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 1-10 of 11 results. Next

A016161 Expansion of g.f. 1/((1-5*x)*(1-7*x)).

Original entry on oeis.org

1, 12, 109, 888, 6841, 51012, 372709, 2687088, 19200241, 136354812, 964249309, 6798573288, 47834153641, 336059778612, 2358521965909, 16540171339488, 115933787267041, 812299450322412, 5689910849522509
Offset: 0

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Author

Keywords

Comments

Also, this is the number of incongruent integer-edged Heron triangles whose circumdiameter is the product of n distinct primes each of shape 4k + 1. Cf. A003462, A109021. - R. K. Guy, Jan 31 2007

Crossrefs

Programs

  • Magma
    [n le 2 select 12^(n-1) else 12*Self(n-1) -35*Self(n-2): n in [1..30]]; // G. C. Greubel, Nov 09 2024
    
  • Mathematica
    CoefficientList[Series[1/((1-5x)(1-7x)),{x,0,30}],x] (* or *) LinearRecurrence[ {12,-35},{1,12},30] (* Harvey P. Dale, Nov 16 2021 *)
  • PARI
    Vec(1/((1-5*x)*(1-7*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 24 2012
    
  • SageMath
    A016161=BinaryRecurrenceSequence(12,-35,1,12)
    [A016161(n) for n in range(31)] # G. C. Greubel, Nov 09 2024

Formula

From R. J. Mathar, Sep 18 2008: (Start)
a(n) = (7^(n+1) - 5^(n+1))/2 = A081200(n+1).
Binomial transform of A080962. (End)
a(n) = 7*a(n-1) + 5^n. - Vincenzo Librandi, Feb 09 2011
E.g.f.: exp(5*x)*(7*exp(2*x) - 5)/2. - Stefano Spezia, Oct 25 2023

A081201 7th binomial transform of (0,1,0,1,0,1,....), A000035.

Original entry on oeis.org

0, 1, 14, 148, 1400, 12496, 107744, 908608, 7548800, 62070016, 506637824, 4113568768, 33271347200, 268347559936, 2159841173504, 17357093552128, 139326933401600, 1117436577120256, 8956419276406784, 71752914167922688, 574632673083392000, 4600717543107198976
Offset: 0

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Author

Paul Barry, Mar 11 2003

Keywords

Comments

Binomial transform of A081200.
Conjecture (verified up to a(8)): Number of collinear 6-tuples of points in a 6 X 6 X 6 X... n-dimensional cubic grid. [R. H. Hardin, May 23 2010]
From Wolfdieter Lang, Jul 17 2017: (Start)
For a combinatorial interpretation of a(n) with special 8-letter words of length n see the comment in A081200 on the 7-letter analog.
The binomial transform of {a(n)}_{n >= 0} is A081202, the 9-letter analog.
(End)

Crossrefs

Apart from offset same as A016170.

Programs

Formula

a(n) = 14*a(n-1) - 48*a(n-2) with n>1, a(0)=0, a(1)=1.
G.f.: x/((1-6*x)*(1-8*x)).
a(n) = (1/2)*(8^n - 6^n).
E.g.f.: exp(7*x)*sinh(x). - G. C. Greubel, Nov 10 2024

Extensions

Name clarified by Pontus von Brömssen, Nov 11 2020

A081199 5th binomial transform of (0,1,0,1,...), A000035.

Original entry on oeis.org

0, 1, 10, 76, 520, 3376, 21280, 131776, 807040, 4907776, 29708800, 179301376, 1080002560, 6496792576, 39047864320, 234555621376, 1408407470080, 8454739787776, 50745618595840, 304542431051776, 1827529464217600, 10966276296933376, 65802055828111360, 394829927154712576
Offset: 0

Views

Author

Paul Barry, Mar 11 2003

Keywords

Comments

Binomial transform of A005059.
Conjecture (verified up to a(9)): Number of collinear 4-tuples of points in a 4 X 4 X 4 X ... n-dimensional cubic grid. - R. H. Hardin, May 24 2010
a(n) is also the total number of words of length n, over an alphabet of six letters, of which one of them appears an odd number of times. See a Lekraj Beedassy, Jul 22 2003, comment on A006516 (4-letter case), and the Balakrishnan reference there. For the 2-, 3-, 5- and 7-letter case analogs see A131577, A003462, A005059 and A081200, respectively. - Wolfdieter Lang, Jul 16 2017

Crossrefs

Cf. A000035, A003462, A005059, A006516, A081200 (binomial transform of a(n), and 7-letter case), A131577.
Apart from offset the same as A016149.

