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 1-10 of 265 results. Next

A255881 Expansion of exp( Sum_{n >= 1} A000364(n)*x^n/n ).

Original entry on oeis.org

1, 1, 3, 23, 371, 10515, 461869, 28969177, 2454072147, 269732425859, 37312477130105, 6342352991066661, 1299300852841580893, 315702973949640373933, 89765549161833322593411, 29526682496433138896248775, 11124674379405792463701519059
Offset: 0

Views

Author

Peter Bala, Mar 09 2015

Keywords

Comments

A000364(n) = (-1)^n*2^(2*n)*Euler(2*n,1/2), where E(n,x) is the n-th Euler polynomial. In general it appears that when k is a nonzero integer, the expansion of exp( Sum_{n >= 1} k^(2*n)*E(2*n,1/k)*(-x)^n/n ) has (positive) integer coefficients. See A255882 (k = 3), A255883(k = 4) and A255884 (k = 6).

Crossrefs

Programs

  • Maple
    #A255881
    k := 2:
    exp(add(k^(2*n)*euler(2*n, 1/k)*(-x)^n/n, n = 1 .. 16)): seq(coeftayl(%, x = 0, n), n = 0 .. 16);
  • Mathematica
    A000364:= Table[Abs[EulerE[2 n]], {n, 0, 80}]; a:= With[{nmax = 70}, CoefficientList[Series[Exp[Sum[A000364[[k + 1]]*x^(k)/(k), {k, 1, 75}]], {x, 0, nmax}], x]]; Table[a[[n]], {n, 1, 50}] (* G. C. Greubel, Aug 26 2018 *)

Formula

O.g.f.: exp( x + 5*x^2/2 + 61*x^3/3 + 1385*x^4/4 + ... ) = 1 + x + 3*x^2 + 23*x^3 + 371*x^4 + ....
a(0) = 1 and for n >= 1, n*a(n) = Sum_{k = 1..n} (-1)^k*2^(2*k)*E(2*k,1/2)*a(n-k).
a(n) ~ 2^(4*n + 3) * n^(2*n - 1/2) / (Pi^(2*n + 1/2) * exp(2*n)). - Vaclav Kotesovec, Jun 08 2019

A060058 Triangle of numbers related to A000330 (sum of squares) and A000364 (Euler numbers).

Original entry on oeis.org

1, 1, 1, 1, 5, 5, 1, 14, 61, 61, 1, 30, 331, 1385, 1385, 1, 55, 1211, 12284, 50521, 50521, 1, 91, 3486, 68060, 663061, 2702765, 2702765, 1, 140, 8526, 281210, 5162421, 49164554, 199360981, 199360981, 1, 204, 18522, 948002, 28862471, 510964090, 4798037791, 19391512145, 19391512145
Offset: 0

Views

Author

Wolfdieter Lang, Mar 16 2001

Keywords

Examples

			Triangle T(n, k) starts:
  [0] 1;
  [1] 1,   1;
  [2] 1,   5,    5;
  [3] 1,  14,   61,     61;
  [4] 1,  30,  331,   1385,    1385;
  [5] 1,  55, 1211,  12284,   50521,    50521;
  [6] 1,  91, 3486,  68060,  663061,  2702765,   2702765;
  [7] 1, 140, 8526, 281210, 5162421, 49164554, 199360981, 199360981;
  ...
		

Crossrefs

Cf. A060059 (row sums), A000364 (main diagonal Euler numbers).
Columns: A000012 (powers of 1), A000330 (sum of squares), A060060-2 for m=0,...,4.
See triangle A060074.

Programs

  • Maple
    T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1) else (n - k + 1)^2 * T(n, k - 1) + T(n - 1, k) fi fi end:
    seq(print(seq(T(n, k), k=0..n)), n=0..7);  # Peter Luschny, Sep 30 2023
  • Mathematica
    a[, -1] = 0; a[0, 0] = 1; a[n, m_] /; n < m = 0; a[n_, m_] := a[n, m] = a[n-1, m] + (n+1-m)^2*a[n, m-1]; Table[a[n, m], {n, 0, 8}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jul 09 2013 *)

