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.

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A274181 Decimal expansion of Phi(1/2, 2, 2), where Phi is the Lerch transcendent.

Original entry on oeis.org

3, 2, 8, 9, 6, 2, 1, 0, 5, 8, 6, 0, 0, 5, 0, 0, 2, 3, 6, 1, 0, 6, 2, 5, 2, 8, 0, 6, 3, 8, 7, 2, 0, 4, 3, 4, 9, 7, 6, 7, 9, 3, 8, 9, 9, 2, 2, 4, 5, 0, 5, 7, 0, 1, 7, 3, 7, 3, 8, 8, 1, 9, 1, 4, 9, 2, 6, 8, 4, 1, 7, 6, 2, 8, 6, 7, 3, 2, 8, 0, 3, 2, 6, 7, 3, 6, 1, 2, 7, 4, 3, 5, 1, 6, 6, 3, 4, 2, 8, 7, 4
Offset: 0

Views

Author

Johannes W. Meijer and N. H. G. Baken, Jun 17 2016, Jul 08 2016

Keywords

Comments

The exponential integral distribution is defined by p(x, m, n, mu) = ((n+mu-1)^m * x^(mu-1) / (mu-1)!) * E(x, m, n), see A163931 and the Meijer link. The moment generating function of this probability distribution function is M(a, m, n, mu) = Sum_{k>=0}(((mu+k-1)!/((mu-1)!*k!)) * ((n+mu-1) / (n+mu+k-1))^m * a^k).
In the special case that mu=1 we get p(x, m, n, mu=1) = n^m * E(x, m, n) and M(a, m, n, mu=1) = n^m * Phi(a, m, n), with Phi the Lerch transcendent. If n=1 and mu=1 we get M(a, m, n=1, mu=1) = polylog(m, a)/a = Li_m(a)/a.

Examples

			0.32896210586005002361062528063872043497679389922...
		

References

  • William Feller, An introduction to probability theory and its applications, Vol. 1. p. 285, 1968.

Crossrefs

Cf. A163931, A002162 (Phi(1/2, 1, 1)/2), A076788 (Phi(1/2, 2, 1)/2), A112302, A008276.

Programs

  • Maple
    Digits := 101; c := evalf(LerchPhi(1/2, 2, 2));
  • Mathematica
    N[HurwitzLerchPhi[1/2, 2, 2], 25] (* G. C. Greubel, Jun 19 2016 *)
  • PARI
    Pi^2/3 - 2*log(2)^2 - 2 \\ Altug Alkan, Jul 08 2016
    
  • PARI
    lerchphi(.5,2,2) \\ Charles R Greathouse IV, Jan 30 2025
    
  • Python
    from mpmath import mp, lerchphi
    mp.dps=102
    print([int(d) for d in list(str(lerchphi(1/2, 2, 2))[2:-1])]) # Indranil Ghosh, Jul 04 2017

Formula

Equals Phi(1/2, 2, 2) with Phi the Lerch transcendent.
Equals Sum_{k>=0}(1/((2+k)^2*2^k)).
Equals 4 * polylog(2, 1/2) - 2.
Equals Pi^2/3 - 2*log(2)^2 - 2.
Equals Integral_{x=0..oo} x*exp(-x)/(exp(x)-1/2) dx. - Amiram Eldar, Aug 24 2020

A054649 Triangle T(n, k) giving coefficients in expansion of n! * Sum_{i=0..n} binomial(x - n, i) in powers of x.

Original entry on oeis.org

1, 1, 0, 1, -3, 4, 1, -9, 32, -36, 1, -18, 131, -426, 528, 1, -30, 375, -2370, 7544, -9600, 1, -45, 865, -8955, 52414, -163800, 213120, 1, -63, 1729, -26565, 245854, -1366932, 4220376, -5574240, 1, -84, 3122, -66696, 893249, -7664916, 41096908, -125747664, 167973120
Offset: 0

Views

Author

N. J. A. Sloane, Apr 16 2000

Keywords

Examples

			Triangle begins:
  1;
  1,   0;
  1,  -3,    4;
  1,  -9,   32,    -36;
  1, -18,  131,   -426,    528;
  1, -30,  375,  -2370,   7544,    -9600;
  1, -45,  865,  -8955,  52414,  -163800,  213120;
  1, -63, 1729, -26565, 245854, -1366932, 4220376, -5574240;
  ...
From _Peter Luschny_, Nov 27 2021: (Start)
The row reversed triangle can be seen as the coefficients of a sequence of monic polynomials with monomials sorted in ascending order which start:
[0]     1;
[1]              x;
[2]     4 -    3*x +      x^2;
[3]   -36 +   32*x -    9*x^2 +     x^3;
[4]   528 -  426*x +  131*x^2 -  18*x^3 +    x^4;
[5] -9600 + 7544*x - 2370*x^2 + 375*x^3 - 30*x^4 + x^5; (End)
		

