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 1-10 of 18 results. Next

A140763 A051838 gives numbers m such that the sum of first m primes divides the product of the first m primes. This sequence gives corresponding values of the sum of first m primes.

Original entry on oeis.org

2, 10, 77, 238, 874, 2747, 2914, 3266, 3638, 4661, 5117, 5830, 6601, 6870, 7141, 9523, 10191, 10887, 11966, 13490, 16401, 19113, 21037, 23069, 40313, 41741, 46191, 50887, 53342, 54998, 58406, 60146, 61910, 65534, 68341, 72179, 75130, 76127, 80189, 82253
Offset: 1

Views

Author

Enoch Haga, May 28 2008

Keywords

Comments

Sums (divisors) associated with A140761.

Examples

			a(2)=10 because when 30 is divided by 10, the quotient is 3 and integral.
		

Crossrefs

Programs

  • Mathematica
    Module[{nn=200,s,p},s=Accumulate[Prime[Range[nn]]];p=FoldList[ Times,Prime[ Range[nn]]];Select[Thread[{p,s}],Divisible[#[[1]],#[[2]]]&]][[All,2]] (* Harvey P. Dale, Jun 07 2022 *)

Formula

a(n)=A116536(n)/A159578(n) = A007504(A051838(n)). - R. J. Mathar, Jun 09 2008
Sum_{i=1..A051838(n)} prime(i).

Extensions

Corrected and edited by N. J. A. Sloane, Oct 01 2011

A159580 Integral quotients occur consecutively this many times (in sequence associated with A051838 and A116536).

Original entry on oeis.org

2, 3, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 3, 2, 2, 2, 3, 2, 2, 3, 3, 2, 3, 2, 4, 4, 2, 2, 2, 2, 2, 2, 3, 2, 2, 3, 2, 2, 3, 3, 2, 2, 2, 2, 2, 3, 3, 2, 3, 2, 2, 3, 2, 3, 3, 2, 2, 2, 3, 2, 2, 3, 2, 2, 2, 5, 2, 2, 2, 2, 2, 2, 2, 4, 2
Offset: 1

Views

Author

Enoch Haga, Apr 16 2009

Keywords

Examples

			a(1)=2 because the 1st subsequence (38,39) of consecutive integers has run length 2.
a(2)=3 because the 2nd subsequence is (56,57,58) which has run length 3.
a(3)=2 because the 3rd subsequence is (167,168) which has run length 2.
		

Crossrefs

Extensions

Recomputed based on b-file of A051838. - R. J. Mathar, Aug 28 2018

A002110 Primorial numbers (first definition): product of first n primes. Sometimes written prime(n)#.

Original entry on oeis.org

1, 2, 6, 30, 210, 2310, 30030, 510510, 9699690, 223092870, 6469693230, 200560490130, 7420738134810, 304250263527210, 13082761331670030, 614889782588491410, 32589158477190044730, 1922760350154212639070, 117288381359406970983270, 7858321551080267055879090
Offset: 0

