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 1-8 of 8 results.

A115137 Second diagonal of triangle A113647 (called Y(2,1)).

Original entry on oeis.org

1, 1, 7, 41, 247, 1545, 9975, 66057, 446455, 3067913, 21372919, 150618121, 1071841271, 7691763721, 55600938999, 404488323081, 2959189475319, 21757613309961, 160691417776119, 1191577871450121, 8868160862158839
Offset: 0

Views

Author

Wolfdieter Lang, Jan 13 2006

Keywords

Examples

			41=a(3)= A062992(3) - 2*A062992(2) = 67 - 2*13.

Links

Programs

  • Mathematica
    CoefficientList[Series[(1-2*x)*(2*(1-Sqrt[1-8*x])/(4*x)-1)/(1+x), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 19 2012 *)

Formula

a(n)= b(n) - 2*b(n-1) with b(n):=A062992(n)= A064062(n+1), n>=1. a(0):=1.
G.f.: (1-2*x)*(2*c(2*x)-1)/(1+x) with c(x) g.f. of A000108 (Catalan).
a(n)= A113647(n, n), n>=1.
Recurrence: (n-2)*(n+1)*a(n) = (7*n^2-19*n+14)*a(n-1) + 4*(n-1)*(2*n-3)*a(n-2). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ 2^(3*n+2)/(3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 19 2012

A115150 Third diagonal (M=3) sequence of triangle A113647, called Y(2,1).

Original entry on oeis.org

1, 15, 113, 783, 5361, 36879, 255985, 1794063, 12689393, 90505231, 650379249, 4705157135, 34244198385, 250572963855, 1842382110705, 13605619630095, 100872203796465, 750556607938575, 5602962592235505, 41952165966643215
Offset: 0

Views

Author

Wolfdieter Lang, Jan 13 2006

Keywords

Links

Crossrefs

a(n)=A115138(n+2), n>=0.

Programs

  • Mathematica
    CoefficientList[Series[((4*x-2+x^2)+2*(1-4*x)*(1-Sqrt[1-8*x])/(4*x))/((x^2)*(1+x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 19 2012 *)

Formula

a(n)= A113647(n+2, n+1), n>=0.
G.f.: ((4*x-2+x^2) + 2*(1-4*x)*c(2*x))/((x^2)*(1+x)), with the o.g.f. c(x):=(1-sqrt(1-4*x))/(2*x) of A000108 (Catalan numbers).
Recurrence: (n-1)*(n+3)*a(n) = (7*n^2 + 2*n + 3)*a(n-1) + 4*n*(2*n+1)*a(n-2). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ 2^(3*n+9)/(9*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 19 2012

A115153 Sixth diagonal (M=6) sequence of triangle A113647, called Y(2,1).

Original entry on oeis.org

1, 127, 1665, 16255, 141441, 1163135, 9273473, 72613759, 562430081, 4327407487, 33161347201, 253517365119, 1935665528961, 14771256557439, 112715410440321, 860346088685439, 6570305359184001, 50209563856600959, 383989436028813441
Offset: 0

Views

Author

Wolfdieter Lang, Jan 13 2006

Keywords

Links

Programs

  • Mathematica
    CoefficientList[Series[((-2+16*x-24*x^2+x^5)+2*(1-10*x+24*x^2-8*x^3)*(1-Sqrt[1-8*x])/(4*x))/((x^5)*(1+x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 19 2012 *)

Formula

a(n)= A113647(n+5, n+1), n>=0.
G.f.: ((-2 + 16*x - 24*x^2 + x^5) + 2*(1 - 10*x + 24*x^2 - 8*x^3)*c(2*x))/((x^5)*(1+x)), with the o.g.f. c(x):=(1-sqrt(1-4*x))/(2*x) of A000108 (Catalan numbers).
Recurrence: (n-1)*(n+6)*a(n) = (7*n^2+23*n+30)*a(n-1) + 4*(n+2)*(2*n+3)*a(n-2). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ 7*2^(3n+13)/(9*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 19 2012

A115136 Row sums of triangle A113647.

Original entry on oeis.org

1, 4, 21, 124, 773, 4988, 33029, 223228, 1533957, 10686460, 75309061, 535920636, 3845881861, 27800469500, 202244161541, 1479594737660, 10878806654981, 80345708888060, 595788935725061, 4434080431079420, 33109442115403781
Offset: 0

Views

Author

Wolfdieter Lang, Jan 13 2006

Keywords

Formula

a(n)=sum(A113647(n, m), m=1..n+1), n>=0.
Conjecture: (n+2)*a(n) +(-11*n-4)*a(n-1) +(25*n-42)*a(n-2) +5*(-n+12)*a(n-3) +2*(-13*n+20)*a(n-4) +8*(2*n-7)*a(n-5)=0. - R. J. Mathar, Nov 10 2013

A115151 Fourth diagonal (M=4) sequence of triangle A113647, called Y(2,1).

Original entry on oeis.org

1, 31, 289, 2271, 16929, 123871, 901153, 6553567, 47759393, 349143007, 2561474593, 18860670943, 139371085857, 1033405464543, 7687240679457, 57356977176543, 429173772386337, 3219806849335263, 24215844242325537
Offset: 0

Views

Author

Wolfdieter Lang, Jan 13 2006

Keywords

Links

Crossrefs

Cf. A000108, A113647.

Programs

  • Mathematica
    CoefficientList[Series[((-2+8*x+x^3)+2*(1-6*x+4*x^2)*(1-Sqrt[1-8*x])/(4*x))/((x^3)*(1+x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 19 2012 *)

Formula

a(n) = A113647(n+3, n+1), n>=0.
G.f.: ((-2 + 8*x +x^3) + 2*(1-6*x+4*x^2)*c(2*x))/((x^3)*(1+x)), with the o.g.f. c(x):=(1-sqrt(1-4*x))/(2*x) of A000108 (Catalan numbers).
Recurrence: (n-1)*(n+4)*a(n) = (7*n^2 + 9*n + 8)*a(n-1) + 4*(n+1)*(2*n+1)*a(n-2). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ 5*2^(3*n+9)/(9*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 19 2012

A115152 Fifth diagonal (M=5) sequence of triangle A113647, called Y(2,1).

Original entry on oeis.org

1, 63, 705, 6207, 50113, 389183, 2965441, 22380607, 168132545, 1260716095, 9450356673, 70882689087, 532259536833, 4002476458047, 30145737916353, 227429364793407, 1718693633458113, 13009919057854527, 98641252341252033
Offset: 0

Views

Author

Wolfdieter Lang, Jan 13 2006

Keywords

Links

Programs

  • Mathematica
    CoefficientList[Series[((-2+12*x-8*x^2+x^4)+2*(1-8*x+12*x^2)*(1-Sqrt[1-8*x])/(4*x))/((x^4)*(1+x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 19 2012 *)

Formula

a(n)= A113647(n+4, n+1), n>=0.
G.f.: ((-2 + 12*x - 8*x^2 + x^4) + 2*(1-8*x+12*x^2)*c(2*x))/((x^4)*(1+x)), with the o.g.f. c(x):=(1-sqrt(1-4*x))/(2*x) of A000108 (Catalan numbers).
Recurrence: (n-1)*(n+5)*a(n) = (7*n^2+16*n+17)*a(n-1) + 4*(n+1)*(2*n+3)*a(n-2). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ 2^(3*n+12)/(3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 19 2012

A333565 O.g.f.: (1 + 4*x)/((1 + x)*sqrt(1 - 8*x)).

Original entry on oeis.org

1, 7, 33, 223, 1537, 11007, 80385, 595455, 4456449, 33615871, 255148033, 1946337279, 14908784641, 114597822463, 883479412737, 6828492980223, 52895475040257, 410544577183743, 3191929428770817, 24855137310736383, 193811815161921537, 1513167009951514623, 11827298001565515777
Offset: 0

Views

Author

Peter Bala, Apr 11 2020

Keywords

Comments

This sequence satisfies the Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ), for all prime p and positive integers n and k, since the power series E(x) := exp( Sum_{n >= 1} a(n)*x^n/n ) has integer coefficients. See Stanley, Ex. 5.2 (a), p. 72, and its solution on p. 104.
We conjecture that this sequence satisfies the stronger congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 3 and positive integers n and k. The particular case when n = k = 1 follows from the corresponding result for A333564. Some examples of these congruences are given below.

Examples

			Examples of congruences:
a(11) - a(1) = 1946337279 - 7 = (2^3)*(11^3)*182789 == 0 ( mod 11^3 ).
a(2*11) - a(2) = 11827298001565515777 - 33 = (2^5)*(3^2)*(11^3)*107* 288357478039 == 0 ( mod 11^3 ).
a(5^2) - a(5) = 5680983691406772011007 - 11007 = (2^8)*(3^3)*(5^6)*7* 19*1123*352183001 == 0 ( mod 5^6 ).

References

  • R. P. Stanley. Enumerative combinatorics. Vol. 2, (volume 62 of Cambridge Studies in Advanced Mathematics). Cambridge University Press, Cambridge, 1999.

Crossrefs

Cf. A000984, A113647, A115137, A119259, A333564.

Programs

  • Maple
    a := proc (n) option remember; `if`(n = 0, 1, `if`(n = 1, 7, `if`(n = 2, 33, ((3*n+4)*a(n-1)+(36*n-76)*a(n-2)+(32*n-80)*a(n-3))/n)))
end proc:
seq(a(n), n = 0..25);
  • Mathematica
    a[n_] := (-1)^n - 2^(n+2) Binomial[2n, n-1] Hypergeometric2F1[1, 2n +1, n + 2, 2];
Table[Simplify[a[n]], {n, 0, 22}] (* Peter Luschny, Apr 13 2020 *)
CoefficientList[Series[(1+4x)/((1+x)Sqrt[1-8x]),{x,0,30}],x] (* Harvey P. Dale, Jan 24 2021 *)

Formula

a(n) = (2^n)*binomial(2*n,n) + 3*sum_{k = 0..n-1} (-1)^(n+k+1)*2^k* binomial(2*k,k).
a(n) = 4*A333564(n) + (-1)^n for n >= 1.
a(n) = 2*A119259(n) - (-1)^n.
a(n) = (-1)^n + 4*Sum_{k = 1..n} (3*k-1)*2^(k-1)*A000108(k-1).
a(n) ~ 8^n * 4/(3*sqrt(Pi*n)).
Congruences: a(p) == 7 ( mod p^3 ) for all prime p >= 3.
O.g.f. A(x) = 1 + 7*x + 33*x^2 + ... satisfies the differential equation (x + 1)*(4*x + 1)*(8*x - 1)*A'(x) + (16*x^2 - 4*x + 7)*A(x) = 0. Cf. A333564.
P-recursive: n*(3*n - 4)*a(n) = (21*n^2 - 40*n + 12)*a(n-1) + 4*(3*n - 1)*(2*n - 3)*a(n-2) with a(0) = 1 and a(1) = 7.
Alternative form: (a(n) + a(n-1))/(a(n) - a(n-2)) = P(n)/Q(n), where P(n) = 4*(3*n - 1)*(2*n - 3) and Q(n) = (21*n^2 - 40*n + 12).
Also, n*a(n) = (3*n + 4)*a(n-1) + 4*(9*n - 19)*a(n-2) + 16*(2*n - 5)*a(n-3) with a(0) = 1, a(1) = 7 and a(2) = 33.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 7*x + 41*x^2 + 247*x^3 + ... is the o.g.f. of the second diagonal of triangle A113647. See also A115137.

A115138 A sequence related to Catalan numbers A000108.

Original entry on oeis.org

1, -1, 1, 15, 113, 783, 5361, 36879, 255985, 1794063, 12689393, 90505231, 650379249, 4705157135, 34244198385, 250572963855, 1842382110705, 13605619630095, 100872203796465, 750556607938575, 5602962592235505, 41952165966643215, 314983352736153585
Offset: 0

Views

Author

Wolfdieter Lang, Jan 13 2006

Keywords

Comments

See also A115150, the third diagonal of triangle A113647 (called Y(2,1)).

Examples

			15= a(3) = A062992(3) - 4*A062992(2) = 67 - 4*13.

Formula

(1-4*x)*(2*c(2*x)-1)/(1+x) with c(x) g.f. of A000108 (Catalan).
a(n)= A113647(n, n-1), n>=2.
a(n)= b(n)-4*b(n-1) with b(n):=A062992(n), n>=2; a(0)=1, a(1)=-1.
Showing 1-8 of 8 results.

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