Crossrefs

Programs

  • Maple
    # Some older Maple versions are known to have a bug in the hypergeom function.
    with(ListTools): with(PolynomialTools):
    CoeffList := p -> op(Reverse(CoefficientList(simplify(p), x))):
    p := k -> k!*hypergeom([-k, -x + k], [-k], -1):
    seq(CoeffList(p(k)), k = 0..8); # Peter Luschny, Nov 27 2021
  • Mathematica
    c[n_, k_] := Product[n-i, {i, 0, k-1}]/k!; row[n_] := CoefficientList[ n!*Sum[c[x-n, k], {k, 0, n}], x] // Reverse; Table[ row[n], {n, 0, 8}] // Flatten  (* Jean-François Alcover, Oct 04 2012 *)
  • PARI
    row(n) = Vec(n!*sum(k=0, n, binomial(x-n, k))); \\ Seiichi Manyama, Sep 24 2021

Formula

T(n, k) = n! * [x^(n - k)] hypergeom([-n, -x + n], [-n], -1). - Peter Luschny, Nov 27 2021

A102411 Even triangle !n. This table read by rows gives the coefficients of sum formulas of n-th Left factorials (A003422). The k-th row (6>=k>=1) contains T(i,k) for i=1 to k+2, where k=[2*n+1+(-1)^(n-1)]/4 and T(i,k) satisfies !n = Sum_{i=1..k+2} T(i,k) * n^(i-1) / (2*k-2)!.

Original entry on oeis.org

0, 1, 0, -16, 5, 1, 0, 5256, -3068, 276, 32, 0, 2070720, 2367420, -912150, 53220, 3510, 0, -36031524480, 15327895296, -40587120, -387492840, 21414120, 758184, 840, -212459319878400, -75473246681280, 38182549456800, -2562251680800, -195611371200, 13639812480, 285616800, 453600
Offset: 1

Views

Author

André F. Labossière, Jan 07 2005

Keywords

Comments

The sum of signed coefficients for each k-th row is divisible by (2*k-2)!. Moreover, another variant (but an incomplete one, and sorted differently) of the above sequence is presented in A101752.

Examples

			Triangle starts:
0, 1, 0;
-16, 5, 1, 0;
5256, -3068, 276, 32, 0;
2070720, 2367420, -912150, 53220, 3510, 0;
-36031524480, 15327895296, -40587120, -387492840, 21414120, 758184, 840;
...
!11=4037914; substituting n=11 in the formula of the k-th row we obtain k=6 and the coefficients T(i,6) are those needed for computing !11.
=> !11 = [ -212459319878400 -75473246681280*11 +38182549456800*11^2 -2562251680800*11^3 -195611371200*11^4 +13639812480*11^5 +285616800*11^6 +453600*11^7 ]/10! = 4037914.
		

Crossrefs

A102412 Odd triangle !n. This table read by rows gives the coefficients of sum formulas of n-th Left factorials (A003422). The k-th row (6>=k>=1) contains T(i,k) for i=1 to k+1, where k=[2*n+3+(-1)^n]/4 and T(i,k) satisfies !n = Sum_{i=1..k+1} T(i,k) * n^(i-1) / (2*k-2)!.

Original entry on oeis.org

0, 1, -4, 4, 0, 96, -396, 108, 0, 1012320, -192900, -64890, 11460, 90, -2038014720, 1977810240, -304486560, -12131280, 2792160, 21840, -33190735737600, 4445760574080, 2334485260800, -394554283200, 2330344800, 1198048320, 8215200
Offset: 1

Views

Author

André F. Labossière, Jan 07 2005

Keywords

Comments

Incidentally, the sum of signed coefficients for each k-th row is divisible by (2*k-2)!.

Examples

			Triangle starts:
0, 1;
-4, 4, 0;
96, -396, 108, 0;
1012320, -192900, -64890, 11460, 90;
-2038014720, 1977810240, -304486560, -12131280, 2792160, 21840;
...
!11=4037914; substituting n=11 in the formula of the k-th row we obtain k=6 and the coefficients T(i,6) are those needed for computing !11.
=> !11 = [ -33190735737600 +4445760574080*11 +2334485260800*11^2 -394554283200*11^3 +2330344800*11^4 +1198048320*11^5 +8215200*11^6 ]/10! = 4037914.
		

Crossrefs

A115298 Row sums of A115297.