Views

Author

Keywords

Comments

See A034386 for the second definition of primorial numbers: product of primes in the range 2 to n.
a(n) is the least number N with n distinct prime factors (i.e., omega(N) = n, cf. A001221). - Lekraj Beedassy, Feb 15 2002
Phi(n)/n is a new minimum for each primorial. - Robert G. Wilson v, Jan 10 2004
Smallest number stroked off n times after the n-th sifting process in an Eratosthenes sieve. - Lekraj Beedassy, Mar 31 2005
Apparently each term is a new minimum for phi(x)*sigma(x)/x^2. 6/Pi^2 < sigma(x)*phi(x)/x^2 < 1 for n > 1. - Jud McCranie, Jun 11 2005
Let f be a multiplicative function with f(p) > f(p^k) > 1 (p prime, k > 1), f(p) > f(q) > 1 (p, q prime, p < q). Then the record maxima of f occur at n# for n >= 1. Similarly, if 0 < f(p) < f(p^k) < 1 (p prime, k > 1), 0 < f(p) < f(q) < 1 (p, q prime, p < q), then the record minima of f occur at n# for n >= 1. - David W. Wilson, Oct 23 2006
Wolfe and Hirshberg give ?, ?, ?, ?, ?, 30030, ?, ... as a puzzle.
Records in number of distinct prime divisors. - Artur Jasinski, Apr 06 2008
For n >= 2, the digital roots of a(n) are multiples of 3. - Parthasarathy Nambi, Aug 19 2009 [with corrections by Zak Seidov, Aug 30 2015]
Denominators of the sum of the ratios of consecutive primes (see A094661). - Vladimir Joseph Stephan Orlovsky, Oct 24 2009
Where record values occur in A001221. - Melinda Trang (mewithlinda(AT)yahoo.com), Apr 15 2010
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
The above comment from Parthasarathy Nambi follows from the observation that digit summing produces a congruent number mod 9, so the digital root of any multiple of 3 is a multiple of 3. prime(n)# is divisible by 3 for n >= 2. - Christian Schulz, Oct 30 2013
The peaks (i.e., local maximums) in a graph of the number of repetitions (i.e., the tally of values) vs. value, as generated by taking the differences of all distinct pairs of odd prime numbers within a contiguous range occur at regular periodic intervals given by the primorial numbers 6 and greater. Larger primorials yield larger (relative) peaks, however the range must be >50% larger than the primorial to be easily observed. Secondary peaks occur at intervals of those "near-primorials" divisible by 6 (e.g., 42). See A259629. Also, periodicity at intervals of 6 and 30 can be observed in the local peaks of all possible sums of two, three or more distinct odd primes within modest contiguous ranges starting from p(2) = 3. - Richard R. Forberg, Jul 01 2015
If a number k and a(n) are coprime and k < (prime(n+1))^b < a(n), where b is an integer, then k has fewer than b prime factors, counting multiplicity (i.e., bigomega(k) < b, cf. A001222). - Isaac Saffold, Dec 03 2017
If n > 0, then a(n) has 2^n unitary divisors (A034444), and a(n) is a record; i.e., if k < a(n) then k has fewer unitary divisors than a(n) has. - Clark Kimberling, Jun 26 2018
Unitary superabundant numbers: numbers k with a record value of the unitary abundancy index, A034448(k)/k > A034448(m)/m for all m < k. - Amiram Eldar, Apr 20 2019
Psi(n)/n is a new maximum for each primorial (psi = A001615) [proof in link: Patrick Sole and Michel Planat, proposition 1 page 2]; compare with comment 2004: Phi(n)/n is a new minimum for each primorial. - Bernard Schott, May 21 2020
The term "primorial" was coined by Harvey Dubner (1987). - Amiram Eldar, Apr 16 2021
a(n)^(1/n) is approximately (n log n)/e. - Charles R Greathouse IV, Jan 03 2023
Subsequence of A267124. - Frank M Jackson, Apr 14 2023

Examples

			a(9) = 23# = 2*3*5*7*11*13*17*19*23 = 223092870 divides the difference 5283234035979900 in the arithmetic progression of 26 primes A204189. - _Jonathan Sondow_, Jan 15 2012
		

References

  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 50.
  • G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, Cambridge, University Press, 1940, p. 49.
  • P. Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 4.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 114.
  • D. Wolfe and S. Hirshberg, Underspecified puzzles, in Tribute to A Mathemagician, Peters, 2005, pp. 73-74.

Crossrefs

A034386 gives the second version of the primorial numbers.
Subsequence of A005117 and of A064807. Apart from the first term, a subsequence of A083207.
Cf. A001615, A002182, A002201, A003418, A005235, A006862, A034444 (unitary divisors), A034448, A034387, A033188, A035345, A035346, A036691 (compositorial numbers), A049345 (primorial base representation), A057588, A060735 (and integer multiples), A061742 (squares), A072938, A079266, A087315, A094348, A106037, A121572, A053589, A064648, A132120, A260188.
Cf. A061720 (first differences), A143293 (partial sums).
Cf. also A276085, A276086.
The following fractions are all related to each other: Sum 1/n: A001008/A002805, Sum 1/prime(n): A024451/A002110 and A106830/A034386, Sum 1/nonprime(n): A282511/A282512, Sum 1/composite(n): A250133/A296358.

