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.

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A000178 Superfactorials: product of first n factorials.

Original entry on oeis.org

1, 1, 2, 12, 288, 34560, 24883200, 125411328000, 5056584744960000, 1834933472251084800000, 6658606584104736522240000000, 265790267296391946810949632000000000, 127313963299399416749559771247411200000000000, 792786697595796795607377086400871488552960000000000000
Offset: 0

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Author

Keywords

Comments

a(n) is also the Vandermonde determinant of the numbers 1,2,...,(n+1), i.e., the determinant of the (n+1) X (n+1) matrix A with A[i,j] = i^j, 1 <= i <= n+1, 0 <= j <= n. - Ahmed Fares (ahmedfares(AT)my-deja.com), May 06 2001
a(n) = (1/n!) * D(n) where D(n) is the determinant of order n in which the (i,j)-th element is i^j. - Amarnath Murthy, Jan 02 2002
Determinant of S_n where S_n is the n X n matrix S_n(i,j) = Sum_{d|i} d^j. - Benoit Cloitre, May 19 2002
Appears to be det(M_n) where M_n is the n X n matrix with m(i,j) = J_j(i) where J_k(n) denote the Jordan function of row k, column n (cf. A059380(m)). - Benoit Cloitre, May 19 2002
a(2n+1) = 1, 12, 34560, 125411328000, ... is the Hankel transform of A000182 (tangent numbers) = 1, 2, 16, 272, 7936, ...; example: det([1, 2, 16, 272; 2, 16, 272, 7936; 16, 272, 7936, 353792; 272, 7936, 353792, 22368256]) = 125411328000. - Philippe Deléham, Mar 07 2004
Determinant of the (n+1) X (n+1) matrix whose i-th row consists of terms 1 to n+1 of the Lucas sequence U(i,Q), for any Q. When Q=0, the Vandermonde matrix is obtained. - T. D. Noe, Aug 21 2004
Determinant of the (n+1) X (n+1) matrix A whose elements are A(i,j) = B(i+j) for 0 <= i,j <= n, where B(k) is the k-th Bell number, A000110(k) [I. Mezo, JIS 14 (2011) # 11.1.1]. - T. D. Noe, Dec 04 2004
The Hankel transform of the sequence A090365 is A000178(n+1); example: det([1,1,3; 1,3,11; 3,11,47]) = 12. - Philippe Deléham, Mar 02 2005
Theorem 1.3, page 2, of Polynomial points, Journal of Integer Sequences, Vol. 10 (2007), Article 07.3.6, provides an example of an Abelian quotient group of order (n-1) superfactorial, for each positive integer n. The quotient is obtained from sequences of polynomial values. - E. F. Cornelius, Jr. (efcornelius(AT)comcast.net), Apr 09 2007
Starting with offset 1 this is a 'Matryoshka doll' sequence with alpha=1, the multiplicative counterpart to the additive A000292. seq(mul(mul(i,i=alpha..k), k=alpha..n),n=alpha..12). - Peter Luschny, Jul 14 2009
For n>0, a(n) is also the determinant of S_n where S_n is the n X n matrix, indexed from 1, S_n(i,j)=sigma_i(j), where sigma_k(n) is the generalized divisor sigma function: A000203 is sigma_1(n). - Enrique Pérez Herrero, Jun 21 2010
a(n) is the multiplicative Wiener index of the (n+1)-vertex path. Example: a(4)=288 because in the path on 5 vertices there are 3 distances equal to 2, 2 distances equal to 3, and 1 distance equal to 4 (2*2*2*3*3*4=288). See p. 115 of the Gutman et al. reference. - Emeric Deutsch, Sep 21 2011
a(n-1) = Product_{j=1..n-1} j! = V(n) = Product_{1 <= i < j <= n} (j - i) (a Vandermondian V(n), see the Ahmed Fares May 06 2001 comment above), n >= 1, is in fact the determinant of any n X n matrix M(n) with entries M(n;i,j) = p(j-1,x = i), 1 <= i, j <= n, where p(m,x), m >= 0, are monic polynomials of exact degree m with p(0,x) = 1. This is a special x[i] = i choice in a general theorem given in Vein-Dale, p. 59 (written for the transposed matrix M(n;j,x_i) = p(i-1,x_j) = P_i(x_j) in Vein-Dale, and there a_{k,k} = 1, for k=1..n). See the Aug 26 2013 comment under A049310, where p(n,x) = S(n,x) (Chebyshev S). - Wolfdieter Lang, Aug 27 2013
a(n) is the number of monotonic magmas on n elements labeled 1..n with a symmetric multiplication table. I.e., Product(i,j) >= max(i,j); Product(i,j) = Product(j,i). - Chad Brewbaker, Nov 03 2013
The product of the pairwise differences of n+1 integers is a multiple of a(n) [and this does not hold for any k > a(n)]. - Charles R Greathouse IV, Aug 15 2014
a(n) is the determinant of the (n+1) X (n+1) matrix M with M(i,j) = (n+j-1)!/(n+j-i)!, 1 <= i <= n+1, 1 <= j <= n+1. - Stoyan Apostolov, Aug 26 2014
All terms are in A064807 and all terms after a(2) are in A005101. - Ivan N. Ianakiev, Sep 02 2016
Empirical: a(n-1) is the determinant of order n in which the (i,j)-th entry is the (j-1)-th derivative of x^(x+i-1) evaluated at x=1. - John M. Campbell, Dec 13 2016
Empirical: If f(x) is a smooth, real-valued function on an open neighborhood of 0 such that f(0)=1, then a(n) is the determinant of order n+1 in which the (i,j)-th entry is the (j-1)-th derivative of f(x)/((1-x)^(i-1)) evaluated at x=0. - John M. Campbell, Dec 27 2016
Also the automorphism group order of the n-triangular honeycomb rook graph. - Eric W. Weisstein, Jul 14 2017
Is the zigzag Hankel transform of A000182. That is, a(2*n+1) is the Hankel transform of A000182 and a(2*n+2) is the Hankel transform of A000182(n+1). - Michael Somos, Mar 11 2020
Except for n = 0, 1, superfactorial a(n) is never a square (proof in link Mabry and Cormick, FFF 4 p. 349); however, when k belongs to A349079 (see for further information), there exists m, 1 <= m <= k such that a(k) / m! is a square. - Bernard Schott, Nov 29 2021

