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.

Previous Showing 11-17 of 17 results.

A104698 Triangle read by rows: T(n,k) = Sum_{j=0..n-k} binomial(k, j)*binomial(n-j+1, k+1).

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 9, 6, 1, 5, 16, 19, 8, 1, 6, 25, 44, 33, 10, 1, 7, 36, 85, 96, 51, 12, 1, 8, 49, 146, 225, 180, 73, 14, 1, 9, 64, 231, 456, 501, 304, 99, 16, 1, 10, 81, 344, 833, 1182, 985, 476, 129, 18, 1, 11, 100, 489, 1408, 2471, 2668, 1765, 704, 163, 20, 1, 12
Offset: 0

Views

Author

Gary W. Adamson, Mar 19 2005

Keywords

Comments

The n-th column of the triangle is the binomial transform of the n-th row of A081277, followed by zeros. Example: column 3, (1, 6, 19, 44, ...) = binomial transform of row 3 of A081277: (1, 5, 8, 4, 0, 0, 0, ...). A104698 = reversal by rows of A142978. - Gary W. Adamson, Jul 17 2008
This sequence is jointly generated with A210222 as an array of coefficients of polynomials u(n,x): initially, u(1,x)=v(1,x)=1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) + 1 and v(n,x) = 2x*u(n-1,x) + v(n-1,x) + 1. See the Mathematica section at A210222. - Clark Kimberling, Mar 19 2012
This Riordan triangle T appears in a formula for A001100(n, 0) = A002464(n), for n >= 1. - Wolfdieter Lang, May 13 2025

Examples

			The Riordan triangle T begins:
  n\k  0   1   2    3    4    5    6   7   8  9 10 ...
  ----------------------------------------------------
  0:   1
  1:   2   1
  2:   3   4   1
  3:   4   9   6    1
  4:   5  16  19    8    1
  5:   6  25  44   33   10    1
  6:   7  36  85   96   51   12    1
  7:   8  49 146  225  180   73   14   1
  8:   9  64 231  456  501  304   99  16   1
  9:  10  81 344  833 1182  985  476 129  18  1
  10: 11 100 489 1408 2471 2668 1765 704 163 20  1
  ... reformatted and extended by _Wolfdieter Lang_, May 13 2025
From _Wolfdieter Lang_, May 13 2025: (Start)
Zumkeller recurrence (adapted for offset [0,0]): 19 = T(4, 2) = T(2, 1) + T(3, 1) + T(3,3) = 4 + 9 + 6 = 19.
A-sequence recurrence: 19 = T(4, 2) = 1*T(3. 1) + 2*T(3. 2) - 2*T(3, 3) = 9 + 12 - 2 = 19.
Z-sequence recurrence: 5 = T(4, 0) = 2*T(3, 0) - 1*T(3, 1) + 2*T(3, 2) - 6*T(3, 3) = 8 - 9 + 12 + 6 = 5.
Boas-Buck recurrence: 19 = T(4, 2) = (1/2)*((2 + 0)*T(2, 2) + (2 + 2*2)*T(3, 2)) = (1/2)*(2 + 36) = 19. (End)
		

Crossrefs

Diagonal sums are A008937(n+1).
Cf. A048739 (row sums), A008288, A005900 (column 3), A014820 (column 4)
Cf. A081277, A142978 by antidiagonals, A119328, A110271 (matrix inverse).