Programs

  • Haskell
    a002110 n = product $ take n a000040_list
    a002110_list = scanl (*) 1 a000040_list
    -- Reinhard Zumkeller, Feb 19 2012, May 03 2011
    
  • Magma
    [1] cat [&*[NthPrime(i): i in [1..n]]: n in [1..20]]; // Bruno Berselli, Oct 24 2012
    
  • Magma
    [1] cat [&*PrimesUpTo(p): p in PrimesUpTo(60)]; // Bruno Berselli, Feb 08 2015
    
  • Maple
    A002110 := n -> mul(ithprime(i),i=1..n);
  • Mathematica
    FoldList[Times, 1, Prime[Range[20]]]
    primorial[n_] := Product[Prime[i], {i, n}]; Array[primorial,20] (* José María Grau Ribas, Feb 15 2010 *)
    Join[{1}, Denominator[Accumulate[1/Prime[Range[20]]]]] (* Harvey P. Dale, Apr 11 2012 *)
  • PARI
    a(n)=prod(i=1,n, prime(i)) \\ Washington Bomfim, Sep 23 2008
    
  • PARI
    p=1; for (n=0, 100, if (n, p*=prime(n)); write("b002110.txt", n, " ", p) )  \\ Harry J. Smith, Nov 13 2009
    
  • PARI
    a(n) = factorback(primes(n)) \\ David A. Corneth, May 06 2018
    
  • Python
    from sympy import primorial
    def a(n): return 1 if n < 1 else primorial(n)
    [a(n) for n in range(51)]  # Indranil Ghosh, Mar 29 2017
    
  • Sage
    [sloane.A002110(n) for n in (1..20)] # Giuseppe Coppoletta, Dec 05 2014
    
  • Scheme
    ; with memoization-macro definec
    (definec (A002110 n) (if (zero? n) 1 (* (A000040 n) (A002110 (- n 1))))) ;; Antti Karttunen, Aug 30 2016

Formula

Asymptotic expression for a(n): exp((1 + o(1)) * n * log(n)) where o(1) is the "little o" notation. - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 08 2001
a(n) = A054842(A002275(n)).
Binomial transform = A136104: (1, 3, 11, 55, 375, 3731, ...). Equals binomial transform of A121572: (1, 1, 3, 17, 119, 1509, ...). - Gary W. Adamson, Dec 14 2007
a(0) = 1, a(n+1) = prime(n)*a(n). - Juri-Stepan Gerasimov, Oct 15 2010
a(n) = Product_{i=1..n} A000040(i). - Jonathan Vos Post, Jul 17 2008
a(A051838(n)) = A116536(n) * A007504(A051838(n)). - Reinhard Zumkeller, Oct 03 2011
A000005(a(n)) = 2^n. - Carlos Eduardo Olivieri, Jun 16 2015
a(n) = A035345(n) - A005235(n) for n > 0. - Jonathan Sondow, Dec 02 2015
For all n >= 0, a(n) = A276085(A000040(n+1)), a(n+1) = A276086(A143293(n)). - Antti Karttunen, Aug 30 2016
A054841(a(n)) = A002275(n). - Michael De Vlieger, Aug 31 2016
a(n) = A270592(2*n+2) - A270592(2*n+1) if 0 <= n <= 4 (conjectured for all n by Alon Kellner). - Jonathan Sondow, Mar 25 2018
Sum_{n>=1} 1/a(n) = A064648. - Amiram Eldar, Oct 16 2020
Sum_{n>=1} (-1)^(n+1)/a(n) = A132120. - Amiram Eldar, Apr 12 2021
Theta being Chebyshev's theta function, a(0) = exp(theta(1)), and for n > 0, a(n) = exp(theta(m)) for A000040(n) <= m < A000040(n+1) where m is an integer. - Miles Englezou, Nov 26 2024

A007504 Sum of the first n primes.

Original entry on oeis.org

0, 2, 5, 10, 17, 28, 41, 58, 77, 100, 129, 160, 197, 238, 281, 328, 381, 440, 501, 568, 639, 712, 791, 874, 963, 1060, 1161, 1264, 1371, 1480, 1593, 1720, 1851, 1988, 2127, 2276, 2427, 2584, 2747, 2914, 3087, 3266, 3447, 3638, 3831, 4028, 4227, 4438, 4661, 4888
Offset: 0

