 1 k = n
T(3,n,k) = T(1,n,k)^3 + T(1,n,k)*T(2,n,n-1-k), 0 <= k < n
T(3,n,n) = 1 k = n
T(2,n,k) = T(1,n,k)^2 + T(1,n,k)*T(1,n,n-1-k), 0 <= k < n
T(2,n,n) = 1 k = n
T(3,n,k) = A194595
T(2,n,k) = A103371
T(1,n,k) = A007318 (Pascal's Triangle)
closed formula (conjectured): T(n,n) = 1, k = n
T(n,k) = A + B + C + D, k < n
A = (C(n,k))^4
B = (C(n,k))^3 * C(n,n-1-k)
C = (C(n,k))^2 *(C(n,n-1-k))^2
D = C(n,k) *(C(n,n-1-k))^3
[Susanne Wienand]
Let S(n,k) = binomial(2*n,n)^(k+1)*((n+1)^(k+1)-n^(k+1))/(n+1)^k. Then T(2*n,n) = S(n,3). - Peter Luschny, Oct 20 2011
T(n,k) = A198063(n+1,k+1)*C(n,k)^4/(k+1)^3. - Peter Luschny, Oct 29 2011
T(n,k) = h(n,k)*binomial(n,k)^4, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^4)/(1+2*k-n) if 1+2*k-n <> 0, otherwise h(n,k) = 4. - Peter Luschny, Nov 24 2011

A197654 Triangle by rows T(n,k), showing the number of meanders with length 5(n+1) and containing 5(k+1) L's and 5(n-k) R's, where L's and R's denote arcs of equal length and a central angle of 72 degrees which are positively or negatively oriented.

Original entry on oeis.org

1, 5, 1, 31, 62, 1, 121, 1215, 363, 1, 341, 13504, 20256, 1364, 1, 781, 96875, 500000, 193750, 3905, 1, 1555, 501066, 7321875, 9762500, 1252665, 9330, 1, 2801, 2033647, 72656661, 262609375, 121094435, 6100941, 19607, 1
Offset: 0

Views

Author

Susanne Wienand, Oct 19 2011

Keywords

Comments

Definition of a meander:
A binary curve C is a triple (m, S, dir) such that:
(a) S is a list with values in {L,R} which starts with an L,
(b) dir is a list of m different values, each value of S being allocated a value of dir,
(c) consecutive L's increment the index of dir,
(d) consecutive R's decrement the index of dir,
(e) the integer m>0 divides the length of S.
Then C is a meander if each value of dir occurs length(S)/m times.
Let T(m,n,k) = number of meanders (m, S, dir) in which S contains m(k+1) L's and m(n-k) R's, so that length(S) = m(n+1).
For this sequence, m = 5, T(n,k) = T(5,n,k).
The values in the triangle were proved by brute force for 0 <= n <= 6. The formulas have not yet been proved in general.
The number triangle can be calculated recursively by the number triangles and A007318, A103371, A194595 and A197653. The first column seems to be A053699. The diagonal right hand is A000012. The diagonal with k = n-1 seems to be A152031 and to start with the second number of A152031. Row sums are in A198257.
The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 29. - Susanne Wienand, Jul 01 2015

Examples

			For n = 5 and k = 2, T(n,k) = 500000
Example for recursive formula:
T(1,5,2) = 10
T(4,5,5-1-2) = T(4,5,2) = 40000
T(5,5,2) = 10^5 + 10*40000 = 500000
Example for closed formula:
T(5,2) = A + B + C + D + E
A = 10^5
B = 10^4 * 10
C = 10^3 * 10^2
D = 10^2 * 10^3
E = 10   * 10^4
T(5,2) = 5 * 10^5 = 500000
Some examples of list S and allocated values of dir if n = 5 and k = 2:
Length(S) = (5+1)*5 = 30 and S contains (2+1)*5 = 15 Ls.
  S: L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,R,R,R,R,R,R,R,R,R,R,R,R,R,R,R
dir: 1,2,3,4,0,1,2,3,4,0,1,2,3,4,0,0,4,3,2,1,0,4,3,2,1,0,4,3,2,1
  S: L,L,L,L,L,L,L,R,R,L,L,R,R,R,L,R,R,R,L,R,L,L,L,R,R,L,R,R,R,R
dir: 1,2,3,4,0,1,2,2,1,1,2,2,1,0,0,0,4,3,3,3,3,4,0,0,4,4,4,3,2,1
  S: L,L,L,L,L,R,L,L,L,L,L,R,L,R,R,R,R,R,R,R,R,R,R,L,L,L,L,R,R,R
dir: 1,2,3,4,0,0,0,1,2,3,4,4,4,4,3,2,1,0,4,3,2,1,0,0,1,2,3,3,2,1
Each value of dir occurs 30/5 = 6 times.
The triangle begins:
1,
5, 1,
31, 62, 1,
121, 1215, 363, 1,
341, 13504, 20256, 1364, 1,
781, 96875, 500000, 193750, 3905, 1,
...

Links

Crossrefs

Cf. A000012, A007318, A053699, A103371, A152031, A194595, A197653, A197655, A198257.

Programs

  • Maple
    A197654 := (n,k)->(k^4+2*k^3*(1-n)+2*k^2*(2+n+2*n^2)+k*(3+n-n^2-3*n^3)+ n^4+n^3+n^2+n+1)*binomial(n,k)^5/(1+k)^4;
seq(print(seq(A197654(n, k), k=0..n)), n=0..7);  # Peter Luschny, Oct 20 2011
  • Mathematica
    T[n_, k_] := (k^4 + 2*k^3*(1-n) + 2*k^2*(2+n+2*n^2) + k*(3+n-n^2-3*n^3) + n^4+n^3+n^2+n+1)*Binomial[n, k]^5/(1+k)^4; Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 20 2017, after Peter Luschny *)
  • PARI
    A197654(n,k) = {if(n ==1+2*k,5,(1+k)*(1-((n-k)/(1+k))^5)/(1+2*k-n))*binomial(n,k)^5} \\ Peter Luschny, Nov 24 2011
  • Sage
    def S(N,n,k) : return binomial(n,k)^(N+1)*sum(sum((-1)^(N-j+i)*binomial(N-i,j)*((n+1)/(k+1))^j for i in (0..N) for j in (0..N)))
def A197654(n,k) : return S(4,n,k)
for n in (0..5) : print([A197654(n,k) for k in (0..n)])  # Peter Luschny, Oct 24 2011

Formula

Recursive formula (conjectured):
T(n,k) = T(5,n,k) = T(1,n,k)^5 + T(1,n,k)*T(4,n,n-1-k), 0 <= k < n
T(5,n,n) = 1 k = n
T(4,n,k) = T(1,n,k)^4 + T(1,n,k) * T(3,n,n-1-k), 0 <= k < n
T(4,n,n) = 1 k = n
T(3,n,k) = T(1,n,k)^3 + T(1,n,k) * T(2,n,n-1-k), 0 <= k < n
T(3,n,n) = 1 k = n
T(2,n,k) = T(1,n,k)^2 + T(1,n,k) * T(1,n,n-1-k), 0<= k < n
T(2,n,n) = 1 k = n
T(4,n,k) = A197653
T(3,n,k) = A194595
T(2,n,k) = A103371
T(1,n,k) = A007318 (Pascal's Triangle)
Closed formula (conjectured): T(n,n) = 1, k = n
T(n,k) = A + B + C + D + E, k < n
A = (C(n,k))^5
B = (C(n,k))^4 * C(n,n-1-k)
C = (C(n,k))^3 *(C(n,n-1-k))^2
D = (C(n,k))^2 *(C(n,n-1-k))^3
E = C(n,k) *(C(n,n-1-k))^4 [Susanne Wienand]
Let S(n,k) = binomial(2*n,n)^(k+1)*((n+1)^(k+1)-n^(k+1))/(n+1)^k. Then T(2*n,n) = S(n,4). [Peter Luschny, Oct 20 2011]
T(n,k) = A198064(n+1,k+1)C(n,k)^5/(k+1)^4. [Peter Luschny, Oct 29 2011]
T(n,k) = h(n,k)*binomial(n,k)^5, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^5)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 5. [Peter Luschny, Nov 24 2011]

A198065 Triangle read by rows (n >= 0, 0 <= k <= n, m = 5); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)*C(i,j)*n^j*k^(m-j).

Original entry on oeis.org

0, 1, 1, 32, 6, 32, 243, 63, 63, 243, 1024, 364, 192, 364, 1024, 3125, 1365, 665, 665, 1365, 3125, 7776, 3906, 2016, 1458, 2016, 3906, 7776, 16807, 9331, 5187, 3367, 3367, 5187, 9331, 16807, 32768, 19608, 11648, 7448, 6144, 7448, 11648, 19608, 32768, 59049
Offset: 0

Views

Author

Peter Luschny, Oct 26 2011

Keywords

Examples

			[0]                        0
[1]                       1, 1
[2]                    32, 6, 32
[3]                 243, 63, 63, 243
[4]            1024, 364, 192, 364, 1024
[5]         3125, 1365, 665, 665, 1365, 3125
[6]     7776, 3906, 2016, 1458, 2016, 3906, 7776
[7] 16807, 9331, 5187, 3367, 3367, 5187, 9331, 16807

Crossrefs

Cf. A057427, A003056, A073254, A198063, A198064.
Cf. A000584, A022522, A197655.

Programs

  • Magma
    &cat[[n*(k^2-k*n+n^2)*(3*k^2-3*k*n+n^2): k in [0..n]]: n in [0..9]];  // Bruno Berselli, Nov 02 2011
  • Maple
    A198065 := (n,k) -> 3*n*k^4-6*k^3*n^2+7*k^2*n^3-4*k*n^4+n^5:

Formula

T(n,k) = 3*n*k^4-6*k^3*n^2+7*k^2*n^3-4*k*n^4+n^5.
T(n,0) = T(n,n) = n^m = n^5 = A000584(n).
T(2n,n) = (m+1)n^m = 6n^5.
T(2n+1,n+1) = (n+1)^(m+1)-n^(m+1) = (n+1)^6-n^6 = A022522(n).
Sum{k=0..n} T(n,k) = (13n^6+30n^5+20n^4-3n^2)/30.
T(n+1,k+1)C(n,k)^6/(k+1)^5 = A197655(n,k).

A198062 Array read by antidiagonals, m>=0, n>=0, k>=0, A(m, n, k) = sum{j=0..m} sum{i=0..m} (-1)^(j+i)*C(i,j)*n^j*k^(m-j).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 4, 2, 1, 0, 1, 1, 8, 3, 2, 1, 0, 1, 1, 16, 4, 4, 3, 1, 0, 1, 1, 32, 5, 8, 9, 3, 1, 0, 1, 1, 64, 6, 16, 27, 7, 3, 1, 0, 1, 1, 128, 7, 32, 81, 15, 7, 3, 1, 0, 1, 1, 256, 8, 64, 243, 31, 15, 9, 4, 1, 0, 1, 1, 512, 9
Offset: 0

Views

Author

Peter Luschny, Nov 02 2011

Keywords

Examples

			   [0] [1] [2]  [3] [4]  [5]  [6]  [7]  [8]  [9]
-------------------------------------------------
[0]  1   1   1    1   1    1    1    1    1    1    A000012
[1]  0   1   1    2   2    2    3    3    3    3    A003056
[2]  0   1   1    4   3    4    9    7    7    9    A073254
[3]  0   1   1    8   4    8   27   15   15   27    A198063
[4]  0   1   1   16   5   16   81   31   31   81    A198064
[5]  0   1   1   32   6   32  243   63   63  243    A198065

Links

Crossrefs

Cf. A198060, A198061.

Programs

  • Maple
    A198062_RowAsTriangle := proc(m) local pow; pow :=(a,b)->`if`(a=0 and b=0,1,a^b): proc(n, k) local i, j; add(add((-1)^(j + i)*binomial(i, j)*pow(n, j)* pow(k, m-j), i=0..m),j=0..m) end: end:
for m from 0 to 2 do seq(print(seq(A198062_RowAsTriangle(m)(n,k),k=0..n)),n=0..5) od;
  • Mathematica
    max = 9; RowAsTriangle[m_][n_, k_] := Module[{pow}, pow[a_, b_] := If[a == 0 && b == 0, 1, a^b]; Module[{i, j}, Sum[Sum[(-1)^(j+i)*Binomial[i, j]*pow[n, j]*pow[k, m-j], {i, 0, m}], {j, 0, m}]]]; t = Flatten /@ Table[RowAsTriangle[m][n, k], {m, 0, max}, {n, 0, max}, {k, 0, n}]; Table[t[[n-k+1, k+1]], {n, 0, max}, {k, 0, n }] // Flatten (* Jean-François Alcover, Feb 25 2014, after Maple *)

Formula

A007318(n,k) = A(0,n+1,k+1)*C(n,k)^1/(k+1)^0,
A103371(n,k) = A(1,n+1,k+1)*C(n,k)^2/(k+1)^1,
A194595(n,k) = A(2,n+1,k+1)*C(n,k)^3/(k+1)^2,
A197653(n,k) = A(3,n+1,k+1)*C(n,k)^4/(k+1)^3,
A197654(n,k) = A(4,n+1,k+1)*C(n,k)^5/(k+1)^4,
A197655(n,k) = A(5,n+1,k+1)*C(n,k)^6/(k+1)^5.
Showing 1-6 of 6 results.

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