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 1-5 of 5 results.

A123619 a(n) = A123610(2*n+2,n)/(n+1) = A123618(n)/(n+1).

Original entry on oeis.org

1, 2, 13, 98, 884, 8712, 92033, 1022450, 11819620, 141052808, 1727897780, 21634496072, 275950213712, 3576314656800, 46995009879033, 625082413914450, 8403885788094500, 114069363868845000, 1561609591376307572
Offset: 0

Views

Author

Paul D. Hanna, Oct 03 2006

Keywords

Comments

Related sequences: A123610(2n,n) = A123617(n); A123610(2n+1,n) = A000891(n); A123610(2n+2,n) = A123618(n).

Links

Crossrefs

Cf. A123610 (triangle); A123617, A000891, A123618.

Programs

  • Mathematica
    T[, 0] = 1; T[n, k_] := 1/n DivisorSum[n, If[GCD[k, #] == #, EulerPhi[#]*Binomial[n/#, k/#]^2, 0] &];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* A123610 *)
Table[T[2*n, n], {n, 0, 50}] (* A123617 *)
Table[T[2*n + 2, n], {n, 0, 50}] (* A123618 *)
Table[T[2*n + 2,n]/(n+1), {n, 0, 50}] (* A123619 *)
(* G. C. Greubel, Oct 26 2017 *)
  • PARI
    {a(n)=if(n==0,1,(1/(2*(n+1)^2))*sumdiv(2*n+2,d,if(gcd(n,d)==d, eulerphi(d)*binomial((2*n+2)/d,n/d)^2,0)))}

A123610 Triangle read by rows, where T(n,k) = (1/n)*Sum_{d|(n,k)} phi(d) * binomial(n/d,k/d)^2 for n >= k > 0, with T(n,0) = 1 for n >= 0.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 10, 4, 1, 1, 5, 20, 20, 5, 1, 1, 6, 39, 68, 39, 6, 1, 1, 7, 63, 175, 175, 63, 7, 1, 1, 8, 100, 392, 618, 392, 100, 8, 1, 1, 9, 144, 786, 1764, 1764, 786, 144, 9, 1, 1, 10, 205, 1440, 4420, 6352, 4420, 1440, 205, 10, 1, 1, 11, 275, 2475, 9900
Offset: 0

Views

Author

Paul D. Hanna, Oct 03 2006

Keywords

Comments

A variant of the triangle A047996 of circular binomial coefficients.

Examples

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
  1;
  1,  1;
  1,  2,   1;
  1,  3,   3,    1;
  1,  4,  10,    4,    1;
  1,  5,  20,   20,    5,    1;
  1,  6,  39,   68,   39,    6,    1;
  1,  7,  63,  175,  175,   63,    7,    1;
  1,  8, 100,  392,  618,  392,  100,    8,   1;
  1,  9, 144,  786, 1764, 1764,  786,  144,   9,  1;
  1, 10, 205, 1440, 4420, 6352, 4420, 1440, 205, 10, 1;
  ...
Example of column g.f.s are:
column 1: 1/(1 - x)^2;
column 2: Ser([1, 1, 3, 1]) / ((1 - x)^2*(1 - x^2)^2) = g.f. of A005997;
column 3: Ser([1, 2, 11, 26, 30, 26, 17, 6, 1]) / ((1 - x)^2*(1 - x^2)^2*(1 -x^3)^2);
column 4: Ser([1, 3, 28, 94, 240, 440, 679, 839, 887, 757, 550, 314, 148, 48, 11, 1]) / ((1 - x)^2*(1 - x^2)^2*(1 - x^3)^2*(1 - x^4)^2);
where Ser() denotes a polynomial in x with the given coefficients, as in Ser([1, 1, 3, 1]) = (1 + x + 3*x^2 + x^3).

Links

Crossrefs

Cf. Columns: A005997, A123613, A123614, A123615, A123616.
Cf. A123611 (row sums), A123612 (antidiagonal sums), A123617 (central terms).
Cf. A123618, A123619, A047996 (variant), A128545.

