

A006093


a(n) = prime(n)  1.
(Formerly M1006)


315



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
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OFFSET

1,2


COMMENTS

The values of k for which Sum_{j=0..n} ((1)^j*binomial(k, j)*binomial(k1j, nj)/(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^(n1) + x^(n2) + ... + x + 1 is irreducible.  Robert G. Wilson v, Jun 22 2002
Records for Euler totient function phi.
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
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)
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
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
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.
Harvey Dubner, 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

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.
J. R. Rickard and J. J. Hitchcock, Problem Drive 4, Archimedeans Problems Drive, Eureka, 40 (1979), 2829, 40. (Annotated scanned copy)


FORMULA

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


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
# alternative
ithprime(n)1 ;
end proc:


MATHEMATICA

a[ n_] := If[ n < 1, 0, 1 + Prime @ n] (* Michael Somos, Jul 17 2011 *)


PROG

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


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


KEYWORD

nonn,easy,nice


AUTHOR



EXTENSIONS

Obfuscating comments removed by Joerg Arndt, Mar 11 2010


STATUS

approved