Original entry on oeis.org

1, 3, 7, 13, 21, 31, 43, 55, 73, 91, 115, 139, 165, 193, 227, 263, 301, 339, 381, 423, 471, 517, 569, 625, 685, 745, 809, 871, 937, 1011, 1089, 1167, 1247, 1335, 1425, 1515, 1611, 1707, 1809, 1915, 2023, 2135, 2249, 2363, 2479, 2601, 2735, 2865, 2997, 3129
Offset: 1

Views

Author

André F. Labossière, Jan 19 2006

Keywords

Crossrefs

Programs

  • Maple
    a:=n->add(ithprime(k)-ithprime(ceil(k/2)),k=2..n+1): seq(a(n),n=1..60); # Muniru A Asiru, Jan 04 2019
  • Mathematica
    Accumulate[Table[Prime[n] - Prime[Ceiling[n/2]], {n, 2, 51}]] (* Jon Maiga, Jan 04 2019 *)

Formula

a(n) = Sum_{k=2..n+1} (A000040(k) - A000040(ceiling(k/2))). - Jon Maiga, Jan 04 2019

Extensions

More terms from André F. Labossière, Mar 27 2006

A087748 Triangle formed by reading triangle of Stirling numbers of the first kind (A048994) mod 2.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1
Offset: 0

Views

Author

Philippe Deléham, Oct 02 2003

Keywords

Examples

			Triangle begins:
1,
0, 1,
0, 1, 1,
0, 0, 1, 1,
0, 0, 1, 0, 1,
0, 0, 0, 1, 0, 1,
0, 0, 0, 1, 1, 1, 1,
0, 0, 0, 0, 1, 1, 1, 1,
0, 0, 0, 0, 1, 0, 0, 0, 1,
0, 0, 0, 0, 0, 1, 0, 0, 0, 1,
0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1,
0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1,
0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1,
...
		

References

  • Brand, Neal; Das, Sajal; Jacob, Tom. The number of nonzero entries in recursively defined tables modulo primes. Proceedings of the Twenty-first Southeastern Conference on Combinatorics, Graph Theory, and Computing (Boca Raton, FL, 1990). Congr. Numer. 78 (1990), 47--59. MR1140469 (92h:05004). - From N. J. A. Sloane, Jun 03 2012

Crossrefs

Also parity of triangles A049444, A049459, A051338, A051379, A051523.

Formula

T(n, k) = A087755(n, k) = A048994(n, k) mod 2 = A047999([n/2], k-[(n+1)/2]) = T(n-2, k-2) XOR T(n-2, k-1) with T(0, 0) = T(1, 1) = 1 and T(1, 0) = 0; T(2n, k) = T(2n-1, k-1) XOR T(2n-1, k); T(2n+1, k) = T(2n, k-1). - Henry Bottomley, Dec 01 2003

Extensions

Edited and extended by Henry Bottomley, Dec 01 2003

A101751 Table (read by rows) giving the coefficients of sum formulas of n-th Factorials (A000142). The k-th row (k>=1, n>=2) contains T(i,k) for i=1 to k+1, where k=[2*n+1+(-1)^(n-1)]/4 and T(i,k) satisfies Fact(n) = Sum_{i=1..k+1} T(i,k) * (n-1)^(k-i+1) / (2*k-2)!.

Original entry on oeis.org

1, 0, 1, 3, -6, 32, 264, -2024, 2400, 3420, 55800, -666540, 909720, 2570400, 90440, 13101144, 72406040, -3757930680, 13117344800, 72965762016, -261763004160
Offset: 1

Views

Author

André F. Labossière, Dec 17 2004

Keywords

Examples

			Fact(8) = 5040; substituting n=8 in the formula of the k-th row we obtain k=4 and the coefficients
T(i,4) will be the following: 3420,55800,-666540,909720,2570400, => Fact(8) = [ 3420*7^4 +55800*7^3 -666540*7^2 +909720*7 +2570400 ]/6! = 7! =5040.
		

Crossrefs

A213167 a(n) = n! - (n-2)!.

Original entry on oeis.org

1, 5, 22, 114, 696, 4920, 39600, 357840, 3588480, 39553920, 475372800, 6187104000, 86699289600, 1301447347200, 20835611596800, 354379753728000, 6381450915840000, 121289412980736000, 2426499634470912000
Offset: 2

Views

Author

Olivier Gérard, Nov 02 2012

Keywords

Comments

Row sums of A134433 starting from k=3.
a(n) = sum( (-1)^k*k*A008276(n,k), k=1..n-1).
a(n) = sum( (-1)^k*k*A054654(n,k), k=1..n-2).
For n >= 3, a(n) = number whose factorial base representation (A007623) begins with digits {n-1} and {n-2} followed by n-3 zeros. Viewed in that base, this sequence looks like this: 1, 21, 320, 4300, 54000, 650000, 7600000, 87000000, 980000000, A900000000, BA000000000, ... (where "digits" A and B stand for placeholder values 10 and 11 respectively). - Antti Karttunen, May 07 2015.