Programs

  • Haskell
    a104698 n k = a104698_tabl !! (n-1) !! (k-1)
    a104698_row n = a104698_tabl !! (n-1)
    a104698_tabl = [1] : [2,1] : f [1] [2,1] where
       f us vs = ws : f vs ws where
         ws = zipWith (+) ([0] ++ us ++ [0]) $
              zipWith (+) ([1] ++ vs) (vs ++ [0])
    -- Reinhard Zumkeller, Jul 17 2015
  • Maple
    A104698 := proc(n, k) add(binomial(k, j)*binomial(n-j+1, n-k-j), j=0..n-k) ; end proc:
    seq(seq(A104698(n, k), k=0..n), n=0..15); # R. J. Mathar, Sep 04 2011
    T := (n, k) -> binomial(n + 1, k + 1)*hypergeom([-k, k - n], [-n - 1], -1):
    for n from 0 to 9 do seq(simplify(T(n, k)), k = 0..n) od;
    T := proc(n, k) option remember; if k = 0 then n + 1 elif k = n then 1 else T(n-2, k-1) + T(n-1, k-1) + T(n-1, k) fi end: # Peter Luschny, May 13 2025
  • Mathematica
    u[1, ] = 1; v[1, ] = 1;
    u[n_, x_] := u[n, x] = x u[n-1, x] + v[n-1, x] + 1;
    v[n_, x_] := v[n, x] = 2 x u[n-1, x] + v[n-1, x] + 1;
    Table[CoefficientList[u[n, x], x], {n, 1, 11}] // Flatten (* Jean-François Alcover, Mar 10 2019, after Clark Kimberling *)
  • PARI
    T(n,k)=sum(j=0,n-k,binomial(k,j)*binomial(n-j+1,k+1)) \\ Charles R Greathouse IV, Jan 16 2012
    

Formula

The triangle is extracted from the product A * B; A = [1; 1, 1; 1, 1, 1; ...], B = [1; 1, 1; 1, 3, 1; 1, 5, 5, 1; ...] both infinite lower triangular matrices (rest of the terms are zeros). The triangle of matrix B by rows = A008288, Delannoy numbers.
From Paul Barry, Jul 18 2005: (Start)
Riordan array (1/(1-x)^2, x(1+x)/(1-x)) = (1/(1-x), x)*(1/(1-x), x(1+x)/(1-x)).
T(n, k) = Sum_{j=0..n} Sum_{i=0..j-k} C(j-k, i)*C(k, i)*2^i.
T(n, k) = Sum_{j=0..k} Sum_{i=0..n-k-j} (n-k-j-i+1)*C(k, j)*C(k+i-1, i). (End)
T(n, k) = binomial(n+1, k+1)*2F1([-k, k-n], [-n-1], -1) where 2F1 is a Gaussian hypergeometric function. - R. J. Mathar, Sep 04 2011
T(n, k) = T(n-2, k-1) + T(n-1, k-1) + T(n-1, k) for 1 < k < n; T(n, 0) = n + 1; T(n, n) = 1. - Reinhard Zumkeller, Jul 17 2015
From Wolfdieter Lang, May 13 2025: (Start)
The Riordan triangle T = (1/(1 - x)^2, x*(1 + x)/(1 - x)) has the o.g.f. G(x, y) = 1/((1 - x)*(1 - x - y*x*(1+x))) for the row polynomials R(n, y) = Sum_{k=0..n} T(n, k)*y^k.
The o.g.f. for column k is G(k, x) = (1/(1 - x)^2)*(x*(1 + x)/(1 - x))^k, for k >= 0.
The o.g.f. for the diagonal m is D(m, x) = N(m, x)/(1 - x)^(m+1), with the numerator polynomial N(m, x) = Sum_{k=0..floor(m/2)} A034867(m, k)*x^(2*k) for m >= 0.
The row sums with o.g.f. R(x) = 1/((1 -x)*(1 - 2*x -x^2) give A048739.
The alternating row sums with o.g.f. 1/((1 - x)(1 + x^2)) give A133872.
The A-sequence for this Riordan triangle has o.g.f. A(x) = 1 + x + sqrt(1 + 6*x + x^2))/2 giving A112478(n). Hence T(n, k) = Sum_{j=0..n-k} A112478(j)*T(n-1, k-1+j), for n >= 1, k >= 1, T(n, k) = 0 for n < k, and T(0, 0) = 1.
The Z-sequence has o.g.f. (5 + x - sqrt(1 + 6*x + x^2))/2 = 3 + x - A(x) giving Z(n) = {2, -1, -A112478(n >= 2)}. Hence T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1. For A- and Z-sequences of Riordan triangles see a W. Lang link at A006232 with references.
The Boas-Buck sequences alpha and beta for the Riordan triangle T (see A046521 for the Aug 10 2017 comment and reference) are alpha(n) = A040000(n+1) = repeat{2} and beta(n) = A010673(n+1) = repeat{2,0}. Hence the recurrence for column T(n, k){n>=k}, with input T(k, k) = 1, for k >= 0, is T(n, k) = (1/(n-k)) * Sum{j=k..n-1} (2 + k*(1 + (-1)^(n-1-j))) *T(j,k), for n >= k+1. (End)