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Author

Keywords

Comments

It appears that a(n)^2 - a(n-1)^2 = A034960(n). - Gary Detlefs, Dec 20 2011
This is true. Proof: By definition we have A034960(n) = Sum_{k = (a(n-1)+1)..a(n)} (2*k-1). Since Sum_{k = 1..n} (2*k-1) = n^2, it follows A034960(n) = a(n)^2 - a(n-1)^2, for n > 1. - Hieronymus Fischer, Sep 27 2012 [formulas above adjusted to changed offset of A034960 - Hieronymus Fischer, Oct 14 2012]
Row sums of the triangle in A037126. - Reinhard Zumkeller, Oct 01 2012
Ramanujan noticed the apparent identity between the prime parts partition numbers A000607 and the expansion of Sum_{k >= 0} x^a(k)/((1-x)...(1-x^k)), cf. A046676. See A192541 for the difference between the two. - M. F. Hasler, Mar 05 2014
For n > 0: row 1 in A254858. - Reinhard Zumkeller, Feb 08 2015
a(n) is the smallest number that can be partitioned into n distinct primes. - Alonso del Arte, May 30 2017
For a(n) < m < a(n+1), n > 0, at least one m is a perfect square.
Proof: For n = 1, 2, ..., 6, the proposition is clear. For n > 6, a(n) < ((prime(n) - 1)/2)^2, set (k - 1)^2 <= a(n) < k^2 < ((prime(n) + 1)/2)^2, then k^2 < (k - 1)^2 + prime(n) <= a(n) + prime(n) = a(n+1), so m = k^2 is this perfect square. - Jinyuan Wang, Oct 04 2018
For n >= 5 we have a(n) < ((prime(n)+1)/2)^2. This can be shown by noting that ((prime(n)+1)/2)^2 - ((prime(n-1)+1)/2)^2 - prime(n) = (prime(n)+prime(n-1))*(prime(n)-prime(n-1)-2)/4 >= 0. - Jianing Song, Nov 13 2022
Washington gives an oscillation formula for |a(n) - pi(n^2)|, see links. - Charles R Greathouse IV, Dec 07 2022

References

  • E. Bach and J. Shallit, §2.7 in Algorithmic Number Theory, Vol. 1: Efficient Algorithms, MIT Press, Cambridge, MA, 1996.
  • H. L. Nelson, "Prime Sums", J. Rec. Math., 14 (1981), 205-206.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

See A122989 for the value of Sum_{n >= 1} 1/a(n).

Programs

  • GAP
    P:=Filtered([1..250],IsPrime);;
    a:=Concatenation([0],List([1..Length(P)],i->Sum([1..i],k->P[k]))); # Muniru A Asiru, Oct 07 2018
    
  • Haskell
    a007504 n = a007504_list !! n
    a007504_list = scanl (+) 0 a000040_list
    -- Reinhard Zumkeller, Oct 01 2014, Oct 03 2011
    
  • Magma
    [0] cat [&+[ NthPrime(k): k in [1..n]]: n in [1..50]]; // Bruno Berselli, Apr 11 2011 (adapted by Vincenzo Librandi, Nov 27 2015 after Hasler's change on Mar 05 2014)
    
  • Maple
    s1:=[2]; for n from 2 to 1000 do s1:=[op(s1),s1[n-1]+ithprime(n)]; od: s1;
    A007504 := proc(n)
        add(ithprime(i), i=1..n) ;
    end proc: # R. J. Mathar, Sep 20 2015
  • Mathematica
    Accumulate[Prime[Range[100]]] (* Zak Seidov, Apr 10 2011 *)
    primeRunSum = 0; Table[primeRunSum = primeRunSum + Prime[k], {k, 100}] (* Zak Seidov, Apr 16 2011 *)
  • PARI
    A007504(n) = sum(k=1,n,prime(k)) \\ Michael B. Porter, Feb 26 2010
    
  • PARI
    a(n) = vecsum(primes(n)); \\ Michel Marcus, Feb 06 2021
    
  • Python
    from itertools import accumulate, count, islice
    from sympy import prime
    def A007504_gen(): return accumulate(prime(n) if n > 0 else 0 for n in count(0))
    A007504_list = list(islice(A007504_gen(),20)) # Chai Wah Wu, Feb 23 2022