Formula

a(n, m) = a(n-1, m) + ((n+1-m)^2)*a(n, m-1), a(n, -1) := 0, a(0, 0) = 1, a(n, m) = 0 if n < m.
a(n, m) = ay(n-m+1, m) if n >= m >= 0, with the rectangular array ay(n, m) := Sum_{j=1..n} (j^2)*ay(j+1, m-1), n >= 0, m >= 1; input: ay(n, 0)=1 (iterated sums of squares).
G.f. for m-th column: 1/(1-x) for m=0, (x^m)*(Sum_{k=0..m} A060063(m, k)*x^k)/(1-x)^(3*m+1), m >= 1.
Recursion for g.f.s for m-th column: (1-x)*G(m, x) = x*G''(m-1, x) - G'(m-1, x) + G(m-1, x)/x, m >= 2; G(1, x) = x*(1+x)/(1-x)^4; the apostrophe denotes differentiation w.r.t. x. G(0, x) = 1/(1-x). - Wolfdieter Lang, Feb 13 2004

A002437 a(n) = A000364(n) * (3^(2*n+1) + 1)/4.

Original entry on oeis.org

1, 7, 305, 33367, 6815585, 2237423527, 1077270776465, 715153093789687, 626055764653322945, 698774745485355051847, 968553361387420436695025, 1632180870878422847476890007, 3286322019402928956112227932705, 7791592461957309952817483706344167, 21485762937086358457367440231243675985
Offset: 0

Views

Author

Keywords

Comments

The terms are multiples of the Euler numbers (A000364).

Examples

			a(4) = A000364(4) * (3^(2*4+1)+1)/4 = 1385 * (3^9+1)/4 = 1385 * 4921 = 6815585.
		

References

  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 75.
  • J. W. L. Glaisher, Messenger of Math., 28 (1898), 36-79, see esp. p. 51.
  • L. B. W. Jolley, Summation of Series, Dover, 2nd ed. (1961)
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Bisections: A156168, A156169.
Cf. other sequences with a g.f. of the form cos(x)/(1 - k*sin^2(x)): A012494 (k=-1), A001209 (k=1/2), A000364 (k=1), A000281 (k=2), A156134 (k=3).

Programs

  • Maple
    Q:=proc(n) option remember; if n=0 then RETURN(1); else RETURN(expand((u^2+1)*diff(Q(n-1),u)+u*Q(n-1))); fi; end;
    [seq(subs(u=sqrt(3),Q(2*n)),n=0..25)];
  • Mathematica
    Table[Abs[EulerE[2 n]] (3^(2 n + 1) + 1) / 4, {n, 0, 30}] (* Vincenzo Librandi, Feb 07 2017 *)

Formula

A000364(n) * (3^(2*n+1) + 1)/4.
Q_2n(sqrt(3)), where the polynomials Q_n() are defined in A104035. - N. J. A. Sloane, Nov 06 2009
a(n) = (-1)^n*Sum_{k = 0..2*n-1} w^(2*n+k)*Sum_{j = 1..2*n-1} (-1)^(k-j)*binomial(2*n-1,k-j)*(2*j - 1)^(2*n-2), where w = exp(2*Pi*i/6) (i = sqrt(-1)). Cf. A002439. - Peter Bala, Jan 21 2011
Sum_{n>=1} (-1)^floor((n-1)/2) 1/A007310(n)^s = r_s with r_{2s+1} = 2 *(Pi/6)^(2s+1) *a(s) /(2s)!. [Jolley eq (315)]. - R. J. Mathar, Mar 24 2011
From Peter Bala, Feb 06 2017: (Start)
E.g.f.: cos(x)^2/cos(3*x) = cos(x)/(1 - 4*sin(x)^2) = 1 + 7*x^2/2! + 305*x^4/4! + 33367*x^6/6! + .... This is the even part of (1/2)*sec(x + Pi/3). Cf. A000191. (End)
a(n) = (1/2)*Integral_{x = 0..inf} x^(2*n)*cosh(Pi*x/3)/cosh(Pi*x/2) dx. - Cf. A000281. - Peter Bala, Nov 08 2019

Extensions

More terms from Herman P. Robinson
Further terms from N. J. A. Sloane, Nov 06 2009

A086646 Triangle, read by rows, of numbers T(n,k), 0 <= k <= n, given by T(n,k) = A000364(n-k)*binomial(2*n, 2*k).

