A257505 -id:A257505 - cp's OEIS frontend

A255566 a(0) = 0; after which, a(2n) = A255411(a(n)), a(2n+1) = A256450(a(n)).

0, 1, 4, 2, 18, 6, 12, 3, 96, 24, 48, 8, 72, 15, 16, 5, 600, 120, 240, 30, 360, 56, 60, 10, 480, 87, 88, 20, 90, 21, 22, 7, 4320, 720, 1440, 144, 2160, 270, 288, 36, 2880, 416, 420, 67, 432, 73, 66, 13, 3600, 567, 568, 107, 570, 109, 108, 26, 576, 111, 112, 27, 114, 28, 52, 9, 35280, 5040, 10080, 840, 15120, 1584, 1680, 168

Offset: 0

Antti Karttunen, May 05 2015

This sequence can be represented as a binary tree. Each left hand child is produced as A255411(n), and each right hand child as A256450(n), when parent contains n >= 1:
                                      0
                                      |
                   ...................1...................
                  4                                       2
       18......../ \........6                  12......../ \........3
       / \                 / \                 / \                 / \
      /   \               /   \               /   \               /   \
     /     \             /     \             /     \             /     \
   96       24         48       8          72       15         16       5
600  120 240  30    360  56   60 10     480  87   88  20     90  21   22 7
etc.
Because all terms of A255411 are even it means that odd terms can occur only in odd positions (together with some even terms, for each one of which there is a separate infinite cycle), while terms in even positions are all even.
After its initial 1, A255567 seems to give all the terms like 2, 3, 12, ... where the left hand child of the right hand child is one more than the right hand child of the left hand child (as for 2: 16 = 15+1, as for 3: 22 = 21+1, as for 12: 88 = 87+1).

Inverse: A255565.
Cf. A000142, A001511, A001563, A083318, A255411, A256450, A257679.
Cf. also A255567 and arrays A257503, A257505.
Related or similar permutations: A273666, A273667.

a(0) = 0; after which, a(2n) = A255411(a(n)), a(2n+1) = A256450(a(n)).
Other identities:
For all n >= 0, a(2^n) = A001563(n+1). [The leftmost branch of the binary tree is given by n*n!]
For all n >= 0, a(A083318(n)) = A000142(n+1). [And the next innermost vertices by (n+1)! This follows because A256450(n*n! - 1) = (n+1)! - 1.]
For all n >= 1, A257679(a(n)) = A001511(n).

Formula changed because of the changed starting offset of A256450 - Antti Karttunen, May 30 2016

A130744 a(n) = n(n+2)n!.

Original entry on oeis.org

0, 3, 16, 90, 576, 4200, 34560, 317520, 3225600, 35925120, 435456000, 5708102400, 80472268800, 1214269056000, 19527937228800, 333456963840000, 6025763487744000, 114887039275008000, 2304854534062080000

Offset: 0

Paul Curtz, Jul 12 2007

For n >= 1, a(n) = number whose factorial base representation (A007623) begins with a double digit {n}{n}, which is followed by n-1 zeros. Viewed in that base, this sequence looks like this: 0, 11, 220, 3300, 44000, 550000, 6600000, 77000000, 880000000, 9900000000, AA000000000, BB0000000000, ... (where "digits" A and B stand for placeholder values 10 and 11 respectively). - Antti Karttunen, May 07 2015

			G.f. = 3*x + 16*x^2 + 90*x^3 + 576*x^4 + 4200*x^5 + 34560*x^6 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Index entries for sequences related to factorial base representation.

Cf. A000142, A001048, A005563, A007623.
Column 3 of A257503 (apart from initial zero. Equally, row 3 of A257505).
Subsequence of both A227130 and A227148.
Cf. A001113, A001620, A091725, A099285.

