A000142 -id:A000142 - cp's OEIS frontend

A080568 Sum of the Fibonacci numbers A000045 and the factorials A000142.

1, 2, 3, 8, 27, 125, 728, 5053, 40341, 362914, 3628855, 39916889, 479001744, 6227021033, 87178291577, 1307674368610, 20922789888987, 355687428097597, 6402373705730584, 121645100408836181, 2432902008176646765, 51090942171709450946, 1124000727777607697711

Offset: 0

Alasdair Kergon (agk(AT)oxlug.org), Feb 21 2003

Original puzzle was to find the next four terms after 2, 3, 8, 27, 125. The cubes were intended to mislead - providing one further term would have made the intended sequence too obvious.

Cambridge Archimedeans' Problems Drive, 2001.

Vincenzo Librandi, Table of n, a(n) for n = 0..100

Cf. A000045, A000142.

[Factorial(n)+Fibonacci(n): n in [0..25]]; // Vincenzo Librandi, May 03 2011

with(combstruct): with(combinat): a:=proc(m) [ZL, {ZL=Set(Cycle(Z, card>=m))}, labeled]; end: ZLL:=a(1): seq(count(ZLL, size=n)+fibonacci(n), n=0..19); # Zerinvary Lajos, Jun 11 2008

Table[n!+Fibonacci[n],{n,0,100}] (* Vladimir Joseph Stephan Orlovsky, May 02 2011 *)

vector(30, n, n--; n! + fibonacci(n)) \\ Michel Marcus, Jul 05 2015

E.g.f.: (2/sqrt(5))*exp(x/2)*sinh(sqrt(5)*x/2)+1/(1-x). - Alois P. Heinz, Jul 05 2015

A100685 Powers of factorials A000142.

Original entry on oeis.org

1, 2, 4, 6, 8, 16, 24, 32, 36, 64, 120, 128, 216, 256, 512, 576, 720, 1024, 1296, 2048, 4096, 5040, 7776, 8192, 13824, 14400, 16384, 32768, 40320, 46656, 65536, 131072, 262144, 279936, 331776, 362880, 518400, 524288, 1048576, 1679616, 1728000

Offset: 1

Kyle Schalm and Jonathan Sondow, Dec 08 2004

Subsequence of A001013. Supersequence of A036740 without its first term.

Supersequence also of A046882 and A055209 without their first terms. - Jonathan Sondow and Robert G. Wilson v, Dec 19 2004

			24 = (4!)^1 and 36 = (3!)^2.

G. C. Greubel, Table of n, a(n) for n = 1..9660
Index entries for sequences related to factorial numbers.

Cf. A000142, A001013, A036740, A331373.
Cf. also A046882 and A055209.
Subsequences: A000079, A000400, A009968.

With[{ln = Log[10!]}, Table[With[{f = m!}, Table[f^j, {j, 0, Floor[ln/Log[f]]}]], {m, 2, 10}]] //Flatten //Union

Sum_{n>=1} 1/a(n) = 1 + A331373. - Amiram Eldar, Nov 21 2021

A126787 G.f.: B(x)B(2!x^2)B(3!x^3)*..., where B(x) is g.f. of A000142.

Original entry on oeis.org

1, 1, 4, 14, 66, 308, 1888, 12240, 95640, 827904, 8106960, 87387264, 1035645312, 13316300928, 184988692800, 2756878875648, 43888205438208, 742943286892800, 13326434312808960, 252448071959572992, 5036116692383428608, 105523926692032447488

Offset: 0

Vladeta Jovovic, Feb 18 2007

Take each Ferrers diagram of the partitions of n, label(linearly order) the dots within each row, then linearly order any of the rows that are of equal length. - Geoffrey Critzer, Mar 21 2009

Alois P. Heinz, Table of n, a(n) for n = 0..450 (terms n=176..300 from Vaclav Kotesovec)

Cf. A096161, A110143.

