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 29 results. Next

A007917 Version 1 of the "previous prime" function: largest prime <= n.

Original entry on oeis.org

2, 3, 3, 5, 5, 7, 7, 7, 7, 11, 11, 13, 13, 13, 13, 17, 17, 19, 19, 19, 19, 23, 23, 23, 23, 23, 23, 29, 29, 31, 31, 31, 31, 31, 31, 37, 37, 37, 37, 41, 41, 43, 43, 43, 43, 47, 47, 47, 47, 47, 47, 53, 53, 53, 53, 53, 53, 59, 59, 61, 61, 61, 61, 61, 61, 67, 67, 67, 67, 71, 71, 73, 73, 73, 73
Offset: 2

Views

Author

R. Muller

Keywords

Comments

Version 2 of the "previous prime" function (see A151799) is "largest prime < n". This produces the same sequence of numerical values, except the offset (or indexing) starts at 3 instead of 2.
Maple's "prevprime" function uses version 2.
Also the largest prime dividing n! or lcm(1,...,n). - Labos Elemer, Jun 22 2000
Also largest prime among terms of (n+1)st row of Pascal's triangle. - Jud McCranie, Jan 17 2000
Also largest integer k such that A000203(k) <= n+1. - Benoit Cloitre, Mar 17 2002. - Corrected by Antti Karttunen, Nov 07 2017
Also largest prime factor of A061355(n) (denominator of Sum_{k=0..n} 1/k!). - Jonathan Sondow, Jan 09 2005
Also prime(pi(x)) where pi(x) is the prime counting function = number of primes <= x. - Cino Hilliard, May 03 2005
Also largest prime factor, occurring to the power p, in denominator of Sum_{k=1..n} 1/k^p, for any positive integer p. - M. F. Hasler, Nov 10 2006
For n > 10, these values are close to the most negative eigenvalues of A191898 (conjecture). - Mats Granvik, Nov 04 2011

References

  • K. Atanassov, On the 37th and the 38th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 2, 83-85.
  • J. Castillo, Other Smarandache Type Functions: Inferior/Superior Smarandache f-part of x, Smarandache Notions Journal, Vol. 10, No. 1-2-3, 1999, 202-204.

Crossrefs

Programs

Formula

Equals A006530(A000142(n)). - Jonathan Sondow, Jan 09 2005
Equals A006530(A056040(n)). - Peter Luschny, Mar 04 2011
a(n) = A000040(A049084(A007918(n)) + 1 - A010051(n)). - Reinhard Zumkeller, Jul 26 2012
From Wesley Ivan Hurt, May 22 2013: (Start)
omega( Product_{i=2..n} a(i) ) = pi(n).
Omega( Product_{i=2..n} a(i) ) = n - 1. (End)
For n >= 2, a(A000203(n)) = A070801(n). - Antti Karttunen, Nov 07 2017
a(n) = n + 1 - Sum_{i=1..n} floor(pi(i)/pi(n)) = n + 1 - A175851(n). - Ridouane Oudra, Jun 24 2024
Conjecture: a(n) = floor(log(Sum_{k=2..n} exp(A000010(k)+1))). - Joseph M. Shunia, Aug 09 2024
a(n) = A000040(A000720(n)). - Ridouane Oudra, Oct 04 2024

Extensions

Edited by N. J. A. Sloane, Apr 06 2008

A061354 Numerator of Sum_{k=0..n} 1/k!.

Original entry on oeis.org

1, 2, 5, 8, 65, 163, 1957, 685, 109601, 98641, 9864101, 13563139, 260412269, 8463398743, 47395032961, 888656868019, 56874039553217, 7437374403113, 17403456103284421, 82666416490601, 6613313319248080001, 69439789852104840011
Offset: 0

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Author

Amarnath Murthy, Apr 28 2001

Keywords

Comments

p divides a(p-1) for prime p = {2, 5, 13, 37, 463, ...} which apparently coincides with A064384(n) = {2, 5, 13, 37, 463, ...} Primes p such that p divides 0!-1!+2!-3!+...+(-1)^{p-1}(p-1)!. - Alexander Adamchuk, Jun 14 2007
GCD(a(n), a(n+2)) = A124779(n) is either 1 or a prime 2, 5, 13, 37, 463, ... = A064384. - Jonathan Sondow, Jun 12 2007
For proofs of Adamchuk's and my Comments, see the link "The Taylor series for e ...". - Jonathan Sondow, Jun 18 2007

Examples

			1, 2, 5/2, 8/3, 65/24, 163/60, 1957/720, 685/252, ...
		

