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 1-6 of 6 results.

A166914 a(n) = 20*a(n-1) - 64*a(n-2) for n > 1; a(0) = 21, a(1) = 340.

Original entry on oeis.org

21, 340, 5456, 87360, 1398016, 22369280, 357912576, 5726617600, 91625947136, 1466015416320, 23456247709696, 375299967549440, 6004799497568256, 96076792028200960, 1537228672719650816, 24595658764588154880
Offset: 0

Views

Author

Klaus Brockhaus, Oct 27 2009

Keywords

Comments

Related to Reverse and Add trajectory of 318 in base 4: A075153(6*n+2) = 240*a(n).

Links

Crossrefs

Cf. A075153, A166912, A166913, A166915, A166916, A166917, A166927, A166965, A166984.

Programs

  • Magma
    [Binomial(4^(n+3), 2)/96: n in [0..30]]; // G. C. Greubel, Oct 02 2024
  • Mathematica
    CoefficientList[Series[(21-80x)/((1-4x)(1-16x)),{x,0,20}],x]  (* or *) LinearRecurrence[{20,-64},{21,340},20] (* Harvey P. Dale, Feb 23 2011 & Mar 30 2012 *)
  • PARI
    {m=15; v=concat([21, 340], vector(m-2)); for(n=3, m, v[n]=20*v[n-1]-64*v[n-2]); v}
  • SageMath
    A166914=BinaryRecurrenceSequence(20,-64,21,340)
[A166914(n) for n in range(31)] # G. C. Greubel, Oct 02 2024

Formula

a(n) = (64*16^n - 4^n)/3.
G.f.: (21 - 80*x)/((1-4*x)*(1-16*x)).
Limit_{n -> infinity} a(n)/a(n-1) = 16.
From G. C. Greubel, May 28 2016: (Start)
a(n) = 20*a(n-1) - 64*a(n-2).
E.g.f.: (1/3)*(-exp(4*x) + 64*exp(16*x)). (End)

A166917 a(n) = 20*a(n-1) - 64*a(n-2) for n > 1; a(0) = 85, a(1) = 1364.

Original entry on oeis.org

85, 1364, 21840, 349504, 5592320, 89478144, 1431654400, 22906486784, 366503854080, 5864061927424, 93824991887360, 1501199874392064, 24019198007050240, 384307168179912704, 6148914691147038720, 98382635059426361344, 1574122160955116748800, 25185954575299047849984
Offset: 0

Views

Author

Klaus Brockhaus, Oct 27 2009

Keywords

Comments

Related to Reverse and Add trajectory of 318 in base 4: A075153(6*n+5) = 240*a(n).

Links

Crossrefs

Cf. A075153, A166912, A166913, A166914, A166915, A166916, A166927, A166965, A166984.

Programs

  • Magma
    [Binomial(4^(n+4), 2)/384: n in [0..30]]; // G. C. Greubel, Oct 02 2024
  • Mathematica
    LinearRecurrence[{20,-64}, {85, 1364}, 50] (* G. C. Greubel, May 28 2016 *)
  • PARI
    {m=15; v=concat([85, 1364], vector(m-2)); for(n=3, m, v[n]=20*v[n-1]-64*v[n-2]); v}
  • SageMath
    A166917=BinaryRecurrenceSequence(20,-64,85,1364)
[A166917(n) for n in range(31)] # G. C. Greubel, Oct 02 2024

Formula

a(n) = (256*16^n - 4^n)/3.
G.f.: (85 - 336*x)/((1-4*x)*(1-16*x)).
Limit_{n -> infinity} a(n)/a(n-1) = 16.
E.g.f.: (1/3)*(256*exp(16*x) - exp(4*x)). - G. C. Greubel, May 28 2016

A278080 Expansion of e.g.f. (1/4!)*sin^4(x)/cos(x) (coefficients of even powers only).

Original entry on oeis.org

0, 0, 1, -5, 126, 1490, 118151, 8256885, 808428076, 100199284180, 15432169163901, 2889536106161375, 646438926423519626, 170294687860735726470, 52177485058722877649251, 18397662218707151323777465, 7396641315814156362154666776
Offset: 0

Views

Author

Peter Bala, Nov 10 2016

Keywords

Comments

This sequence gives the coefficients in an asymptotic expansion of a series related to the constant Pi. It can be shown that (1/4!)*Pi/4 = Sum_{k >= 1} (-1)^(k-1)/((2*k - 5)*(2*k - 3)*(2*k - 1)*(2*k + 1)*(2*k + 3)). Using Proposition 1 of Borwein et al. it can be shown that the following asymptotic expansion holds for the tails of the series: for N divisible by 4, 2*( (1/4!)*Pi/4 - Sum_{k = 1..N/2} (-1)^(k-1)/((2*k - 5)*(2*k - 3)*(2*k - 1)*(2*k + 1)*(2*k + 3)) ) ~ 1/N^5 - (-5)/N^7 + 126/N^9 - 1490/N^11 + 118151/N^13 - .... An example is given below. Cf. A024235 and A278195.

