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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A241095 a(n) = (5^n - 1)^n.

Original entry on oeis.org

1, 4, 576, 1906624, 151613669376, 297546691796890624, 14546328186912540283109376, 17761976839391142146587243652890624, 542089984111981439129333008468155364987109376, 413588400456285638417956135979678948104381423950212890624
Offset: 0

Views

Author

Hannah Ko, Apr 15 2014

Keywords

Comments

Number of n X n {-2,-1,0,1,2} matrices with no zero rows.

Crossrefs

Programs

  • Mathematica
    Join[{1},Table[(5^n-1)^n,{n,10}]] (* Harvey P. Dale, May 24 2023 *)

Extensions

More terms from Alois P. Heinz, Apr 23 2014

A241098 (4^n - 1)^n.

Original entry on oeis.org

1, 3, 225, 250047, 4228250625, 1120413075641343, 4715453174592516890625, 316777275155162685106909462527, 340240830764391036687105719527812890625, 5845805845679338940092384222659294205213252255743, 1606922719369349023077959288770050490334792256259918212890625
Offset: 0

Views

Author

Hannah Ko, Apr 15 2014

Keywords

Examples

			a(3) = 63^3 = 250047.
		

Crossrefs

Programs

Formula

(4^n - 1)^n

A251657 a(n) = (2^n + 3)^n.

Original entry on oeis.org

1, 5, 49, 1331, 130321, 52521875, 90458382169, 662062621900811, 20248745068443234721, 2548385124666493326171875, 1305282261160894865367626964649, 2701607566979638625212777041914285051, 22497539334127167666989016452232087989410801, 751859086636251929847496735809485838154930419921875
Offset: 0

Views

Author

Paul D. Hanna, Jan 29 2015

Keywords

Comments

This is a special case of the more general statement:
Sum_{n>=0} m^n * F(q^n*x)^b * log( F(q^n*x) )^n / n! = Sum_{n>=0} x^n * [y^n] F(y)^(m*q^n + b) where F(x) = exp(x), q=2, m=1, b=3.

Examples

			E.g.f.: A(x) = 4^0 + 5^1*x + 7^2*x^2/2! + 11^3*x^3/3! + 19^4*x^4/4! + 35^5*x^5/5! + 67^6*x^6/6! + 131^7*x^7/7! +...+ (2^n+3)^n*x^n/n! +...
such that
A(x) = exp(3*x) + 2*exp(3*2*x) + 2^4*exp(3*4*x)*x^2/2! + 2^9*exp(3*8*x)*x^3/3! + 2^16*exp(3*16*x)*x^4/4! +...+ 2^(n^2)*exp(3*2^n*x)*x^n/n! +...
		

Crossrefs

Programs

  • Mathematica
    Table[(2^n+3)^n,{n,0,20}] (* Harvey P. Dale, Mar 16 2016 *)
  • PARI
    {a(n,q=2,m=1,b=3) =( m*q^n + b)^n}
    for(n=0,15,print1(a(n,q=2,m=1,b=3),", "))
    
  • PARI
    {a(n,q=2,m=1,b=3) = sum(k=0,n, binomial(n,k) * b^k * m^(n-k) * (q^n)^(n-k))}
    for(n=0,15,print1(a(n,q=2,m=1,b=3),", "))
    
  • PARI
    {a(n,q=2,m=1,b=3) = n!*polcoeff(sum(k=0, n, m^k * q^(k^2) * exp(b*q^k*x +x*O(x^n)) * x^k/k!), n)}
    for(n=0,15,print1(a(n,q=2,m=1,b=3),", "))
    
  • PARI
    {a(n,q=2,m=1,b=3) = polcoeff(sum(k=0, n, m^k * q^(k^2) * x^k / (1 - b*q^k*x  +x*O(x^n))^(k+1) ), n)}
    for(n=0,15,print1(a(n,q=2,m=1,b=3),", "))

Formula

E.g.f.: Sum_{n>=0} 2^(n^2) * exp(3*2^n*x) * x^n/n! = Sum_{n>=0} (2^n + 3)^n * x^n/n!.
O.g.f.: Sum_{n>=0} 2^(n^2) * x^n / (1 + 3*2^n*x)^(n+1).
a(n) = Sum_{k=0..n} binomial(n, k) * 3^k * (2^n)^(n-k).
a(n) = Sum_{k=0..n} A155810(k)*3^k.

