

A006093


Primes minus 1.
(Formerly M1006)


231



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

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((1)^j*binomial(k, j)*binomial(k1j, nj)/(j+1), j=0..n) 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, .5, .5] 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^(n1) + x^(n2) + ... + 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(n1) = Sum_(1...n) A075526(n1) = 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. [JuriStepan Gerasimov, Oct 22 2009]
A171400(a(n)) = 1 for n <> 2: subsequence of A171401, except a(2) = 2. [Reinhard Zumkeller, Dec 08 2009]
Nominator of (1  1/prime(n)). [JuriStepan 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. [From 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 nth 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).)  ZhiWei 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


REFERENCES

Archimedeans Problems Drive, Eureka, 40 (1979), 28.
Dubner, Harvey. "Generalized Fermat primes." J. Recreational Math., 18 (1985): 279280.
M. Gardner, The Colossal Book of Mathematics, pp. 31, W. W. Norton & Co., NY, 2001.
M. Gardner, Mathematical Circus, pp. 2512, Alfred A. Knopf, NY, 1979.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

T. D. Noe, Table of n, a(n) for n=1..10000
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.
Index entries for sequences generated by sieves [Reinhard Zumkeller, Jul 10 2009]


FORMULA

a(n) = (p1)! mod p where p is the nth prime, by Wilson's theorem. [From Jonathan Sondow, Jul 13 2010]
a(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


MAPLE

for n from 2 to 271 do if (n! mod n^2 = n*(n1) and (n<>4) then print(n1) fi od; # Gary Detlefs, Sep 10 2010


MATHEMATICA

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 *)


PROG

(PARI) isA006093(n) = isprime(n+1) \\ Michael B. Porter, Apr 09 2010
(PARI) A006093(n) = prime(n)1 \\ Michael B. Porter, Apr 09 2010
(Haskell)
a006093 = (subtract 1) . a000040  Reinhard Zumkeller, Mar 06 2012
(MAGMA) [NthPrime(n)1: n in [1..100]]; // Vincenzo Librandi, Nov 17 2015


CROSSREFS

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, A034693, 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).
Sequence in context: A249427 A075728 A146886 * A127965 A117891 A262935
Adjacent sequences: A006090 A006091 A006092 * A006094 A006095 A006096


KEYWORD

nonn,easy,nice


AUTHOR

N. J. A. Sloane.


EXTENSIONS

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


STATUS

approved



