cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

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

Original entry on oeis.org

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

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Author

Keywords

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_{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
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

References

  • 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).

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, A034694. Different from A075728.
Complement of A072668 (composite numbers minus 1), A072670(a(n))=0.
Essentially the same as A039915.
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.

Programs

Formula

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

Extensions

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

A005867 a(0) = 1; for n > 0, a(n) = (prime(n)-1)*a(n-1).

Original entry on oeis.org

1, 1, 2, 8, 48, 480, 5760, 92160, 1658880, 36495360, 1021870080, 30656102400, 1103619686400, 44144787456000, 1854081073152000, 85287729364992000, 4434961926979584000, 257227791764815872000, 15433667505888952320000
Offset: 0

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Author

Keywords

Comments

Local minima of Euler's phi function. - Walter Nissen
Number of potential primes in a modulus primorial(n+1) sieve. - Robert G. Wilson v, Nov 20 2000
Let p=prime(n) and let p# be the primorial (A002110), then it can be shown that any p# consecutive numbers have exactly a(n-1) numbers whose lowest prime factor is p. For a proof, see the "Proofs Regarding Primorial Patterns" link. For example, if we let p=7 and consider the interval [101,310] containing 210 numbers, we find the 8 numbers 119, 133, 161, 203, 217, 259, 287, 301. - Dennis Martin (dennis.martin(AT)dptechnology.com), Jul 16 2006
From Gary W. Adamson, Apr 21 2009: (Start)
Equals (-1)^n * (1, 1, 1, 2, 8, 48, ...) dot (-1, 2, -3, 5, -7, 11, ...).
a(6) = 480 = (1, 1, 1, 2, 8, 48) dot (-1, 2, -3, 5, -7, 11) = (-1, 2, -3, 10, -56, 528). (End)
It can be proved that there are at least T prime numbers less than N, where the recursive function T is: T = N - N*Sum_{i=0..T(sqrt(N))} A005867(i)/A002110(i). This can show for example that at least 0.16*N numbers are primes less than N for 29^2 > N > 23^2. - Ben Paul Thurston, Aug 23 2010
First column of A096294. - Eric Desbiaux, Jun 20 2013
Conjecture: The g.f. for the prime(n+1)-rough numbers (A000027, A005408, A007310, A007775, A008364, A008365, A008366, A166061, A166063) is x*P(x)/(1-x-x^a(n)+x^(a(n)+1)), where P(x) is an order a(n) polynomial with symmetric coefficients (i.e., c(0)=c(n), c(1)=c(n-1), ...). - Benedict W. J. Irwin, Mar 18 2016
a(n)/A002110(n+1) (primorial(n+1)) is the ratio of natural numbers whose smallest prime factor is prime(n+1); i.e., prime(n+1) coprime to A002110(n). So the ratio of even numbers to natural numbers = 1/2; odd multiples of 3 = 1/6; multiples of 5 coprime to 6 (A084967) = 2/30 = 1/15; multiples of 7 coprime to 30 (A084968) = 8/210 = 4/105; etc. - Bob Selcoe, Aug 11 2016
The 2-adic valuation of a(n) is A057773(n), being sum of the 2-adic valuations of the product terms here. - Kevin Ryde, Jan 03 2023
For n > 1, a(n) is the number of prime(n+1)-rough numbers in [1, primorial(prime(n))]. - Alexandre Herrera, Aug 29 2023

Examples

			a(3): the mod 30 prime remainder set sieve representation yields the remainder set: {1, 7, 11, 13, 17, 19, 23, 29}, 8 elements.
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A057773 (2-adic valuation).
Column 1 of A281890.

Programs

  • Haskell
    a005867 n = a005867_list !! n
    a005867_list = scanl (*) 1 a006093_list
    -- Reinhard Zumkeller, May 01 2013
  • Maple
    A005867 := proc(n)
        mul(ithprime(j)-1,j=1..n) ;
    end proc: # Zerinvary Lajos, Aug 24 2008, R. J. Mathar, May 03 2017
  • Mathematica
    Table[ Product[ EulerPhi[ Prime[ j ] ], {j, 1, n} ], {n, 1, 20} ]
    RecurrenceTable[{a[0]==1,a[n]==(Prime[n]-1)a[n-1]},a,{n,20}] (* Harvey P. Dale, Dec 09 2013 *)
    EulerPhi@ FoldList[Times, 1, Prime@ Range@ 18] (* Michael De Vlieger, Mar 18 2016 *)
  • PARI
    for(n=0, 22, print1(prod(k=1,n, prime(k)-1), ", "))
    