Programs

  • Mathematica
    T[, 0] = 1; T[n, k_] := 1/n DivisorSum[n, If[GCD[k, #] == #, EulerPhi[#]* Binomial[n/#, k/#]^2, 0]&]; Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 06 2015, adapted from PARI *)
  • PARI
    {T(n,k)=if(k==0,1,(1/n)*sumdiv(n,d,if(gcd(k,d)==d, eulerphi(d)*binomial(n/d,k/d)^2,0)))}

Formula

T(2*n+1, n) = (2*n + 1)*A000108(n)^2 = (2*n + 1)*((2*n)!/(n!(n+1)!))^2 = A000891(n) for n >= 0.
Row sums are 2*A047996(2*n,n) = 2*A003239(n) for n > 0.
Row sums equal the row sums of triangle A128545.
For n >= 1, the g.f. of column n has the form: P_n(x)/(Product_{m=1..n} (1 - x^m)^2), where P_n(x) is a polynomial with n^2 coefficients such that the sum of the coefficients is P_n(1) = (2*n - 1)!.
From Petros Hadjicostas, Oct 24 2017: (Start)
Proofs of the following formulae can be found in the links.
G.f.: Sum_{n>=1, k>=0} T(n,k)*x^n*y^k = -Sum_{s>=1} (phi(s)/s)*log(f(x^s,y^s)), where phi(s) is Euler's totient function at s, f(x,y) = (sqrt(g(x,y)) + 1 -(1 + y)*x)/2, and g(x,y) = 1 - 2*(1 + y)*x + (1 - y)^2*x^2. (Term T(0,0) is not used in this g.f.)
Row g.f.: Sum_{k>=0} T(n,k)*y^k = (1/n)*Sum_{d|n} phi(d)*R(n/d, y^d), where R(m, y) = [z^m] (1 + (1 + y)*z + y*z^2)^m. (End)

A123617 Central terms of triangle A123610: a(n) = A123610(2*n,n).

Original entry on oeis.org

1, 2, 10, 68, 618, 6352, 71188, 841332, 10352618, 131328068, 1706742160, 22619741212, 304685855700, 4160480013848, 57476485976388, 802048167035968, 11290551106506218, 160168176177137896, 2287724464324213972
Offset: 0

Views

Author

Paul D. Hanna, Oct 03 2006

Keywords

Comments

Related sequences: A123610(2n+1,n) = A000891(n); A123610(2n+2,n) = A123618(n); A123610(2n+2,n)/(n+1) = A123619(n).

Links

Crossrefs

Cf. A123610 (triangle); A000891, A123618.

Programs

  • Mathematica
    T[, 0] = 1; T[n, k_] := 1/n DivisorSum[n, If[GCD[k, #] == #, EulerPhi[#]*Binomial[n/#, k/#]^2, 0] &];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* A123610 *)
Table[T[2*n, n], {n, 0, 50}] (* A123617 *)
Table[T[2*n + 2, n], {n, 0, 50}] (* A123618 *)
Table[T[2*n + 2,n]/(n+1), {n, 0, 50}] (* A123619 *)
(* G. C. Greubel, Oct 26 2017 *)
  • PARI
    {a(n)=if(n==0,1,(1/2/n)*sumdiv(2*n,d,if(gcd(n,d)==d, eulerphi(d)*binomial(2*n/d,n/d)^2,0)))}

A123611 Row sums of triangle A123610.

Original entry on oeis.org

1, 2, 4, 8, 20, 52, 160, 492, 1620, 5408, 18504, 64132, 225440, 800048, 2865720, 10341208, 37568340, 137270956, 504176992, 1860277044, 6892335720, 25631327688, 95640894056, 357975249028, 1343650267296, 5056424257552
Offset: 0

Views

Author

Paul D. Hanna, Oct 03 2006

Keywords

Links

Crossrefs

Cf. A047996, A003239; A123610 (triangle), A123612 (antidiagonal sums); central terms: A123617, A123618.
Cf. A128545.

