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 1-4 of 4 results.

A154283 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(2*n+1,i) * binomial(k+2-i,2)^n, 0 <= k <= 2*(n-1).

Original entry on oeis.org

1, 1, 4, 1, 1, 20, 48, 20, 1, 1, 72, 603, 1168, 603, 72, 1, 1, 232, 5158, 27664, 47290, 27664, 5158, 232, 1, 1, 716, 37257, 450048, 1822014, 2864328, 1822014, 450048, 37257, 716, 1, 1, 2172, 247236, 6030140, 49258935, 163809288, 242384856, 163809288, 49258935, 6030140, 247236, 2172, 1
Offset: 1

Views

Author

Roger L. Bagula, Jan 06 2009

Keywords

Comments

From Yahia Kahloune, Jan 30 2014: (Start)
In general, let b(k,e,p) = Sum_{i=0..k} (-1)^i*binomial(e*p+1,i)*binomial(k+e-i,e)^p. Then T(n,k) = b(k,2,n).
With these coefficients we can calculate: Sum_{i=1..n} binomial(i+e-1,e)^p = Sum_{k=0..e*(p-1)} b(k,e,p)*binomial(n+e+k,e*p+k).
For example, A085438(n) = Sum_{i=1..n} binomial(1+i,2)^3 = T(3,0)*binomial(2+n,7) + T(3,1)*binomial(3+n,7) + T(3,2)*binomial(4+n,7) + T(3,3)*binomial(5+n,7) + T(3,4)*binomial(6+n,7) = (1/5040)*(90*n^7 + 630*n^6 + 1638*n^5 + 1890*n^4 + 840*n^3 - 48*n).
(End)
T(n,k) is the number of permutations of 2 indistinguishable copies of 1..n with exactly k descents. A descent is a pair of adjacent elements with the second element less than the first. - Andrew Howroyd, May 06 2020

Examples

			Triangle begins:
  1;
  1,     4,       1;
  1,    20,      48,        20,           1;
  1,    72,     603,      1168,         603,           72,           1;
  1,   232,    5158,     27664,       47290,        27664,        5158,  232, 1;
  1,   716,   37257,    450048,     1822014,      2864328,     1822014, ...;
  1,  2172,  247236,   6030140,    49258935,    163809288,   242384856, ...;
  1,  6544, 1568215,  72338144,  1086859301,   6727188848, 19323413187, ...;
  1, 19664, 9703890, 811888600, 21147576440, 225167210712, ... ;
  ...
The T(2,1) = 4 permutations of 1122 with 1 descent are 1212, 1221, 2112, 2211. - _Andrew Howroyd_, May 15 2020

Links

Crossrefs

Columns k=0..9 are A000012, A061981, A151624, A151625, A151626, A151627, A151628, A151629, A151630, A151631.
Row sums are A000680.
Similar triangles for e=1..6: A173018 (or A008292), this sequence, A174266, A236463, A237202, A237252.
Cf. A000680, A005799, A085438, A334781.

Programs

  • Magma
    [(&+[(-1)^j*Binomial(2*n+1,j)*Binomial(k-j+2,2)^n: j in [0..k]]): k in [0..2*n-2], n in [1..12]]; // G. C. Greubel, Jun 13 2022
  • Maple
    A154283 := proc(n,k)
        (1-x)^(2*n+1)*add( (l*(l+1)/2)^n*x^(l-1),l=0..k+1) ;
        coeftayl(%,x=0,k) ;
end proc: # R. J. Mathar, Feb 01 2013
  • Mathematica
    p[x_, n_]= (1-x)^(2*n+1)*Sum[(k*(k+1)/2)^n*x^k, {k, 0, Infinity}]/x;
Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n,10}]//Flatten
  • PARI
    T(n,k)={sum(i=0, k, (-1)^i*binomial(2*n+1, i)*binomial(k+2-i, 2)^n)} \\ Andrew Howroyd, May 09 2020
  • SageMath
    def A154283(n,k): return sum((-1)^j*binomial(2*n+1, j)*binomial(k-j+2, 2)^n for j in (0..k))
flatten([[A154283(n,k) for k in (0..2*n-2)] for n in (1..12)]) # G. C. Greubel, Jun 13 2022

Formula

T(n,k) = (-1) times coefficient of x^k in (x-1)^(2*n+1) * Sum_{k>=0} (k*(k+1)/2)^n *x^(k-1).
From Yahia Kahloune, Jan 29 2014: (Start)
Sum_{i=1..n} binomial(1+i,2)^p = Sum_{k=0..2*p-2} T(p,k)*binomial(n+2+k,2*p+1).
binomial(n,2)^p = Sum_{k=0..2*p-2} T(p,k)*binomial(n+k,2*p). (End)
From Peter Bala, Dec 21 2019: (Start)
E.g.f. as a continued fraction: (1-x)/(1-x + ( 1-exp((1-x)^2*t))*x/(1-x + (1-exp(2*(1-x)^2*t))*x/(1-x + (1-exp(3*(1-x)^2*t))*x/(1-x + ... )))) = 1 + x*t + x*(x^2 + 4*x + 1)*t^2/2! + x*(x^4 + 20*x^3 + 48*x^2 + 20*x + 1)*t^3/3! + ... (use Prodinger equation 1.1).
The sequence of alternating row sums (unsigned) [1, 1, 2, 10, 104, 1816,...] appears to be A005799. (End)

Extensions

Edited by N. J. A. Sloane, Jan 30 2014 following suggestions from Yahia Kahloune (among other things, the signs of all terms have been reversed).
Edited by Andrew Howroyd, May 09 2020

A151583 Number of permutations of 2 indistinguishable copies of 1..n arranged in a circle with exactly 2 adjacent element pairs in decreasing order.