Programs

  • Magma
    [6^n/2-4^n/2: n in [0..25]]; // Vincenzo Librandi, Aug 07 2013
  • Maple
    seq(add(2^(2*n-k)*binomial(n,k)/2,k=1..n),n=0..20); # Zerinvary Lajos, Apr 18 2009
  • Mathematica
    CoefficientList[Series[x / ((1 - 4 x) (1 - 6 x)), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 07 2013 *)
    LinearRecurrence[{10, -24}, {0, 1}, 21] (* Michael De Vlieger, Jul 16 2017 *)

Formula

a(n) = 10*a(n-1) - 24*a(n-2) with n>1, a(0)=0, a(1)=1.
G.f.: x/((1-4*x)*(1-6*x)).
a(n) = 6^n/2 - 4^n/2.
E.g.f.: exp(4*x)*(exp(2*x) - 1)/2. - Stefano Spezia, Jul 23 2024

A081202 8th binomial transform of (0,1,0,1,0,1,....), A000035.

Original entry on oeis.org

0, 1, 16, 193, 2080, 21121, 206896, 1979713, 18640960, 173533441, 1602154576, 14701866433, 134294124640, 1222488408961, 11099284691056, 100571785292353, 909893629141120, 8222275592839681, 74233110849544336, 669726411243809473, 6038936596379658400, 54430221633714537601
Offset: 0

Views

Author

Paul Barry, Mar 11 2003

Keywords

Comments

Binomial transform of A081201.
From Wolfdieter Lang, Jul 17 2017: (Start)
For a combinatorial interpretation of a(n) with special 9-letter words of length n see the comment in A081200 on the 7-letter analog.
The binomial transform of {a(n)}_{n >=0} is A081203, the 10-letter analog.
(End)

Crossrefs

Apart from offset same as A016178.

Programs

Formula

a(n) = 16*a(n-1) - 63*a(n-2), a(0)=0, a(1)=1.
G.f.: x/((1-7*x)*(1-9*x)).
a(n) = (9^n - 7^n)/2.
E.g.f.: exp(7*x)*(exp(2*x) - 1)/2. - Stefano Spezia, Jul 23 2024

A081203 9th binomial transform of (0,1,0,1,0,1,.....), A000035.

Original entry on oeis.org

0, 1, 18, 244, 2952, 33616, 368928, 3951424, 41611392, 432891136, 4463129088, 45705032704, 465640261632, 4725122093056, 47800976744448, 482407813955584, 4859262511644672, 48874100093157376, 490992800745259008
Offset: 0

Views

Author

Paul Barry, Mar 11 2003

Keywords

Comments

Binomial transform of A081202.
From Wolfdieter Lang, Jul 17 2017: (Start)
For a combinatorial interpretation of a(n) with special 10-letter words of length n see the comment in A081200 on the 7-letter analog.
The binomial transform of {a(n)}_{n >= 0} is {0, A016190}, the 11-letter analog.
(End)

Crossrefs

Apart from the first term, identical to A016186.

Programs

  • Magma
    [10^n/2 - 8^n/2: n in [0..25]]; // Vincenzo Librandi, Aug 07 2013
  • Mathematica
    CoefficientList[Series[x / ((1 - 8 x) (1 - 10 x)), {x, 0, 20}], x] (* Vincenzo Librandi, Aug 07 2013 *)
    LinearRecurrence[{18,-80},{0,1},20] (* Harvey P. Dale, Aug 05 2018 *)

Formula

a(n) = 18*a(n-1)-80*a(n-2), a(0)=0, a(1)=1.
G.f.: x/((1-8*x)*(1-10*x)).
a(n) = 10^n/2 - 8^n/2.

A162590 Polynomials with e.g.f. exp(x*t)/csch(t), triangle of coefficients read by rows.

Original entry on oeis.org

0, 1, 0, 0, 2, 0, 1, 0, 3, 0, 0, 4, 0, 4, 0, 1, 0, 10, 0, 5, 0, 0, 6, 0, 20, 0, 6, 0, 1, 0, 21, 0, 35, 0, 7, 0, 0, 8, 0, 56, 0, 56, 0, 8, 0, 1, 0, 36, 0, 126, 0, 84, 0, 9, 0, 0, 10, 0, 120, 0, 252, 0, 120, 0, 10, 0, 1, 0, 55, 0, 330, 0, 462, 0, 165, 0, 11, 0, 0, 12, 0, 220, 0, 792, 0, 792, 0
Offset: 0

