A007778 -id:A007778 - cp's OEIS frontend

A062276 a(n) = floor(n^(n+1) / (n+1)^n).

0, 0, 0, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 16, 16, 16, 17, 17, 17, 18, 18, 18, 19, 19, 20, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 24, 24, 24, 25, 25, 25, 26, 26, 27, 27, 27, 28, 28

Offset: 0

Henry Bottomley, Jul 02 2001

nonn

a(n) is close to n/e (cf. A032634).

			a(2) = floor(2^3/3^2) = floor(8/9) = 0.

Harry J. Smith, Table of n, a(n) for n = 0..1000

Cf. A032634. Cf. A000169, A007778, A007925.

List([0..90],n->Int(n^(n+1)/(n+1)^n)); # Muniru A Asiru, Jul 01 2018

seq(floor(n^(n+1)/(n+1)^n),n=0..90); # Muniru A Asiru, Jul 01 2018

Array[Floor[#^(# + 1)/(# + 1)^#] &, 78, 0] (* Michael De Vlieger, Jul 01 2018 *)

{ default(realprecision, 50); for (n=0, 1000, write("b062276.txt", n, " ", floor(n^(n + 1) / (n + 1)^n)) ) } \\ Harry J. Smith, Aug 03 2009

A085283 a(n) = nn^n - (n-1)(n-1)^n.

Original entry on oeis.org

1, 1, 7, 65, 781, 11529, 201811, 4085185, 93864121, 2413042577, 68618940391, 2138428376721, 72470493235141, 2653457921150425, 104382202543721467, 4390455017903519489, 196621779843659466481, 9340717969198079777313

Offset: 0

Paul Barry, Jun 26 2003

The system of equations
x(0) = n*x(1) + 1,
(n-1)*x(1) = n*x(2) + 1,
...
(n-1)*x(n) = n*x(n+1) + 1.
relates to the Monkey-And-Coconuts problem and reduces to the single equation
A007778(n-1)*x(0) = A007778(n)*x(n+1) + a(n),
whose solutions {x(0),x(n+1)} are given by {A014293(n), A085606(n)=A007778(n-1) - 1}. - Lekraj Beedassy, Jul 15 2003
For n >= 1, a(n) is equal to the number of functions f: {1,2,...,n+1}->{1,2,...,n} such that Im(f) contains a fixed element. - Aleksandar M. Janjic and Milan Janjic, Feb 27 2007

Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
BBC, The Monkey and the Coconuts
K. Belcourt, How Many Coconuts
Santo D'Agostino, "The Coconut Problem"; Updated With Solution, May 2011.
J. Dean, Sailors, monkey and coconuts
A. K. Dewdney, The Monkey and the Coconuts
G. Lewandrowski, The Monkey Problem
R. Raja, Monkeys and Coconuts
A. Rishi, Nemo's sailors and the monkey
Dr. Rob, The MathForum, Coconut piles
D. J. Wright, Five Pirates and a Monkey

Cf. A045531, A055869.

Join[{1},Table[n*n^n-(n-1)(n-1)^n,{n,20}]] (* Harvey P. Dale, Sep 08 2016 *)

E.g.f.: -(x + 2*x*W(-x) + W(-x)^2)/(W(-x)*(1 + W(-x))^3), where W(x) is the Lambert W function. - Fabian Pereyra, Sep 26 2023

A110676 Number of prime factors with multiplicity of 1 + n^(n+1).

Original entry on oeis.org

1, 2, 2, 3, 3, 4, 3, 6, 3, 5, 4, 5, 4, 9, 3, 4, 9, 3, 6, 10, 6, 7, 6, 11, 5, 11, 10, 5, 10, 8, 3, 12, 6, 10, 8, 5, 6, 13, 8, 6, 11, 6, 10, 16, 4, 4, 6, 9, 6, 11, 8, 4, 10, 10, 5, 13, 10, 7, 11, 6, 6, 21, 4, 23, 8, 6, 8, 16, 15, 7, 12, 7, 8, 19, 8, 13, 14, 5, 6, 20, 6, 10, 13, 12, 7, 9, 9, 6, 21

Offset: 1

Jonathan Vos Post, Sep 14 2005

As also noticed by T. D. Noe, for odd n: 2 | a(n), for even n: (n+1)^2 | a(n). Coincidentally, a(74) includes 13 multidigit prime factors all of which end with the digit 1. There is no upper limit to this sequence, which rapidly becomes slow to compute. The derived sequences of n such that a(n) = k for any constant k > 2 do not yet appear in the OEIS. For instance, a(n) = 3 for n = 4, 5, 7, 9, 15, 18, 31, ... Is each such derived sequence finite?

			a(1) = 1 because 1+1^2 = 2 is prime (and the only such prime).
a(2) = 2 because 1 + 2^3 = 9 = 3^2 which has (with multiplicity) two prime factors.
a(3) = 2 because 1 + 3^4 = 82 = 2 * 41 (the last such semiprime?).
a(4) = 3 because 1 + 4^5 = 1025 = 5^2 * 41 which has (with multiplicity) 3 prime factors.
a(8) = 6 because 1 + 8^9 = 134217729 = 3^4 * 19 * 87211.
a(14) = 9 because 1 + 14^15 = 155568095557812225 = 3^2 * 5^2 * 61 * 71 * 101 * 811 * 1948981.
a(1000) > 52.