B:= proc(n) option remember; local x; unapply(`if`(n<=0, 1, B(n-1)(x)+ n! *x^n), x) end: BB:= proc(n) local x, d; unapply(convert(series(mul(B(floor(n/d))(d!*x^d), d=1..n), x, n+1), polynom), x) end: a:= n-> coeff(BB(n)(x), x, n): seq(a(n), n=0..25); # Alois P. Heinz, Sep 25 2008
# second Maple program:
b:= proc(n, i) option remember; `if`(n=0 or i=1, n!,
      add(b(n-i*j, i-1)*j!*i!^j, j=0..n/i))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..30);  # Alois P. Heinz, Oct 02 2017

CoefficientList[Series[Product[Sum[x^(n*k) n!^k*k!, {k, 0, 20}], {n, 1, 20}], {x, 0, 20}], x] (* Geoffrey Critzer, Mar 21 2009 *)

a(n) ~ 2*n! * (1 + 1/(2*n) + 3/n^2 + 13/n^3 + 82/n^4 + 587/n^5 + 4966/n^6). - Vaclav Kotesovec, Mar 16 2015

More terms from Alois P. Heinz, Sep 25 2008

A139003 Number of operations A000142 (i.e., x!) or A000196 (i.e., floor(sqrt(x))) needed to get n, starting with 3.

Original entry on oeis.org

1, 2, 0, 20, 4, 1, 14, 17, 31, 6, 26, 41, 35, 20, 31, 31, 19, 28, 27, 38, 21, 33, 21, 21, 26, 3, 51, 38, 28, 26, 20, 35, 36, 36, 13, 23, 27, 62, 45, 50, 45, 40, 9, 15, 31, 8, 32, 52, 36, 13, 68, 69, 57, 33, 54, 36, 46, 34, 49, 63, 56, 68, 14, 63, 23, 33, 36, 47, 43, 16, 38, 66, 38

Offset: 1

M. F. Hasler, Apr 09 2008

nonn

Knuth conjectured that any number can be obtained in that way, starting from 4.
This seems also to be true using 3 as the starting value. Since 3 is the minimal possible choice, this variant could be considered to be more natural.
To ensure the sequence is well-defined, define a(n)=-1 if it is not possible to get n in the given way.
See A139004 for references and links.
In fact a single 2 is enough to get any positive integer, if Knuth's conjecture that one 4 is enough is true. From 2, (((-tan(2.))!)!)! = 5.592..., then floor, factorial gets 120, then sqrt, sqrt gives 3.162..., and floor gives 3, or negate, floor, negate gives 4. - N. J. A. Sloane, Feb 26 2025

			Representing the operation x -> floor(sqrt(x)) by "s" and x -> x! by "f",
we have:
a(1) = 1 since 1 = s3 is clearly the shortest way to obtain 1 from 3.
a(2) = 2 since 2 = sf3 is clearly the shortest way to obtain 2 from 3.
a(3) = 0 since no operation is required to get 3 which is there at the beginning.
a(5) = 4 since 5 = ssff3 is the shortest way to obtain 5 from 3.
a(6) = 1 since 6 = f3 is certainly the shortest way to get 6 from 3.
a(4) = 20 = 7+9+a(5) since 4 = ssssssfsssssssffssff3 = floor(35!^(1/2^6)), 35 = floor((5!)!^(1/2^7)).

Jon E. Schoenfield, Table of n, a(n) for n = 1..1000
Jon E. Schoenfield, Table of n, a(n), and shortest path for n = 1..1000

Cf. A139004.

A139003( n, S=Set(3), LIM=10^5 )={ for( i=0,LIM, setsearch( S, n) & return(i); S=setunion( S, setunion( Set( vector( #S, j, sqrtint(eval(S[j])))), Set( vector( #S, j, if( LIM > j=eval(S[j]), j!))))))}

a(3) = 0; a(n) = min { a(k)+1 ; n^2 <= k < (n+1)^2 or k! = n }

a(9)-a(11) from Max Alekseyev, Nov 03 2008
Corrected formula, added terms from a(12) onward. - Jon E. Schoenfield, Nov 17 2008, Nov 19 2008
Comments and example edited by Jon E. Schoenfield, Sep 15 2013