A217873 a(n) = 4*n*(n^2 + 2)/3.

Original entry on oeis.org

0, 4, 16, 44, 96, 180, 304, 476, 704, 996, 1360, 1804, 2336, 2964, 3696, 4540, 5504, 6596, 7824, 9196, 10720, 12404, 14256, 16284, 18496, 20900, 23504, 26316, 29344, 32596, 36080, 39804, 43776, 48004, 52496, 57260, 62304, 67636, 73264, 79196, 85440, 92004
Offset: 0

Views

Author

M. F. Hasler, Oct 13 2012

Keywords

Comments

Occurs as 4th column in the table A142978 of figurate numbers for n-dimensional cross polytope.

Crossrefs

Programs

  • Magma
    [4*n*(n^2+2)/3: n in [0..45]]; // Vincenzo Librandi, Nov 08 2012
  • Mathematica
    Table[4n(n^2 + 2)/3, {n, 0, 39}] (* Alonso del Arte, Oct 22 2012 *)
    LinearRecurrence[{4,-6,4,-1},{0,4,16,44},50] (* Harvey P. Dale, Mar 16 2015 *)
  • Maxima
    makelist(4*n*(n^2+2)/3, n, 0, 41); /* Martin Ettl, Oct 15 2012 */
    
  • PARI
    a(n)=(n^2+2)*n/3*4
    

Formula

a(n) = 4*A006527(n).
From Peter Luschny, Oct 14 2012: (Start)
a(n) = A008412(n)/2.
a(n) = A174794(n+1) - 1.
First differences are in A112087.
Second differences are in A008590 and A022144.
Binomial transformation of (a(n), n > 0) is A082138. (End)
G.f.: 4*x*(1 + x^2)/(x - 1)^4. - R. J. Mathar, Oct 15 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), a(0)=0, a(1)=4, a(2)=16, a(3)=44. - Harvey P. Dale, Mar 16 2015
From Elmo R. Oliveira, Aug 09 2025: (Start)
E.g.f.: 4*exp(x)*x*(3 + 3*x + x^2)/3.
a(n) = A292022(n)/3. (End)

A072410 Number of iterations of the map k -> A000001(k) needed to reach 1 starting at n, or -1 if no such number exists.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 3, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 2, 3, 1, 3, 1, 3, 2, 2, 1, 3, 2, 2, 1, 2, 1, 2, 2, 2, 2, 2, 1, 2, 1, 2, 3, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 2, 3, 1, 3, 1, 3, 2, 2, 1, 2, 1, 2, 1, 3, 1, 3, 1, 3, 2, 2, 1, 3, 1, 2, 2, 4, 1, 3, 1, 3, 2
Offset: 1

Views

Author

N. J. A. Sloane, Oct 03 2008

Keywords

Comments

The old entry with this sequence number was a duplicate of A052409.
It appears that a(n) is the number of times n appears in A142978, excluding the first column of infinitely many 1's. - Ron Wolf, Dec 16 2020
Preceding comment is incorrect. The first counterexample is a(19) = 1, whereas 19 appears twice in A142978. - Eric M. Schmidt, Mar 22 2021