Examples

			Let N = 100000. The truncated series 2*Sum_{k = 1..N/2} (-1)^(k-1)/((2*k - 5)*(2*k - 3)*(2*k - 1)*(2*k + 1)*(2*k + 3)) = 0.065449846949787359134638(3)038183229(2)6754107(569)820314(8536)0364.... The bracketed digits show where this decimal expansion differs from that of Pi/48. The numbers 1, 5, 126, -1490 must be added to the bracketed numbers to give the correct decimal expansion to 60 digits: Pi/48 = 0.065449846949787359134638(4) 038183229(7)6754107(695)820314(7046)0364.. ..

Links

Crossrefs

Cf. A000364, A004174, A024235, A166984, A278079, A278194, A278195.

Programs

  • Maple
    A000364 := n -> abs(euler(2*n)):
seq(1/4!*(A000364(n) + (-1)^n*(9^n - 5)/4), n = 0..20);
  • Mathematica
    With[{nn=40},Take[CoefficientList[Series[1/4! Sin[x]^4/Cos[x],{x,0,nn}],x] Range[0,nn]!,{1,-1,2}]] (* Harvey P. Dale, Feb 09 2025 *)

Formula

a(n) = [x^(2*n)/(2*n)!] ( (1/4!)*sin^4(x)/cos(x) ).
a(n) = (1/4!)*( A000364(n) + (-1)^n*(9^(n) - 5)/4 ).
a(n) = (-1)^n/(2^4*4!) * 2^(2*n)*( E(2*n,5/2) - 4*E(2*n,3/2) + 6*E(2*n,1/2) - 4*E(2*n,-1/2) + E(2*n,-3/2) ), where E(n,x) is the Euler polynomial of order n.
E.g.f. (1/4!)*sin^4(x)/cos(x) = x^4/4! - 5*x^6/6! + 126*x^8/8! + 1490*x^10/10! + ....
O.g.f. for a signed version of the sequence: Sum_{n >= 0} ( (1/2^n) * Sum_{k = 0..n} (-1)^k*binomial(n, k)/((1 - (2*k - 3)*x)*(1 - (2*k - 1)*x)*(1 - (2*k + 1)*x)*(1 - (2*k + 3)*x)*(1 - (2*k + 5)*x)) ) = 1 + 5*x^2 + 126*x^4 - 1490*x^6 + 118151*x^8 - ....

A278195 Expansion of e.g.f. (1/6!)*sin^6(x)/cos(x) (coefficients of even powers only).

Original entry on oeis.org

0, 0, 0, 1, -28, 882, -17116, 803803, 13713336, 3671012164, 506128123928, 96524822605365, 21542790273363260, 5676618945053498806, 1739246268204447115932, 613255488134158250903887, 246554708506039690689322544, 112115693433705109495581088008
Offset: 0

Views

Author

Peter Bala, Nov 15 2016

Keywords

Comments

This sequence gives the coefficients in an asymptotic expansion of a series related to the constant Pi. It can be shown that (1/6!)*Pi/4 = Sum_{k >= 1} (-1)^k/((2*k - 7)*(2*k - 5)*(2*k - 3)*(2*k - 1)*(2*k + 1)*(2*k + 3)*(2*k + 5)). Using Proposition 1 of Borwein et al. it can be shown that the following asymptotic expansion holds for the tails of this series: for N divisible by 4, 2*{ (1/6!)*Pi/4 - Sum_{k = 1..N/2} (-1)^k/((2*k - 7)*(2*k - 5)*(2*k - 3)*(2*k - 1)*(2*k + 1)*(2*k + 3)*(2*k + 5)) } ~ -1/N^7 + (-28)/N^9 - 882/N^11 + (-17116)/N^13 - 803803/N^15 + .... An example is given below.

Examples

			Expansion of (1/6!)*sin^6(x)/cos(x) starts x^6/6! - 28*x^8/8! + 882*x^10/10! - 17116*x^12/12! + ....
Let N = 100000. The truncated series 2*Sum_{k = 1..N/2} (-1)^k/( (2*k - 7)*(2*k - 5)*(2*k - 3)*(2*k - 1)*(2*k + 1)*(2*k + 3)*(2*k + 5) ) = 0.0021816615649929119711546134606107(7)5891803(617)860677(2450)20121.... The bracketed digits show where this decimal expansion differs from that of Pi/1440. The numbers -1, -28, -882 must be added to the bracketed numbers to give the correct decimal expansion: Pi/1440 = 0.0021816615649929119711546134606107(6)5891803(589)860677(1568)20121....

