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.

A107711 Triangle read by rows: T(0,0)=1, T(n,m) = binomial(n,m) * gcd(n,m)/n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 5, 10, 5, 1, 1, 1, 1, 3, 5, 5, 3, 1, 1, 1, 1, 7, 7, 35, 7, 7, 1, 1, 1, 1, 4, 28, 14, 14, 28, 4, 1, 1, 1, 1, 9, 12, 42, 126, 42, 12, 9, 1, 1, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 1, 1, 11, 55, 165, 66, 462, 66, 165, 55, 11, 1, 1
Offset: 0

Views

Author

Leroy Quet, Jun 10 2005

Keywords

Comments

T(0,0) is an indeterminate, but 1 seems a logical value to assign it. T(n,0) = T(n,1) = T(n,n-1) = T(n,n) = 1.
T(2n,n) = A001700(n-1) (n>=1). - Emeric Deutsch, Jun 13 2005

Examples

			T(6,2)=5 because binomial(6,2)*gcd(6,2)/6 = 15*2/6 = 5.
The triangle T(n,m) begins:
n\m 0  1  2   3   4    5   6   7  8  9  10...
0:  1
1:  1  1
2:  1  1  1
3:  1  1  1   1
4:  1  1  3   1   1
5:  1  1  2   2   1    1
6:  1  1  5  10   5    1   1
7:  1  1  3   5   5    3   1   1
8:  1  1  7   7  35    7   7   1  1
9:  1  1  4  28  14   14  28   4  1  1
10: 1  1  9  12  42  126  42  12  9  1   1
n\m 0  1  2   3   4    5   6   7  8  9  10...
... reformatted - _Wolfdieter Lang_, Feb 23 2014

Links

Crossrefs

Row sums A159554.
Cf. A007318, A001700, A000108, A050169, A234040, A026741, A234042.
Cf. A109004.

Programs

  • Haskell
    a107711 n k = a107711_tabl !! n !! k
a107711_row n = a107711_tabl !! n
a107711_tabl = [1] : zipWith (map . flip div) [1..]
               (tail $ zipWith (zipWith (*)) a007318_tabl a109004_tabl)
-- Reinhard Zumkeller, Feb 28 2014
  • Maple
    a:=proc(n,k) if n=0 and k=0 then 1 elif k<=n then binomial(n,k)*gcd(n,k)/n else 0 fi end: for n from 0 to 13 do seq(a(n,k),k=0..n) od; # yields sequence in triangular form. - Emeric Deutsch, Jun 13 2005
  • Mathematica
    T[0, 0] = 1; T[n_, m_] := Binomial[n, m] * GCD[n, m]/n;
Table[T[n, m], {n, 1, 13}, {m, 1, n}] // Flatten (* Jean-François Alcover, Nov 16 2017 *)

Formula

From Wolfdieter Lang, Feb 28 2014 (Start)
T(n, m) = T(n-1,m)*(n-1)*gcd(n,m)/((n-m)*gcd(n-1,m)), n > m >= 1, T(n, 0) = 1, T(n, n) = 1, otherwise 0.
T(n, m) = binomial(n-1,m-1)*gcd(n,m)/m for n >= m >= 1, T(n,0) = 1, otherwise 0 (from iteration of the preceding recurrence).
T(n, m) = T(n-1, m-1)*(n-1)*gcd(n,m)/(m*gcd(n-1,m-1)) for n >= m >= 2, T(n, 0) = 1, T(n, 1) = 0, otherwise 0 (from the preceding formula).
T(2*n, n) = A001700(n-1) (n>=1) (see the Emeric Deutsch comment above), T(2*n, n-1) = A234040(n), T(2*n+1,n) = A000108(n), n >= 0 (Catalan numbers).
Column sequences: T(n+2, 2) = A026741(n+1), T(n+3, 3) = A234041(n), T(n+4, 4) = A208950(n+2), T(n+5, 5) = A234042, n >= 0. (End)

Extensions

More terms from Emeric Deutsch, Jun 13 2005

A234042 a(n) = binomial(n+4,4)*gcd(n,5)/5.

Original entry on oeis.org

1, 1, 3, 7, 14, 126, 42, 66, 99, 143, 1001, 273, 364, 476, 612, 3876, 969, 1197, 1463, 1771, 10626, 2530, 2990, 3510, 4095, 23751, 5481, 6293, 7192, 8184, 46376, 10472, 11781, 13209, 14763, 82251, 18278, 20254, 22386, 24682, 135751, 29799, 32637, 35673, 38916
Offset: 0

Views

Author

Wolfdieter Lang, Feb 24 2014

Keywords

Comments

The sixth column of the triangle A107711.