Examples

			a(3) = (1/6)* | 1 1 1 | 2 4 8 | 3 9 27 |
a(7) = n! * a(n-1) = 7! * 24883200 = 125411328000.
a(12) = 1! * 2! * 3! * 4! * 5! * 6! * 7! * 8! * 9! * 10! * 11! * 12!
= 1^12 * 2^11 * 3^10 * 4^9 * 5^8 * 6^7 * 7^6 * 8^5 * 9^4 * 10^3 * 11^2 * 12^1
= 2^56 * 3^26 * 5^11 * 7^6 * 11^2.
G.f. = 1 + x + 2*x^2 + 12*x^3 + 288*x^4 + 34560*x^5 + 24883200*x^6 + ...
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 545.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 135-145.
  • 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. 50.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 231.
  • H. J. Ryser, Combinatorial Mathematics. Mathematical Association of America, Carus Mathematical Monograph 14, 1963, p. 53.
  • 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).
  • R. Vein and P. Dale, Determinants and Their Applications in Mathematical Physics, Springer, 1999.

Crossrefs

Programs

  • Magma
    [&*[Factorial(k): k in [0..n]]: n in [0..20]]; // Bruno Berselli, Mar 11 2015
    
  • Maple
    A000178 := proc(n)
        mul(i!,i=1..n) ;
    end proc:
    seq(A000178(n),n=0..10) ; # R. J. Mathar, Oct 30 2015
  • Mathematica
    a[0] := 1; a[1] := 1; a[n_] := n!*a[n - 1]; Table[a[n], {n, 1, 12}] (* Stefan Steinerberger, Mar 10 2006 *)
    Table[BarnesG[n], {n, 2, 14}] (* Zerinvary Lajos, Jul 16 2009 *)
    FoldList[Times,1,Range[20]!] (* Harvey P. Dale, Mar 25 2011 *)
    RecurrenceTable[{a[n] == n! a[n - 1], a[0] == 1}, a, {n, 0, 12}] (* Ray Chandler, Jul 30 2015 *)
    BarnesG[Range[2, 20]] (* Eric W. Weisstein, Jul 14 2017 *)
  • Maxima
    A000178(n):=prod(k!,k,0,n)$ makelist(A000178(n),n,0,30); /* Martin Ettl, Oct 23 2012 */
    