[n*(n+2)*Factorial(n): n in [0..25]]; // Vincenzo Librandi, Aug 11 2011

a[n_]:=(n+2)!-(n+1)!-n!; (* Vladimir Joseph Stephan Orlovsky, Dec 05 2008 *)

a(n)=n!*(n*(n+2)) \\ Charles R Greathouse IV, Aug 11 2011

(define (A130744 n) (* n (+ 2 n) (A000142 n))) ;; Antti Karttunen, May 07 2015

0 = +a(n) * (+a(n+1) + 2*a(n+2) - 6*a(n+3) + a(n+4)) + a(n+1) * (+5*a(n+2) - 6*a(n+3) + a(n+4)) + a(n+2) * (+3*a(n+2) - a(n+4)) + a(n+3) * (+a(n+3)) if n>=0. - Michael Somos, Mar 26 2014
From Antti Karttunen, May 07 2015: (Start)
a(n) = n * (n! + (n+1)!) = n * A001048(n+1).
a(n) = A005563(n) * A000142(n).
a(n) = (n+2)! - (n+1)! - n! [from Orlovsky's Mathematica-code].
(End)
From Amiram Eldar, May 17 2022: (Start)
Sum_{n>=1} 1/a(n) = (Ei(1) - gamma)/2 - 1/4, where Ei(1) = A091725 and gamma = A001620.
Sum_{n>=1} (-1)^(n+1)/a(n) = (gamma - Ei(-1))/2 - 1/e + 1/4, where Ei(-1) = -A099285 and e = A001113. (End)

More terms from Vladimir Joseph Stephan Orlovsky, Dec 05 2008

A255565 a(0) = 0; for n >= 1: if n = A255411(k) for some k, then a(n) = 2a(k), otherwise, n = A256450(h) for some h, and a(n) = 1 + 2a(h).

Original entry on oeis.org

0, 1, 3, 7, 2, 15, 5, 31, 11, 63, 23, 127, 6, 47, 255, 13, 14, 95, 4, 511, 27, 29, 30, 191, 9, 1023, 55, 59, 61, 383, 19, 2047, 111, 119, 123, 767, 39, 4095, 223, 239, 247, 1535, 79, 8191, 447, 479, 495, 3071, 10, 159, 16383, 895, 62, 959, 991, 6143, 21, 319, 32767, 1791, 22, 125, 1919, 1983, 126, 12287, 46, 43, 639, 65535, 254, 3583, 12

Offset: 0

Antti Karttunen, May 05 2015

Because all terms of A255411 are even it means that even terms can only occur in even positions (together with some odd terms, for each one of which there is a separate infinite cycle), while terms in odd positions are all odd.

Inverse: A255566.
Cf. A000079, A000142, A001511, A001563, A083318, A255411, A256450, A257679, A257680, A257682, A257685.
Cf. also arrays A257503, A257505.
Related or similar permutations: A273665, A273668.

a(0) = 0; for n >= 1: if A257680(n) = 0 [i.e., n is one of the terms of A255411], then a(n) = 2*a(A257685(n)), otherwise [when n is one of the terms of A256450], a(n) = 1 + 2*a(A273662(n)).
Other identities:
For all n >= 1, A001511(a(n)) = A257679(n).
For all n >= 1, a(A001563(n)) = A000079(n-1) = 2^(n-1).
For all n >= 1, a(A000142(n)) = A083318(n-1).

Formula changed because of the changed starting offset of A256450 - Antti Karttunen, May 30 2016

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

Olivier Gérard, Nov 02 2012

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.

Index entries for sequences related to factorial base representation

Column 4 of A257503 (apart from initial 1. Equally, row 4 of A257505).
Cf. A000142, A007623, A134433.
Cf. A008276, A094638, A008275, A130534.
Cf. A054654, A048994, A132393.
Cf. A067318.

Table[n! - (n - 2)!, {n, 2, 20}]
#[[3]]-#[[1]]&/@Partition[Range[0,20]!,3,1] (* Harvey P. Dale, Aug 10 2023 *)

A213167(n):=n!-(n-2)!$
makelist(A213167(n),n,2,30); /* Martin Ettl, Nov 03 2012 */

(define (A213167 n) (- (A000142 n) (A000142 (- n 2)))) ;; Antti Karttunen, May 07 2015

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

cp's OEIS Frontend

A255566 a(0) = 0; after which, a(2n) = A255411(a(n)), a(2n+1) = A256450(a(n)).

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A130744 a(n) = n*(n+2)*n!.

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A255565 a(0) = 0; for n >= 1: if n = A255411(k) for some k, then a(n) = 2*a(k), otherwise, n = A256450(h) for some h, and a(n) = 1 + 2*a(h).

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

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Mathematica

Maxima

Scheme

Formula

A130744 a(n) = n(n+2)n!.

A255565 a(0) = 0; for n >= 1: if n = A255411(k) for some k, then a(n) = 2a(k), otherwise, n = A256450(h) for some h, and a(n) = 1 + 2a(h).