Links

Crossrefs

Cf. A000364, A004174, A166984, A278079, A278080, A278194.

Programs

  • Maple
    A000364 := n -> abs(euler(2*n)):
seq((1/6!)*(A000364(n) - (1/16)*((-25)^n - 7*(-9)^n + 22*(-1)^n) ), n = 0..20);
  • Mathematica
    With[{nn=40},Take[CoefficientList[Series[Sin[x]^6/Cos[x] 1/6!,{x,0,nn}],x] Range[0,nn]!,{1,-1,2}]] (* Harvey P. Dale, Sep 12 2019 *)

Formula

a(n) = [x^(2*n)/(2*n)!] ( 1/6!*sin^6(x)/cos(x) ).
a(n) = (1/6!)*( A000364(n) - 1/16*((-25)^n - 7*(-9)^n + 22*(-1)^n) ).
a(n) = (-1)^(n+1)/(2^6*6!) * 2^(2*n)*( E(2*n,7/2) - 6*E(2*n,5/2) + 15*E(2*n,3/2) - 20*E(2*n,1/2) + 15*E(2*n,-1/2) - 6*E(2*n,-3/2) + E(2*n,-5/2) ), where E(n,x) is the Euler polynomial of order n.

A381512 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where A(n,k) = (2*n+k)!/k! * [x^(2*n+k)] sinh(x)^k.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 4, 1, 0, 1, 10, 16, 1, 0, 1, 20, 91, 64, 1, 0, 1, 35, 336, 820, 256, 1, 0, 1, 56, 966, 5440, 7381, 1024, 1, 0, 1, 84, 2352, 24970, 87296, 66430, 4096, 1, 0, 1, 120, 5082, 90112, 631631, 1397760, 597871, 16384, 1, 0, 1, 165, 10032, 273988, 3331328, 15857205, 22368256, 5380840, 65536, 1, 0
Offset: 0

Views

Author

Seiichi Manyama, May 11 2025

Keywords

Examples

			Square array begins:
  1, 1,    1,     1,       1,        1, ...
  0, 1,    4,    10,      20,       35, ...
  0, 1,   16,    91,     336,      966, ...
  0, 1,   64,   820,    5440,    24970, ...
  0, 1,  256,  7381,   87296,   631631, ...
  0, 1, 1024, 66430, 1397760, 15857205, ...

Crossrefs

Columns k=0..7 give A000007, A000012, A000302, A002452(n+1), A166984, A002453, 4^n * A002451(n), A381513.
Main diagonal gives A383837.
Cf. A136630, A160562, A286899.

Programs

  • PARI
    a(n, k) = (2*n+k)!/k!*polcoef(sinh(x+x*O(x^(2*n+k)))^k, 2*n+k);

Formula

G.f. of column k: 1/Product_{j=0..floor(k/2)} (1 - (k-2*j)^2*x).
A(n,k) = k^2 * A(n-1,k) + A(n,k-2) for k > 1.
A(n,k) = (1/(2^k*k!)) * Sum_{j=0..k} (-1)^j * (k-2*j)^(2*n+k) * binomial(k,j).

A026337 a(n) = 4^n*(4^n - 1)/2.

Original entry on oeis.org

0, 6, 120, 2016, 32640, 523776, 8386560, 134209536, 2147450880, 34359607296, 549755289600, 8796090925056, 140737479966720, 2251799780130816, 36028796884746240, 576460751766552576, 9223372034707292160, 147573952581086478336, 2361183241400462868480
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Cf. A166984.

Programs

  • Magma
    [4^n*(4^n-1)/2: n in [0..30]]; // Vincenzo Librandi, May 01 2011
  • Maple
    seq(binomial(4^n,2),n=0..18); # Zerinvary Lajos, Jan 07 2008
  • Mathematica
    Table[4^n (4^n-1)/2,{n,0,30}] (* or *) LinearRecurrence[{20,-64},{0,6},30] (* Harvey P. Dale, Nov 05 2023 *)
  • SageMath
    [binomial(4^n, 2) for n in range(21)] # G. C. Greubel, Oct 02 2024

Formula

a(n) = binomial(4^n, 2), n >= 0. - Zerinvary Lajos, Jan 07 2008
From R. J. Mathar, Mar 20 2009: (Start)
a(n) = 20*a(n-1) - 64*a(n-2).
G.f.: 6*x/((1-4*x)*(1-16*x)). (End)
a(n) = 6*A166984(n-1). - R. J. Mathar, Jun 23 2013
E.g.f.: exp(10*x)*sinh(6*x). - G. C. Greubel, Oct 02 2024
Showing 1-6 of 6 results.

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

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