Examples

			Conway et al. remark that every number less than 2048 reaches 1 after at most 5 steps and give the following examples:
672 -> 1280 -> 1116461 -> 1
1024 -> 49487367289 -> 1
720 -> 840 -> 186 -> 6 -> 2 -> 1
320 -> 1640 -> 68 -> 5 -> 1
384 -> 20169 -> 67 -> 1
128 -> 2328 -> 64 -> 267 -> 1
960 -> 11394 -> 60 -> 13 -> 1
864 -> 4725 -> 51 -> 1
1344 -> 11720 -> 49 -> 2 -> 1
1440 -> 5958 -> 16 -> 14 -> 2 -> 1
1248 -> 1460 -> 15 -> 1
256 -> 56092 -> 11 -> 1
1728 -> 47937 -> 6 -> 2 -> 1
512 -> 10494213 -> 5 -> 1
1536 -> 408641062 -> 4 -> 2 -> 1
1664 -> 21507 -> 2 -> 1
1280 -> 1116461 -> 1
		

Crossrefs

Cf. A000001, A066952 (indices of records).

A142977 Table of coefficients in the expansion of the rational function 1/{(1-x)^2 - y*(1+x)^2}.

Original entry on oeis.org

1, 1, 2, 1, 6, 3, 1, 10, 19, 4, 1, 14, 51, 44, 5, 1, 18, 99, 180, 85, 6, 1, 22, 163, 476, 501, 146, 7, 1, 26, 243, 996, 1765, 1182, 231, 8, 1, 30, 339, 1804, 4645, 5418, 2471, 344, 9, 1, 34, 451, 2964, 10165, 17718, 14407, 4712, 489, 10
Offset: 0

Views

Author

Peter Bala, Jul 15 2008

Keywords

Comments

The row entries are the figurate numbers of the odd dimensional cross polytopes. See A142978 for the complete table of figurate numbers of n-dimensional cross polytopes. The rows are the partial sums of the even-numbered rows of the square array of Delannoy numbers A008288.

Examples

			The square array begins
 n\k| 0...1....2.....3.....4.......5
------------------------------------
 .0.| 1...2....3.....4......5......6 ... A000027
 .1.| 1...6...19....44.....85....146 ... A005900
 .2.| 1..10...51...180....501...1182 ... A069038
 .3.| 1..14...99...476...1765...5418 ... A099193
 .4.| 1..18..163...996...4645..17718 ... A099196
 .5.| 1..22..243..1804..10165..46530 ... A300624
 ...
		

Crossrefs

Cf. A005900 (row 1), A008288, A069038 (row 2), A099193 (row 3), A099196 (row 4), A300624 (row 5), A142978, A142983.

Programs

  • Maple
    with(combinat): T:=(n,k) -> add(binomial(2n,k-j)*binomial(2n+j+1,j), j = 0..k): for n from 0 to 9 do seq(T(n,k), k = 0..9) end do;

Formula

T(n,k) = Sum_{j = 0..k} C(2*n, k-j)*C(2*n+j+1, j).
O.g.f.: 1/{(1 - x)^2 - y*(1 + x)^2} = Sum_{n, k >= 0} T(n,k)*x^k*y^n = 1/(1 - y) * Sum_{m >= 0} U(m, (1 + y)/(1 - y))*x^m, where U(m, y) denotes the m-th Chebyshev polynomial of the second kind.
O.g.f. row n: (1 + x)^(2*n)/(1 - x)^(2*n+2).
O.g.f. column k: 1/(1 - y)*U(k, (1 + y)/(1 - y)).
The entries in the n-th row appear in the series acceleration formula for the constant log(2): Sum_{k >= 1} (-1)^(k+1)/(T(n,k)*T(n,k+1)) = 1 + (4*n + 2)*( log(2) - (1 - 1/2 + 1/3 - ... + 1/(2*n + 1)) ).
For example, n = 1 gives log(2) = 4/6 + (1/6)*( 1/(1*6) - 1/(6*19) + 1/(19*44) - 1/(44*85) + ... ). See A142983 for further details.