Cf. A001222, A007778, A110567.

a(n) = bigomega(1+(n^(n+1))) \\ Georg Fischer, Jun 21 2024

a(n) = A001222(A110567(n)) = A001222(1 + A007778(n)) = A001222(1 + (n^(n+1))).

Trivially a(n) << n log n. At most n^(n+1) + 1 is of the form 2*3^k. - Charles R Greathouse IV, Jun 24 2024

a(13) and 3 other terms corrected by Georg Fischer, Jun 21 2024

A110874 Number of prime factors of 2 + n^(n+1) counted with multiplicity.

Original entry on oeis.org

1, 2, 1, 5, 2, 2, 4, 5, 2, 5, 4, 4, 5, 3, 1, 4, 5, 3, 4, 6, 3, 8, 4, 5, 4, 4, 2, 6, 3, 6, 5, 5, 5, 6, 6, 8, 6, 6, 4, 5, 4, 6, 4, 5, 3, 8, 4, 3, 5, 5, 5, 7, 7, 11, 4, 5, 4, 13, 4, 6, 2, 5, 2, 6, 6, 5, 8, 9, 5, 9, 4, 7, 4, 4, 5, 7, 6, 7, 6, 9, 4, 9, 5, 8, 5, 8

Offset: 1

Jonathan Vos Post, Sep 18 2005

nonn

Compared with A110676, number of prime factors with multiplicity of 2 + n^(n+1), this seems to have an unlimited number of primes (n = 1, 3, 15, ...) and semiprimes (n = 2, 5, 6, 9, 27, ...). Of course, n even gives n | a(n).

			a(1) = 1 because 2 + 1^2 = 3 is prime (one prime factor).
a(2) = 2 because 2 + 2^3 = 10 = 2 * 5 is semiprime (two prime factors).
a(3) = 1 because 2 + 3^4 = 83 is prime.
a(4) = 5 because 2 + 4^5 = 1026 = 2 * 3^3 * 19 has five prime factors (3 has multiplicity of 3).
a(5) = 2 because 2 + 5^6 = 15627 = 3 * 5209 is semiprime (two prime factors).
a(6) = 2 because 2 + 6^7 = 279938 = 2 * 139969 is semiprime (two prime factors).
a(15) = 1 because 2 + 15^16 = 6568408355712890627 is prime. What is the next prime?

Sean A. Irvine, Table of n, a(n) for n = 1..100

Cf. A001222, A007778, A110567, A110676.

Table[PrimeOmega[2+n^(n+1)],{n,41}] (* Harvey P. Dale, Nov 08 2020 *)

a(n) = A001222(1 + A110567(n)) = A001222(2 + A007778(n)) = A001222(2 + n^(n+1)).

More terms from Sean A. Irvine, Sep 17 2023

A138748 a(n) = (n+(n+1)) + (n*(n+1)) + (n^(n+1)).

Original entry on oeis.org

1, 6, 19, 100, 1053, 15666, 279991, 5764872, 134217817, 3486784510, 100000000131, 3138428376876, 106993205379253, 3937376385699498, 155568095557812463, 6568408355712890896, 295147905179352826161, 14063084452067724991350, 708235345355337676358011, 37589973457545958193356020, 2097152000000000000000000461

Offset: 0

Anthony J. DeFusco II, Mar 28 2008

Sum of three arithmetic operations (sum, product, powers) of two consecutive integers.

			a(0) = (0+1) + (0*1) + (0^1) = 1;
a(1) = (1+2) + (1*2) + (1^2) = 6;
a(2) = (2+3) + (2*3) + (2^3) = 19; etc.