Formula

a(n) ~ n^2 * log(n) / 2. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 24 2001 (see Bach & Shallit 1996)
a(n) = A014284(n+1) - 1. - Jaroslav Krizek, Aug 19 2009
a(n+1) - a(n) = A000040(n+1). - Jaroslav Krizek, Aug 19 2009
a(A051838(n)) = A002110(A051838(n)) / A116536(n). - Reinhard Zumkeller, Oct 03 2011
a(n) = min(A068873(n), A073619(n)) for n > 1. - Jonathan Sondow, Jul 10 2012
a(n) = A033286(n) - A152535(n). - Omar E. Pol, Aug 09 2012
For n >= 3, a(n) >= (n-1)^2 * (log(n-1) - 1/2)/2 and a(n) <= n*(n+1)*(log(n) + log(log(n))+ 1)/2. Thus a(n) = n^2 * log(n) / 2 + O(n^2*log(log(n))). It is more precise than in Fares's comment. - Vladimir Shevelev, Aug 01 2013
a(n) = (n^2/2)*(log n + log log n - 3/2 + (log log n - 3)/log n + (2 (log log n)^2 - 14 log log n + 27)/(4 log^2 n) + O((log log n/log n)^3)) [Sinha]. - Charles R Greathouse IV, Jun 11 2015
G.f: (x*b(x))/(1-x), where b(x) is the g.f. of A000040. - Mario C. Enriquez, Dec 10 2016
a(n) = A008472(A002110(n)), for n > 0. - Michel Marcus, Jul 16 2020

Extensions

More terms from Stefan Steinerberger, Apr 11 2006
a(0) = 0 prepended by M. F. Hasler, Mar 05 2014

A116536 Numbers that can be expressed as the ratio of the product and the sum of consecutive prime numbers starting from 2.

Original entry on oeis.org

1, 3, 125970, 1278362451795, 305565807424800745258151050335, 2099072522743338791053378243660769678400212601239922213271230, 330455532167461882998265688366895823334392289157931734871641555
Offset: 1

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Author

Keywords

Comments

Let prime(i) denote the i-th prime (A000040). Let F(m) = (Product_{i=1..m} prime(i)) / (Sum_{i=1..m} prime(i)). Sequence gives integer values of F(m) and A051838 gives corresponding values of m. - N. J. A. Sloane, Oct 01 2011

Examples

			a(1) = 1 because 2/2 = 1.
a(2) = 3 because (2*3*5)/(2+3+5) = 30/10 = 3.
a(3) = 125970 because (2*3*5*7*11*13*17*19)/(2+3+5+7+11+13+17+19) = 9699690/77 = 125790.
		

References

  • G. Balzarotti and P. P. Lava, Le sequenze di numeri interi, Hoepli, 2008, p. 158.

Crossrefs

Programs

  • Haskell
    import Data.Maybe (catMaybes)
    a116536 n = a116536_list !! (n-1)
    a116536_list = catMaybes $ zipWith div' a002110_list a007504_list where
       div' x y | m == 0    = Just x'
                | otherwise = Nothing where (x',m) = divMod x y
    -- Reinhard Zumkeller, Oct 03 2011
  • Magma
    [p/s: n in [1..40] | IsDivisibleBy(p,s) where p is &*[NthPrime(i): i in [1..n]] where s is &+[NthPrime(i): i in [1..n]]];  // Bruno Berselli, Sep 30 2011
    
  • Maple
    P:=proc(n) local i,j, pp,sp; pp:=1; sp:=0; for i from 1 by 1 to n do pp:=pp*ithprime(i); sp:=sp+ithprime(i); j:=pp/sp; if j=trunc(j) then print(j); fi; od; end: P(100);
  • Mathematica
    seq = {}; sum = 0; prod = 1; p = 1; Do[p = NextPrime[p]; prod *= p; sum += p; If[Divisible[prod, sum], AppendTo[seq, prod/sum]], {50}]; seq (* Amiram Eldar, Nov 02 2020 *)

Formula

a(n) = A002110(A051838(n)) / A007504(A051838(n)). - Reinhard Zumkeller, Oct 03 2011
a(n) = A159578(n)/A001414(A159578(n)). - Amiram Eldar, Nov 02 2020

A159578 Dividend associated with A116536.

Original entry on oeis.org

2, 30, 9699690, 304250263527210, 267064515689275851355624017992790, 5766152219975951659023630035336134306565384015606066319856068810, 962947420735983927056946215901134429196419130606213075415963491270, 29819592777931214269172453467810429868925511217482600306406141434158090
Offset: 1

Views

Author

Enoch Haga, Apr 16 2009

Keywords

Comments

a(2)-a(4) are mentioned by Alladi and Erdős (1977). They conjectured that this sequence is infinite. - Amiram Eldar, Nov 02 2020

Examples

			a(2) = 30 because 2*3*5 = 30, 2+3+5 = 10, and 30/10 = 3 in A116536.
		

Crossrefs

Intersection of A002110 and A036844.

Programs

  • Maple
    # First define t1, the sequence A051838.
    t1:=[1,3,8,13,23,38,39,41,43,48,50,53,56,57,58,66,68,
    70,73,77,84,90,94,98,126,128,134,140,143,145,149,
    151,153,157,160,164,167,168,172,174,176,182,191,
    194,196,200,210,212,215,217,218,219,222,225,228,
    229];
    p:=ithprime;
    num:=n->mul(p(i),i=1..t1[n]);
    s:=[num(i),i=1..11)];
  • Mathematica
    seq = {}; sum = 0; prod = 1; p = 1; Do[p = NextPrime[p]; prod *= p; sum += p; If[Divisible[prod, sum], AppendTo[seq, prod]], {50}]; seq (* Amiram Eldar, Nov 02 2020 *)
    Module[{nn=50,s,p},s=Accumulate[Prime[Range[nn]]];p=FoldList[Times,Prime[Range[ nn]]]; Select[Thread[{p,s}],Divisible[#[[1]],#[[2]]]&]][[All,1]] (* Harvey P. Dale, Jun 07 2022 *)

Formula

a(n) = A002110(A051838(n)). - Amiram Eldar, Nov 02 2020

Extensions

Corrected by N. J. A. Sloane, Oct 02 2011 (all the terms were wrong).

A196415 Values of n such that (product of first n composite numbers) / (sum of first n composite numbers) is an integer.

Original entry on oeis.org

1, 4, 7, 10, 13, 15, 16, 21, 32, 33, 56, 57, 60, 70, 77, 80, 83, 84, 88, 92, 93, 97, 112, 114, 115, 120, 122, 130, 134, 141, 147, 153, 155, 164, 165, 188, 191, 196, 201, 202, 213, 222, 225, 226, 229, 243, 245, 248, 252, 260, 264, 265, 268, 273, 274, 281
Offset: 1

Views

Author

N. J. A. Sloane, Oct 02 2011

Keywords

Comments

A036691(a(n)) mod A053767(a(n)) = 0, A141092(n) = A036691(a(n)) / A053767(a(n)). [Reinhard Zumkeller, Oct 03 2011]

Crossrefs

Programs

  • Haskell
    import Data.List (elemIndices)
    a196415 n = a196415_list !! (n-1)
    a196415_list =
       map (+ 1) $ elemIndices 0 $ zipWith mod a036691_list a053767_list
    -- Reinhard Zumkeller, Oct 03 2011
  • Maple
    # First define list of composite numbers:
    tc:=[4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,
    28,30,32,33,34,35,36,38,39,40,42,44,45,46,48,49,
    50,51,52,54,55,56,57,58,60,62,63,64,65,66,68,69,
    70,72,74,75,76,77,78,80,81,82,84,85,86,87,88];
    a1:=n->mul(tc[i],i=1..n);
    a2:=n->add(tc[i],i=1..n);
    sn:=[];
    s0:=[];
    s1:=[];
    s2:=[];
    for n from 1 to 40 do
      t1:=a1(n)/a2(n);
      if whattype(t1) = integer then
       sn:= [op(sn),n];
       s0:= [op(s0),t1];
       s1:= [op(s1),a1(n)];
       s2:= [op(s2),a2(n)];
    fi;
    od:
    sn; s0; s1; s2;
    # alternatively
    for n from 1 to 1000 do
            if type(A036691(n)/A053767(n),'integer') then
                    printf("%d,",n);
            end if;
    end do: # R. J. Mathar, Oct 03 2011
  • Mathematica
    c = Select[Range[2,355], ! PrimeQ@# &]; p = 1; s = 0; Select[Range@ Length@c, Mod[p *= c[[#]], s += c[[#]]] == 0 &] (* Giovanni Resta, Apr 03 2013 *)

Extensions

More terms from Arkadiusz Wesolowski, Oct 03 2011

A140761 Primes p(j) = A000040(j), j>=1, such that p(1)*p(2)*...*p(j) is an integral multiple of p(1)+p(2)+...+p(j).

Original entry on oeis.org

2, 5, 19, 41, 83, 163, 167, 179, 191, 223, 229, 241, 263, 269, 271, 317, 337, 349, 367, 389, 433, 463, 491, 521, 701, 719, 757, 809, 823, 829, 859, 877, 883, 919, 941, 971, 991, 997, 1021, 1033, 1049, 1091, 1153, 1181, 1193, 1223, 1291, 1301, 1319, 1327, 1361
Offset: 1

Views

Author

Enoch Haga, May 28 2008

Keywords

Examples

			a(2) = 5 because it is the last consecutive prime in the run 2*3*5 = 30 and 2+3+5 = 10; since 30/10 = 3, it is the first integral quotient.
		

