A060073 -id:A060073 - cp's OEIS frontend

A128466 Primes of the form ((k+1)^k - 1)/k^2 = A060073(k+1).

2, 7, 311, 7563707819165039903

Offset: 1

Alexander Adamchuk, Mar 09 2007

nonn

Corresponding numbers k are listed in A127837.
Terms are the primes in A060073.
Next term has 15850 = 1 + floor((4357*log(4358) - 2*log(4357))/log(10)) digits and is too large to include. - M. F. Hasler, May 22 2007

Cf. A127837, A037205, A060072, A060073, A058128, A128456

Select[Table[((n+1)^n-1)/n^2,{n,500}],PrimeQ]  (* Harvey P. Dale, Apr 30 2011 *)

A128466(n)=A060073(A127837(n)+1) /* see there. --- or: */ forprime(p=1,10^5,if(ispseudoprime(n=((p+1)^p-1)/p^2),print1(n,", "))); \\ M. F. Hasler, May 22 2007

a(n) = ((A127837(n) + 1)^A127837(n) - 1) / A127837(n)^2.

A023811 Largest metadrome (number with digits in strict ascending order) in base n.

Original entry on oeis.org

0, 1, 5, 27, 194, 1865, 22875, 342391, 6053444, 123456789, 2853116705, 73686780563, 2103299351334, 65751519677857, 2234152501943159, 81985529216486895, 3231407272993502984, 136146740744970718253, 6106233505124424657789, 290464265927977839335179

Offset: 1

Olivier Gérard

Also smallest zeroless pandigital number in base n. - Franklin T. Adams-Watters, Nov 15 2006

The smallest permutational number in A134640 in the n-positional system. - Artur Jasinski, Nov 07 2007

			a(5) = 1234[5] (in base 5) = 1*5^3 + 2*5^2 + 3*5 + 4 = 125 + 50 + 15 + 4 = 194.
a(10) = 123456789 (in base 10).

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Christian Perfect, Integer sequence reviews on Numberphile (or vice versa), 2013.
Chai Wah Wu, Pandigital and penholodigital numbers, arXiv:2403.20304 [math.GM], 2024. See p. 1.

Cf. A049363, A051846, A058128, A060073.

Cf. A062813, A134640.

a023811 n = foldl (\val dig -> val * n + dig) 0 [0 .. n - 1]
-- Reinhard Zumkeller, Aug 29 2014

[0] cat [(n^n-n^2+n-1)/(n-1)^2: n in [2..20]]; // Vincenzo Librandi, May 22 2012

0, seq((n^n-n^2+n-1)/(n-1)^2, n=2..100); # Robert Israel, Dec 13 2015

Table[Total[(#1 n^#2) & @@@ Transpose@ {Range[n - 1], Reverse@ (Range[n - 1] - 1)}], {n, 20}] (* Michael De Vlieger, Jul 24 2015 *)
Table[Sum[(b - k)*b^(k - 1), {k, b - 1}], {b, 30}] (* Clark Kimberling, Aug 22 2015 *)
Table[FromDigits[Range[0, n - 1], n], {n, 20}] (* L. Edson Jeffery, Dec 13 2015 *)

{for(i=1,18,cuo=0; for(j=1,i-1,cuo=cuo+j*i^(i-j-1)); print1(cuo,", "))} \\\ Douglas Latimer, May 16 2012

A023811(n)=if(n>1,(n^n-n^2)\(n-1)^2+1)  \\ M. F. Hasler, Jan 22 2013

def a(n): return (n**n - n**2 + n - 1)//((n - 1)**2) if n > 1 else 0
print([a(n) for n in range(1, 21)]) # Michael S. Branicky, Apr 24 2023

a(n) = Sum_{j=1...n-1} j*n^(n-1-j).
lim_{n->infinity} a(n)/a(n-1) - a(n-1)/a(n-2) = exp(1). - Conjectured by Gerald McGarvey, Sep 26 2004. Follows from the formula below and lim_{n->infinity} (1+1/n)^n = e. - Franklin T. Adams-Watters, Jan 25 2010
a(n) = (n^n-n^2+n-1)/(n-1)^2 = A058128(n)-1 = n*A060073(n)-1 (for n>=2). - Henry Bottomley, Feb 21 2001

Edited by M. F. Hasler, Jan 22 2013

A060072 a(n) = (n^(n-1) - 1)/(n-1) for n>1, a(1) = 0.