A141476 Triangle T(n,k) = A000142(n-k)*A003319(k+1) read by rows.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 6, 2, 3, 13, 24, 6, 6, 13, 71, 120, 24, 18, 26, 71, 461, 720, 120, 72, 78, 142, 461, 3447, 5040, 720, 360, 312, 426, 922, 3447, 29093, 40320, 5040, 2160, 1560, 1704, 2766, 6894, 29093, 273343, 362880, 40320, 15120, 9360, 8520, 11064, 20682

Offset: 0

Paul Curtz, Aug 09 2008

			The triangle starts at row n=0 with columns 0 <= k <= n:
    1;
    1,   1;
    2,   1,   3;
    6,   2,   3,  13;
   24,   6,   6,  13,  71;
  120,  24,  18,  26,  71, 461;

(* b = A003319 *) b[0]=0; b[n_] := b[n] = n! - Sum[k!*b[n-k], {k, 1, n-1}];
T[n_, k_] := (n-k)! b[k+1]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 18 2018 *)

Row sums: Sum_{k=0..n} T(n,k) = A000142(n+1) = (n+1)!.

Edited and extended by R. J. Mathar, Dec 05 2008

A226718 n! mod tetrahedral(n), that is A000142(n) mod A000292(n).

Original entry on oeis.org

0, 2, 6, 4, 15, 48, 0, 0, 45, 120, 66, 168, 0, 0, 120, 288, 153, 360, 0, 0, 231, 528, 0, 0, 0, 0, 378, 840, 435, 960, 0, 0, 0, 0, 630, 1368, 0, 0, 780, 1680, 861, 1848, 0, 0, 1035, 2208, 0, 0, 0, 0, 1326, 2808, 0, 0, 0, 0, 1653, 3480, 1770, 3720, 0, 0, 0, 0, 2145, 4488

Offset: 1

Alex Ratushnyak, Jun 15 2013

Ivan Neretin, Table of n, a(n) for n = 1..10000

Cf. A000142, A000292, A119690.

A226718 := proc(n)
    n! mod ( n*(n+1)*(n+2)/6) ;
end proc: # R. J. Mathar, Jun 18 2013

Table[Mod[n!, n (n + 1) (n + 2)/6], {n, 66}] (* Ivan Neretin, May 18 2015 *)

f = 1
for i in range(1, 100):
    f *= i
    print(f % (i*(i+1)*(i+2)//6), end=', ')

a(n) = n! mod (n*(n+1)*(n+2)/6).

For n>4: if neither n+1 nor n+2 is prime, then a(n)=0. Otherwise, a(n)=n(n+1)/2 for odd n and a(n)=n(n+2) for even n. - Ivan Neretin, May 18 2015

A248652 Union of the factorial numbers (A000142) and the double factorials of odd numbers (A001147).

Original entry on oeis.org

1, 2, 3, 6, 15, 24, 105, 120, 720, 945, 5040, 10395, 40320, 135135, 362880, 2027025, 3628800, 34459425, 39916800, 479001600, 654729075, 6227020800, 13749310575, 87178291200, 316234143225, 1307674368000, 7905853580625, 20922789888000, 213458046676875, 355687428096000

Offset: 1

Olivier Gérard, Oct 10 2014

nonn

Douglas Hoftstadter, Keynote lecture, DIMACS Workshop on Recognition of Integer Sequences, Oct. 10, 2014.

Abstract of D. Hoftstadter in the Workshop's program

Cf. A000142, A001147, A005214.

See A268645 for another version.

max = 10^15;
A000142 = Reap[For[k = 0, k!  <= max, k++, Sow[k!]]][[2, 1]];
A001147 = Reap[For[k = 1, k!! <= max, k = k+2, Sow[k!!]]][[2, 1]];
Union[A000142, A001147] (* Jean-François Alcover, Jul 18 2022 *)

Revised by N. J. A. Sloane, Feb 09 2016

A277362 Self-convolution of a(n)/4^n gives factorials (A000142).