Programs

  • Mathematica
    Total /@ Table[If[k == 0, 1, 1/n DivisorSum[n, If[Divisible[k, #], EulerPhi[#] Binomial[n/#, k/#]^2, 0] &]], {n, 0, 25}, {k, 0, n}] (* Michael De Vlieger, Apr 03 2017, after Jean-François Alcover at A123610 *)
  • PARI
    {a(n)=sum(k=0,n,if(k==0,1,(1/n)*sumdiv(n,d,if(gcd(k,d)==d, eulerphi(d)*binomial(n/d,k/d)^2,0))))}

Formula

a(n) = 2*A047996(2*n,n) = 2*A003239(n) for n > 0, where A047996 is the triangle of circular binomial coefficients and A003239(n) = number of rooted planar trees with n non-root nodes.
Also equals the row sums of triangle A128545, where A128545(n,k) is the coefficient of q^(n*k) in the q-binomial coefficient [2n,n] for n >= k >= 0.
a(n) = (1/n) * Sum_{d | n} phi(n/d) * binomial(2*d, d) for n>0. - Andrew Howroyd, Apr 02 2017
G.f.: 1 - Sum_{n>=1} (phi(n)/n) * log((1-2*x^n + sqrt(1-4*x^n))/2) = 1 - 2*Sum_{n>=1} (phi(n)/n) * log((1+sqrt(1-4*x^n))/2). (Except for the term a(0) = 1, the first g.f. follows from the g.f. in A123610 by setting y=1, as suggested by P. D. Hanna.) - Petros Hadjicostas, Oct 24 2017

A123612 Antidiagonal sums of triangle A123610.

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 17, 31, 68, 145, 325, 728, 1685, 3891, 9140, 21565, 51311, 122666, 295037, 712477, 1728262, 4207027, 10276693, 25178708, 61866141, 152397945, 376309596, 931239093, 2309219447, 5737078442, 14278587533, 35595622719
Offset: 0

Views

Author

Paul D. Hanna, Oct 03 2006

Keywords

Comments

The g.f. was suggested by P. D. Hanna. It can be proved either by letting y=x in the bivariate g.f. for sequence A123610 or by using the formula of A. Howroyd (below) for this sequence and the l.g.f. for sequence A167539. The second proof proceeds as follows: Sum_{n>=1} a(n)*x^n = Sum_{n>=1} (1/n)*Sum_{d|n} phi(n/d)*g(d), where g(d) = A167539(d). Then Sum_{n>=1} a(n)*x^n = Sum_{m>=1} (phi(m)/m)*Sum_{d>=1} g(d)*(x^m)^d/d = Sum_{m>=1} (phi(m)/m)*G(x^m), where G(x) = l.g.f. of sequence g(n) = A167539(n). - Petros Hadjicostas, Oct 25 2017

Links

Crossrefs

Cf. A123610 (triangle), A123611 (row sums); central terms: A123617, A123618, A167539.

Programs

  • Mathematica
    Total /@ Table[Function[m, If[k == 0, 1, 1/m DivisorSum[m, If[GCD[k, #] == #, EulerPhi[#] Binomial[m/#, k/#]^2, 0] &]]][n - k + 1], {n, -1, 30}, {k, 0, Ceiling[n/2]}] (* Michael De Vlieger, Apr 03 2017, after Jean-François Alcover at A123610 *)
  • PARI
    {a(n)=sum(k=0,n\2,if(k==0,1,(1/(n-k))*sumdiv(n-k,d,if(gcd(k,d)==d, eulerphi(d)*binomial((n-k)/d,k/d)^2,0))))}

Formula

a(n) = (1/n) * Sum_{d | n} phi(n/d) * A167539(d) for n>0. - Andrew Howroyd, Apr 02 2017
G.f.: 1-Sum_{n>=1} (phi(n)/n)*f(x^n), where f(x) = log((1-x-x^2+sqrt((1+x+x^2)*(1-3*x+x^2)))/2) = -log((1-x-x^2-sqrt((1+x+x^2)*(1-3*x+x^2)))/(2*x^3)). - Petros Hadjicostas, Oct 25 2017
Showing 1-5 of 5 results.

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

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