  • PARI
    A000178(n)=prod(k=2,n,k!) \\ M. F. Hasler, Sep 02 2007
    
  • PARI
    a(n)=polcoeff(1-sum(k=0, n-1, a(k)*x^k/prod(j=1, k+1, (1+j!*x+x*O(x^n)) )), n) \\ Paul D. Hanna, Oct 02 2013
    
  • PARI
    for(j=1,13, print1(prod(k=1,j,k^(j-k)),", ")) \\ Hugo Pfoertner, Apr 09 2020
    
  • Python
    A000178_list, n, m = [1], 1,1
    for i in range(1,100):
        m *= i
        n *= m
        A000178_list.append(n) # Chai Wah Wu, Aug 21 2015
    
  • Python
    from math import prod
    def A000178(n): return prod(i**(n-i+1) for i in range(2,n+1)) # Chai Wah Wu, Nov 26 2023
  • Ruby
    def mono_choices(a,b,n)
        n - [a,b].max
    end
    def comm_mono_choices(n)
        accum =1
        0.upto(n-1) do |i|
            i.upto(n-1) do |j|
                accum = accum * mono_choices(i,j,n)
            end
        end
        accum
    end
    1.upto(12) do |k|
        puts comm_mono_choices(k)
    end # Chad Brewbaker, Nov 03 2013
    

Formula

a(0) = 1, a(n) = n!*a(n-1). - Lee Hae-hwang, May 13 2003, corrected by Ilya Gutkovskiy, Jul 30 2016
a(0) = 1, a(n) = 1^n * 2^(n-1) * 3^(n-2) * ... * n = Product_{r=1..n} r^(n-r+1). - Amarnath Murthy, Dec 12 2003 [Formula corrected by Derek Orr, Jul 27 2014]
a(n) = sqrt(A055209(n)). - Philippe Deléham, Mar 07 2004
a(n) = Product_{i=1..n} Product_{j=0..i-1} (i-j). - Paul Barry, Aug 02 2008
log a(n) = 0.5*n^2*log n - 0.75*n^2 + O(n*log n). - Charles R Greathouse IV, Jan 13 2012
Asymptotic: a(n) ~ exp(zeta'(-1) - 3/4 - (3/4)*n^2 - (3/2)*n)*(2*Pi)^(1/2 + (1/2)*n)*(n+1)^((1/2)*n^2 + n + 5/12). For example, a(100) is approx. 0.270317...*10^6941. (See A213080.) - Peter Luschny, Jun 23 2012
G.f.: 1 + x/(U(0) - x) where U(k) = 1 + x*(k+1)! - x*(k+2)!/U(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 02 2012
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 1/(1 + 1/((k+1)!*x*G(k+1)))); (continued fraction). - Sergei N. Gladkovskii, Jun 14 2013
G.f.: 1 = Sum_{n>=0} a(n)*x^n / Product_{k=1..n+1} (1 + k!*x). - Paul D. Hanna, Oct 02 2013
A203227(n+1)/a(n) -> e, as n -> oo. - Daniel Suteu, Jul 30 2016
From Ilya Gutkovskiy, Jul 30 2016: (Start)
a(n) = G(n+2), where G(n) is the Barnes G-function.
a(n) ~ exp(1/12 - n*(3*n+4)/4)*n^(n*(n+2)/2 + 5/12)*(2*Pi)^((n+1)/2)/A, where A is the Glaisher-Kinkelin constant (A074962).
Sum_{n>=0} (-1)^n/a(n) = A137986. (End)
0 = a(n)*a(n+2)^3 + a(n+1)^2*a(n+2)^2 - a(n+1)^3*a(n+3) for all n in Z (if a(-1)=1). - Michael Somos, Mar 11 2020
Sum_{n>=0} 1/a(n) = A287013 = 1/A137987. - Amiram Eldar, Nov 19 2020
a(n) = Wronskian(1, x, x^2, ..., x^n). - Mohammed Yaseen, Aug 01 2023
From Andrea Pinos, Apr 04 2024: (Start)
a(n) = e^(Sum_{k=1..n} (Integral_{x=1..k+1} Psi(x) dx)).
a(n) = e^(Integral_{x=1..n+1} (log(sqrt(2*Pi)) - (x-1/2) + x*Psi(x)) dx).
a(n) = e^(Integral_{x=1..n+1} (log(sqrt(2*Pi)) - (x-1/2) + (n+1)*Psi(x) - log(Gamma(x))) dx).
Psi(x) is the digamma function. (End)

A087316 a(n) = Sum_{k=1..n} prime(k)^prime(n-k+1).

