The On-Line Encyclopedia of Integer Sequences (OEIS)

Revisions by Peter Bala (See also Peter Bala's wiki page)

(Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
Showing entries 1-10 | older changes

A058806

a(n) = n! * H_n(n) where H_0(n) = 1/n, H_m(n) = Sum_{k=1..n} H_{m-1}(k).
(history; published version)

#29 by Peter Bala at Fri Mar 07 09:16:08 EST 2025

STATUS

editing

proposed

A008548

Quintuple factorial numbers: Product_{k=0..n-1} (5*k+1).
(history; published version)

#103 by Peter Bala at Fri Mar 07 09:14:52 EST 2025

STATUS

editing

proposed

A007696

Quartic (or 4-fold) factorial numbers: a(n) = Product_{k = 0..n-1} (4*k + 1).
(history; published version)

#136 by Peter Bala at Fri Mar 07 09:14:45 EST 2025

STATUS

editing

proposed

A008548

Quintuple factorial numbers: Product_{k=0..n-1} (5*k+1).
(history; published version)

#102 by Peter Bala at Fri Mar 07 09:13:31 EST 2025

FORMULA

Let T(x) = 1/(1 - 4*x)^(1/4) be the e.g.f. for the sequence of triple factorial numbers A007696. Then the e.g.f. A(x) for the quintuple factorial numbers satisfies T( int Integral_{t = 0..x} A(t) dt ) = A(x). Cf. A007559 and A007696. - Peter Bala, Jan 02 2015

STATUS

approved

editing

A058806

a(n) = n! * H_n(n) where H_0(n) = 1/n, H_m(n) = Sum_{k=1..n} H_{m-1}(k).
(history; published version)

#28 by Peter Bala at Fri Mar 07 09:11:30 EST 2025

FORMULA

a(1) = 1; a(n) = 2*(2*n-1)*a(n-1)* - (2*(2n-1) - (2nn-3)!/(n-1)!.

a(n) = n!*Sum_{k=1..n} (binomial(2*n-k-1, n-k)/k). - Vladimir Kruchinin, Mar 17 2016

From _Peter Bala, Mar 07 2025: (Start)

a(n+1) = - (2*n+1)!^2/n!^3 * Integral_{x = 0..1} log(x)*x^n*(1 - x)^n = (2*n+1)!^2/n!^3 * Sum_{k = 0..n} (-1)^k * binomial(n, k)/(n+k+1)^2.

(n-1)*a(n) = 2*(4*n^2-10*n+7)*a(n-1) - 4*(n-2)*(2*n-3)^2*a(n-2) with a(1) = 1, a(2) = 5. (End)

EXAMPLE

a(3) = 3! *(1 + (1 + (1 + 1/2)) + (1 + (1 + 1/2) + (1 + 1/2 + 1/3))) = 47.

MAPLE

a := proc(n) option remember; if n = 1 then 1 else 2*(2*n-1)*a(n-1) - (2*n-3)!/(n-1)! fi; end:

seq(a(n), n = 1..20); # Peter Bala, Mar 07 2025

STATUS

approved

editing

A007696

Quartic (or 4-fold) factorial numbers: a(n) = Product_{k = 0..n-1} (4*k + 1).
(history; published version)

#135 by Peter Bala at Thu Mar 06 16:59:20 EST 2025

FORMULA

Let T(x) = 1/(1 - 3*x)^(1/3) be the e.g.f. for the sequence of triple factorial numbers A007559. Then the e.g.f. A(x) for the quartic factorial numbers satisfies T(int_Integral_{t = 0..x} A(t) dt) = A(x). (Cf. A007559 and A008548.) - Peter Bala, Jan 02 2015

KEYWORD

nonn,easy

STATUS

approved

editing

Discussion

Fri Mar 07

09:14

Peter Bala: minor edit

A113476

Decimal expansion of (log(2) + Pi/sqrt(3))/3.
(history; published version)

#55 by Peter Bala at Thu Mar 06 12:25:26 EST 2025

STATUS

editing

proposed

A342316

Decimal expansion of Pi/2 - log(2).
(history; published version)

#23 by Peter Bala at Thu Mar 06 12:25:07 EST 2025

STATUS

editing

proposed

A196521

Decimal expansion of Pi/4-log(2)/2.
(history; published version)

#56 by Peter Bala at Thu Mar 06 12:23:57 EST 2025

STATUS

editing

proposed

A106516

A Pascal-like triangle based on 3^n.
(history; published version)

#42 by Peter Bala at Thu Mar 06 12:23:43 EST 2025

STATUS

editing

proposed