A039915 -id:A039915 - cp's OEIS frontend

A006093 a(n) = prime(n) - 1.

1, 2, 4, 6, 10, 12, 16, 18, 22, 28, 30, 36, 40, 42, 46, 52, 58, 60, 66, 70, 72, 78, 82, 88, 96, 100, 102, 106, 108, 112, 126, 130, 136, 138, 148, 150, 156, 162, 166, 172, 178, 180, 190, 192, 196, 198, 210, 222, 226, 228, 232, 238, 240, 250, 256, 262, 268, 270

Offset: 1

N. J. A. Sloane

These are also the numbers that cannot be written as i*j + i + j (i,j >= 1). - Rainer Rosenthal, Jun 24 2001; Henry Bottomley, Jul 06 2002
The values of k for which Sum_{j=0..n} (-1)^j*binomial(k, j)*binomial(k-1-j, n-j)/(j+1) produces an integer for all n such that n < k. Setting k=10 yields [0, 1, 4, 11, 19, 23, 19, 11, 4, 1, 0] for n = [-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9], so 10 is in the sequence. Setting k=3 yields [0, 1, 1/2, 1/2] for n = [-1, 0, 1, 2], so 3 is not in the sequence. - Dug Eichelberger (dug(AT)mit.edu), May 14 2001
n such that x^n + x^(n-1) + x^(n-2) + ... + x + 1 is irreducible. - Robert G. Wilson v, Jun 22 2002
Records for Euler totient function phi.
Together with 0, n such that (n+1) divides (n!+1). - Benoit Cloitre, Aug 20 2002; corrected by Charles R Greathouse IV, Apr 20 2010
n such that phi(n^2) = phi(n^2 + n). - Jon Perry, Feb 19 2004
Numbers having only the trivial perfect partition consisting of a(n) 1's. - Lekraj Beedassy, Jul 23 2006
Numbers n such that the sequence {binomial coefficient C(k,n), k >= n } contains exactly one prime. - Artur Jasinski, Dec 02 2007
Record values of A143201: a(n) = A143201(A001747(n+1)) for n > 1. - Reinhard Zumkeller, Aug 12 2008
From Reinhard Zumkeller, Jul 10 2009: (Start)
The first N terms can be generated by the following sieving process:
  start with {1, 2, 3, 4, ..., N - 1, N};
  for i := 1 until SQRT(N) do
  (if (i is not striked out) then
  (for j := 2 * i + 1 step i + 1 until N do
  (strike j from the list)));
  remaining numbers = {a(n): a(n) <= N}. (End)
a(n) = partial sums of A075526(n-1) = Sum_{1..n} A075526(n-1) = Sum_{1..n} (A008578(n+1) - A008578(n)) = Sum_{1..n} (A158611(n+2) - A158611(n+1)) for n >= 1. - Jaroslav Krizek, Aug 04 2009
A006093 U A072668 = A000027. - Juri-Stepan Gerasimov, Oct 22 2009
A171400(a(n)) = 1 for n <> 2: subsequence of A171401, except for a(2) = 2. - Reinhard Zumkeller, Dec 08 2009
Numerator of (1 - 1/prime(n)). - Juri-Stepan Gerasimov, Jun 05 2010
Numbers n such that A002322(n+1) = n. This statement is stronger than repeating the property of the entries in A002322, because it also says in reciprocity that this sequence here contains no numbers beyond the Carmichael numbers with that property. - Michel Lagneau, Dec 12 2010
a(n) = A192134(A095874(A000040(n))); subsequence of A192133. - Reinhard Zumkeller, Jun 26 2011
prime(a(n)) + prime(k) < prime(a(k) + k) for at least one k <= a(n): A212210(a(n),k) < 0. - Reinhard Zumkeller, May 05 2012
Except for the first term, numbers n such that the sum of first n natural numbers does not divide the product of first n natural numbers; that is, n*(n + 1)/2 does not divide n!. - Jayanta Basu, Apr 24 2013
BigOmega(a(n)) equals BigOmega(a(n)*(a(n) + 1)/2), where BigOmega = A001222. Rationale: BigOmega of the product on the right hand side factorizes as BigOmega(a/2) + Bigomega(a+1) = BigOmega(a/2) + 1 because a/2 and a + 1 are coprime, because BigOmega is additive, and because a + 1 is prime. Furthermore Bigomega(a/2) = Bigomega(a) - 1 because essentially all 'a' are even. - Irina Gerasimova, Jun 06 2013
Record values of A060681. - Omar E. Pol, Oct 26 2013
Deficiency of n-th prime. - Omar E. Pol, Jan 30 2014
Conjecture: All the sums Sum_{k=s..t} 1/a(k) with 1 <= s <= t are pairwise distinct. In general, for any integers d >= -1 and m > 0, if Sum_{k=i..j} 1/(prime(k)+d)^m = Sum_{k=s..t} 1/(prime(k)+d)^m with 0 < i <= j and 0 < s <= t then we must have (i,j) = (s,t), unless d = m = 1 and {(i,j),(s,t)} = {(4,4),(8,10)} or {(4,7),(5,10)}. (Note that 1/(prime(8)+1)+1/(prime(9)+1)+1/(prime(10)+1) = 1/(prime(4)+1) and Sum_{k=5..10} 1/(prime(k)+1) = 1/(prime(4)+1) + Sum_{k=5..7} 1/(prime(k)+1).) - Zhi-Wei Sun, Sep 09 2015
Numbers n such that (prime(i)^n + n) is divisible by (n+1), for all i >= 1, except when prime(i) = n+1. - Richard R. Forberg, Aug 11 2016
a(n) is the period of Fubini numbers (A000670) over the n-th prime. - Federico Provvedi, Nov 28 2020