A337851 a(n) = (2^n + 2)^n.

Original entry on oeis.org

1, 4, 36, 1000, 104976, 45435424, 82653950016, 627485170000000, 19631688197463081216, 2504194578379511247798784, 1292628144912333835229805413376, 2687153475176994340820312500000000000, 22431765115399782718874449007331506546282496
Offset: 0

Views

Author

Paul D. Hanna, Sep 26 2020

Keywords

Comments

In general, we have the o.g.f. identity:
Sum_{n>=0} m^n * q^(n^2) * x^n/(1 - b*q^n*x)^(n+1) = Sum_{n>=0} (m*q^n + b)^n * x^n ; here, q=2, m=1, b=2.
In general, we have the e.g.f. identity:
Sum_{n>=0} m^n * q^(n^2) * exp(b*q^n*x) * x^n / n! = Sum_{n>=0} (m*q^n + b)^n * x^n / n! ; here, q=2, m=1, b=2.

Examples

			O.g.f.: A(x) = 1 + 4*x + 36*x^2 + 1000*x^3 + 104976*x^4 + 45435424*x^5 + 82653950016*x^6 + 627485170000000*x^7 + 19631688197463081216*x^8 + ...
where
A(x) = 1/(1 - 2*x) + 2*x/(1 - 2^2*x)^2 + 2^4*x^2/(1 - 2^3*x)^3 + 2^9*x^3/(1 - 2^4*x)^4 + 2^16*x^4/(1 - 2^5*x)^5 + 2^25*x^5/(1 - 2^6*x)^6 + ...
		

Crossrefs

Programs

  • PARI
    {a(n,q,m,b) = (m*q^n + b)^n}
    for(n=0,15, print1(a(n,q=2,m=1,b=2),", "))
    
  • PARI
    /* E.g.f. formula: */
    {a(n,q,m,b) = polcoeff( sum(k=0,n, m^k * q^(k^2) * x^k / (1 - b*q^k*x +x*O(x^n))^(k+1)), n)}
    for(n=0,15, print1(a(n,q=2,m=1,b=2),", "))
    
  • PARI
    /* E.g.f. formula: */
    {a(n,q,m,b) = n! * polcoeff( sum(k=0,n, m^k * q^(k^2) * exp(b*q^k*x +x*O(x^n)) * x^k/k!), n)}
    for(n=0,15, print1(a(n,q=2,m=1,b=2),", "))

Formula

O.g.f.: Sum_{n>=0} 2^(n^2) * x^n/(1 - 2^(n+1)*x)^(n+1) = Sum_{n>=0} (2^n + 2)^n * x^n.
E.g.f.: Sum_{n>=0} 2^(n^2) * exp(2^(n+1)*x) * x^n / n! = Sum_{n>=0} (2^n + 2)^n * x^n / n!.
a(n) = 2^n * A165327(n) for n >= 0.

A202990 E.g.f: Sum_{n>=0} 3^n * 2^(n^2) * exp(-2*2^n*x) * x^n/n!.

Original entry on oeis.org

1, 4, 100, 10648, 4477456, 7339040224, 47045881000000, 1186980379913527168, 118530511097526559703296, 47035767668340696232372862464, 74367598058372171073462490000000000, 469253945833810205185008441288962454059008
Offset: 0

Views

Author

Paul D. Hanna, Dec 26 2011

Keywords

Comments

E.g.f. series identity: Sum_{n>=0} m^n * q^(n^2) * exp(b*q^n*x) * x^n/n! = Sum_{n>=0} (m*q^n + b)^n * x^n/n! for all q, m, b.
O.g.f. series identity: Sum_{n>=0} m^n * q^(n^2) * x^n/(1-b*q^n*x)^(n+1) = Sum_{n>=0} (m*q^n + b)^n * x^n for all q, m, b.