Links

Crossrefs

Cf. A107711, A208950 (fifth column of A107711), A109009 (gcd(n,5)).

Programs

  • Mathematica
    a[n_] := Binomial[n + 4, 4] * GCD[n, 5]/5; Table[a[n], {n, 0, 40}] (* Amiram Eldar, Sep 20 2022 *)
  • PARI
    a(n) = binomial(n+4,4)*gcd(n,5)/5 \\ Charles R Greathouse IV, Feb 16 2017

Formula

a(n) = A107711(n+5,5) = binomial(n+5,5)*gcd(n,5)/(n+5), with n >= 0.
O.g.f.: ((1+x^20) + x*(1+x^18) + 3*x^2*(1+x^16) + 7*x^3*(1+x^14) + 14*x^4*(1+x^12) + 121*x^5*(1+x^10)+37*x^6*(1+x^8) + 51*x^7*(1+x^6) + 64*x^8*(1+x^4) + 73*x^9*(1+x^2) + 381*x^10)/(1-x^5)^5. From the 5-section using n = 5*k + j, for j = 0, 1, 2, 3, 4.
Sum_{n>=0} 1/a(n) = 20/3 - 16*sqrt(10-22/sqrt(5))*Pi/5. - Amiram Eldar, Sep 20 2022

A185138 a(4*n) = n*(4*n-1); a(2*n+1) = n*(n+1)/2; a(4*n+2) = (2*n+1)*(4*n+1).

Original entry on oeis.org

0, 0, 1, 1, 3, 3, 15, 6, 14, 10, 45, 15, 33, 21, 91, 28, 60, 36, 153, 45, 95, 55, 231, 66, 138, 78, 325, 91, 189, 105, 435, 120, 248, 136, 561, 153, 315, 171, 703, 190, 390, 210, 861, 231, 473, 253, 1035, 276, 564, 300, 1225, 325
Offset: 0

Views

Author

Paul Curtz, Mar 12 2012

Keywords

Comments

a(n) is divisible by the n-th term of the sequence 3, 3, 1, 1, 3, 3 (periodically repeated with period 6).
a(n) is divisible by b(floor((n-1)/3)), where b(n) = 1, 3, 2, 3, 7, 3, 5, 3, 13, 3, 8, 3, 19, 3,... , n>=0, is defined by inserting a 3 after each entry of A165355.
(n+1)*(n+2)*(n+3)/2=3*A000292(n+1) is divisible by a(n+2), so there is an integer sequence c(n)= 3*A000292(n+1)/a(n+2) = 3, 12, 10, 20, 7, 28, 18,... with c(2*n)=A123167(n+1) and c(n)/A109613(n+2)=A176895(n).
The sequence of denominators of a(n+2)/n has period length 8: 1, 2, 1, 4, 1, 1, 1, 4.
A table T(k,c) = a(1+c*(1+2k)) of (2*k+1)-sections starts as follows:
0 1 1 3 3 15...
0 3 6 45 21 60...
0 15 15 60 55 325...
0 14 28 231 105 315...
0 45 45 189 171 1035...
The table of T'(k,c) = T(k,c)/(2k+1), columns c>=0, looks as follows, construction similar to A165943:
0 1 1 3 3 15 6 14 k=0
0 1 2 15 7 20 15 77 k=1
0 3 3 12 11 65 24 63 k=2
0 2 4 33 15 45 33 175 k=3
0 5 5 21 19 115 42 112 k=4
0 3 6 51 23 70 51 273 k=5
The entries T'(k,c) are divisible by A060819(c).
Differences are T'(2,c)-T'(0,c) = T'(4,c)-T'(2,c) = 0, 2, 2, 9, 8, 50, 18, 49, 32, ... which is A168077(c) multiplied by the c-th term of the period-4 sequence 2, 2, 2, 1.
Differences are T'(3,c)- T'(1,c) = T'(5,c)-T'(3,c) = 0, 1, 2, 18, 8, 25, 18, 98, 32,... which is A168077(c) multiplied by the period-4 sequence 2, 1, 2, 2.
The reduced fractions T'(0,c)/T'(1,c) = 1, 1/2, 1/5, 3/7, 3/4, 2/5, 2/11, 5/13, 5/7, 3/8, 3/17, 7/19, .., c>=1, have a numerator sequence A026741(floor(c/2)+1). The denominator sequence is f(c) = 1, 2, 5, 7, 4, 5,.. = A001651(c+1)/A130658(c+1), with f(2*c+1) +f(2*c+2) = 3, 12, 9, 24 .. =3*A022998(c).