Extensions

Restored missing program. - Peter Bala, Oct 02 2008

A361745 Square array of circular Delannoy numbers A(i,j) (i >= 0, j >= 0) read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 6, 16, 6, 1, 1, 8, 36, 36, 8, 1, 1, 10, 64, 114, 64, 10, 1, 1, 12, 100, 264, 264, 100, 12, 1, 1, 14, 144, 510, 768, 510, 144, 14, 1, 1, 16, 196, 876, 1800, 1800, 876, 196, 16, 1, 1, 18, 256, 1386, 3648, 5010, 3648, 1386, 256, 18, 1
Offset: 0

Views

Author

Noah Snyder, Mar 22 2023

Keywords

Comments

An (n,m) Delannoy loop is an oriented unbased loop on a toroidal grid with points labeled by Z/n x Z/m composed of steps of the form (1,0), (0,1), and (1,1), and which loops around the torus exactly once in each of the x-direction and the y-direction. The circular Delannoy numbers count the number of (n,m) Delannoy loops. This array is a modification of the ordinary Delannoy numbers A008288.
Dimensions of hom spaces Hom(S^{{i}}, S^{{j}}) in the circular Delannoy category attached to the oligomorphic group of order preserving self-bijections of the circle.

Examples

			The square array A(n,m) (n >= 0, m >= 0) begins:
  1, 1,  1,   1,   1,    1,    1,    1,     1,     1, ...
  1, 2,  4,   6,   8,   10,   12,   14,    16,    18, ...
  1, 4, 16,  36,  64,  100,  144,  196,   256,   324, ...
  1, 6, 36, 114, 264,  510,  876, 1386,  2064,  2934, ...
  1, 8, 64, 264, 768, 1800, 3648, 6664, 11264, 17928, ...
.
The triangle T(n,m) (0 <= m <= n) begins:
  [0] 1;
  [1] 1,  1;
  [2] 1,  2,   1;
  [3] 1,  4,   4,   1;
  [4] 1,  6,  16,   6,    1;
  [5] 1,  8,  36,  36,    8,    1;
  [6] 1, 10,  64, 114,   64,   10,   1;
  [7] 1, 12, 100, 264,  264,  100,  12,   1;
  [8] 1, 14, 144, 510,  768,  510, 144,  14,  1;
  [9] 1, 16, 196, 876, 1800, 1800, 876, 196, 16, 1;
		

Crossrefs

Circular analog of A008288.
Main diagonal: A361743.
Row sums: A361758.

Programs

  • Maple
    A := (n, k) -> `if`(n*k=0, 1, 2*n*k*hypergeom([1 - n, 1 - k], [2], 2)):
    seq(print(seq(simplify(A(n, k)), k = 0..9)), n=0..4); # Peter Luschny, Mar 23 2023
  • Mathematica
    a[n_Integer?Positive, m_Integer?Positive] := Sum[k Binomial[n, k] Binomial[m, k] 2^k, {k, 1, Min[n,m]}]
  • Python
    from math import comb
    def A361745_A(n,m): # compute square array A(n,m)
        return 1 if not(m and n) else sum(comb(n-1,i)*comb(m+i,n) for i in range(max(n-m,0),n))*n<<1 # Chai Wah Wu, Mar 23 2023

Formula

A(n,m) = A(m,n).
A(n,m) = Sum_{k=0..min(n,m)} binomial(n,k)*binomial(m,k)*k*2^k for n >= 1.
A(n,m) = n*(D(n,m-1) + D(n-1,m-1)) = n*(D(n,m) - D(n-1,m)) for n,m >= 1, where D(i,j) = A008288(i,j) are the Delannoy numbers.
G.f.: 2*x*y/(1-x-y-x*y)^2 (valid for n,m > 1).
For n,m >= 1, A(n,m) = 2*n*A142978(n,m).
A(n,m) = 2*n*m*hypergeom([1-n, 1-m], [2], 2) for n,m >= 1. - Peter Luschny, Mar 23 2023

A156136 A triangle of polynomial coefficients related to Mittag-Leffler polynomials: p(x,n)=Sum[Binomial[n, k]*Binomial[n - 1, n - k]*2^k*x^k, {k, 0, n}]/(2*x).