Original entry on oeis.org

1, 2, 14, 164, 2646, 53852, 1316364, 37467080, 1215510118, 44249471916, 1785942489700, 79150848980216, 3821494523507708, 199668288426274968, 11225643465179779544, 675769562728962818448, 43370783734391689628294, 2956329387192674856638668

Offset: 0

Vladimir Reshetnikov, Oct 10 2016

nonn

Self-convolution of a(n) gives A047053.

Vaclav Kotesovec, Table of n, a(n) for n = 0..360

Cf. A000142, A000302, A047053.

a:= proc(n) option remember; `if`(n=0, 1,
     (n!*4^n-add(a(k)*a(n-k), k=1..n-1))/2)
    end:
seq(a(n), n=0...20);  # Alois P. Heinz, Oct 12 2016

With[{n = 20}, Sqrt[Sum[k! (4 x)^k, {k, 0, n - 1}] + O[x]^n][[3]]]
CoefficientList[Series[Sqrt[-Gamma[0, -1/(4*x)]/(x*E^(1/(4*x)))]/2, {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 27 2021 *)

Sum_{k=0..n} a(k)/4^k * a(n-k)/4^(n-k) = n!.

a(n) ~ 2^(2*n-1) * n!. - Vaclav Kotesovec, Oct 27 2021

A280531 a(n) = A049501(A000142(n)).

Original entry on oeis.org

0, 0, 1, 2, 2, 4, 11, 16, 16, 25, 30, 64, 39, 83, 111, 139, 139, 165, 205, 242, 283, 320, 336, 474, 440, 395, 637, 655, 886, 842, 1019, 1159, 1159, 1153, 1327, 1366, 1328, 1243, 1487, 1756, 1623, 2362, 2394, 2274, 2487, 2642, 2907, 2843, 3211, 3049, 3736

Offset: 0

Indranil Ghosh, Jan 04 2017

nonn

a(n) is the major index (1st definition) of n!.

			For n=4, A000142(n) = 24 and A049501(24) = 2. So a(n) = 2.

Indranil Ghosh, Table of n, a(n) for n = 0..10000

Cf. A000142, A049501.

Cf. A280062 (Major index (2nd definition) of n!).

import math
def M(n):
    x=bin(int(n))[2:]
    s=0
    for i in range(1,len(x)):
        if x[i-1]=="1" and x[i]=="0":
            s+=i
    return s
a=lambda n: M(math.factorial(n))

A289924 p-INVERT of (n!), n >= 1 (A000142, shifted), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 4, 17, 79, 402, 2253, 14037, 98152, 774973, 6911131, 69225314, 771593257, 9470565513, 126755983488, 1834510979193, 28511931874423, 473179672441090, 8346048191981797, 155838573499885229, 3069991622444141848, 63618933765102190149, 1383222300396890185731

Offset: 0

Clark Kimberling, Aug 14 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A289780 for a guide to related sequences.

Michael De Vlieger, Table of n, a(n) for n = 0..448
Mikhail Khovanov, Victor Ostrik and Yakov Kononov, Two-dimensional topological theories, rational functions and their tensor envelopes, arXiv:2011.14758 [math.QA], 2020.

Cf. A000142.

z = 60; s = Sum[k! x^k, {k, 1, z}]; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000142 shifted *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x] , 1]  (* A289924 *)

cp's OEIS Frontend

A080568 Sum of the Fibonacci numbers A000045 and the factorials A000142.

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A100685 Powers of factorials A000142.

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A126787 G.f.: B(x)*B(2!*x^2)*B(3!*x^3)*..., where B(x) is g.f. of A000142.

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A139003 Number of operations A000142 (i.e., x!) or A000196 (i.e., floor(sqrt(x))) needed to get n, starting with 3.

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A141476 Triangle T(n,k) = A000142(n-k)*A003319(k+1) read by rows.

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Mathematica

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A226718 n! mod tetrahedral(n), that is A000142(n) mod A000292(n).

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A248652 Union of the factorial numbers (A000142) and the double factorials of odd numbers (A001147).

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Mathematica

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A126787 G.f.: B(x)B(2!x^2)B(3!x^3)*..., where B(x) is g.f. of A000142.