Original entry on oeis.org

1, 1, 1, 5, 6, 1, 61, 75, 15, 1, 1385, 1708, 350, 28, 1, 50521, 62325, 12810, 1050, 45, 1, 2702765, 3334386, 685575, 56364, 2475, 66, 1, 199360981, 245951615, 50571521, 4159155, 183183, 5005, 91, 1, 19391512145, 23923317720, 4919032300, 404572168, 17824950, 488488, 9100, 120, 1
Offset: 0

Views

Author

Philippe Deléham, Jul 26 2003

Keywords

Comments

The elements of the matrix inverse are apparently given by T^(-1)(n,k) = (-1)^(n+k)*A086645(n,k). - R. J. Mathar, Mar 14 2013
Let E(y) = Sum_{n >= 0} y^n/(2*n)! = cosh(sqrt(y)). Then this triangle is the generalized Riordan array (1/E(-y), y) with respect to the sequence (2*n)! as defined in Wang and Wang. - Peter Bala, Aug 06 2013
Let P_n be the poset of even size subsets of [2n] ordered by inclusion. Then Sum_{k=0..n}(-1)^(n-k)*T(n,k)*x^k is the characteristic polynomial of P_n. - Geoffrey Critzer, Feb 24 2021

Examples

			Triangle begins:
      1;
      1,     1;
      5,     6,     1;
     61,    75,    15,    1;
   1385,  1708,   350,   28,  1;
  50521, 62325, 12810, 1050, 45, 1;
  ...
From _Peter Bala_, Aug 06 2013: (Start)
Polynomial  |        Real zeros to 5 decimal places
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
R(5,-x)     | 1, 9.18062, 13.91597
R(10,-x)    | 1, 9.00000, 25.03855,  37.95073
R(15,-x)    | 1, 9.00000, 25.00000,  49.00895, 71.83657
R(20,-x)    | 1, 9.00000, 25.00000,  49.00000, 81.00205, 114.87399
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
(End)
		

Crossrefs

Cf. A000281.
Cf. A000795 (row sums).
Cf. A055133, A086645 (unsigned matrix inverse), A103364, A104033.
T(2n,n) give |A214445(n)|.

Programs

  • Maple
    A086646 := proc(n,k)
        if k < 0 or k > n then
            0 ;
        else
            A000364(n-k)*binomial(2*n,2*k) ;
        end if;
    end proc: # R. J. Mathar, Mar 14 2013
  • Mathematica
    R[0, _] = 1;
    R[n_, x_] := R[n, x] = x^n - Sum[(-1)^(n-k) Binomial[2n, 2k] R[k, x], {k, 0, n-1}];
    Table[CoefficientList[R[n, x], x], {n, 0, 8}] // Flatten (* Jean-François Alcover, Dec 19 2019 *)
    T[0, 0] := 1; T[n_, 0] := -Sum[(-1)^k T[n, k], {k, 1, n}]; T[n_, k_]/;0Oliver Seipel, Jan 11 2025 *)

Formula

cosh(u*t)/cos(t) = Sum_{n>=0} S_2n(u)*(t^(2*n))*(1/(2*n)!). S_2n(u) = Sum_{k>=0} T(n,k)*u^(2*k). Sum_{k>=0} (-1)^k*T(n,k) = 0. Sum_{k>=0} T(n,k) = 2^n*A005647(n); A005647: Salie numbers.
Triangle T(n,k) read by rows; given by [1, 4, 9, 16, 25, 36, 49, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is the operator defined in A084938.
Sum_{k=0..n} (-1)^k*T(n,k)*4^(n-k) = A000281(n). - Philippe Deléham, Jan 26 2004
Sum_{k=0..n} T(n,k)*(-4)^k*9^(n-k) = A002438(n+1). - Philippe Deléham, Aug 26 2005
Sum_{k=0..n} (-1)^k*9^(n-k)*T(n,k) = A000436(n). - Philippe Deléham, Oct 27 2006
From Peter Bala, Aug 06 2013: (Start)
Let E(y) = Sum_{n >= 0} y^n/(2*n)! = cosh(sqrt(y)). Generating function: E(x*y)/E(-y) = 1 + (1 + x)*y/2! + (5 + 6*x + x^2)*y^2/4! + (61 + 75*x + 15*x^2 + x^3)*y^3/6! + .... The n-th power of this array has a generating function E(x*y)/E(-y)^n. In particular, the matrix inverse is a signed version of A086645 with a generating function E(-y)*E(x*y).
Recurrence equation for the row polynomials: R(n,x) = x^n - Sum_{k = 0..n-1} (-1)^(n-k)*binomial(2*n,2*k)*R(k,x) with initial value R(0,x) = 1.
It appears that for arbitrary complex x we have lim_{n -> infinity} R(n,-x^2)/R(n,0) = cos(x*Pi/2). A stronger result than pointwise convergence may hold: the convergence may be uniform on compact subsets of the complex plane. This would explain the observation that the real zeros of the polynomials R(n,-x) seem to converge to the odd squares 1, 9, 25, ... as n increases. Some numerical examples are given below. Cf. A055133, A091042 and A103364.
R(n,-1) = 0; R(n,-9) = (-1)^n*2*4^n; R(n,-25) = (-1)^n*2*(16^n - 4^n);
R(n,-49) = (-1)^n*2*(36^n - 16^n + 4^n). (End)