Original entry on oeis.org

1, 2, 2, 3, 12, 4, 4, 36, 48, 8, 5, 80, 240, 160, 16, 6, 150, 800, 1200, 480, 32, 7, 252, 2100, 5600, 5040, 1344, 64, 8, 392, 4704, 19600, 31360, 18816, 3584, 128, 9, 576, 9408, 56448, 141120, 150528, 64512, 9216, 256, 10, 810, 17280, 141120, 508032
Offset: 0

Views

Author

Roger L. Bagula, Feb 04 2009

Keywords

Examples

			1;
2, 2;
3, 12, 4;
4, 36, 48, 8;
5, 80, 240, 160, 16;
6, 150, 800, 1200, 480, 32;
7, 252, 2100, 5600, 5040, 1344, 64;
8, 392, 4704, 19600, 31360, 18816, 3584, 128;
9, 576, 9408, 56448, 141120, 150528, 64512, 9216, 256;
10, 810, 17280, 141120, 508032, 846720, 645120, 207360, 23040, 512;
		

References

  • Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), pp. 75-76

Crossrefs

A142983, A142978, A047781 (row sums).

Programs

  • Mathematica
    Clear[t0, p, x, n, m];
    p[x_, n_] = Sum[Binomial[n, k]*Binomial[n - 1, n - k]*2^k*x^k, {k, 0, n}]/(2*x);
    Table[FullSimplify[ExpandAll[p[x, n]]], {n, 1, 10}];
    Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 1, 10}];
    Flatten[%]

Formula

p(x,n)=Sum[Binomial[n, k]*Binomial[n - 1, n - k]*2^k*x^k, {k, 0, n}]/(2*x);
p(x,n)=n Hypergeometric2F1[1 - n, 1 - n, 2, 2 x];
t(n,m)=coefficiemts(p(x,n))
T(n,m) = 2^m*A103371(n,m). - R. J. Mathar, Dec 05 2017

A364429 a(0) = 1, a(n) = (2*n^5 + 20*n^3 + 23*n) * 2/15 for n>=1.

Original entry on oeis.org

1, 6, 36, 146, 456, 1182, 2668, 5418, 10128, 17718, 29364, 46530, 71000, 104910, 150780, 211546, 290592, 391782, 519492, 678642, 874728, 1113854, 1402764, 1748874, 2160304, 2645910, 3215316, 3878946, 4648056, 5534766, 6552092, 7713978, 9035328, 10532038, 12221028
Offset: 0

Views

Author

Steven Lu, Jul 24 2023

Keywords

Comments

a(n) is the 6th n-orthoplex (n-dimensional cross-polytope) number.

Examples

			a(3) = 146 since the 6th octahedral number is 146; A005900(6) = 146.
a(4) = 456 since the 6th 16-cell number is 456; A014820(5) = 456.
		

Crossrefs

Cf. A142978 (column 6 with an initial 1).

Programs

  • Mathematica
    Prepend[Table[2/15 (2 x^5 + 20 x^3 + 23 x), {x, 100}], 1]
  • Python
    print([1]+[(2*i**5+20*i**3+23*i)*2//15 for i in range(1,101)])

Formula

a(0) = 1, a(n) = (2*n^5 + 20*n^3 + 23*n) * 2/15 for n>=1.
G.f.: (1 + 15*x^2 + 15*x^4 + x^6)/(1 - x)^6. - Stefano Spezia, Jul 24 2023
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