Formula

a(n) = phi(product of first n primes) = A000010(A002110(n)).
a(n) = Product_{k=1..n} (prime(k)-1) = Product_{k=1..n} A006093(n).
Sum_{n>=0} a(n)/A002110(n+1) = 1. - Bob Selcoe, Jan 09 2015
a(n) = A002110(n)-((1/A000040(n+1) - A038110(n+1)/A038111(n+1))*A002110(n+1)). - Jamie Morken, Mar 27 2019
a(n) = |Sum_{k=0..n} A070918(n,k)|. - Alois P. Heinz, Aug 18 2019
a(n) = A058251(n)/A060753(n+1). - Jamie Morken, Apr 25 2022
a(n) = A002110(n) - A016035(A002110(n)) - 1 for n >= 1. - David James Sycamore, Sep 07 2024
Sum_{n>=0} 1/a(n) = A345974. - Amiram Eldar, Jun 26 2025

Extensions

Offset changed to 0, Name changed, and Comments and Examples sections edited by T. D. Noe, Apr 04 2010

A060735 a(1)=1, a(2)=2; thereafter, a(n) is the smallest number m not yet in the sequence such that every prime that divides a(n-1) also divides m.

Original entry on oeis.org

1, 2, 4, 6, 12, 18, 24, 30, 60, 90, 120, 150, 180, 210, 420, 630, 840, 1050, 1260, 1470, 1680, 1890, 2100, 2310, 4620, 6930, 9240, 11550, 13860, 16170, 18480, 20790, 23100, 25410, 27720, 30030, 60060, 90090, 120120, 150150, 180180, 210210
Offset: 1

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Author

Robert G. Wilson v, Apr 23 2001

Keywords

Comments

Also, numbers k at which k / (phi(k) + 1) increases.
Except for the initial 1, this sequence is a primorial (A002110) followed by its multiples until the next primorial, then the multiples of that primorial and so on. - Wilfredo Lopez (chakotay147138274(AT)yahoo.com), Dec 28 2006
a(1)=1, a(2)=2. For n >= 3, a(n) is the smallest integer > a(n-1) that is divisible by every prime which divides lcm(a(1), a(2), a(3), ..., a(n)). - Leroy Quet, Feb 23 2010
Numbers n for which A053589(n) = A260188(n), thus numbers with only one nonzero digit when written in primorial base A049345. - Antti Karttunen, Aug 30 2016
Lexicographically earliest infinite sequence of distinct positive numbers with property that every prime that divides a(n-1) also divides a(n). - N. J. A. Sloane, Apr 08 2022

Examples

			After a(2)=2 the next term must be even, so a(3)=4.
Then a(4) must be even so a(4) = 6.
Now a(5) must be a multiple of 2*3=6, so a(5)=12.
Then a(6)=18, a(7)=24, a(8)=30.
Now a(9) must be a multiple of 2*3*5 = 30, so a(9)=60. And so on.
		

Crossrefs

Indices of ones in A276157 and A267263.
One more than A343048.

Programs

  • Maple
    seq(seq(k*mul(ithprime(i),i=1..n-1),k=1..ithprime(n)-1),n=1..10); # Vladeta Jovovic, Apr 08 2004
    a := proc(n) option remember; if n=1 then return 1 fi; a(n-1);
    % + convert(numtheory:-factorset(%), `*`) end:
    seq(a(n), n=1..42); # after Zumkeller, Peter Luschny, Aug 30 2016
  • Mathematica
    a = 0; Do[ b = n/(EulerPhi[ n ] + 1); If[ b > a, a = b; Print[ n ] ], {n, 1, 10^6} ]
    f[n_] := Range[Prime[n + 1] - 1] Times @@ Prime@ Range@ n;  Array[f, 7, 0] // Flatten (* Robert G. Wilson v, Jul 22 2015 *)
  • PARI
    first(n)=my(v=vector(n),k=1,p=1,P=1); v[1]=1; for(i=2,n, v[i]=P*k++; if(k>p && isprime(k), p=k; P=v[i]; k=1)); v \\ Charles R Greathouse IV, Jul 22 2015
    
  • PARI
    is_A060735(n,P=1)={forprime(p=2,,n>(P*=p)||return(1);n%P&&return)} \\ M. F. Hasler, Mar 14 2017
    
  • Python
    from functools import cache;
    from sympy import primefactors, prod
    @cache
    def a(n): return 1 if n == 0 else a(n-1) + prod(primefactors(a(n-1)))
    print([a(n) for n in range(42)]) # Trey Deitch, Jun 08 2024

Formula

a(1) = 1, a(n) = a(n-1) + rad(a(n-1)) with rad=A007947, squarefree kernel. - Reinhard Zumkeller, Apr 10 2006
a(A101301(n)+1) = A002110(n). - Enrique Pérez Herrero, Jun 10 2012
a(n) = 1 + A343048(n). - Antti Karttunen, Nov 14 2024

Extensions

Definition corrected by Franklin T. Adams-Watters, Apr 16 2009
Simpler definition, comments, examples from N. J. A. Sloane, Apr 08 2022

A377566 Lexicographically earliest infinite sequence of distinct positive integers such that if j = a(n-1) is primorial, a(n) is the smallest prime not already a term, whereas if j is not primorial a(n) is the smallest novel number > j divisible by rad(j).

Original entry on oeis.org

1, 2, 3, 6, 5, 10, 20, 30, 7, 14, 28, 42, 84, 126, 168, 210, 11, 22, 44, 66, 132, 198, 264, 330, 660, 990, 1320, 1650, 1980, 2310, 13, 26, 52, 78, 156, 234, 312, 390, 780, 1170, 1560, 1950, 2340, 2730, 5460, 8190, 10920, 13650, 16380, 19110, 21840, 24570, 27300, 30030, 17
Offset: 1

Views

Author

David James Sycamore, Nov 03 2024

Keywords

Comments

In other words: j in A002110 implies a(n) = p, next missing prime; j not in A002110 implies a(n) = m*rad(j), with minimal novel m.
Immediately following odd prime term p = prime(n), 2*p occurs, and as the sequence extends, multiples of intervening primes q; 2A101301(n) gives the number of steps (terms) from prime(n) to A002110(n).
Sequence can be generated by the following recursion: If a(t) = prime(n), n > 1 then a(t+k-1) = A060735(k)*prime(n); k = 1,2...A101301(n)+1; see Example.

Examples

			If j = a(n-1) is squarefree then a(n) = 2*j.
a(9) = prime(4) = 7, A101301(4) = 7, so there are 7+1 = 8 terms from 7 to A002110(4) = 210, namely: A060735(7+k-1)*7, k = 1,2,...8; so: 1*7,2*7,4*7,6*7,12*7,18*7,24*7,30*7 = 7,14,28,42,84,126,168,210.
		

Crossrefs

Programs

  • Mathematica
    {{1, 2, 3, 6}}~Join~Table[Prime[m + 2]*If[n == 0, 1, Product[Prime[i], {i, n}]]*k, {m, 10}, {n, 0, m}, {k, 1 + Boole[n > 1], If[n == 0, 1, Prime[n + 1]]}] // Flatten
    (* faster for large datasets, or *)
    nn = 1000; c[] := False; m[] := 1; f[x_] := FactorInteger[x][[All, 1]]; Array[Set[{a[#], c[#], m[#]}, {#, True, 2}] &, 2]; j = 2; u = v = 3;
    Do[If[Or[IntegerQ@ Log2[j],
         And[EvenQ[j], Union@ Differences@ PrimePi[#] == {1}] ],
         k = v, k = Times @@ #;
         While[c[k m[k]], m[k]++]; k *= m[k]] &[f[j]];
       Set[{a[n], c[k], j}, {k, True, k}];
       If[k == u, While[c[u], u++]];
       If[k == v, While[c[v], v = NextPrime[v] ] ], {n, 3, nn}];
    Array[a, nn] (* Michael De Vlieger, Nov 04 2024 *)

A235431 The smallest positive number that must be added to or subtracted from the sum of the first n primes in order to get a prime.

Original entry on oeis.org

1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 4, 3, 4, 1, 2, 5, 2, 1, 4, 1, 4, 1, 2, 3, 4, 5, 2, 3, 2, 5, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 10, 1, 4, 11, 2, 1, 6
Offset: 1

Views

Author

R. J. Cano, Jan 17 2014

Keywords

Comments

The primes in A013918 would have associated a(n)=0 if not for the qualifier "positive" in the definition.
The sum of the first n primes appears to be close to a prime. For illustration, the maximum for a(n) among the first 5 million terms is a(808500) = 218.
See A013916 for the above mentioned indices, numbers n such that the sum of the first n primes is prime. - M. F. Hasler, Jan 20 2014

Examples

			The sum of the first 9 primes is 100, and by adding 1 we get 101. Since 101 is a prime, a(9) = 1.
The sum of the first 10 primes is 129, since 129 - 2 = prime(31) = 127 or 129 + 2 = prime(32) = 131, a(10) = 2.
The sum of the first 129 primes minus 1 is a prime, this is 42468 - 1 = prime(4443), so a(129) = 1.
		

Crossrefs

Programs

  • PARI
    a(n)=my(u=sum(j=1,n,prime(j)),k=1);while(!(isprime(u+k)||isprime(u-k)),k++);k

Formula

Algorithm:
Let S be the sum of the first n primes;
initially, let k=1;
increment k while neither S-k nor S+k is prime;
return a(n)=k.
a(n) = min(A013632(A007504(n)), A049711(A007504(n))). - M. F. Hasler, Jan 20 2014
Showing 1-5 of 5 results.