Crossrefs

Cf. A061355 (denominators), A093101, A064384, A064384, A124779, A129924.

Programs

  • Mathematica
    exp[n_]:=Apply[Plus,1/Range[0,n]!];Numerator[Table[exp[n],{n,0,21}]]  (* Geoffrey Critzer, May 05 2013 *)
    A061354[n_] := Numerator[Sum[1/k!, {k, 0, n}]]; Array[A061354, 22, 0] (* JungHwan Min, Nov 08 2016 *)
    Accumulate[1/Range[0,30]!]//Numerator (* Harvey P. Dale, Apr 13 2018 *)
  • PARI
    { default(realprecision, 500); e=exp(1); for (n=0, 200, a=numerator(floor(n!*e)/n!); if (n==0, a=1); write("b061354.txt", n, " ", a) ) } \\ Harry J. Smith, Jul 21 2009

Formula

a(n) = A000522(n)/A093101(n).
Numerators of floor(n!*exp(1))/n!, n>=1. Numerators of coefficients in expansion of exp(x)/(1-x). - Vladeta Jovovic, Aug 11 2002
a(n) = (1+n+n(n-1)+...+n!)/GCD(n!,1+n+n(n-1)+...+n!). - Jonathan Sondow, Aug 18 2006

A093101 Cancellation factor in reducing Sum_{k=0...n} 1/k! to lowest terms.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 20, 1, 10, 1, 8, 5, 2, 5, 4, 1, 130, 1, 4000, 1, 2, 5, 52, 5, 494, 1, 40, 1, 10, 13, 4, 25, 38, 5, 16, 13, 230, 13, 20, 1, 46, 5, 104, 475, 62, 1, 20, 1, 130, 31, 832, 2755, 74, 5, 4, 13, 50, 1, 40, 23, 2, 2795, 76, 34385, 2, 1, 80, 1, 650, 1, 2812, 5, 74, 5
Offset: 0

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Author

Jonathan Sondow, May 10 2004, Oct 18 2006

Keywords

Comments

Same as n!/A061355(n) and (1+n+n(n-1)+n(n-1)(n-2)+...+n!)/A061354(n).
a(n) is relatively prime to n.
gcd(a(n),a(n+1)) = 1.

Examples

			E.g. 1/0!+1/1!+1/2!+1/3!=16/6=(2*8)/(2*3) so a(3)=2.
		

Crossrefs

(n+1)!/(a(n)*a(n+1)) = A123899(n).
(n+3)!/(a(n)*a(n+1)*a(n+2)) = A123900(n).
(n+3)/GCD(a(n), a(n+2)) = A123901(n).
Cf. also A000522, A061354, A061355.

Programs

  • Mathematica
    f[n_] := n! / Denominator[ Sum[1/k!, {k, 0, n}]]; Table[ f[n], {n, 0, 74}] (* Robert G. Wilson v *)
    (* Second program: *)
    A[n_] := If[n==0,1,n*A[n-1]+1]; Table[GCD[A[n],n! ], {n, 0, 74}]
  • PARI
    A000522(n) = sum(k=0, n, binomial(n, k)*k!); \\ This function from Joerg Arndt, Dec 14 2014
    A093101(n) = gcd(n!,A000522(n)); \\ Antti Karttunen, Jul 12 2017

Formula

a(n) = gcd(n!, 1+n+n(n-1)+n(n-1)(n-2)+...+n!).
a(n) = gcd(n!, A(n)) where A(0) = 1, A(n) = n*A(n-1)+1.