Original entry on oeis.org

4, 17, 84, 545, 7824, 281771, 51540600, 3347558057, 1146374959980, 288113965730819, 529172633067826888, 283453407513524913023, 4122282265785671687518812, 1586581830624893452605127040309, 412109111737176949907195758658736
Offset: 1

Views

Author

Amarnath Murthy, Sep 03 2003

Keywords

Examples

			Examples from _Jonathan Vos Post_, Jan 06 2006: (Start)
a(1) = 4 because prime(1)^prime(1) = 2^2 = 4.
a(2) = 17 because prime(1)^prime(2) + prime(2)^prime(1) = 2^3 + 3^2 = 17.
a(3) = 84 because 2^5 + 3^3 + 5^2 = 84.
a(4) = 545 = 2^7 + 3^5 + 5^3 + 7^2.
a(5) = 7824 = 2^11 + 3^7 + 5^5 + 7^3 + 11^2.
a(6) = 281771 = 2^13 + 3^11 + 5^7 + 7^5 + 11^3 + 13^2.
a(7) = 51540600 = 2^17 + 3^13 + 5^11 + 7^7 + 11^5 + 13^3 + 17^2.
a(8) = 3347558057 = 2^19 + 3^17 + 5^13 + 7^11 + 11^7 + 13^5 + 17^3 + 19^2.
a(9) = 1146374959980 = 2^23 + 3^19 + 5^17 + 7^13 + 11^11 + 13^7 + 17^5 + 19^3 + 23^2. (End)
		

Crossrefs

Programs

  • Maple
    a:=n->sum(ithprime(k)^ithprime(n-k+1),k=1..n): seq(a(n),n=1..16); # Emeric Deutsch, Apr 13 2005
  • Mathematica
    Table[Sum[Prime[k]^Prime[n - k + 1], {k, 1, n}], {n, 1, 15}] (* Vaclav Kotesovec, Jun 08 2025 *)
  • PARI
    a(n) = sum(k=1, n, prime(k)^prime(n-k+1)); \\ Michel Marcus, Aug 20 2019
    
  • Python
    from sympy import prime
    def a(n): return sum(prime(k)**prime(n-k+1) for k in range(1, n+1))
    print([a(n) for n in range(1, 16)]) # Michael S. Branicky, Apr 17 2021

Extensions

More terms from Sam Alexander, Oct 20 2003
Further terms from Emeric Deutsch, Apr 13 2005
Edited by N. J. A. Sloane, Aug 19 2008 at the suggestion of R. J. Mathar

A113231 Ascending descending base exponent transform of triangular numbers (A000217).

Original entry on oeis.org

1, 4, 34, 956, 106721, 75818480, 490656737694, 22960404169011552, 7141530219670856270919, 20319415706020976355219258316, 1104797870481014132439711155738991604
Offset: 1

Views

Author

Jonathan Vos Post, Jan 07 2006

Keywords

Comments

A003101 is the ascending descending base exponent transform of natural numbers A000027. The ascending descending base exponent transform applied to the Fibonacci numbers is A113122; applied to the tribonacci numbers is A113153; applied to the Lucas numbers is A113154. Since the parity of the triangular numbers cycles odd, odd, even, even; the parity of this sequence cycles odd, even, even, even. The smallest prime in this sequence is a(5) = 127601. What is the next prime? What is the first triangular value?

Examples

			a(1) = 1 because T(1)^T(1) = 1^1 = 1.
a(2) = 4 because T(1)^T(2) + T(2)^T(1) = 1^3 + 3^1 = 4.
a(3) = 34 = 1^6 + 3^3 + 6^1.
a(4) = 956 = 1^10 + 3^6 + 6^3 + 10^1.
a(5) = 106721 = 1^15 + 3^10 + 6^6 + 10^3 + 15^1.
a(6) = 75818480 = 1^21 + 3^15 + 6^10 + 10^6 + 15^3 + 21^1.
a(7) = 490656737694 = 1^28 + 3^21 + 6^15 + 10^10 + 15^6 + 21^3 + 28^1.
a(8) = 22960404169011552 = 1^36 + 3^28 + 6^21 + 10^15 + 15^10 + 21^6 + 28^3 + 36^1.
a(9) = 7141530219670856270919 = 1^45 + 3^36 + 6^28 + 10^21 + 15^15 + 21^10 + 28^6 + 36^3 + 45^1.
		