Crossrefs

Programs

  • Mathematica
    seq = {}; sum = 0; prod = 1; p = 1; Do[p = NextPrime[p]; prod *= p; sum += p; If[Divisible[prod, sum], AppendTo[seq, p]], {200}]; seq (* Amiram Eldar, Nov 02 2020 *)

Formula

Find integral quotients of products of consecutive primes divided by their sum.
a(n) = A000040(A051838(n)). - R. J. Mathar, Jun 09 2008

Extensions

Edited by R. J. Mathar, Jun 09 2008
a(1) added by Amiram Eldar, Nov 02 2020

A255217 Primorial mod sum-of-primes.

Original entry on oeis.org

0, 1, 0, 6, 14, 18, 52, 0, 70, 90, 50, 98, 0, 148, 82, 150, 110, 453, 450, 213, 262, 637, 0, 69, 530, 129, 1106, 339, 1110, 1416, 1290, 1443, 994, 30, 2274, 933, 646, 0, 0, 168, 0, 3234, 0, 786, 2014, 3270, 1680, 0, 1222, 0, 1070, 690, 0, 2934, 416, 0, 0, 0, 708
Offset: 1

Views

Author

Walter Carlini, Apr 25 2015

Keywords

Comments

Does 0 appear infinitely often in this sequence? See A051838.

Examples

			For n = 4, a(4) = (2*3*5*7) mod (2+3+5+7) = 210 mod 17 = 6.
		

Crossrefs

Cf. A002110 (Primorial numbers), A007504 (Sum of first n primes)

Programs

  • Mathematica
    Table[Mod[Product[Prime[i],{i,n}],Sum[Prime[i],{i,n}]],{n,60}] (* Ivan N. Ianakiev, Apr 25 2015 *)
    With[{pr=Prime[Range[60]]},Mod[#[[1]],#[[2]]]&/@Thread[{FoldList[ Times, pr], Accumulate[pr]}]] (* Harvey P. Dale, Jan 22 2016 *)
  • PARI
    a(n) = my(vp=primes(n)); vecprod(vp) % vecsum(vp); \\ Michel Marcus, Dec 05 2021
    
  • PARI
    lista(nn) = {my(s=0, p=1); forprime(q=2, nn, s += q; p *= q; print1(p%s, ", "););} \\ Michel Marcus, Dec 05 2021

Formula

a(n) = prime(n)# mod A007504(n).

Extensions

More terms from Michel Marcus, Apr 25 2015

A322608 Values of k such that (product of squarefree numbers <= k) / (sum of squarefree numbers <= k) is an integer.

Original entry on oeis.org

1, 3, 11, 14, 17, 21, 23, 33, 34, 37, 46, 47, 55, 58, 59, 61, 62, 67, 69, 73, 82, 83, 87, 94, 95, 97, 101, 106, 107, 109, 114, 115, 119, 123, 127, 133, 134, 141, 146, 151, 157, 158, 159, 161, 165, 166, 173, 181, 187, 197, 202, 203, 210, 218, 219, 223, 226, 230
Offset: 1

Views

Author

Paolo P. Lava, Dec 20 2018

Keywords

Examples

			3 is in the sequence because (1*2*3)/(1+2+3) = 1.
11 is in the sequence because (1*2*3*5*6*7*10*11)/(1+2+3+5+6+7+10+11) = 138600/45 = 3080.
		

Crossrefs

Cf. A005117, A051838, A111059, A116536, A141092, A173143, A322607 (values of the quotient), A347690 (numbers of terms in the numerators).

Programs

  • Maple
    with(numtheory): P:=proc(q) local a,b,c,n; a:=1; b:=0; c:=[];
    for n from 1 to q do if issqrfree(n) then a:=a*n; b:=b+n;
    if frac(a/b)=0 then c:=[op(c),n];
    fi; fi; od; op(c); end: P(60);
  • Mathematica
    seq = {}; sum = 0; prod = 1; Do[If[SquareFreeQ[n], sum += n; prod *= n; If[Divisible[prod, sum], AppendTo[seq, n]]], {n, 1, 230}]; seq (* Amiram Eldar, Mar 05 2021 *)

Extensions

Definition corrected by N. J. A. Sloane, Sep 19 2021 at the suggestion of Harvey P. Dale.
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