Extensions

More terms from Robert G. Wilson v, May 14 2004

A137452 Triangular array of the coefficients of the sequence of Abel polynomials A(n,x) := x*(x-n)^(n-1).

Original entry on oeis.org

1, 0, 1, 0, -2, 1, 0, 9, -6, 1, 0, -64, 48, -12, 1, 0, 625, -500, 150, -20, 1, 0, -7776, 6480, -2160, 360, -30, 1, 0, 117649, -100842, 36015, -6860, 735, -42, 1, 0, -2097152, 1835008, -688128, 143360, -17920, 1344, -56, 1, 0, 43046721, -38263752, 14880348, -3306744, 459270, -40824, 2268, -72, 1
Offset: 0

Views

Author

Roger L. Bagula, Apr 18 2008

Keywords

Comments

Row sums give A177885.
The Abel polynomials are associated with the Abel operator t*exp(y*t)*p(x) = t*p(x+y).
From Peter Luschny, Jan 14 2009: (Start)
Abs(T(n,k)) is the number of rooted labeled trees on n+1 vertices with a root degree k (Clarke's formula).
The row sums in the unsigned case, Sum_{k=0..n} abs(T(n,k)), count the trees on n+1 labeled nodes, A000272(n+1). (End)
Exponential Riordan array [1, W(x)], W(x) the Lambert W-function. - Paul Barry, Nov 19 2010
The inverse array is the exponential Riordan array [1, x*exp(x)], which is A059297. - Peter Bala, Apr 08 2013
The inverse Bell transform of [1,2,3,...]. See A264428 for the Bell transform and A264429 for the inverse Bell transform. - Peter Luschny, Dec 20 2015
Also the Bell transform of (-1)^n*(n+1)^n. - Peter Luschny, Jan 18 2016

Examples

			Triangle begins:
  1;
  0,        1;
  0,       -2,       1;
  0,        9,      -6,       1;
  0,      -64,      48,     -12,      1;
  0,      625,    -500,     150,    -20,      1;
  0,    -7776,    6480,   -2160,    360,    -30,    1;
  0,   117649, -100842,   36015,  -6860,    735,  -42,   1;
  0, -2097152, 1835008, -688128, 143360, -17920, 1344, -56, 1;
		

References

  • Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), pp. 14 and 29

Crossrefs

Row sums A177885.
Cf. A000272, A061356, A059297 (inverse array), A264429.

Programs

  • Maple
    T := proc(n,k) if n = 0 and k = 0 then 1 else binomial(n-1,k-1)*(-n)^(n-k) fi end; seq(print(seq(T(n,k),k=0..n)),n=0..7); # Peter Luschny, Jan 14 2009
    # The function BellMatrix is defined in A264428.
    BellMatrix(n -> (-n-1)^n, 9); # Peter Luschny, Jan 27 2016
  • Mathematica
    a0 = 1 a[x, 0] = 1; a[x, 1] = x; a[x_, n_] := x*(x - a0*n)^(n - 1); Table[Expand[a[x, n]], {n, 0, 10}]; a1 = Table[CoefficientList[a[x, n], x], {n, 0, 10}]; Flatten[a1]
    (* Second program: *)
    BellMatrix[f_, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
    B = BellMatrix[Function[n, (-n-1)^n], rows = 12];
    Table[B[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 28 2018, after Peter Luschny *)
  • Sage
    # uses[inverse_bell_transform from A264429]
    def A137452_matrix(dim):
        nat = [n for n in (1..dim)]
        return inverse_bell_transform(dim, nat)
    A137452_matrix(10) # Peter Luschny, Dec 20 2015