A060164 Number of orbits of length n under the map whose periodic points are counted by A000364.

Original entry on oeis.org

1, 2, 20, 345, 10104, 450450, 28480140, 2423938845, 267208852820, 37037118818700, 6304443126648900, 1292877846962865230, 314390193022547991720, 89447117243116404721950, 29436259549934873636908816, 11094961973721205588579579845, 4748429366816935180127543967840
Offset: 1

Views

Author

Thomas Ward, Mar 13 2001

Keywords

Comments

The sequence A000364 seems to record the number of points of period n under a map. The number of orbits of length n for this map gives the sequence above.

Examples

			u(3) = 20 since the conjectured map whose periodic points are counted by A000364 would have 1 fixed point and 61 points of period 3, so it must have 20 orbits of length 3.
		

Crossrefs

Formula

a(n) = (1/n)* Sum_{d|n} mu(d)*A000364(n/d).

A125053 Variant of triangle A008301, read by rows of 2*n+1 terms, such that the first column is the secant numbers (A000364).

Original entry on oeis.org

1, 1, 3, 1, 5, 15, 21, 15, 5, 61, 183, 285, 327, 285, 183, 61, 1385, 4155, 6681, 8475, 9129, 8475, 6681, 4155, 1385, 50521, 151563, 247065, 325947, 378105, 396363, 378105, 325947, 247065, 151563, 50521, 2702765, 8108295, 13311741, 17908935
Offset: 0

Views

Author

Paul D. Hanna, Nov 21 2006, Dec 20 2006

Keywords

Comments

Foata and Han refer to this as the triangle of Poupard numbers h_n(k). - N. J. A. Sloane, Feb 17 2014
Central terms (A125054) equal the binomial transform of the tangent numbers (A000182).

Examples

			If we write the triangle like this:
......................... ...1;
................... ...1, ...3, ...1;
............. ...5, ..15, ..21, ..15, ...5;
....... ..61, .183, .285, .327, .285, .183, ..61;
. 1385, 4155, 6681, 8475, 9129, 8475, 6681, 4155, 1385;
then the first nonzero term is the sum of the previous row:
1385 = 61 + 183 + 285 + 327 + 285 + 183 + 61,
the next term is 3 times the first:
4155 = 3*1385,
and the remaining terms in each row are obtained by the rule illustrated by:
6681 = 2*4155 - 1385 - 4*61;
8475 = 2*6681 - 4155 - 4*183;
9129 = 2*8475 - 6681 - 4*285;
8475 = 2*9129 - 8475 - 4*327;
6681 = 2*8475 - 9129 - 4*285;
4155 = 2*6681 - 8475 - 4*183;
1385 = 2*4155 - 6681 - 4*61.
An alternate recurrence is illustrated by:
4155 = 1385 + 2*(61 + 183 + 285 + 327 + 285 + 183 + 61);
6681 = 4155 + 2*(183 + 285 + 327 + 285 + 183);
8475 = 6681 + 2*(285 + 327 + 285);
9129 = 8475 + 2*(327);
and then for k>n, T(n,k) = T(n,2*n-k).
		