A028387 := proc(n) n+(n+1)^2 ; end: A007778 := proc(n) n^(n+1) ; end: A138748 := proc(n) A028387(n)+A007778(n) ; end: seq(A138748(n),n=0..20) ; # R. J. Mathar, Apr 03 2008

Table[2n+1+n(n+1)+n^(n+1),{n,0,20}] (* Harvey P. Dale, Sep 13 2024 *)

a(n) = A028387(n) + A007778(n). - R. J. Mathar, Apr 03 2008

More terms from R. J. Mathar, Apr 03 2008

A166974 Number of single-component graphs where the product of the valences of the nodes is n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 4, 2, 2, 1, 6, 1, 2, 2, 8, 1, 6, 1, 6, 2, 2, 1, 16, 2, 2, 4, 6, 1, 8, 1, 16, 2, 2, 2, 25, 1, 2, 2, 16, 1, 8, 1, 6, 6, 2, 1, 46, 2, 6, 2, 6, 1, 18, 2, 16, 2, 2, 1, 36, 1, 2, 6, 40, 2, 8, 1, 6, 2, 8, 1, 84, 1, 2, 6, 6, 2, 8, 1, 49, 12, 2, 1, 36, 2, 2, 2, 16, 1, 38, 2, 6, 2, 2, 2, 137

Offset: 0

Franklin T. Adams-Watters, Oct 26 2009

A single-component graph is any nonempty connected graph. If the empty graph was allowed, a(1) would be 2 instead of 1.
The sequence can be computed for n > 1 by looking at the graph that results when all valence 1 nodes are removed. This will be a connected graph, and labeling each node with its original valence, the product of the labels will be the original product. Each node must be labeled with at least its valence, and at least 2. Each such labeling, up to graph equivalence, uniquely defines the original graph, so we need only count the labelings for connected graphs with up to BigOmega(n) nodes.
Note, in particular, that a(n) = 1 for any prime, and 2 for any semiprime.
This product for the complete graph on n points is (n-1)^n. For the complete bipartite graph with n and m points in the parts the product is n^m*m^n. For the cyclic graph with n nodes it is 2^n.

Andrew Weimholt, Table of n, a(n) for n = 0..255

Cf. A000079, A001222 (BigOmega), A001349, A002905, A007778, A062275.

Corrected and extended by Andrew Weimholt, Oct 26 2009

A193637 a(n) = a(n-1)^2 - n^(n+1).

Original entry on oeis.org

0, -1, -7, -32, 0, -15625, 243860689, 59468035633789920, 3536447262141707692104062559388672, 12506459237909580203511583184455022770672120296396568887010875139183

Offset: 0

Arkadiusz Wesolowski, Aug 01 2011

Example of a recursive sequence which produces a table containing two zeros.

			a(2) = -7 because a(1) = -1 and (-1)^2 - 2^(2+1) = -7.

Vincenzo Librandi, Table of n, a(n) for n = 0..12
Eric Weisstein's World of Mathematics, Recursive Sequence

Cf. A003095, A007778.

RecurrenceTable[{a[n] == a[n - 1]^2 - n^(n + 1), a[0] == 0}, a, {n, 10}]

a=0; for(n=0, 10, print1(a=a^2-n^(n+1), ", "));

a(0) = 0, a(n) = a(n-1)^2 - n^(n+1).

A242749 G.f. satisfies: A(x) = G(x/A(x)) such that A(xG(x)) = G(x) = Sum_{n>=0} (n+1)^(n+1)x^n.

Original entry on oeis.org

1, 4, 11, 60, 611, 8632, 151538, 3132140, 73883667, 1949844168, 56785116742, 1806695366616, 62314198956510, 2315470815127792, 92214156916779444, 3918743752606940812, 177018691811732542595, 8471087431826716955880, 428141645771934036086942, 22791557465710675500959688

Offset: 0

Paul D. Hanna, May 21 2014

nonn

			G.f.: A(x) = 1 + 4*x + 11*x^2 + 60*x^3 + 611*x^4 + 8632*x^5 + 151538*x^6 +...
such that A(x*G(x)) = G(x) where:
G(x) = 1 + 4*x + 27*x^2 + 256*x^3 + 3125*x^4 +...+ (n+1)^(n+1)*x^n +...
also, A(x) = G(x/A(x)):
A(x) = 1 + 4*x/A(x) + 27*x^2/A(x)^2 + 256*x^3/A(x)^3 + 3125*x^4/A(x)^4 +...+ (n+1)^(n+1)*x^n/A(x)^n +...
If we form a table of coefficients of x^k in A(x)^n like so:
[1,  4,  11,   60,   611,   8632,  151538,   3132140, ...];
[1,  8,  38,  208,  1823,  23472,  389174,   7739808, ...];
[1, 12,  81,  508,  4164,  48852,  759407,  14463624, ...];
[1, 16, 140, 1024,  8418,  91920, 1335712,  24248640, ...];
[1, 20, 215, 1820, 15625, 163664, 2232620,  38498580, ...];
[1, 24, 306, 2960, 27081, 279936, 3623894,  59297664, ...];
[1, 28, 413, 4508, 44338, 462476, 5764801,  89716400, ...];
[1, 32, 536, 6528, 69204, 739936, 9018480, 134217728, ...]; ...
then the main diagonal forms the sequence A007778:
[1, 8, 81, 1024, 15625, 279936, 5764801, 134217728, ..., (n+1)^(n+2), ...].