Formula

Row n gives the coefficients of the expansion of x*(x-n)^(n-1).
Abs(T(n,k)) = C(n-1,k-1)*n^(n-k). - Peter Luschny, Jan 14 2009
From Wolfdieter Lang, Nov 08 2022: (Start)
From the exponential Riordan (also Sheffer of Jabotinsky) type (1, LambertW) array (see comments).
E.g.f. of column sequence k, LambertW(x)^k/k!, for k >= 0.
E.g.f. of row polynomials P_n(y) = Sum_{k=0..n} T(n, k)*y^k: exp(y*LambertW(x)).
Recurrence for T: T(n, k) = 0 for n < k; T(n, 0) = 1 for n = 0 otherwise 0; T(n, k) = (n/k)*Sum_{j=0..n-k} binomial(k-1+j,k-1)*(-1)^j*T(n-1, k-1+j). (Jabotinsky type convolution triangle, the e.g.f.s for the a- and z-sequences are exp(-x), and 0. See the link in A006232.)
Recurrence for column k of T: T(n, k) = 0 for n < k, T(k, k) = 1, for k >= 0 otherwise T(n, k) = (n!*k/(n-k))*Sum_{j=k..n-1} (1/j!)*beta(n-1-j)*T(j, k), where beta(n) = A264234(n+1)/A095996(n+1) = {-1, 2, -9/2, 32/3, -625/24, ...} with o.g.f. d/dx(log(LambertW(x)/x)). See the Boas-Buck or Rainville references given in A046521, and my Aug 10 2017 comment there.
Recurrence for the row polynomials P_0(x) = 1, and P_n(x) = x*substitute(z=d/dx, exp(-z)/(1+z)) P_(n-1)(x), for n >= 1, with coefficient z^k of exp(-z)/(1+z) given by (-1)^k*A061354(k)/A061355(k). See the Roman reference Corollary 3.7.2., p. 50. (End)
The column sequences for the unsigned triangle Abs(T(n, k)), for k >= 2, are also given by {n^(n-k)*(n-1)*s(k-2, n)/(k-1)!}A049444.%20-%20_Wolfdieter%20Lang">{n>=k} with the row polynomials s(n, x) = risingfactorial(x - (n+1), n) of A049444. - _Wolfdieter Lang, Nov 21 2022

Extensions

Better name by Peter Bala, Apr 08 2013
Edited by Joerg Arndt, Apr 08 2013

A120265 a(n) = numerator(Sum_{k=1..n} 1/k!).

Original entry on oeis.org

1, 3, 5, 41, 103, 1237, 433, 69281, 62353, 6235301, 8573539, 164611949, 5349888343, 29959374721, 561738276019, 35951249665217, 4701317263913, 11001082397556421, 52255141388393, 4180411311071440001, 43894318766250120011, 386270005143001056097
Offset: 1

Views

Author

Alexander Adamchuk, Jun 30 2006

Keywords

Examples

			1, 3/2, 5/3, 41/24, 103/60, 1237/720, 433/252, 69281/40320, 62353/36288, 6235301/3628800, 8573539/4989600, 164611949/
95800320, 5349888343/3113510400, ...
		

Crossrefs

Cf. A061354, A061355 (denominator).

Programs

  • Maple
    a:= n-> numer(add(1/i!, i=1..n)): seq(a(n), n=1..23); # Zerinvary Lajos, Mar 28 2007
  • Mathematica
    Numerator[Table[Sum[1/k!,{k,1,n}],{n,1,30}]]
  • PARI
    a(n) = numerator(sum(k=1, n, 1/k!)); \\ Michel Marcus, Jun 01 2018

Formula

A061355(n) = denominator(Sum_{k=1..n} 1/k!).
a(n) = A061354(n) - A061355(n).
a(n) = numerator(exp(1)*gamma(n + 1,1)/gamma(n + 1) - 1). - Gerry Martens, May 31 2018
(exp(x)-1) / (1-x) is the o.g.f. for the sequence of fractions. - Joerg Arndt, Jun 01 2018

A124779 a(n) = gcd(A(n), A(n+2))/gcd(d(n), d(n+2)) where A(n) = Sum_{k=0..n} n!/k! and d(n) = gcd(A(n), n!).