Links

Crossrefs

Cf. A000217, A014634, A026741, A033991, A064038, A060819, A165355, A208950.

Programs

  • Maple
    A185138 := proc(n)
        if n mod 4 = 0 then
                return n/4*(n-1) ;
        elif n mod 2 = 1 then
                return (n-1)*(n+1)/8 ;
        else
                return (n-1)*n/2 ;
        end if;
end proc: # R. J. Mathar, Apr 05 2012
  • Mathematica
    Clear[b];b[1] = 0; b[2] = 0; b[3] = 1; b[4] = 1; b[5] = 3; b[6] = 3; b[7] = 15;b[8] = 6;b[n_Integer] := b[n] = ((-2 + n) (-4 (-4 + n) (-3 + n) (-2 + n) (8 + n (-9 + 2 n)) b[-3 + n] + (-5 + n) ((-3 +n) ((-4 + n) (211 + 2 n (-215 + n (147 + n (-41 + 4 n)))) - 4 (-1 + n) (19 + n (-13 + 2 n)) b[-2 + n]) - 4 (-4 + n)^2 (8 + n (-9 + 2 n)) b[-1 + n])))/(4 (-5 + n) (-4 + n) (-3 + n)^2 (19 + n (-13 + 2 n)))
a = Table[b[n], {n, 1, 52}] (* Roger L. Bagula, Mar 14 2012 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{0,0,1,1,3,3,15,6,14,10,45,15},60] (* Harvey P. Dale, Nov 23 2015 *)
  • PARI
    x='x+O('x^50); concat([0,0], Vec(-x^2*(3*x^8+x^7+5*x^6+3*x^5+12*x^4+3*x^3+3*x^2+x+1)/ ((x-1)^3*(x+1)^3*(x^2+1)^3))) \\ G. C. Greubel, Jun 23 2017

Formula

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(2*n) = A064038(2*n), a(2*n+1) = A000217(n).
a(n) = 3*A208950(n)/A109613(n).
a(n+1) = A060819(n) * A026741(n+2)(floor(n/2)).
G.f.: -x^2*(3*x^8+x^7+5*x^6+3*x^5+12*x^4+3*x^3+3*x^2+x+1)/ ((x-1)^3*(x+1)^3*(x^2+1)^3). - R. J. Mathar, Mar 22 2012
a(n) = (4*n^2-3*n-1+(2*n^2-3*n+1)*(-1)^n + n*(n-1)*(1+(-1)^n)*(-1)^((2*n-3-(-1)^n)/4))/16. - Luce ETIENNE, May 13 2016
Sum_{n>=2} 1/a(n) = 2 - Pi/4 + 7*log(2)/2. - Amiram Eldar, Aug 12 2022

A218289 Denominator of the sixth increasing diagonal of the autosequence of the second kind from (-1)^n/(n+1).

Original entry on oeis.org

6, 12, 12, 12, 210, 168, 504, 72, 198, 660, 1716, 1092, 546, 336, 4080, 2448, 5814, 684, 1596, 4620, 10626, 6072, 2760, 1560, 17550, 9828, 21924, 2436, 5394, 14880, 32736, 17952, 7854, 4284, 46620, 25308, 54834
Offset: 0

Views

Author

Paul Curtz, Oct 25 2012

Keywords

Comments

See A194767. a(n) is a multiple of 6. The terms 6, 210, 504, 1716, 4080, 5814, ... have the form k*(k+1)*(k+2), for k = 1, 5, 7, 11, 15, 17, 21, 25, ... .

Links

Crossrefs

Cf. A208950(n+2).

Formula

a(n) = A007531(n+3)/s(n) = (n+1)*(n+2)*(n+3)/s(n) where s(n) repeats 1, 2, 5, 10, 1, 2, 1, 10, 5, 2.
a(n) = (n+1)*(n+2)*(n+3)*a(n-10)/((n-7)*(n-8)*(n-9)) for n>9 (empirical). - Jean-François Alcover, Nov 29 2016
Showing 1-4 of 4 results.

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