Crossrefs

Cf. A008301, A000364 (secant numbers, which are the row sums), A125054 (central terms), A125055, A000182, A008282.
Cf. A210111 (left half).

Programs

  • Haskell
    a125053 n k = a125053_tabf !! n !! k
    a125053_row n = a125053_tabf !! n
    a125053_tabf = iterate f [1] where
    f zs = zs' ++ reverse (init zs') where
    zs' = (sum zs) : g (map (* 2) zs) (sum zs)
    g [x] y = [x + y]
    g xs y = y' : g (tail $ init xs) y' where y' = sum xs + y
    -- Reinhard Zumkeller, Mar 17 2012
  • Maple
    T := proc(n, k) option remember; local j;
      if n = 1 then 1
    elif k = 1 then add(T(n-1, j), j=1..2*n-3)
    elif k = 2 then 3*T(n, 1)
    elif k > n then T(n, 2*n-k)
    else 2*T(n, k-1) - T(n, k-2) - 4*T(n-1, k-2)
      fi end:
    seq(print(seq(T(n,k), k=1..2*n-1)), n=1..5); # Peter Luschny, May 11 2014
  • Mathematica
    t[n_, k_] := t[n, k] = If[2*n < k || k < 0, 0, If[n == 0 && k == 0, 1, If[k == 0, Sum[t[n-1, j], {j, 0, 2*n-2}], If[k <= n, t[n, k-1] + 2*Sum[t[n-1, j], {j, k-1, 2*n-1-k}], t[n, 2*n-k]]]]]; Table[t[n, k], {n, 0, 6}, {k, 0, 2*n}] // Flatten (* Jean-François Alcover, Dec 06 2012, translated from Pari *)
  • PARI
    T(n,k)=if(2*n
    				

Formula

Sum_{k=0..2n} C(2n,k)*T(n,k) = 4^n * A000182(n), where A000182 are the tangent numbers.
Sum_{k=0..2n} (-1)^n*C(2n,k)*T(n,k) = (-4)^n.

A034971 a(n) = floor(E_(n+1)/E_(n)) where E_n is n-th Euler number (see A028296 and A000364).

Original entry on oeis.org

5, 12, 22, 36, 53, 73, 97, 124, 154, 187, 223, 263, 306, 352, 402, 454, 510, 569, 632, 697, 766, 838, 914, 992, 1074, 1159, 1248, 1339, 1434, 1532, 1634, 1738, 1846, 1957, 2071, 2189, 2310, 2434, 2561, 2691, 2825, 2962, 3102, 3246, 3393, 3542, 3696, 3852
Offset: 1

Views

Author

Keywords

Comments

First nine terms are identical to those of A025740.

References

  • J. Peters J. and J. Stein, Mathematische Tafeln. Revised Russian Edition in 1968, Moscow, Table 9b.

Crossrefs

Programs

  • Mathematica
    Wolfram: ae[ x_ ] := Abs[ EulerE[ 2*x ] ] and fo[ x_ ] := Floor[ N[ ae[ x ]/ae[ x-1 ], 80 ] ]
    Table[Floor[Abs[EulerE[2n+2]]/Abs[EulerE[2n]]],{n,60}] (* Harvey P. Dale, May 01 2011 *)

Formula

a(n) = floor(E(n+1)/E(n)) where E(n) is the n-th coefficient in the expansion of sec(x).

A047893 Number of decimal digits of Euler (Zig) numbers A000364.

Original entry on oeis.org

1, 1, 2, 4, 5, 7, 9, 11, 13, 15, 17, 20, 22, 25, 27, 30, 32, 35, 38, 41, 44, 46, 49, 52, 55, 58, 61, 64, 68, 71, 74, 77, 80, 84, 87, 90, 94, 97, 100, 104, 107, 111, 114, 118, 121, 125, 128, 132, 135, 139, 143, 146, 150, 154, 157, 161, 165, 168, 172, 176, 180, 183
Offset: 1

Views

Author

Keywords

Examples

			a(4) = floor(log_10(1385)) + 1 = 4, E(4) = 1385, the 4th Euler number.
		

References

  • J. Peters and J. Stein, Mathematische Tafeln, Revised Russian Edition, Moscow, 1968.

Crossrefs

Cf. A000364, A034971, A034972, A000182 (tangent numbers), A047894.

Programs

  • Mathematica
    a[n_] := IntegerLength[EulerE[2*n]]; Array[a, 100] (* Amiram Eldar, Mar 20 2025 *)
  • PARI
    a(n) = #Str(subst(bernpol(2*n+1), 'x, 1/4)*4^(2*n+1)*(-1)^(n+1)/(2*n+1));

Formula

a(n) = A055642(A000364(n)) = floor(log_10(|E(2n)|)) + 1 for n >=2, where E is Euler's E function.

Extensions

a(1) corrected by Amiram Eldar, Mar 20 2025

A173226 Partial sums of A000364.

Original entry on oeis.org

1, 2, 7, 68, 1453, 51974, 2754739, 202115720, 19593627865, 2424473303306, 372795661540831, 69721670054678732, 15584255833611765637, 4102656765126735657998, 1256362298168756601126283, 442800255547191861154809104, 177962191835086481297819598769
Offset: 0

Views

Author

Jonathan Vos Post, Feb 13 2010

Keywords

Comments

Partial sums of Euler numbers. Partial sums of secant or "Zig" numbers. The subsequence of prime partial sum of Euler numbers begins 2, 7, 1453, no more through a(17). What is the next such prime?

Crossrefs

Programs

  • Mathematica
    Accumulate[Table[Abs[EulerE[2n]],{n,0,20}]] (* Harvey P. Dale, Aug 10 2024 *)
  • Python
    from sympy import euler
    def A173226(n): return sum(abs(euler(i)) for i in range(0,(n<<1)+1,2)) # Chai Wah Wu, Apr 16 2023

Formula

a(n) = Sum_{i=0..n} A000364(i).
G.f.: 1/U(0)/(1-x) where U(k)=1 + x - x*(2*k+1)*(2*k+2)/(1 - x*(2*k+1)*(2*k+2)/U(k+1)) ; (continued fraction, 2-step). - Sergei N. Gladkovskii, Oct 15 2012
G.f.: 1/(1-x)/Q(0), where Q(k)= 1 - x*(2*k+1)^2/(1 - x*(2*k+2)^2/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 27 2013
G.f.: Q(0)/(1-x), where Q(k) = 1 - x*(k+1)^2/( x*(k+1)^2 - 1/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 22 2013

A255895 O.g.f.: exp( Sum_{n>=1} A000364(n+1)*x^n/n ), where A000364 is the Euler numbers.

Original entry on oeis.org

1, 5, 43, 635, 15811, 626543, 36670645, 2968583909, 316359049075, 42804896729263, 7156058577755345, 1447112885406569905, 347972935518441794285, 98096607566761160791345, 32036179679728008237066395, 11996707788374286511611298267, 5105803366761640532801681583635
Offset: 0

Views

Author

Paul D. Hanna, Mar 09 2015

Keywords

Comments

Inspired by Peter Bala's sequence A255881, which suggests the following conjecture: exp( Sum_{n>=1} A000364(n+m)*x^n/n ) is an integer series for integer m>=0.

Examples

			O.g.f.: A(x) = 1 + 5*x + 43*x^2 + 635*x^3 + 15811*x^4 + 626543*x^5 +...
where
log(A(x)) = 5*x + 61*x^2/2 + 1385*x^3/3 + 50521*x^4/4 + 2702765*x^5/5 + 199360981*x^6/6 + 19391512145*x^7/7 + 2404879675441*x^8/8 +...+ A000364(n+1)*x^n/n +...
		

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[Exp[Sum[Abs[EulerE[2*k+2]]*x^k/k, {k, 1, n}]], {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Nov 05 2021 *)
  • PARI
    /* By the Ordinary Generating Function: */
    {A000364(n)=polcoeff(sum(m=0, n, (2*m)!/2^m * x^m/prod(k=1, m, 1+k^2*x+x*O(x^n))), n)}
    {a(n)=local(A=1);A=exp(sum(m=1,n,A000364(m+1)*x^m/m) +x*O(x^n));polcoeff(A,n)}
    for(n=0,20,print1(a(n),", "))

Formula

a(n) ~ 2^(4*n + 7) * n^(2*n + 3/2) / (exp(2*n) * Pi^(2*n + 5/2)). - Vaclav Kotesovec, Nov 05 2021
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