Original entry on oeis.org

1, 2, 5, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 37, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Jonathan Sondow, Nov 07 2006

Keywords

Comments

The next term > 1 is a(460) = 463. The primes 2, 5, 13, 37, 463 are the only terms > 1 up to n = 600000. If a(n) > 1 with n > 1, then a(n) = n+3 is prime. This uses A(n+2) = (n+2)(n+1)*A(n) + n+3. The terms > 1 are A064384 = primes p such that p divides 0!-1!+2!-3!+...+(-1)^{p-1}(p-1)!. The proof uses (n-1)!/(n-k-1)! = (n-1)(n-2)...(n-k) == (-1)^k k! (mod n). Cf. Cloitre's comment in A064383.
An integer p > 1 is in the sequence if and only if p is prime and p|A(p-1), where A(0) = 1 and A(n) = n*A(n-1)+1 for n > 0. - Jonathan Sondow, Dec 22 2006
Michael Mossinghoff has calculated that there are only five primes in the sequence up to 150 million. Heuristics suggest it contains infinitely many. - Jonathan Sondow, Jun 12 2007

Examples

			a(2) = gcd(A(2), A(4))/gcd(d(2), d(4)) = gcd(5, 65)/gcd(1, 1) = 5/1 = 5.
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, Springer-Verlag, 3rd edition, 2004, B43.

Crossrefs

A(n) = A000522, d(n) = A093101, gcd(A(n), A(n+2)) = A124780, gcd(d(n), d(n+2)) = A124781, (n+3)/gcd(A(n), A(n+2)) = A124782, (n+3)/gcd(d(n), d(n+2)) = A123901. Cf. A061354, A061355, A123899, A123900.
Cf. A129924.

Programs

  • Mathematica
    (A[n_] := Sum[n!/k!, {k, 0, n}]; d[n_] := GCD[A[n],n! ]; Table[GCD[A[n],A[n+2]]/GCD[d[n],d[n+2]], {n,0,100}])
  • PARI
    A124779(n)={my(An=A000522(n),A2=A000522(n+2));gcd(An, A2)/gcd([An,n!,A2,(n+2)!])} \\ M. F. Hasler, Jun 04 2019

Formula

a(n) = A124780(n)/A124781(n) = A124782(n)/A123901(n).
a(n) = gcd(A(n), A(n+2))/gcd(A(n), A(n+2), n!) where A(n)=1+n+n(n-1)+...+n!. - Jonathan Sondow, Nov 10 2006
a(n) = gcd(N(n), N(n+2)), where N(n) = A061354(n) = numerator of Sum[1/k!,{k,0,n}]. - Jonathan Sondow, Jun 12 2007

A069880 Number of terms in the simple continued fraction for Sum_{k=1..n} 1/k!.

Original entry on oeis.org

1, 2, 3, 5, 6, 9, 9, 13, 14, 18, 19, 20, 24, 24, 23, 29, 33, 36, 31, 38, 41, 42, 46, 50, 53, 58, 56, 57, 70, 73, 77, 69, 76, 76, 78, 77, 80, 85, 89, 101, 101, 105, 106, 104, 106, 112, 115, 124, 113, 126, 124, 124, 130, 144, 144, 148, 140, 149, 141, 151, 157, 158, 172
Offset: 1

Views

Author

Benoit Cloitre, May 04 2002

Keywords

Examples

			For n=4, Sum_{k=1..n} 1/k! = 1/1! + 1/2! + 1/3! + 1/4! = 1/1 + 1/2 + 1/6 + 1/24 = 41/24 = 1 + 1/(1 + 1/(2 + 1/(2 + 1/3))) = CF[1;1,2,2,1], so a(4) = 5.
		

Crossrefs

Programs

  • Mathematica
    lcf[f_] := Length[ContinuedFraction[f]]; lcf /@ Accumulate[Table[1/k!, {k, 1, 100}]] (* Amiram Eldar, Apr 30 2022 *)

Formula

Does lim_{n->infinity} a(n)/(n * log(log(n))) = C = 2.XXX...?

A354331 a(n) is the denominator of Sum_{k=0..n} 1 / (2*k+1)!.