Crossrefs

Column 4 of A257503 (apart from initial 1. Equally, row 4 of A257505).
Cf. A067318.

Programs

Formula

a(n) = n! - (n-2)!.
G.f.: (1/G(0) - 1 - x)/x^2 where G(k) = 1 - x/(x - 1/(x - (k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Dec 13 2012
G.f.: (1+x)/x^2*(1/Q(0)-1), where Q(k)= 1 - 2*k*x - x^2*(k + 1)^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 08 2013
G.f.: 2*Q(0), where Q(k)= 1 - 1/( (k+1)*(k+2) - x*(k+1)^2*(k+2)^2*(k+3)/(x*(k+1)*(k+2)*(k+3) - 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 08 2013

A102409 Even triangle n!. This table read by rows gives the coefficients of sum formulas of n-th factorials (A000142). The k-th row (6>=k>=1) contains T(i,k) for i=1 to k+3, where k=[2*n+1+(-1)^(n-1)]/4 and T(i,k) satisfies n! = Sum_{i=1..k+3} T(i,k) * n^(i-1) / (2*k-2)!.

Original entry on oeis.org

0, 1, 0, 0, 0, -20, 8, 0, 0, 20280, -6530, -1275, 362, 3, 0, -8749440, 21627600, -4871940, -66510, 48300, 390, 0, -261763004160, 72965762016, 13117344800, -3757930680, 72406040, 13101144, 90440, 0, -974260634054400, -1140185248443360, 353509119454680, -8136128999880, -3234018579750
Offset: 1

Views

Author

André F. Labossière, Jan 07 2005

Keywords

Comments

Incidentally, the sum of signed coefficients for each k-th row is divisible by (2*k-2)!. Moreover, another variant (but an incomplete one, and sorted differently) of the above sequence is presented in A101751.

Examples

			Triangle starts:
0, 1, 0, 0;
0, -20, 8, 0, 0;
20280, -6530, -1275, 362, 3, 0;
-8749440, 21627600, -4871940, -66510, 48300, 390, 0;
-261763004160, 72965762016, 13117344800, -3757930680, 72406040, 13101144, 90440, 0;
...
11!=39916800; substituting n=11 in the formula of the k-th row we obtain k=6 and the coefficients T(i,6) are those needed for computing 11!.
=> 11! = [ -974260634054400 -1140185248443360*11 +353509119454680*11^2 -8136128999880*11^3 -3234018579750*11^4 +109743298560*11^5 +6053880420*11^6 +34067880*11^7 +9450*11^8 ]/10! = 39916800.
		

Crossrefs

A102410 Odd triangle n!. This table read by rows gives the coefficients of sum formulas of n-th Factorials (A000142). The k-th row (6>=k>=1) contains T(i,k) for i=1 to k+2, where k=[2*n+3+(-1)^n]/4 and T(i,k) satisfies n! = Sum_{i=1..k+2} T(i,k) * n^(i-1) / (2*k-2)!.

Original entry on oeis.org

1, 0, 0, -6, 3, 1, 0, 2400, -2024, 264, 32, 0, 2570400, 909720, -666540, 55800, 3420, 0, -19071521280, 12195884736, -762499920, -282106440, 22425480, 741384, 840, -219303218534400, -11953192930560, 27128332828800, -2808016545600, -125442525600, 14164990560, 280576800
Offset: 1

Views

Author

André F. Labossière, Jan 07 2005

Keywords

Comments

Incidentally, the sum of signed coefficients for each k-th row is divisible by (2*k-2)!.

Examples

			Triangle starts:
1, 0, 0;
-6, 3, 1, 0;
2400, -2024, 264, 32, 0;
2570400, 909720, -666540, 55800, 3420, 0;
-19071521280, 12195884736, -762499920, -282106440, 22425480, 741384, 840;
...
11!=39916800; substituting n=11 in the formula of the k-th row we obtain k=6 and the coefficients T(i,6) are those needed for computing 11!.
=> 11! = [ -219303218534400 -11953192930560*11 +27128332828800*11^2 -2808016545600*11^3 -125442525600*11^4 +14164990560*11^5 +280576800*11^6 +453600*11^7 ]/10! = 39916800.
		

Crossrefs

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