Views

Author

Peter Luschny, Jul 07 2009

Keywords

Comments

Comment from Peter Bala (Dec 06 2011): "Let P denote Pascal's triangle A070318 and put M = 1/2*(P-P^-1). M is A162590 (see also A131047). Then the first column of (I-t*M)^-1 (apart from the initial 1) lists the row polynomials for" A196776(n,k), which gives the number of ordered partitions of an n set into k odd-sized blocks. - Peter Luschny, Dec 06 2011
The n-th row of the triangle is formed by multiplying by 2^(n-1) the elements of the first row of the limit as k approaches infinity of the stochastic matrix P^(2k-1) where P is the stochastic matrix associated with the Ehrenfest model with n balls. The elements of a stochastic matrix P give the probability of arriving in a state j given the previous state i. In particular the sum of every row of the matrix must be 1, and so the sum of the terms in the n-th row of this triangle is 2^(n-1). Furthermore, by the properties of Markov chains, we can interpret P^(2k) as the (2k)-step transition matrix of the Ehrenfest model and its limit exists and it is again a stochastic matrix. The rows of the triangle divided by 2^(n-1) are the even rows (second, fourth, ...) and the odd rows (first, third, ...) of the limit matrix P^(2k). - Luca Onnis, Oct 29 2023

Examples

			Triangle begins:
  0
  1,  0
  0,  2,  0
  1,  0,  3,  0
  0,  4,  0,  4,  0
  1,  0, 10,  0,  5,  0
  0,  6,  0, 20,  0,  6,  0
  1,  0, 21,  0, 35,  0,  7,  0
  ...
  p[0](x) = 0;
  p[1](x) = 1
  p[2](x) = 2*x
  p[3](x) = 3*x^2 +  1
  p[4](x) = 4*x^3 +  4*x
  p[5](x) = 5*x^4 + 10*x^2 +  1
  p[6](x) = 6*x^5 + 20*x^3 +  6*x
  p[7](x) = 7*x^6 + 35*x^4 + 21*x^2 + 1
  p[8](x) = 8*x^7 + 56*x^5 + 56*x^3 + 8*x
.
Cf. the triangle of odd-numbered terms in rows of Pascal's triangle (A034867).
p[n] (k), n=0,1,...
k=0:  0, 1,  0,   1,    0,     1, ... A000035, (A059841)
k=1:  0, 1,  2,   4,    8,    16, ... A131577, (A000079)
k=2:  0, 1,  4,  13,   40,   121, ... A003462
k=3:  0, 1,  6,  28,  120,   496, ... A006516
k=4:  0, 1,  8,  49,  272,  1441, ... A005059
k=5:  0, 1, 10,  76,  520,  3376, ... A081199, (A016149)
k=6:  0, 1, 12, 109,  888,  6841, ... A081200, (A016161)
k=7:  0, 1, 14, 148, 1400, 12496, ... A081201, (A016170)
k=8:  0, 1, 16, 193, 2080, 21121, ... A081202, (A016178)
k=9:  0, 1, 18, 244, 2952, 33616, ... A081203, (A016186)
k=10: 0, 1, 20, 301, 4040, 51001, ... ......., (A016190)
.
p[n] (k), k=0,1,...
p[0]: 0,  0,   0,    0,    0,     0, ... A000004
p[1]: 1,  1,   1,    1,    1,     1, ... A000012
p[2]: 0,  2,   4,    6,    8,    10, ... A005843
p[3]: 1,  4,  13,   28,   49,    76, ... A056107
p[4]: 0,  8,  40,  120,  272,   520, ... A105374
p[5]: 1, 16, 121,  496, 1441,  3376, ...
p[6]: 0, 32, 364, 2016, 7448, 21280, ...
		

Crossrefs

Cf. A119467.

Programs

  • Maple
    # Polynomials: p_n(x)
    p := proc(n,x) local k;
    pow := (n,k) -> `if`(n=0 and k=0,1,n^k);
    add((k mod 2)*binomial(n,k)*pow(x,n-k),k=0..n) end;
    # Coefficients: a(n)
    seq(print(seq(coeff(i!*coeff(series(exp(x*t)/csch(t), t,16),t,i),x,n), n=0..i)), i=0..8);
  • Mathematica
    p[n_, x_] := Sum[Binomial[n, 2*k-1]*x^(n-2*k+1), {k, 0, n+2}]; row[n_] := CoefficientList[p[n, x], x] // Append[#, 0]&; Table[row[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Jun 28 2013 *)
    n = 15; "n-th row"
    mat = Table[Table[0, {j, 1, n + 1}], {i, 1, n + 1}];
    mat[[1, 2]] = 1;
    mat[[n + 1, n]] = 1;
    For[i = 2, i <= n, i++, mat[[i, i - 1]] = (i - 1)/n ];
    For[i = 2, i <= n, i++, mat[[i, i + 1]] = (n - i + 1)/n];
    mat // MatrixForm;
    P2 = Dot[mat, mat];
    R1 = Simplify[
      Eigenvectors[Transpose[P2]][[1]]/
       Total[Eigenvectors[Transpose[P2]][[1]]]]
    R2 = Table[Dot[R1, Transpose[mat][[k]]], {k, 1, n + 1}]
    even = R1*2^(n - 1) (* Luca Onnis, Oct 29 2023 *)

Formula

p_n(x) = Sum_{k=0..n} (k mod 2)*binomial(n,k)*x^(n-k).
E.g.f.: exp(x*t)/csch(t) = 0*(t^0/0!) + 1*(t^1/1!) + (2*x)*(t^2/2!) + (3*x^2+1)*(t^3/3!) + ...
The 'co'-polynomials with generating function exp(x*t)*sech(t) are the Swiss-Knife polynomials (A153641).

A121213 a(n) = 7^n - 5^n.

Original entry on oeis.org

0, 2, 24, 218, 1776, 13682, 102024, 745418, 5374176, 38400482, 272709624, 1928498618, 13597146576, 95668307282, 672119557224, 4717043931818, 33080342678976, 231867574534082, 1624598900644824, 11379821699045018
Offset: 0

Views

Author

Mohammad K. Azarian, Aug 19 2006

Keywords

Crossrefs

Programs

Formula

a(n) = 12*a(n-1) - 35*a(n-2) with a(0)=0, a(1)=2. - Vincenzo Librandi, Jul 21 2010
a(n) = 2*A081200(n). - Reinhard Zumkeller, Aug 01 2010
G.f.: 2*x/((5*x-1)*(7*x-1)). - Colin Barker, Nov 05 2012
E.g.f.: 2*exp(6*x)*sinh(x). - Elmo R. Oliveira, Mar 31 2025

A327316 Triangular array read by rows: row n shows the coefficients of this polynomial of degree n: p(x,n) = ((x+r)^n - (x+s)^n)/(r - s), where r = 3 and s = 2.

Original entry on oeis.org

1, 5, 2, 19, 15, 3, 65, 76, 30, 4, 211, 325, 190, 50, 5, 665, 1266, 975, 380, 75, 6, 2059, 4655, 4431, 2275, 665, 105, 7, 6305, 16472, 18620, 11816, 4550, 1064, 140, 8, 19171, 56745, 74124, 55860, 26586, 8190, 1596, 180, 9, 58025, 191710, 283725, 247080
Offset: 1

Views

Author

Clark Kimberling, Nov 01 2019

Keywords

Comments

For every choice of integers r and s, the polynomials p(n,x) form a strong divisibility sequence. Thus, if r, s, and x are integers, then p(x,n) is a strong divisibility sequence. That is, gcd(p(x,h),p(x,k)) = p(x,gcd(h,k)).

Examples

			First seven rows:
     1
     5      2
    19     15     3
    65     76    30     4
   211    325   190    50    5
   665   1266   975   380   75    6
  2059   4655  4431  2275  665  105   7
		

Crossrefs

Cf. A001047 (x=0), A005061 (x=1), A005060 (x=2), A005062 (x=3), A081200 (x=1/2).

Programs

  • Mathematica
    f[x_, n_] := ((x + r)^n - (x + s)^n)/(r - s);
    r = 3; s = 2;
    Column[Table[Expand[f[x, n]], {n, 1, 5}]]
    c[x_, n_] := CoefficientList[Expand[f[x, n]], x]
    TableForm[Table[c[x, n], {n, 1, 10}]] (* A327316 array *)
    Flatten[Table[c[x, n], {n, 1, 12}]]   (* A327316 sequence *)

A102728 Array read by antidiagonals: T(n, k) = ((n+1)^k-(n-1)^k)/2.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 1, 2, 1, 0, 1, 4, 4, 0, 0, 1, 6, 13, 8, 1, 0, 1, 8, 28, 40, 16, 0, 0, 1, 10, 49, 120, 121, 32, 1, 0, 1, 12, 76, 272, 496, 364, 64, 0, 0, 1, 14, 109, 520, 1441, 2016, 1093, 128, 1, 0, 1, 16, 148, 888, 3376, 7448, 8128, 3280, 256, 0, 0, 1, 18, 193, 1400, 6841
Offset: 0