Original entry on oeis.org

0, 1, 4, 21, 156, 1555, 19608, 299593, 5380840, 111111111, 2593742460, 67546215517, 1941507093540, 61054982558011, 2085209001813616, 76861433640456465, 3041324492229179280, 128583032925805678351, 5784852794328402307380, 275941052631578947368421

Offset: 1

Henry Bottomley, Feb 21 2001

nonn

(n-1)-digit repunits in base n written in decimal.

			a(10)=111111111; i.e., just nine 1's (converted from base 10 to decimal).

Harry J. Smith, Table of n, a(n) for n=1..200

Cf. A000169, A037205, A055869, A060073, A125118, A228275.

Cf. other sequences of generalized repunits, such as A053696, A055129, A031973, A125598, A173468, A023037, A119598, A085104, and A162861.

[0] cat [ (n^(n-1) -1)/(n-1) : n in [2..25]]; // G. C. Greubel, Aug 15 2022

Join[{0},Array[(#^(#-1)-1)/(#-1)&,20,2]] (* Harvey P. Dale, Jun 04 2013 *)

a(n) = if (n==1, 0, (n^(n - 1) - 1)/(n - 1)); \\ Harry J. Smith, Jul 01 2009

[0]+[(n^(n-1) -1)/(n-1) for n in (2..25)] # G. C. Greubel, Aug 15 2022

a(n+1) = Sum_{k=1..n} n^(k-1)*C(n, k). - Olivier Gérard, Jun 26 2001 [Corrected by Mathew Englander, Dec 15 2020]
a(n) = Sum_{j=2..n} n^(n-j). - Zerinvary Lajos, Sep 11 2006
a(n+1) = A125118(n,n). - Reinhard Zumkeller, Nov 21 2006
a(n) = Integral_{x=1/n..1} 1/x^n dx. - Francesco Daddi, Aug 01 2011
a(n) = A037205(n-1)/(n-1) = A060073(n)*(n-1) = A023037(n) - A000169(n).
a(n) = [x^n] x^2/((1 - x)*(1 - n*x)). - Ilya Gutkovskiy, Oct 04 2017
a(n) = 1 + A228275(n, n-2) for n >= 2. - Mathew Englander, Dec 14 2020

Name edited by Michel Marcus, Dec 14 2020

A198410 a(n) = ((3^(n-1) + 1)^3 -1)/3^n.

Original entry on oeis.org

7, 37, 271, 2269, 19927, 177877, 1596511, 14355469, 129159847, 1162320517, 10460530351, 94143710269, 847290203767, 7625602267957, 68630391713791, 617673439330669, 5559060695695687, 50031545486420197, 450283907053258831, 4052555156505760669, 36472996387631139607

Offset: 2

Michel Lagneau, Oct 24 2011

This sequence is generalizable :
Proposition: p^n divide (p^(n-1) + 1)^ p - 1.
Proof: Let a and p be two integers such that p>=2, and k = gcd(a, p). Then ak divides (a + 1)^p - 1 because (a+1)^p - 1 = [a^p + binomial(p,1)*a^(p-1) + … + binomial(p,p-2)*a^2] + pa == binomial(p,1)*n^(n-1) == 0 (mod p^n) with a = p^(n-1) and k = p.
a(n) is the least k such that k*3^n+1 is a cube. Thus, the cube is given by (3^(n-1)+1)^3. - Derek Orr, Mar 23 2014

			a(2) = ((3 + 1)^3 - 1)/3^2 = 63/9 = 7.

Vincenzo Librandi, Table of n, a(n) for n = 2..200
Index entries for linear recurrences with constant coefficients, signature (13,-39,27).

Cf. A060073.