Archimedeans Problems Drive, Eureka, 40 (1979), 28.
Harvey Dubner, Generalized Fermat primes, J. Recreational Math., 18 (1985): 279-280.
M. Gardner, The Colossal Book of Mathematics, pp. 31, W. W. Norton & Co., NY, 2001.
M. Gardner, Mathematical Circus, pp. 251-2, Alfred A. Knopf, NY, 1979.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..10000
Thomas F. Bloom, Unit fractions with shifted prime denominators, arXiv:2305.02689 [math.NT], 2023.
R. P. Boas & N. J. A. Sloane, Correspondence, 1974
Harvey Dubner, Generalized Fermat primes, J. Recreational Math. 18.4 (1985-1986), 279. (Annotated scanned copy)
Armel Mercier, Problem E 3065, American Mathematical Monthly, 1984, p. 649.
Armel Mercier, S. K. Rangarajan, J. C. Binz and Dan Marcus, Problem E 3065, American Mathematical Monthly, No. 4, 1987, pp. 378.
Poo-Sung Park, Additive uniqueness of PRIMES-1 for multiplicative functions, arXiv:1708.03037 [math.NT], 2017.
J. R. Rickard and J. J. Hitchcock, Problem Drive 4, Archimedeans Problems Drive, Eureka, 40 (1979), 28-29, 40. (Annotated scanned copy)
Index entries for sequences generated by sieves

a(n) = K(n, 1) and A034693(K(n, 1)) = 1 for all n. The subscript n refers to this sequence and K(n, 1) is the index in A034693. - Labos Elemer
Cf. A000040, A034694. Different from A075728.
Complement of A072668 (composite numbers minus 1), A072670(a(n))=0.
Essentially the same as A039915.
Cf. A084920, A006093, A050997, A008864, A060800, A131991, A131992, A131993.
Cf. A101301 (partial sums), A005867 (partial products).
Column 1 of the following arrays/triangles: A087738, A249741, A352707, A378979, A379010.
The last diagonal of A162619, and of A174996, the first diagonal in A131424.
Row lengths of irregular triangles A086145, A124223, A212157.

Filtered([1..280],IsPrime)-1; # Muniru A Asiru, Nov 25 2018

a006093 = (subtract 1) . a000040  -- Reinhard Zumkeller, Mar 06 2012

[NthPrime(n)-1: n in [1..100]]; // Vincenzo Librandi, Nov 17 2015

for n from 2 to 271 do if (n! mod n^2 = n*(n-1) and (n<>4) then print(n-1) fi od; # Gary Detlefs, Sep 10 2010
# alternative
A006093 := proc(n)
    ithprime(n)-1 ;
end proc:
seq(A006093(n),n=1..100) ; # R. J. Mathar, Feb 06 2019

Table[Prime[n] - 1, {n, 1, 30}] (* Vladimir Joseph Stephan Orlovsky, Apr 27 2008 *)
a[ n_] := If[ n < 1, 0, -1 + Prime @ n] (* Michael Somos, Jul 17 2011 *)
Prime[Range[60]] - 1 (* Alonso del Arte, Oct 26 2013 *)

isA006093(n) = isprime(n+1) \\ Michael B. Porter, Apr 09 2010

A006093(n) = prime(n)-1 \\ Michael B. Porter, Apr 09 2010

\\ Sieve as described in Rainer Rosenthal's comment.
m=270;s=vector(m);for(i=1,m,for(j=i,m,k=i*j+i+j;if(k<=m,s[k]=1)));for(k=1,m,if(s[k]==0,print1(k,", "))); \\ Hugo Pfoertner, May 14 2019

from sympy import prime
for n in range(1,100): print(prime(n)-1, end=', ') # Stefano Spezia, Nov 30 2018

a(n) = (p-1)! mod p where p is the n-th prime, by Wilson's theorem. - Jonathan Sondow, Jul 13 2010
a(n) = A000010(prime(n)) = A000010(A006005(n)). - Antti Karttunen, Dec 16 2012
a(n) = A005867(n+1)/A005867(n). - Eric Desbiaux, May 07 2013
a(n) = A000040(n) - 1. - Omar E. Pol, Oct 26 2013
a(n) = A033879(A000040(n)). - Omar E. Pol, Jan 30 2014

Correction for change of offset in A158611 and A008578 in Aug 2009 Jaroslav Krizek, Jan 27 2010
Obfuscating comments removed by Joerg Arndt, Mar 11 2010
Edited by Charles R Greathouse IV, Apr 20 2010

A223701 Irregular triangle of numbers k such that prime(n) is the largest prime factor of k^2 - 1.

Original entry on oeis.org

3, 2, 5, 7, 17, 4, 9, 11, 19, 26, 31, 49, 161, 6, 8, 13, 15, 29, 41, 55, 71, 97, 99, 127, 244, 251, 449, 4801, 8749, 10, 21, 23, 34, 43, 65, 76, 89, 109, 111, 197, 199, 241, 351, 485, 769, 881, 1079, 6049, 19601, 12, 14, 25, 27, 51, 53, 64, 79, 129, 131, 155

Offset: 1

T. D. Noe, Apr 03 2013

Note that the first number of each row forms the sequence 3, 2, 4, 6, 10, 12,..., which is A039915. The first 25 rows, except the first, are in A181447-A181470.

			Irregular triangle:
  {3},
  {2, 5, 7, 17},
  {4, 9, 11, 19, 26, 31, 49, 161},
  {6, 8, 13, 15, 29, 41, 55, 71, 97, 99, 127, 244, 251, 449, 4801, 8749}

Andrew Howroyd, Table of n, a(n) for n = 1..16223 (first 25 rows for primes up to 97)
Florian Luca and Filip Najman, On the largest prime factor of x^2-1, arXiv:1005.1533 [math.NT], 2010.
Florian Luca and Filip Najman, On the largest prime factor of x^2-1, Mathematics of Computation 80 (2011), 429-435. (Paper has errata that was posted on the MOC website.)
Filip Najman, List of Publications Page (Adjacent to entry number 4 are links with the data files for the first 25 rows (=16223 terms) of this sequence)

Rows 2..25 with complete data are: A181447, A181448, A181449, A181450, A181451, A181452, A181453, A181454, A181455, A181456, A181457, A181458, A181459, A181460, A181461, A181462, A181463, A181464, A181465, A181466, A181467, A181468, A181469, A181470.
Row 26 is A181568.
Cf. A039915 (first terms), A175607 (last terms), A181471 (row lengths), A379344 (row sums).
Cf. A223702, A223703, A223704 (related tables).

t = Table[FactorInteger[n^2 - 1][[-1,1]], {n, 2, 10^5}]; Table[1 + Flatten[Position[t, Prime[n]]], {n, 6}]

A068499 Numbers m such that m! reduced modulo (m+1) is not zero.

Original entry on oeis.org

1, 2, 3, 4, 6, 10, 12, 16, 18, 22, 28, 30, 36, 40, 42, 46, 52, 58, 60, 66, 70, 72, 78, 82, 88, 96, 100, 102, 106, 108, 112, 126, 130, 136, 138, 148, 150, 156, 162, 166, 172, 178, 180, 190, 192, 196, 198, 210, 222, 226, 228, 232, 238, 240, 250, 256, 262, 268, 270

Offset: 1

Benoit Cloitre, Mar 11 2002

Also n such that tau((n+1)!) = 2* tau(n!)
For n > 2, a(n) is the smallest number such that a(n) !== -1 (mod a(k)+1) for any 1 < k < n. [Franklin T. Adams-Watters, Aug 07 2009]
Also n such that sigma((n+1)!) = (n+2)* sigma(n!), which is the same as A062569(n+1) = (n+2)*A062569(n). - Zak Seidov, Aug 22 2012
This sequence is obtained by the following sieve: keep 1 in the sequence and then, at the k-th step, keep the smallest number, x say, that has not been crossed off before and cross off all the numbers of the form k*(x + 1) - 1 with k > 1. The numbers that are left form the sequence. - Jean-Christophe Hervé, Dec 12 2015
a(n) = A039915(n-1) for 3 < n <= 1000. - Georg Fischer, Oct 19 2018

			Illustration of the sieve: keep 1 = a(1) and then
1st step: take 2 = a(2) and cross off 5, 8, 11, 14, 17, 20, 23, 26, etc.
2nd step: take 3 = a(3) and cross off 7, 11, 15, 19, 23, 27, etc.
3rd step: take 4 = a(4) and cross off 9, 14, 19, 24, etc.
4th step: take 6 = a(5) and cross off 13, 19, 25 etc.
10 is obtained at next step and the smallest crossed off numbers are then 21 and 28. This gives the beginning of the sequence up to 22 = a(10): 1, 2, 3, 4, 6, 10, 12, 16, 18, 22. - _Jean-Christophe Hervé_, Dec 12 2015

Jean-Christophe Hervé, Table of n, a(n) for n = 1..10000
David J. Hemmer and Karlee J. Westrem, Palindrome Partitions and the Calkin-Wilf Tree, arXiv:2402.02250 [math.CO], 2024. See Remark 3.3, p. 6.
Karlee J. Westrem, Schaper numbers, palindrome partitions, and symmetric functions, with applications to characters of the symmetric group, Ph. D. Dissertation, Michigan Tech. Univ. (2025). See p. 49.
Index entries for sequences generated by sieves

Cf. A000040, A039915, A062569, A166460 (almost complement).

Select[Range[300],Mod[#!,#+1]!=0&] (* Harvey P. Dale, Apr 11 2012 *)

{plnt=1 ; nfa=1; mxind=60 ;  for(k=1, 10^7, nfa=nfa*k;
if(nfa % (k+1) != 0 , print1(k, ", "); plnt++ ;
if(mxind <  plnt, break() )))} \\ Douglas Latimer, Apr 25 2012

a(n)=if(n<5,n,prime(n-1)-1) \\ Charles R Greathouse IV, Apr 25 2012

from sympy import prime
def A068499(n): return prime(n-1)-1 if n>3 else n # Chai Wah Wu, Aug 27 2024

For n >= 4, a(n) = prime(n-1) - 1 = A006093(n-1).
For n <> 3, all terms are one less prime. - Zak Seidov, Aug 22 2012
a(n) = Integer part of A078456(n+1)/A078456(n). - Eric Desbiaux, May 07 2013

Theorem: zero does not occur in this sequence. Proof: (p-1)^2-1=(p-2)p. This means that p is greatest prime divisor of (p-1)^2-1 for every p.

An effective abc conjecture (c < rad(abc)^2) would imply that a(24)-a(33) are (2371, 2734, 3360, 4022, 4637, 5575, 6424, 7268, 8351, 9661). - Lucas A. Brown, Oct 01 2022

cp's OEIS Frontend

A006093 a(n) = prime(n) - 1.

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A223701 Irregular triangle of numbers k such that prime(n) is the largest prime factor of k^2 - 1.

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A068499 Numbers m such that m! reduced modulo (m+1) is not zero.

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A181471 a(n) = number of numbers of the form k^2-1 having n-th prime as largest prime divisor.

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A121206 a(n) = (2n)! mod n(2n+1).

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