Examples

			E.g.f.: A(x) = 1 + 4*x + 100*x^2/2! + 10648*x^3/3! + 4477456*x^4/4! +..
By the series identity, the e.g.f.:
A(x) = exp(-2*x) + 3*2*exp(-2*2*x)*x + 3^2*2^4*exp(-2*2^2*x)*x^2/2! + 3^3*2^9*exp(-2*2^3*x)*x^3/3! +...
expands into:
A(x) = 1 + 4*x + 10^2*x^2/2! + 22^3*x^3/3! + 46^4*x^4/4! + 94^5*x^5/5! +...+ (3*2^n-2)^n*x^n/n! +...
		

Crossrefs

Programs

  • Mathematica
    Table[(3*2^n-2)^n,{n,0,12}] (* Harvey P. Dale, Jul 16 2023 *)
  • PARI
    {a(n, q=2, m=3, b=-2)=(m*q^n + b)^n}
    
  • PARI
    {a(n, q=2, m=3, b=-2)=n!*polcoeff(sum(k=0, n, m^k*q^(k^2)*exp(b*q^k*x+x*O(x^n))*x^k/k!), n)}
    
  • PARI
    {a(n, q=2, m=3, b=-2)=polcoeff(sum(k=0, n, m^k*q^(k^2)*x^k/(1-b*q^k*x+x*O(x^n))^(k+1)), n)}

Formula

a(n) = (3*2^n - 2)^n.
O.g.f.: Sum_{n>=0} 3^n * 2^(n^2) * x^n/(1 + 2*2^n*x)^(n+1).

A234811 (6^n - 1)^n.

Original entry on oeis.org

1, 5, 1225, 9938375, 2812412850625, 28412011938974609375, 10313098426045900054366890625, 134710177671603826682045331123397109375, 63339984974231689005132970549727976493235812890625, 1072138503990252055371856714088806945958716851785209488787109375
Offset: 0

Views

Author

Hannah Ko, Apr 19 2014

Keywords

Examples

			a(3) = 215^3 = 9938375.
		

Crossrefs

Programs

A245789 Rectangular array A read by upward antidiagonals: A(k,n) = (2^k-1)^n, n,k >= 1.

Original entry on oeis.org

1, 1, 3, 1, 9, 7, 1, 27, 49, 15, 1, 81, 343, 225, 31, 1, 243, 2401, 3375, 961, 63, 1, 729, 16807, 50625, 29791, 3969, 127, 1, 2187, 117649, 759375, 923521, 250047, 16129, 255, 1, 6561, 823543, 11390625, 28629151, 15752961, 2048383, 65025, 511
Offset: 1

Views

Author

L. Edson Jeffery, Aug 22 2014

Keywords

Comments

A(k,n) is the number of sequences (X_1, X_2, ..., X_k) of subsets of the set {1, 2, ..., n} such that intersect_{j=1..k} X_j = null.

Examples

			Array A begins:
1      3         7           15              31                 63
1      9        49          225             961               3969
1     27       343         3375           29791             250047
1     81      2401        50625          923521           15752961
1    243     16807       759375        28629151          992436543
1    729    117649     11390625       887503681        62523502209
1   2187    823543    170859375     27512614111      3938980639167
1   6561   5764801   2562890625    852891037441    248155780267521
1  19683  40353607  38443359375  26439622160671  15633814156853823
		

References

  • Richard P. Stanley, Enumerative Combinatorics, Cambridge University Press, Vol. 1, Second edition, 2012, p. 14 (Example 1.1.16).

Crossrefs

Cf. A000225, A060867, A128831, etc. (rows 1-3).
Cf. A000012, A000244, A000420, etc. (columns 1-3).
Cf. A055601 (main diagonal).

Programs

  • Mathematica
    (* Array *)
    a[k_, n_] := (2^k - 1)^n; Grid[Table[a[k, n], {n, 12}, {k, 12}]]
    (* Array antidiagonals flattened *)
    Flatten[Table[(2^k - 1)^(n - k + 1), {n, 12}, {k, n}]]

A307232 a(n) is the number of n X n {0,1}-matrices (over the reals) that contain no zeros when squared.

Original entry on oeis.org

1, 1, 3, 73, 6003, 2318521, 4132876803
Offset: 0

Views

Author

Christopher Cormier, Mar 29 2019

Keywords

Comments

For every n, there are trivial solutions where an entire row is filled with 1's and an entire column is filled with 1's, and the column index is equal to the row index. This easily follows from the nature of matrix multiplication. Every matrix that has at least one of these row/column pairs along with any other 1's is also a solution because there are no negative numbers involved here. The number of trivial solutions is given by A307248.

Examples

			For n = 2, the a(2) = 3 solutions are
  1 1    0 1    1 1
  1 0    1 1    1 1
		

Crossrefs

A002416 is the total number of possible square binary matrices.
A307248 gives a lower bound.

Programs

  • MATLAB
    %Exhaustively searches all matrices
    %from n = 1 to 5
    result = zeros(1,5);
    for n = 1:5
    for m = 0:2^(n^2)-1
        p = fliplr(dec2bin(m,n^2) - '0');
        M = reshape(p,[n n]);
        D = M^2;
        if(isempty(find(D==0, 1)))
            result(n) = result(n) + 1;
        end
    end
    end
  • Mathematica
    a[n_] := Module[{b, iter, cnt = 0}, iter = Sequence @@ Table[{b[k], 0, 1}, {k, 1, n^2}]; Do[If[FreeQ[MatrixPower[Partition[Array[b, n^2], n], 2], 0], cnt++], iter // Evaluate]; cnt]; a[0] = 1;
    Do[Print[a[n]], {n, 0, 5}] (* Jean-François Alcover, Jun 23 2019 *)

Extensions

a(6) from Giovanni Resta, May 29 2019

A337852 a(n) = (2^(n+1) + 1)^n.

Original entry on oeis.org

1, 5, 81, 4913, 1185921, 1160290625, 4608273662721, 74051159531521793, 4796659837465472798721, 1248862969947666168212890625, 1304426412609681656861792686592001, 5459157240288132828933147334116110282753, 91477746675481294892349178081259839233191936001
Offset: 0

Views

Author

Paul D. Hanna, Sep 26 2020

Keywords

Comments

In general, we have the o.g.f. identity:
Sum_{n>=0} m^n * q^(n^2) * x^n/(1 - b*q^n*x)^(n+1) = Sum_{n>=0} (m*q^n + b)^n * x^n ; here, q=2, m=2, b=1.
In general, we have the e.g.f. identity:
Sum_{n>=0} m^n * q^(n^2) * exp(b*q^n*x) * x^n / n! = Sum_{n>=0} (m*q^n + b)^n * x^n / n! ; here, q=2, m=2, b=1.

Examples

			O.g.f.: A(x) = 1 + 5*x + 81*x^2 + 4913*x^3 + 1185921*x^4 + 1160290625*x^5 + 4608273662721*x^6 + 74051159531521793*x^7 + 4796659837465472798721*x^8 + ...
where
A(x) = 1/(1 - x) + 2^2*x/(1 - 2*x)^2 + 2^6*x^2/(1 - 2^2*x)^3 + 2^12*x^3/(1 - 2^3*x)^4 + 2^20*x^4/(1 - 2^4*x)^5 + 2^30*x^5/(1 - 2^5*x)^6 + ...
		

Crossrefs

Programs

  • PARI
    {a(n,q,m,b) = (m*q^n + b)^n}
    for(n=0,15, print1(a(n,q=2,m=2,b=1),", "))
    
  • PARI
    /* E.g.f. formula: */
    {a(n,q,m,b) = polcoeff( sum(k=0,n, m^k * q^(k^2) * x^k / (1 - b*q^k*x +x*O(x^n))^(k+1)), n)}
    for(n=0,15, print1(a(n,q=2,m=2,b=1),", "))
    
  • PARI
    /* E.g.f. formula: */
    {a(n,q,m,b) = n! * polcoeff( sum(k=0,n, m^k * q^(k^2) * exp(b*q^k*x +x*O(x^n)) * x^k/k!), n)}
    for(n=0,15, print1(a(n,q=2,m=2,b=1),", "))

Formula

O.g.f.: Sum_{n>=0} 2^(n*(n+1)) * x^n/(1 - 2^n*x)^(n+1) = Sum_{n>=0} (2^(n+1) + 1)^n * x^n.
E.g.f.: Sum_{n>=0} 2^(n*(n+1)) * exp(2^n*x) * x^n / n! = Sum_{n>=0} (2^(n+1) + 1)^n * x^n / n!.