I:=[7,37,271]; [n le 3 select I[n] else 13*Self(n-1)-39*Self(n-2)+27*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Mar 25 2014

A198410 := proc(n)
        (3^(n-1)+1)^3 ;
        (%-1)/3^n ;
end proc:
seq(A198410(n), n=2..20) ; # R. J. Mathar, Oct 25 2011

Table[((3^(n - 1) + 1)^3 - 1)/3^n, {n, 2, 20}] (* Wesley Ivan Hurt, Mar 24 2014 *)
CoefficientList[Series[(7 - 54 x + 63 x^2)/((1 - x) (3 x - 1) (9 x - 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 25 2014 *)
LinearRecurrence[{13,-39,27},{7,37,271},30] (* Harvey P. Dale, Mar 04 2015 *)

a(n)=9^n/27+3^n/3+1 \\ Charles R Greathouse IV, Jun 11 2015

G.f.: -x^2*(7-54*x+63*x^2) / ( (x-1)*(3*x-1)*(9*x-1) ). - R. J. Mathar, Oct 25 2011

a(n) = 13*a(n-1)-39*a(n-2)+27*a(n-3) for n>2. - Vincenzo Librandi, Mar 25 2014

A037205 a(n) = (n+1)^n - 1.

Original entry on oeis.org

0, 1, 8, 63, 624, 7775, 117648, 2097151, 43046720, 999999999, 25937424600, 743008370687, 23298085122480, 793714773254143, 29192926025390624, 1152921504606846975, 48661191875666868480, 2185911559738696531967, 104127350297911241532840, 5242879999999999999999999, 278218429446951548637196400, 15519448971100888972574851071

Offset: 0

N. J. A. Sloane

For n >= 1, a(n) = order of Fibonacci group F(n+1,n).
The terms, written in base n+1, are n digits of value n. For example, a(4) = 624 = 4444 in base 5. - Marc Morgenegg, Nov 30 2016
For n >= 1, in a square grid of side n, this is the number of ways to populate the grid with 1 X 1 blocks (with at least one block) so that no block falls under the effect of gravity. - Paolo Xausa, Apr 12 2021
For n > 1, (n-1)^2 | a(n). - David A. Corneth, Dec 15 2022

D. L. Johnson, Presentation of Groups, Cambridge, 1976, p. 182.
Richard M. Thomas, The Fibonacci groups revisited, in Groups - St. Andrews 1989, Vol. 2, 445-454, London Math. Soc. Lecture Note Ser., 160, Cambridge Univ. Press, Cambridge, 1991.

G. C. Greubel, Table of n, a(n) for n = 0..385
Michael Penn, A divisibility problem., YouTube video, 2021.

A diagonal of A202624.

[(n + 1)^n - 1: n in [0..25]]; // G. C. Greubel, Nov 10 2017

Table[(n + 1)^n - 1, {n, 0, 21}] (* or *)
Table[If[n < 1, Length@ #, FromDigits[#, n + 1]] &@ ConstantArray[n, n], {n, 0, 21}] (* Michael De Vlieger, Nov 30 2016 *)

for(n=0,25, print1((n + 1)^n - 1, ", ")) \\ G. C. Greubel, Nov 10 2017

a(n) = A000169(n+1) - 1 = A060072(n+1)*(n-1) = A060073(n+1)*(n-1)^2.
E.g.f.: 1/(exp(LambertW(-x)) - x) - exp(x). - Ilya Gutkovskiy, Nov 30 2016
E.g.f.: -exp(x) - 1/(x + x/LambertW(-x)). - Vaclav Kotesovec, Dec 05 2016
a(n) = Sum_{k=1..n} binomial(n,k)*n^k [from Paolo Xausa's comment]. - Joerg Arndt, Apr 12 2021

Revised by N. J. A. Sloane, Dec 30 2011

A081215 a(n) = (n^(n+1)+(-1)^n)/(n+1)^2.