Original entry on oeis.org

1, 6, 40, 5040, 362880, 13305600, 6227020800, 1307674368000, 513257472000, 121645100408832000, 51090942171709440000, 8617338912961658880000, 15511210043330985984000000, 10888869450418352160768000000, 2947253997913233984847872000000, 1174691236311131831103651840000000
Offset: 0

Views

Author

Ilya Gutkovskiy, May 24 2022

Keywords

Examples

			1, 7/6, 47/40, 5923/5040, 426457/362880, 15636757/13305600, 7318002277/6227020800, ...
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[1/(2 k + 1)!, {k, 0, n}], {n, 0, 15}] // Denominator
    nmax = 15; CoefficientList[Series[Sinh[Sqrt[x]]/(Sqrt[x] (1 - x)), {x, 0, nmax}], x] // Denominator
  • PARI
    a(n) = denominator(sum(k=0, n, 1/(2*k+1)!)); \\ Michel Marcus, May 24 2022
    
  • Python
    from fractions import Fraction
    from math import factorial
    def A354331(n): return sum(Fraction(1,factorial(2*k+1)) for k in range(n+1)).denominator # Chai Wah Wu, May 24 2022

Formula

Denominators of coefficients in expansion of sinh(sqrt(x)) / (sqrt(x) * (1 - x)).

A354333 a(n) is the denominator of Sum_{k=0..n} (-1)^k / (2*k+1)!.

Original entry on oeis.org

1, 6, 120, 5040, 362880, 39916800, 249080832, 1307674368000, 27360571392000, 121645100408832000, 51090942171709440000, 5170403347776995328000, 15511210043330985984000000, 10888869450418352160768000000, 8841761993739701954543616000000, 432780981798838043038187520000000
Offset: 0

Views

Author

Ilya Gutkovskiy, May 24 2022

Keywords

Examples

			1, 5/6, 101/120, 4241/5040, 305353/362880, 33588829/39916800, 209594293/249080832, ...
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[(-1)^k/(2 k + 1)!, {k, 0, n}], {n, 0, 15}] // Denominator
    nmax = 15; CoefficientList[Series[Sin[Sqrt[x]]/(Sqrt[x] (1 - x)), {x, 0, nmax}], x] // Denominator
  • PARI
    a(n) = denominator(sum(k=0, n, (-1)^k/(2*k+1)!)); \\ Michel Marcus, May 24 2022
    
  • Python
    from fractions import Fraction
    from math import factorial
    def A354333(n): return sum(Fraction(-1 if k % 2 else 1,factorial(2*k+1)) for k in range(n+1)).denominator # Chai Wah Wu, May 24 2022

Formula

Denominators of coefficients in expansion of sin(sqrt(x)) / (sqrt(x) * (1 - x)).

A354335 a(n) is the denominator of Sum_{k=0..n} 1 / (2*k)!.

Original entry on oeis.org

1, 2, 24, 720, 4480, 518400, 479001600, 29059430400, 20922789888000, 6402373705728000, 810967336058880000, 1124000727777607680000, 88635485961891348480000, 14936720782466875392000000, 27717122237428532772864000000, 265252859812191058636308480000000
Offset: 0

Views

Author

Ilya Gutkovskiy, May 24 2022

Keywords

Examples

			1, 3/2, 37/24, 1111/720, 6913/4480, 799933/518400, 739138093/479001600, ...
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[1/(2 k)!, {k, 0, n}], {n, 0, 15}] // Denominator
    nmax = 15; CoefficientList[Series[Cosh[Sqrt[x]]/(1 - x), {x, 0, nmax}], x] // Denominator
  • PARI
    a(n) = denominator(sum(k=0, n, 1/(2*k)!)); \\ Michel Marcus, May 24 2022
    
  • Python
    from fractions import Fraction
    from math import factorial
    def A354335(n): return sum(Fraction(1,factorial(2*k)) for k in range(n+1)).denominator # Chai Wah Wu, May 24 2022

Formula

Denominators of coefficients in expansion of cosh(sqrt(x)) / (1 - x).
Showing 1-10 of 29 results. Next