Original entry on oeis.org

0, 2, 45, 260, 1115, 4230, 15113, 52232, 176823, 590090, 1948133, 6376716, 20725523, 66960782, 215232705, 688746512, 2195381615, 6973567506, 22082966429, 69735686420, 219667415499, 690383309462, 2165293110905, 6778308873240, 21182215233575, 66088511533850
Offset: 1

Views

Author

R. H. Hardin, May 21 2009

Keywords

Links

Crossrefs

With 3..8 descents: A151584, A151585, A151586, A151587, A151588, A151589.
With 3..7 copies of 1..n: A151590, A151597, A151603, A151607, A151610.
Cf. A151576, A151624.

Programs

  • PARI
    a(n) = if(n <= 1, 0, n*(3^n - 4*n)) \\ Andrew Howroyd, May 04 2020
  • PARI
    concat(0, Vec(x^2*(2 + 27*x - 85*x^2 + 33*x^3 - 9*x^4) / ((1 - x)^3*(1 - 3*x)^2) + O(x^30))) \\ Colin Barker, Jul 15 2020

Formula

a(n) = n*(3^n - 4*n) for n > 1. - Andrew Howroyd, May 04 2020
From Colin Barker, Jul 15 2020: (Start)
G.f.: x^2*(2 + 27*x - 85*x^2 + 33*x^3 - 9*x^4) / ((1 - x)^3*(1 - 3*x)^2).
a(n) = 9*a(n-1) - 30*a(n-2) + 46*a(n-3) - 33*a(n-4) + 9*a(n-5) for n>6.
(End)

Extensions

Terms a(13) and beyond from Andrew Howroyd, May 04 2020

A151656 Number of permutations of 7 indistinguishable copies of 1..n with exactly 2 adjacent element pairs in decreasing order.

Original entry on oeis.org

0, 441, 35623, 1561238, 59287158, 2165511047, 78259307721, 2820153607740, 101551366735840, 3656082204395957, 131621033827371963, 4738375497166097906, 170581677602043334746, 6140941779058210902771, 221073915991189916489149, 7958661078139728516094136
Offset: 1

Views

Author

R. H. Hardin, May 29 2009

Keywords

Links

Crossrefs

Cf. A151624.

Programs

  • PARI
    a(n) = {36^n - (7*n + 1)*8^n + 7*n*(7*n + 1)/2} \\ Andrew Howroyd, May 06 2020
  • PARI
    concat(0, Vec(49*x^2*(9 + 232*x - 932*x^2 - 1024*x^3) / ((1 - x)^3*(1 - 8*x)^2*(1 - 36*x)) + O(x^40))) \\ Colin Barker, Jul 18 2020

Formula

a(n) = 36^n - (7*n + 1)*8^n + 7*n*(7*n + 1)/2. - Andrew Howroyd, May 06 2020
From Colin Barker, Jul 18 2020: (Start)
G.f.: 49*x^2*(9 + 232*x - 932*x^2 - 1024*x^3) / ((1 - x)^3*(1 - 8*x)^2*(1 - 36*x)).
a(n) = 55*a(n-1) - 799*a(n-2) + 4381*a(n-3) - 8884*a(n-4) + 7552*a(n-5) - 2304*a(n-6) for n>6.
(End)

Extensions

Terms a(7) and beyond from Andrew Howroyd, May 06 2020

A151658 Number of permutations of 8 indistinguishable copies of 1..n with exactly 2 adjacent element pairs in decreasing order.

Original entry on oeis.org

0, 784, 73200, 3884640, 182107936, 8277726192, 373396825488, 16812327355840, 756652360885056, 34050346486482384, 1532275508306401840, 68952496159266606624, 3102863293076011859040, 139628857613659024861360, 6283298684030318768507856, 282748441663401954476011392
Offset: 1

Views

Author

R. H. Hardin, May 29 2009

Keywords

Links

Crossrefs

Cf. A151624.

Programs

  • PARI
    a(n) = {45^n - (8*n + 1)*9^n + 4*n*(8*n + 1)} \\ Andrew Howroyd, May 06 2020
  • PARI
    concat(0, Vec(16*x^2*(49 + 1341*x - 6093*x^2 - 6561*x^3) / ((1 - x)^3*(1 - 9*x)^2*(1 - 45*x)) + O(x^40))) \\ Colin Barker, Jul 19 2020

Formula

a(n) = 45^n - (8*n + 1)*9^n + 4*n*(8*n + 1). - Andrew Howroyd, May 06 2020
From Colin Barker, Jul 19 2020: (Start)
G.f.: 16*x^2*(49 + 1341*x - 6093*x^2 - 6561*x^3) / ((1 - x)^3*(1 - 9*x)^2*(1 - 45*x)).
a(n) = 66*a(n-1) - 1083*a(n-2) + 6508*a(n-3) - 13671*a(n-4) + 11826*a(n-5) - 3645*a(n-6) for n>6.
(End)

Extensions

Terms a(7) and beyond from Andrew Howroyd, May 06 2020
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.