Views

Author

Lambert Klasen (lambert.klasen(AT)gmx.net) and Gary W. Adamson, Feb 07 2005

Keywords

Comments

Consider a 2 X 2 matrix M = [N, 1] / [1, N]. The n-th row of the array contains the values of the non-diagonal elements of M^k, k=0,1,.... (Corresponding diagonal entry = non-diagonal entry + (N-1)^k.) Table:
N: row sequence g.f. cross references.
0: (1^n-(-1)^n)/2 x/((1+1x)(1-1x)) A000035
1: (2^n-0^n)/2 x/(1-2x) A000079
2: (3^n-1^n)/2 x/((1-1x)(1-3x)) A003462
3: (4^n-2^n)/2 x/((1-2x)(1-4x)) A006516
4: (7^n-3^n)/2 x/((1-3x)(1-5x)) A005059
5: (6^n-4^n)/2 x/((1-4x)(1-6x)) A016149
6: (7^n-5^n)/2 x/((1-5x)(1-7x)) A016161 A081200
7: (8^n-6^n)/2 x/((1-6x)(1-8x)) A016170 A081201
8: (9^n-7^n)/2 x/((1-7x)(1-9x)) A016178 A081202
9: (10^n-8^n)/2 x/((1-8x)(1-10x)) A016186 A081203
10: (11^n-9^n)/2 x/((1-9x)(1-11x)) A016190
11: (12^n-10^n)/2 x/((1-10x)(1-12x)) A016196
...
Characteristic polynomial x^2-2nx+(n^2-1) has roots n+-1, so if r(n) denotes a row sequence, r(n+1)/r(n) converges to n+1.
Columns follow polynomials with certain binomial coefficients:
column: polynomial
0: 0
1: 1
2: 2n
3: 3n^2+ 1 (see A056107)
4: 4n^3+ 4n (= 8*A006003(n))
5: 5n^4+ 10n^2+ 1
6: 6n^5+ 20n^3+ 6n
7: 7n^6+ 35n^4+ 21n^2+ 1
8; 8n^7+ 56n^5+ 56n^3+ 8n
9: 9n^8+ 84n^6+126n^4+ 36n^2+ 1
10: 10n^9+ 120n^7+252n^5+120n^3+ 10n
11: 11n^10+165n^8+462n^6+330n^4+ 55n^2+ 1

Examples

			Array begins:
0,1,0,1,0,1...
0,1,2,4,8,16...
0,1,4,13,40,121...
0,1,6,28,120,496...
0,1,8,49,272,1441...
...
		

Programs

  • PARI
    MM(n,N)=local(M);M=matrix(n,n);for(i=1,n, for(j=1,n,if(i==j,M[i,j]=N,M[i,j]=1)));M for(k=0,12, for(i=0,k,print1((MM(2,k-i)^i)[1,2],","))) T(n, k) = ((n+1)^k-(n-1)^k)/2 for(k=0,10, for(i=0,10,print1(T(k,i),","));print()) for(k=0,10, for(i=0,10,print1(((k+1)^i-(k-1)^i)/2,","));print()) for(k=0,10, for(i=0,10,print1(polcoeff(x/((1-(k-1)*x)*(1-(k+1)*x)),i),","));print())

A105373 Square array by antidiagonals of number of straight lines with n points in a k-dimensional hypercube with n points on each edge.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 28, 8, 1, 1, 120, 49, 10, 1, 1, 496, 272, 76, 12, 1, 1, 2016, 1441, 520, 109, 14, 1, 1, 8128, 7448, 3376, 888, 148, 16, 1, 1, 32640, 37969, 21280, 6841, 1400, 193, 18, 1, 1, 130816, 192032, 131776, 51012, 12496, 2080, 244, 20, 1, 1, 523776
Offset: 1

Views

Author

Henry Bottomley, Apr 02 2005

Keywords

Examples

			Rows start:
  1,  1,   1,   1,    1,     1, ...;
  1,  6,  28, 120,  496,  2016, ...;
  1,  8,  49, 272, 1441,  7448, ...;
  1, 10,  76, 520, 3376, 21280, ...;
  1, 12, 109, 888, 6841, 51012, ...;
  etc.
T(5,3)=109 because in a 5 X 5 X 5 cube there are 25 columns, 25 linear rows in one direction, 25 linear rows in another direction, 5 short diagonals in each of 6 directions and 4 long diagonals; and 3*25 + 6*5 + 4 = 109.
		

Crossrefs

See A102728. Rows essentially include A000012, A006516, A005059, A016149 or A081199, A016161 or A081200, A016170 or A081201, A016178 or A081202 etc. Columns essentially include A000012, A005843, A056107, A105373.

Formula

T(1, k)=1. For n>1: T(n, k) = ((n+2)^k-n^k)/2 = (n+2)*T(n, k-1)+n^(k-1) = A102728(k, n+1).
Showing 1-10 of 11 results. Next