Cf. A180749, A007778.

{a(n)=polcoeff(x/serreverse(x*sum(m=0, n+1, (m+1)^(m+1)*x^m)+x^2*O(x^n)), n)}
for(n=0,20,print1(a(n),", "))

G.f. satisfies: [x^n] A(x)^(n+1) = (n+1)^(n+2).

G.f.: A(x) = x/Series_Reversion(x*G(x)) where G(x) = Sum_{n>=0} (n+1)^(n+1)*x^n.

A258385 a(n) = n^(n+1) * (n-1)^n.

Original entry on oeis.org

0, 8, 648, 82944, 16000000, 4374000000, 1613775332736, 773738492592128, 467988280328060928, 348678440100000000000, 313842837672100000000000, 335790511878017502425382912, 421272520289832237611045879808, 612530145817540311016457192308736

Offset: 1

Daniel Suteu, May 28 2015

			For n=3, a(3) = 3^(3+1) * (3-1)^3 = 3^4 * 2^3 = 81 * 8 = 648.

Daniel Suteu, Table of n, a(n) for n = 1..20

Cf. A007778 (n^(n+1)).

[n^(n+1) * (n-1)^n: n in [1..20]]; // Vincenzo Librandi, May 29 2015

Array[#^(# + 1) (# - 1)^# &, 20] (* Vincenzo Librandi, May 29 2015 *)

a(n)=n^(n+1)*(n-1)^n \\ Charles R Greathouse IV, May 28 2015

func a(n) {
    (n-1)**n * n**(n+1);
};
1.to(Math.inf).each { |n|
    say a(n);
};

a(n) ~ 1/e * n^(2n+1). - Charles R Greathouse IV, May 28 2015

a(n) = A007778(n)*A007778(n-1). - Michel Marcus, Jul 07 2015

More terms from Vincenzo Librandi, May 29 2015

A258388 a(n) = n^(n+1) + (n-1)^n.

Original entry on oeis.org

1, 1, 9, 89, 1105, 16649, 295561, 6044737, 139982529, 3621002129, 103486784401, 3238428376721, 110131633755793, 4044369591078361, 159505471943511513, 6723976451270702849, 301716313535065716481, 14358232357247077816865, 722298429807405401348641

Offset: 0

Daniel Suteu, May 28 2015

			a(3) = 3^(3+1) + (3-1)^3 = 3^4 + 2^3 = 81 + 8 = 89.

Alois P. Heinz, Table of n, a(n) for n = 0..350 (first 20 terms from Daniel Suteu)

Cf. A007778, A084363.

[n^(n+1)+(n-1)^n: n in [0..20]]; // Vincenzo Librandi, May 29 2015

a:= n-> n^(n+1)+(n-1)^n:
seq(a(n), n=0..20);  # Alois P. Heinz, Jun 04 2015

Array[#^(# + 1) + (# - 1)^# &, 20] (* Vincenzo Librandi, May 29 2015 *)

vector(10,n,n^(n+1)+(n-1)^n) \\ Derek Orr, Jun 01 2015

func a(n) {
    (n-1)**n + n**(n+1)
};
1.to(Math.inf).each { |n|
    say a(n);
};

From Robert Israel, Jun 04 2015: (Start)
a(n) = A007778(n-1) + A007778(n) for n>0.
E.g.f.: 1/(1 + W(-x))^3 - 1/(1 + W(-x))^2 - x/(W(-x)*(1+W(-x))) where W is the Lambert W function. (End)

cp's OEIS Frontend

A062276 a(n) = floor(n^(n+1) / (n+1)^n).

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A085283 a(n) = n*n^n - (n-1)*(n-1)^n.

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A110676 Number of prime factors with multiplicity of 1 + n^(n+1).

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A110874 Number of prime factors of 2 + n^(n+1) counted with multiplicity.

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A138748 a(n) = (n+(n+1)) + (n*(n+1)) + (n^(n+1)).

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Maple

Mathematica

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Extensions

A166974 Number of single-component graphs where the product of the valences of the nodes is n.

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Extensions

A193637 a(n) = a(n-1)^2 - n^(n+1).

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A242749 G.f. satisfies: A(x) = G(x/A(x)) such that A(x*G(x)) = G(x) = Sum_{n>=0} (n+1)^(n+1)*x^n.

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A085283 a(n) = nn^n - (n-1)(n-1)^n.

A242749 G.f. satisfies: A(x) = G(x/A(x)) such that A(xG(x)) = G(x) = Sum_{n>=0} (n+1)^(n+1)x^n.