Crossrefs

Programs

  • Mathematica
    A000217[n_] := Binomial[n + 1, 2]; Table[Sum[A000217[k]^(A000217[n - k + 1]), {k, 1, n}], {n, 1, 10}] (* G. C. Greubel, May 18 2017 *)
  • PARI
    for(n=1,10, print1(sum(k=1,n, (binomial(k+1,2))^(binomial(n-k+2,2))), ", ")) \\ G. C. Greubel, May 18 2017

Formula

a(n) = Sum_{i=1..n} (T(i))^(T(n-i+1)), where T(n) are the triangle numbers.
a(n) = Sum_{i=1..n} ((i*(i+1)/2))^((n-i+1)*(n-i+2)/2).
a(n) = Sum_{i=1..n} (A000217(i))^(A000217(n-i+1)).
log(a(n)) ~ n^2 * (-1 + 2*LambertW(2^(-3/2)*exp(1/2)*n))^3 / (8*LambertW(2^(-3/2)*exp(1/2)*n)^2). - Vaclav Kotesovec, Jun 07 2025

A113257 Ascending descending base exponent transform of squares (A000290).

Original entry on oeis.org

1, 5, 266, 268722, 4682453347, 2978988815561863, 722638800922610642480852, 22529984108212742763058965679103268, 57286470055793196612331429228839529219232484069
Offset: 1

Views

Author

Jonathan Vos Post, Jan 07 2006

Keywords

Comments

A003101 is the ascending descending base exponent transform of natural numbers A000027. The ascending descending base exponent transform applied to the Fibonacci numbers is A113122; applied to the tribonacci numbers is A113153; applied to the Lucas numbers is A113154. The smallest prime in this sequence is a(2) = 5. What is the next prime? What is the first square value after 1?

Examples

			a(1) = 1 because (1^2)^(1^2) = 1^1 = 1.
a(2) = 5 because (1^2)^(4^1) + (4^1)^(1^4) = 1^4 + 4^1 = 5.
a(3) = 266 = 1^9 + 4^4 + 9^1.
a(4) = 268722 = 1^16 + 4^9 + 9^4 + 16^1.
a(5) = 4682453347 = 1^25 + 4^16 + 9^9 + 16^4 + 25^1.
a(6) = 2978988815561863 = 1^36 + 4^25 + 9^16 + 16^9 + 25^4 + 36^1.
a(7) = 722638800922610642480852 = 1^49 + 4^36 + 9^25 + 16^16 + 25^9 + 36^4 + 49^1.
a(8) = 22529984108212742763058965679103268 = 1^64 + 4^49 + 9^36 + 16^25 + 25^16 + 36^9 + 49^4 + 64^1.
a(9) = 57286470055793196612331429228839529219232484069 = 1^81 + 4^64 + 9^49 + 16^36 + 25^25 + 36^16 + 49^9 + 64^4 + 81^1.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[(k^2)^((n - k + 1)^2), {k, 1, n}], {n, 1, 10}] (* G. C. Greubel, May 18 2017 *)
  • PARI
    for(n=1,10, print1(sum(k=1,n, (k^2)^((n-k+1)^2) ), ", ")) \\ G. C. Greubel, May 18 2017

Formula

a(n) = Sum_{i=1..n} (i^2)^((n-i+1)^2).
a(n) = Sum_{i=1..n} (A000290(i))^(A000290(n-i+1)).
log(a(n)) ~ n^2 * (-1 + 2*LambertW(exp(1/2)*n/2))^3 / (4*LambertW(exp(1/2)*n/2)^2). - Vaclav Kotesovec, Jun 07 2025

Extensions

a(4) and a(5) corrected by Giovanni Resta, Jun 13 2016

A113271 Ascending descending base exponent transform of 2^n.

Original entry on oeis.org

1, 3, 9, 41, 593, 135457, 8606778433, 36893769626691833985, 680564733921105089459460297630318346497, 231584178474632390853419071752762496470716041121409734167406717963826481922561
Offset: 0

Views

Author

Jonathan Vos Post, Jan 07 2006

Keywords

Comments

A003101 is the ascending descending base exponent transform of natural numbers A000027. The ascending descending base exponent transform applied to the Fibonacci numbers is A113122; applied to the tribonacci numbers is A113153; applied to the Lucas numbers is A113154. The smallest primes in this (always odd) sequence are a(1) = 3, a(3) = 41 and a(5) = 543. What is the next prime?

Examples

			a(0) = 1 because (2^0)^(2^0) = 1^1 = 1.
a(1) = 3 = (2^0)^(2^1) + (2^1)^(2^0) = 1^2 + 2^1.
a(2) = 9 = (2^0)^(2^2) + (2^1)^(2^1) + (2^2)^(2^0) = 1^4 + 2^2 + 4^1.
a(3) = 41 = 1^8 + 2^4 + 4^2 + 8^1.
a(4) = 593 = 1^16 + 2^8 + 4^4 + 8^2 + 16^1
a(5) = 135457 = 1^32 + 2^16 + 4^8 + 8^4 + 16^2 + 32^1.
a(6) = 8606778433 = 1^64 + 2^32 + 4^16 + 8^8 + 16^4 + 32^2 + 64^1.
a(7) = 36893769626691833985 = 1^128 + 2^64 + 4^32 + 8^16 + 16^8 + 32^4 + 64^2 + 128^1.
a(8) = 680564733921105089459460297630318346497 = 1^256 + 2^128 + 4^64 + 8^32 + 16^16 + 32^8 + 64^4 + 128^2 + 256^1.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[(2^k)^(2^(n - k)), {k, 0, n}], {n, 0, 10}] (* G. C. Greubel, May 19 2017 *)
  • PARI
    for(n=0,5, print1(sum(k=0,n, (2^k)^(2^(n-k))), ", ")) \\ G. C. Greubel, May 19 2017

Formula

a(n) = Sum_{i=0..n} (2^i)^(2^(n-i)).
a(n) = Sum_{i=0..n} (2^(n-i))^(2^i).
a(n) = Sum_{i=0..n} (A000079(i))^(A000079(n-i)).
a(n) ~ 2^(2^(n-1) + 1). - Vaclav Kotesovec, Jun 07 2025

Extensions

a(4) corrected by Giovanni Resta, Jun 13 2016
Formulas corrected by G. C. Greubel, May 19 2017

A113208 Half-fixed-point of ascending descending base exponent transform.

Original entry on oeis.org

1, 1, 2, 4, 10, 44, 1426, 17592187106356
Offset: 1

Views

Author

Jonathan Vos Post, Jan 06 2006

Keywords

Comments

a(9) has 429 digits.
The ascending descending base exponent transform applied to the Fibonacci numbers is A113122; applied to the tribonacci numbers is A113153; applied to the Lucas numbers is A113154. There is no nontrivial integer fixed point of the transform.

Examples

			a(2) = 1 because a(1)^a(2) + a(2)^a(1) = 1^1 + 1^1 = 2 and 2/2 = 1.
a(3) = 2 because a(1)^a(3) + a(2)^a(2) + a(3)^a(1) = 1^2 + 1^1 + 2^1 = 4 and 4/2 = 2.
a(4) = 4 because a(1)^a(4) + a(2)^a(3) + a(3)^a(2) + a(4)^a(1) = 1^4 + 1^2 + 2^1 + 4^1 = 8 and 8/2 = 4.
a(5) = 10 because a(1)^a(5) + a(2)^a(4) + a(3)^a(3) + a(4)^a(2) + a(5)^a(1) = 1^10 + 1^4 + 2^2 + 4^1 + 10^1 = 20 and 20/2 = 10.
a(6) = 44 because 1^44 + 1^10 + 2^4 + 4^2 + 10^1 + 44^1 = 88 and 88/2 = 44.
a(7) = (1^1426 + 1^44 + 2^10 + 4^4 + 10^2 + 44^1 + 1426^1)/2 = 1426.
a(8) = (1^17592187106356 + 1^1426 + 2^44 + 4^10 + 10^4 + 44^2 + 1426^1 + 17592187106356^1)/2 = 17592187106356.
		

Crossrefs

Programs

  • Mathematica
    nmax = 9; a[1] = 1; Do[a[n] = a[n] /. Solve[a[n] == Sum[a[i]^a[n - i + 1], {i, 1, n}]/2, a[n]][[1]], {n, 2, nmax}]; Table[a[n], {n, 1, nmax}] (* Vaclav Kotesovec, Jun 08 2025 *)

Formula

a(1) = 1. For n > 1: a(n) = (1/2) * Sum_{i=1..n} a(i)^a(n-i+1).

A113258 Ascending descending base exponent transform of factorials.

Original entry on oeis.org

1, 3, 11, 125, 16824569, 1329227995784915877642188398793079569
Offset: 1

Views

Author

Jonathan Vos Post, Jan 07 2006

Keywords

Comments

A003101 is the ascending descending base exponent transform of natural numbers A000027. The ascending descending base exponent transform applied to the Fibonacci numbers is A113122; applied to the tribonacci numbers is A113153; applied to the Lucas numbers is A113154. The smallest primes in this (always odd) sequence are a(2) = 3 and a(3) = 11. What is the next prime? Is there a nontrivial power after a(4) = 5^3?

Examples

			a(1) = 1 because (1!)^(1!) = 1^1 = 1.
a(2) = 3 because (1!)^(2!) + (2!)^(1!) = 1 + 2 = 3.
a(3) = 11 = (1!)^(3!) + (2!)^(2!) + (3!)^(1!) = 1^6 + 2^2 + 6^1 = 11.
a(4) = 125 = (1!)^(4!) + (2!)^(3!) + (3!)^(2!) + (4!)^(1!).
a(6) = 1329227995784915877642188398793079569 = 1^720 + 2^120 + 6^24 + 24^6 + 120^2 + 720^1.
a(7) = 1!^7! + 2!^6! + 3!^5! + 4!^4! + 5!^3! + 6!^2! + 7!^1! has 217 digits.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[((k)!)^(n - k + 1)!, {k, 1, n}], {n,1,5}] (* G. C. Greubel, May 18 2017 *)
  • PARI
    for(n=1,5, print1(sum(k=1,n, (k!)^((n-k+1)!)), ", ")) \\ G. C. Greubel, May 18 2017

Formula

a(n) = Sum_{i = 1..n} (i!)^((n-i+1)!).
a(n) = Sum_{i = 1..n} (n-i+1)!^i!.
a(n) = Sum_{i = 1..n} (A000142(i))^(A000142(n-i+1)).
a(n) ~ 2^((n-1)!). - Vaclav Kotesovec, Jun 08 2025

A113320 a(1)=1 and a(n) for n>1 has the smallest positive value such that Sum_{i=1..n} a(i)^a(n-i+1) is prime.

Original entry on oeis.org

1, 1, 1, 2, 2, 4, 4, 4, 6, 2, 6, 4, 18, 6, 4, 20, 6, 30, 4, 40, 30, 8, 18, 16, 40, 128, 24, 40, 58, 194, 78, 84, 56, 56, 72, 112, 98, 300, 444, 54, 978, 1938, 120, 126, 6, 1750
Offset: 1

Views

Author

Jonathan Vos Post, Jan 07 2006

Keywords

Comments

Previous name was: Least integers so ascending descending base exponent transforms all prime.
This is the first sequence submitted as a solution to an "ascending descending base exponent transform inverse problem" where the sequence is iteratively defined such that the transform meets a constraint. The sequence is infinite, but it is hard to characterize the asymptotic cost of adding an n-th term. A003101 is the ascending descending base exponent transform of natural numbers A000027. The ascending descending base exponent transform applied to the Fibonacci numbers is A113122; applied to the tribonacci numbers is A113153; applied to the Lucas numbers is A113154.

Examples

			a(1) = 1 by definition.
a(2) = 1 because 1 is the min such that 1^a(2) + a(2)^1 is prime (p=2).
a(3) = 1 because 1 is the min such that 1^a(3) + 1^1 + a(3)^1 is prime (p=5).
a(4) = 2 because 2 is the min such that 1^a(4) + 1^1 + 3^1 + a(4)^1 is prime (p=7).
		

Crossrefs

Programs

  • Mathematica
    inve[w_] := Total[w^Reverse[w]]; a[1] = 1; a[n_] := a[n] = Block[{k = 0}, While[! PrimeQ[ inve@ Append[Array[a, n-1], ++k]]]; k]; Array[a, 46] (* Giovanni Resta, Jun 13 2016 *)
  • PARI
    lista(n)={my(a=vector(n)); a[1]=1; print1(1, ", "); for(n=2, #a, my(t=sum(i=2, n-1, a[i]^a[n-i+1])); my(k=1); while(!ispseudoprime(t+1+k), k++); a[n]=k; print1(k, ", "))} \\ Andrew Howroyd, Jan 03 2020

Formula

a(1) = 1. For n>1, a(n) = min {k>0: a(1)^k + k^a(1) + Sum_{i=2..n-1} a(i)^a(n-i+1) is prime}.

Extensions

Corrected and extended by Giovanni Resta, Jun 13 2016
New name from Giovanni Resta, Jan 03 2020

A113336 Least integers, starting with 2, so ascending descending base exponent transforms all prime.

Original entry on oeis.org

2, 1, 6, 6, 18, 12, 18, 42, 288, 108, 180, 1122, 1458, 660
Offset: 1

Views

Author

Jonathan Vos Post, Jan 07 2006

Keywords

Comments

This is the second sequence submitted as a solution to an "ascending descending base exponent transform inverse problem" where the sequence is iteratively defined such that the transform meets a constraint. The sequence is probably infinite, but it is hard to characterize the asymptotic cost of adding an n-th term (the 9th terms is at least 250). A003101 is the ascending descending base exponent transform of natural numbers A000027. The ascending descending base exponent transform applied to the Fibonacci numbers is A113122; applied to the tribonacci numbers is A113153; applied to the Lucas numbers is A113154.

Examples

			a(1) = 2 by definition.
a(2) = 1 because 1 is the min such that 2^a(2) + a(2)^2 is prime (p=3).
a(3) = 6 because 6 is the min such that 2^a(3) + 1^1 + a(3)^2 is prime (2^6 + 1^1 + 6^1 = 101).
a(4) = 6 because 2^6 + 1^6 + 6^1 + 6^2 = 107 is prime.
a(5) = 18 because 2^18 + 1^6 + 6^6 + 6^1 + 18^2 = 309131 is prime.
a(6) = 12 because 2^12 + 1^18 + 6^6 + 6^6 + 18^1 + 12^2 = 97571 is prime.
a(7) = 18 because 2^18 + 1^12 + 6^18 + 6^6 + 18^6 + 12^1 + 18^2 = 101559990989777 is prime.
a(8) = 42 because 2^42 + 1^18 + 6^12 + 6^18 + 18^6 + 12^6 + 18^1 + 42^2 = 105960216961847 is prime.
a(9) > 250.
		

Crossrefs

Formula

a(1) = 2. For n>1: a(n) = min {n>0: Sum_{i=1..n} a(i)^a(n-i+1) is prime}.

Extensions

a(9)-a(14) from Giovanni Resta, Jun 13 2016

A113498 Ascending descending base exponent transform of omega(n) (A001221).

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 8, 9, 13, 12, 14, 15, 21, 19, 24, 21, 29, 28, 30, 28, 40, 35, 41, 42, 46, 41, 53, 44, 59, 52, 61, 55, 79, 55, 69, 66, 86, 70, 90, 73, 94, 93, 91, 81, 121, 88, 114, 103, 123, 95, 137, 102, 138, 122, 132, 114, 168, 121, 144, 145, 159, 137, 180
Offset: 2

Views

Author

Jonathan Vos Post, Jan 10 2006

Keywords

Examples

			Since omega(n) = A001221(n) = 0, 1, 1, 1, 1, 2, 1, 1, 1, 2 and we skip the initial zero term, we have:
a(1) = 1^1 = 1.
a(2) = 1^1 + 1^1 = 2.
a(3) = 1^1 + 1^1 + 1^1 = 3.
a(4) = 1^1 + 1^1 + 1^1 + 1^1 = 4.
a(5) = 1^1 + 1^1 + 1^1 + 1^1 + 2^1 = 6.
a(9) = 1^1 + 1^1 + 1^1 + 1^1 + 2^2 + 1^1 + 1^1 + 1^1 + 2^1 = 13.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[PrimeNu[k]^(PrimeNu[n - k + 2]), {k, 2, n}], {n, 2, 50}] (* G. C. Greubel, May 18 2017 *)
  • PARI
    for(n=2,25, print1(sum(k=2,n, omega(k)^(omega(n-k+2))), ", ")) \\ G. C. Greubel, May 18 2017

Formula

a(n) = Sum_{i=1..n} (omega(k))^(omega(n-k+2)).
a(n) = Sum_{i=1..n} (A001221(k))^(A001221(n-k+2)).

Extensions

Corrected and extended by Giovanni Resta, Jun 13 2016
Formulas corrected by G. C. Greubel, May 18 2017
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