A355282 Triangle read by rows: T(n, k) = Sum_{i=1..n-k} qStirling1(n-k, i) * qStirling2(n-1+i, n-1) for 0 < k < n with initial values T(n, 0) = 0^n and T(n, n) = 1 for n >= 0, here q = 2.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 9, 4, 1, 0, 343, 79, 11, 1, 0, 50625, 6028, 454, 26, 1, 0, 28629151, 1741861, 68710, 2190, 57, 1, 0, 62523502209, 1926124954, 38986831, 656500, 9687, 120, 1, 0, 532875860165503, 8264638742599, 84816722571, 734873171, 5760757, 40929, 247, 1
Offset: 0

Views

Author

Werner Schulte, Jun 26 2022

Keywords

Comments

We aim at a q-generalization of the Comtet-Lehmer numbers A354794, which are the case q = 1. Here we consider the case q = 2. The generalization is based on the qStirling numbers, for qStirling1 see A342186 and for qStirling2 see A139382. The general construction is as follows:
Let q <> 1 be a fixed integer and f_q(k) = (q^k - 1)/(q - 1) for k >= 0. Define triangle M(q; n, k) for 0 <= k <= n by M(q; n, 0) = 0^n for n >= 0, and M(q; n, k) = 0 for k > n, and M(q; n, k) = M(q; n-1, k-1) + M(q; n-1, k) * f_q(k) for 0 < k <= n. Then M(q; n, n) = 1 for n >= 0 and the matrix inverse I_q = M_q^(-1) exists. Next define the triangle T(q; n, k) for 0 <= k <= n by T(q; n, 0) = 0^n for n >= 0 and T(q; n, k) = Sum_{i=0..n-k} I(q; n-k, i) * M(q; n-1+i, n-1) for 0 < k <= n. Take account of lim_{q->1} (q^n - 1)/(q - 1) = n for n >= 0.
Conjecture: T(q; n+1, 1) = Sum_{i=0..n} I(q; n, i) * M(q; n+i, n) = (f_q(n))^n = ((q^n - 1)/(q - 1))^n for n >= 0.
Conjecture: T(q; n, k) = (Sum_{i=0..n-k} (-1)^i * q-binomial(n-1-i, k-1) * binomial(n-1, i) * q^((n-k)*(n-k-i))) / (q - 1)^(n-k) for 0 < k <= n.

Examples

			Triangle T(n, k) for 0 <= k <= n starts:
n\k : 0               1             2           3         4       5     6   7 8
===============================================================================
  0 : 1
  1 : 0               1
  2 : 0               1             1
  3 : 0               9             4           1
  4 : 0             343            79          11         1
  5 : 0           50625          6028         454        26       1
  6 : 0        28629151       1741861       68710      2190      57     1
  7 : 0     62523502209    1926124954    38986831    656500    9687   120   1
  8 : 0 532875860165503 8264638742599 84816722571 734873171 5760757 40929 247 1
  etc.
		

Crossrefs

Cf. A022166, A139382, A342186, A354794, A055601 (column 1), A125128 (1st subdiagonal).

Programs

  • Maple
    # using qStirling2 from A333143.
    A355282 := proc(n, k) if k = 0 then 0^n elif n = k then 1 else
    add(A342186(n - k, i)*qStirling2(n + i - 2, n - 2, 2), i = 1..n-k) fi end:
    seq(print(seq(A355282(n, k), k = 0..n)), n = 0..8); # Peter Luschny, Jun 28 2022
  • PARI
    mat(nn) = my(m = matrix(nn, nn)); for (n=1, nn, for(k=1, nn, m[n, k] = if (n==1, if (k==1, 1, 0), if (k==1, 1, (2^k-1)*m[n-1, k] + m[n-1, k-1])); ); ); m; \\ A139382
    tabl(nn) = my(m=mat(3*nn), im=1/m); matrix(nn, nn, n, k, n--; k--; if (k==0, 0^n, kMichel Marcus, Jun 27 2022

Formula

Conjecture: T(n+1, 1) = (2^n - 1)^n for n >= 0.
Conjecture: T(n, k) = Sum_{i=0..n-k} (-1)^i * binomial(n-1, i) * [n-1-i, k-1]_2 * 2^((n-k)*(n-k-i)) for 0 < k <= n and T(n, 0) = 0^n for n >= 0, where [x, y]_2 = A022166(x, y).
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