Original entry on oeis.org

1, 0, 1, 5, 41, 434, 5713, 90075, 1657009, 34867844, 826446281, 21794641505, 633095889817, 20088655029078, 691413758034721, 25657845139503479, 1021273028302258913, 43404581642184336392, 1961870762757168078553

Offset: 0

Vladeta Jovovic, Apr 17 2003

From Mathew Englander, Oct 19 2020: (Start)
The sum of two adjacent terms of the sequence cannot be prime.
In base n, a(n) has n-1 digits, which are (beginning from the left): n-2, 2, n-4, 4, and so on, except that if n is even the rightmost digit is 1 instead of 0. In that case, the other digits form a palindrome with every even digit from 2 to n-2 appearing twice. For example, a(14) in base 14 is c2a486684a2c1. If n is odd, then all digits from 1 to n-1 occur exactly once. For example, a(15) in base 15 is d2b496785a3c1e.
For any positive integer k, any prime p, and any positive integer h such that h*p > 2, a(h*p^k - 2) == (-1)^h * (1 - 2^(h-1)) (mod p). For example, a(7*p^k - 2) == 63 (mod p); a(10*p^k - 2) == -511 (mod p).
Suppose k and m are positive integers. If k is even, then a(k*m) == 1, a(k*m+1) == 0, and a(k*m-1) == -1 (all mod m). If k is odd, then a(k*m) == (-1)^m and a(k*m+1) == ceiling(m/2) (both mod m), while a(k*m-1) == m/2 - 1 for m even, and a(k*m-1) == 1 for m odd (mod m).
For proofs of the above, see the Englander link. (End)

Robert Israel, Table of n, a(n) for n = 0..387
Mathew Englander, Comments on OEIS A081215

Cf. A081209, A110567, A076951, A273319, A193746, and A060073.

seq((j^(j+1)+(-1)^j)/(j+1)^2, j=0..50); # Robert Israel, May 19 2016

Array[(#^(# + 1) + (-1)^#)/(# + 1)^2 &, 19, 0] (* Michael De Vlieger, Nov 13 2020 *)

a(n) = (n^(n+1)+(-1)^n)/(n+1)^2; \\ Michel Marcus, Oct 20 2020

a(n) = (-1)^n + Sum_{k=1..n} (-1)^(k+1)*(n+1)^(n-k)*C(n+1,n+2-k). - Gionata Neri, May 19 2016
E.g.f.: (Ei(1,x) - Ei(1,-LambertW(-x)))/x. - Robert Israel, May 19 2016
For n > 1, a(n) = Sum_{k=1..floor(n/2)} (n^(n-2*k) * (2*k/n + n - 2*k)). - Mathew Englander, Oct 19 2020

A127837 Numbers k such that ((k+1)^k-1)/k^2 is a prime.

Original entry on oeis.org

2, 3, 5, 17, 4357

Offset: 1

Farideh Firoozbakht and Alexander Adamchuk, Mar 13 2007

All terms are primes. Corresponding primes of the form ((k+1)^k-1)/k^2 are listed in A128466 = 2, 7, 311, 7563707819165039903, ... .
It seems that if p is in the sequence then the first three numbers k such that k^2 divides (p+1)^k-1 are: 1, p & ((p+1)^p-1)/p. 2 is in the sequence and the first three terms of A127103 are : 1, 2 & ((2+1)^2-1)/2; 3 is in the sequence and the first three terms of A127104 are : 1, 3 & ((3+1)^3-1)/3; 5 is in the sequence and the first three terms of A127106 are : 1, 5 & ((5+1)^5-1)/5.
No other terms below 20000. - Max Alekseyev, Apr 25 2007

			4357 is in the sequence because (4358^4357-1)/4357^2 is prime.

Cf. A128466, A037205, A060072, A060073, A058128, A128456, A127103, A127104, A127106, A128398.

A070189 a(n) = 12345679*n.

Original entry on oeis.org

0, 12345679, 24691358, 37037037, 49382716, 61728395, 74074074, 86419753, 98765432, 111111111, 123456790, 135802469, 148148148, 160493827, 172839506, 185185185, 197530864, 209876543, 222222222, 234567901, 246913580, 259259259, 271604938, 283950617, 296296296, 308641975

Offset: 0

Henry Bottomley, Apr 24 2002

a(82)=1012345678 is the first term which has a digit appearing more than once without an obvious pattern, although a(-82)=-1012345678 might be seen as the concatenation of ten consecutive numbers.

David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See entry 12345679 at p. 188.

Tanya Khovanova, Recursive Sequences.
"TyYann", MegaFavNumbers: Lewis Carroll's Number, video (2020).
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A027889, A057932, A060073, A085959.

Table[12345679*n,{n,0,30}] (* or *) LinearRecurrence[{2,-1},{0,12345679},30] (* Harvey P. Dale, Oct 16 2015 *)

a(n)=12345679*n \\ Charles R Greathouse IV, Jan 09 2012

concat(0, Vec(12345679*x/(1-x)^2 + O(x^26))) \\ Elmo R. Oliveira, Jun 26 2025

a(n) = n*(10^(10-1)-1)/(10-1)^2.
From Elmo R. Oliveira, Jun 26 2025: (Start)
G.f.: 12345679*x/(1-x)^2.
E.g.f.: 12345679*x*exp(x).
a(n) = 333667*A085959(n).
a(n) = 2*a(n-1) - a(n-2). (End)

More terms from Elmo R. Oliveira, Jun 26 2025

A137665 Quotients ((p+1)^p - 1)/p^2 for p = prime(n).

Original entry on oeis.org

2, 7, 311, 42799, 6140565047, 4696537119847, 7563707819165039903, 14523213296398891966759, 105051652240885643072548950287, 8160568057655529131985731272294887039239, 47525417447024678661670292427038339608998847, 20681861558186805237407813095538883147812221153173966103

Offset: 1

Alexander Adamchuk, Feb 04 2008

nonn

p^2 divides a(n) = (p+1)^p - 1, p = prime(n). (p+1)^p - 1 = A137664(n) = {8, 63, 7775, 2097151, 743008370687, 793714773254143, 2185911559738696531967, ...}.
Least prime factors of a(n) are listed in A128456(n) = {2, 7, 311, 127, 23, 157, 7563707819165039903, ...}.
Largest prime factors a(n) are listed in A137666.
a(n) is prime for n = {1, 2, 3, 7, 595, ...} corresponding to p = prime(n) = {2, 3, 5, 17, 4357, ...} = A127837.
Primes in this sequence are A128466.

Cf. A128452, A128456, A128356, A128466, A127837, A128357, A060073, A137664, A137666.

Table[ ((Prime[n] + 1)^Prime[n] - 1)/Prime[n]^2, {n,1,15} ]

a(n) = my(p=prime(n)); polcyclo(p,p+1)/p \\ Hugo Pfoertner, Jul 21 2024

a(n) = ((prime(n) + 1)^prime(n) - 1)/prime(n)^2;

a(n) = A137664(n)/prime(n)^2.

A175771 a(n) = ((n^2 + 1)^n - 1)/n^3.

Original entry on oeis.org

1, 3, 37, 1305, 95051, 11878363, 2277696793, 622353150177, 229930796172439, 110462212541120451, 66954547910007962117, 49988751334503886046233, 45082285083777592171142467, 48321795074001873489007405947

Offset: 1

Michel Lagneau, Sep 01 2010

If we consider the general case ((n^(p-1) + 1)^n - 1)/n^p, we obtain this sequence for p=3, the sequence A060073 for p = 2, and the sequence A000051 for n = 2 and p = 2,3,...

			a(3) = ((3^2 + 1)^3 - 1)/3^3 = 999/27 = 37.

Vincenzo Librandi, Table of n, a(n) for n = 1..200

Cf. A060073, A000051.

[((n^2 + 1)^n - 1)/n^3 : n in [1..20]]; // Vincenzo Librandi, Sep 01 2011

Maple
```
seq(((n^2 + 1)^n - 1)/n^3,n=1..10) ;
```

Table[((n^2 + 1)^n - 1)/n^3, {n,1,20}] (* G. C. Greubel, Nov 11 2018 *)

vector(20, n, ((n^2 + 1)^n - 1)/n^3) \\ G. C. Greubel, Nov 11 2018

cp's OEIS Frontend

A128466 Primes of the form ((k+1)^k - 1)/k^2 = A060073(k+1).

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

Formula

A023811 Largest metadrome (number with digits in strict ascending order) in base n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

PARI

Python

Formula

Extensions

A060072 a(n) = (n^(n-1) - 1)/(n-1) for n>1, a(1) = 0.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

Extensions

A198410 a(n) = ((3^(n-1) + 1)^3 -1)/3^n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A037205 a(n) = (n+1)^n - 1.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A081215 a(n) = (n^(n+1)+(-1)^n)/(n+1)^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs