OFFSET
0,3
COMMENTS
a(n-1) is the number of set partitions of [n] into two or more blocks such that all absolute differences between least elements of consecutive blocks are 1. a(3) = 8: 134|2, 13|24, 14|23, 1|234, 14|2|3, 1|24|3, 1|2|34, 1|2|3|4. - Alois P. Heinz, May 22 2017
Min_{n >= 1} a(n+1)/a(n) = 8/3. This is the answer to the 4th problem proposed during the first day of the final round of the 16th Austrian Mathematical Olympiad in 1985 (see link IMO Compendium). - Bernard Schott, Jan 07 2019
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 0..598
Henry W. Gould, Letters to N. J. A. Sloane, Oct 1973 and Jan 1974.
Mathematics Stack Exchange, Asymptotics of ..., 2011.
Daniel Ropp, Problem 2 - 16th Austrian Mathematical Olympiad (Final round), Crux Mathematicorum, page 7, Vol. 14, Jun. 88.
The IMO compendium, Problem 2, 16th Austrian Mathematical Olympiad, 1985.
FORMULA
a(n) = A026898(n) - 1.
G.f.: G(0)/x-1/(1-x)/x where G(k) = 1 + x*(2*k*x-1)/((2*k*x+x-1) - x*(2*k*x+x-1)^2/(x*(2*k*x+x-1) + (2*k*x+2*x-1)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 26 2013
G.f.: Sum_{k>=1} x^k/(1 - (k + 1)*x). - Ilya Gutkovskiy, Oct 09 2018
a(n) = n^1 + (n-1)^2 + (n-2)^3 + ... + 3^(n-2) + 2^(n-1) + 1^n. - Bernard Schott, Jan 07 2019
log(a(n)) ~ (1 - 1/LambertW(exp(1)*n)) * n * log(1 + n/LambertW(exp(1)*n)). - Vaclav Kotesovec, Jun 15 2021
a(n) ~ sqrt(2*Pi/(n+1 + w(n))) * w(n)^(n+2 - w(n)), where w(n) = (n+1)/LambertW(exp(1)*(n+1)). - Vaclav Kotesovec, Jun 25 2021, after user "leonbloy", see Mathematics Stack Exchange link.
EXAMPLE
For n = 3 we get a(3) = 3^1 + 2^2 + 1^3 = 8. For n = 4 we get a(4) = 4^1 + 3^2 + 2^3 + 1^4 = 22.
MAPLE
A003101 := n->add((n-k+1)^k, k=1..n);
a:= n-> add((n-j+1)^j, j=1..n): seq(a(n), n=0..30); # Zerinvary Lajos, Jun 07 2008
MATHEMATICA
Table[Sum[(n-k+1)^k, {k, n}], {n, 0, 25}] (* Harvey P. Dale, Aug 14 2011 *)
PROG
(PARI) a(n)=sum(k=1, n, (n-k+1)^k) \\ Charles R Greathouse IV, Oct 31 2011
(Haskell)
a003101 n = sum $ zipWith (^) [0 ..] [n + 1, n .. 1]
-- Reinhard Zumkeller, Sep 14 2014
(Magma) [n eq 0 select 0 else (&+[(n-j+1)^j: j in [1..n]]): n in [0..50]]; // G. C. Greubel, Oct 26 2022
(SageMath)
def A003101(n): return sum( (n-k+1)^k for k in range(1, n+1))
[A003101(n) for n in range(50)] # G. C. Greubel, Oct 26 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved