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 1-8 of 8 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

A008864 a(n) = prime(n) + 1.

Original entry on oeis.org

3, 4, 6, 8, 12, 14, 18, 20, 24, 30, 32, 38, 42, 44, 48, 54, 60, 62, 68, 72, 74, 80, 84, 90, 98, 102, 104, 108, 110, 114, 128, 132, 138, 140, 150, 152, 158, 164, 168, 174, 180, 182, 192, 194, 198, 200, 212, 224, 228, 230, 234, 240, 242, 252, 258, 264, 270, 272, 278, 282, 284
Offset: 1

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Author

Keywords

Comments

Sum of divisors of prime(n). - Labos Elemer, May 24 2001
For n > 1, there are a(n) more nonnegative Hurwitz quaternions than nonnegative Lipschitz quaternions, which are counted in A239396 and A239394, respectively. - T. D. Noe, Mar 31 2014
These are the numbers which are in A239708 or in A187813, but excluding the first 3 terms of A187813, i.e., a number m is a term if and only if m is a term > 2 of A187813, or m is the sum of two distinct powers of 2 such that m - 1 is prime. This means that a number m is a term if and only if m is a term > 2 such that there is no base b with a base-b digital sum of b, or b = 2 is the only base for which the base-b digital sum of m is b. a(6) is the only term such that a(n) = A187813(n); for n < 6, we have a(n) > A187813(n), and for n > 6, we have a(n) < A187813(n). - Hieronymus Fischer, Apr 10 2014
Does not contain any number of the format 1 + q + ... + q^e, q prime, e >= 2, i.e., no terms of A060800, A131991, A131992, A131993 etc. [Proof: that requires 1 + p = 1 + q + ... + q^e, or p = q*(1 + ... + q^(e-1)). This is not solvable because the left hand side is prime, the right hand side composite.] - R. J. Mathar, Mar 15 2018
1/a(n) is the asymptotic density of numbers whose prime(n)-adic valuation is odd. - Amiram Eldar, Jan 23 2021

References

  • C. W. Trigg, Problem #1210, Series Formation, J. Rec. Math., 15 (1982), 221-222.

Crossrefs

Column 1 of A341605, column 2 of A286623 and of A328464.
Partial sums of A125266.

Programs

Formula

a(n) = prime(n) + 1 = A000040(n) + 1.
a(n) = A000005(A034785(n)) = A000203(A000040(n)). - Labos Elemer, May 24 2001
a(n) = A084920(n) / A006093(n). - Reinhard Zumkeller, Aug 06 2007
A239703(a(n)) <= 1. - Hieronymus Fischer, Apr 10 2014
From Ilya Gutkovskiy, Jul 30 2016: (Start)
a(n) ~ n*log(n).
Product_{n>=1} (1 + 2/(a(n)*(a(n) - 2))) = 5/2. (End)

A030078 Cubes of primes.

Original entry on oeis.org

8, 27, 125, 343, 1331, 2197, 4913, 6859, 12167, 24389, 29791, 50653, 68921, 79507, 103823, 148877, 205379, 226981, 300763, 357911, 389017, 493039, 571787, 704969, 912673, 1030301, 1092727, 1225043, 1295029, 1442897, 2048383, 2248091, 2571353, 2685619, 3307949
Offset: 1

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Keywords

Comments

Numbers with exactly three factorizations: A001055(a(n)) = 3 (e.g., a(4) = 1*343 = 7*49 = 7*7*7). - Reinhard Zumkeller, Dec 29 2001
Intersection of A014612 and A000578. Intersection of A014612 and A030513. - Wesley Ivan Hurt, Sep 10 2013
Let r(n) = (a(n)-1)/(a(n)+1) if a(n) mod 4 = 1, (a(n)+1)/(a(n)-1) otherwise; then Product_{n>=1} r(n) = (9/7) * (28/26) * (124/126) * (344/342) * (1332/1330) * ... = 48/35. - Dimitris Valianatos, Mar 06 2020
There exist 5 groups of order p^3, when p prime, so this is a subsequence of A054397. Three of them are abelian: C_p^3, C_p^2 X C_p and C_p X C_p X C_p = (C_p)^3. For 8 = 2^3, the 2 nonabelian groups are D_8 and Q_8; for odd prime p, the 2 nonabelian groups are (C_p x C_p) : C_p, and C_p^2 : C_p (remark, for p = 2, these two semi-direct products are isomorphic to D_8). Here C, D, Q mean Cyclic, Dihedral, Quaternion groups of the stated order; the symbols X and : mean direct and semidirect products respectively. - Bernard Schott, Dec 11 2021

Examples

			a(3) = 125; since the 3rd prime is 5, a(3) = 5^3 = 125.
		

References

  • Edmund Landau, Elementary Number Theory, translation by Jacob E. Goodman of Elementare Zahlentheorie (Vol. I_1 (1927) of Vorlesungen über Zahlentheorie), by Edmund Landau, with added exercises by Paul T. Bateman and E. E. Kohlbecker, Chelsea Publishing Co., New York, 1958, pp. 31-32.

Crossrefs

Other sequences that are k-th powers of primes are: A000040 (k=1), A001248 (k=2), this sequence (k=3), A030514 (k=4), A050997 (k=5), A030516 (k=6), A092759 (k=7), A179645 (k=8), A179665 (k=9), A030629 (k=10), A079395 (k=11), A030631 (k=12), A138031 (k=13), A030635 (k=16), A138032 (k=17), A030637 (k=18).
Cf. A060800, A131991, A000578, subsequence of A046099.
Subsequence of A007422 and of A054397.

Programs

Formula

n such that A062799(n) = 3. - Benoit Cloitre, Apr 06 2002
a(n) = A000040(n)^3. - Omar E. Pol, Jul 27 2009
A064380(a(n)) = A000010(a(n)). - Vladimir Shevelev, Apr 19 2010
A003415(a(n)) = A079705(n). - Reinhard Zumkeller, Jun 26 2011
A056595(a(n)) = 2. - Reinhard Zumkeller, Aug 15 2011
A000005(a(n)) = 4. - Wesley Ivan Hurt, Sep 10 2013
a(n) = A119959(n) * A008864(n) -1.- R. J. Mathar, Aug 13 2019
Sum_{n>=1} 1/a(n) = P(3) = 0.1747626392... (A085541). - Amiram Eldar, Jul 27 2020
From Amiram Eldar, Jan 23 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = zeta(3)/zeta(6) (A157289).
Product_{n>=1} (1 - 1/a(n)) = 1/zeta(3) (A088453). (End)

A030514 a(n) = prime(n)^4.

Original entry on oeis.org

16, 81, 625, 2401, 14641, 28561, 83521, 130321, 279841, 707281, 923521, 1874161, 2825761, 3418801, 4879681, 7890481, 12117361, 13845841, 20151121, 25411681, 28398241, 38950081, 47458321, 62742241, 88529281, 104060401, 112550881, 131079601, 141158161
Offset: 1

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Author

Keywords

Comments

Numbers with 5 divisors (1, p, p^2, p^3, p^4, where p is the n-th prime). - Alexandre Wajnberg, Jan 15 2006
Subsequence of A036967. - Reinhard Zumkeller, Feb 05 2008
The n-th number with p divisors is equal to the n-th prime raised to power p-1, where p is prime. - Omar E. Pol, May 06 2008
The general product formula for even s is: product_{p = A000040} (p^s-1)/(p^s+1) = 2*Bernoulli(2s)/( binomial(2s, s)*Bernoulli^2(s)), where the infinite product is over all primes. Here, with s = 4, product_{n = 1, 2, ...} (a(n)-1)/(a(n)+1) = 6/7. In A030516, where s = 6, the product of the ratios is 691/715. For s = 8, the 8th row in A120458, the corresponding product of ratios is 7234/7293. - R. J. Mathar, Feb 01 2009
Except for the first three terms, all others are congruent to 1 mod 240. - Robert Israel, Aug 29 2014

Crossrefs

Programs

Formula

a(n) = A000040(n)^(5-1) = A000040(n)^4, where 5 is the number of divisors of a(n). - Omar E. Pol, May 06 2008
A000005(a(n)) = 5. - Alexandre Wajnberg, Jan 15 2006
A056595(a(n)) = 2. - Reinhard Zumkeller, Aug 15 2011
Sum_{n>=1} 1/a(n) = P(4) = 0.0769931397... (A085964). - Amiram Eldar, Jul 27 2020
From Amiram Eldar, Jan 23 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = zeta(4)/zeta(8) = 105/Pi^4 (A157290).
Product_{n>=1} (1 - 1/a(n)) = 1/zeta(4) = 90/Pi^4 (A215267). (End)

Extensions

Description corrected by Eric W. Weisstein

A060800 a(n) = p^2 + p + 1 where p runs through the primes.

Original entry on oeis.org

7, 13, 31, 57, 133, 183, 307, 381, 553, 871, 993, 1407, 1723, 1893, 2257, 2863, 3541, 3783, 4557, 5113, 5403, 6321, 6973, 8011, 9507, 10303, 10713, 11557, 11991, 12883, 16257, 17293, 18907, 19461, 22351, 22953, 24807, 26733, 28057, 30103, 32221
Offset: 1

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Author

Jason Earls, Apr 27 2001

Keywords

Comments

Terms are divisible by 3 iff p is of the form 6*m+1 (A002476). - Michel Marcus, Jan 15 2017

Examples

			a(3) = 31 because 5^2 + 5 + 1 = 31.
		

Crossrefs

Programs

  • Magma
    [p^2+p+1: p in PrimesUpTo(200)]; // Vincenzo Librandi, Mar 20 2014
  • Maple
    A060800:= n -> map (p -> p^(2)+p+1, ithprime(n)):
    seq (A060800(n), n=1..41); # Jani Melik, Jan 25 2011
  • Mathematica
    #^2 + # + 1&/@Prime[Range[200]] (* Vincenzo Librandi, Mar 20 2014 *)
  • PARI
    { n=0; forprime (p=2, prime(1000), write("b060800.txt", n++, " ", p^2 + p + 1); ) } \\ Harry J. Smith, Jul 13 2009
    

Formula

a(n) = A036690(n) + 1.
a(n) = 1 + A008864(n)*A000040(n) = (A030078(n) - 1)/A006093(n). - Reinhard Zumkeller, Aug 06 2007
a(n) = sigma(prime(n)^2) = A000203(A000040(n)^2). - Zak Seidov, Feb 13 2016
a(n) = A000203(A001248(n)). - Michel Marcus, Feb 15 2016
Product_{n>=1} (1 - 1/a(n)) = zeta(3)/zeta(2) (A253905). - Amiram Eldar, Nov 07 2022

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), May 03 2001

A131992 a(n) = 1 + prime(n) + prime(n)^2 + prime(n)^3 + prime(n)^4.

Original entry on oeis.org

31, 121, 781, 2801, 16105, 30941, 88741, 137561, 292561, 732541, 954305, 1926221, 2896405, 3500201, 4985761, 8042221, 12326281, 14076605, 20456441, 25774705, 28792661, 39449441, 48037081, 63455221, 89451461, 105101005, 113654321
Offset: 1

Views

Author

Reinhard Zumkeller, Aug 06 2007

Keywords

Comments

Thébault shows that a(2) = 121 is the only square in this sequence. - Charles R Greathouse IV, Jul 23 2013
Giovanni Resta has found that 28792661 is the first Sophie Germain prime of this form (and actually of the form p = (n^m-1)/(n-1) for any p-1 > n, m > 1). - M. F. Hasler, Mar 03 2020

Examples

			a(1) = 31 because prime(1) = 2 and 1 + 2 + 2^2 + 2^3 + 2^4 = 1 + 2 + 4 + 8 + 16 = 31.
		

References

  • Victor Thébault, Curiosités arithmétiques, Mathesis 62 (1953), pp. 120-129.

Crossrefs

Equals A053699 restricted to prime indices. Subsequence of primes is A190527.

Programs

Formula

a(n) = 1 + A131991(n)*A000040(n).
a(n) = (A050997(n) - 1)/A006093(n).
a(n) = A000203(prime(n)^4). - R. J. Mathar, Mar 15 2018
a(n) = (prime(n)^5 - 1)/(prime(n) - 1) = A053699(prime(n)). (This is also meant by the 2nd formula.) - M. F. Hasler, Mar 03 2020

A131993 1 + prime(n) + prime(n)^2 + prime(n)^3 + prime(n)^4 + prime(n)^5.

Original entry on oeis.org

63, 364, 3906, 19608, 177156, 402234, 1508598, 2613660, 6728904, 21243690, 29583456, 71270178, 118752606, 150508644, 234330768, 426237714, 727250580, 858672906, 1370581548, 1830004056, 2101864254, 3116505840, 3987077724
Offset: 1

Views

Author

Reinhard Zumkeller, Aug 06 2007

Keywords

Comments

a(n) = 1 + A131992(n)*A000040(n).

Crossrefs

Programs

  • Magma
    [1+(&+[NthPrime(n)^(k): k in [1..5]]): n in [1..100]]; // Berselli - Librandi, Apr 20 2011
  • Mathematica
    Total[#^Range[0,5]]&/@Prime[Range[30]]  (* Harvey P. Dale, Apr 20 2011 *)

Formula

a(n) = (A030516(n) - 1)/A006093(n).

A319076 Square array T(n,k) read by antidiagonal upwards in which column k lists the partial sums of the powers of the k-th prime, n >= 0, k >= 1.

Original entry on oeis.org

1, 3, 1, 7, 4, 1, 15, 13, 6, 1, 31, 40, 31, 8, 1, 63, 121, 156, 57, 12, 1, 127, 364, 781, 400, 133, 14, 1, 255, 1093, 3906, 2801, 1464, 183, 18, 1, 511, 3280, 19531, 19608, 16105, 2380, 307, 20, 1, 1023, 9841, 97656, 137257, 177156, 30941, 5220, 381, 24, 1, 2047, 29524, 488281, 960800, 1948717
Offset: 0

Views

Author

Omar E. Pol, Sep 09 2018

Keywords

Comments

T(n,k) is also the sum of the divisors of the n-th nonnegative power of the k-th prime, n >= 0, k >= 1.

Examples

			The corner of the square array is as follows:
         A126646 A003462 A003463  A023000    A016123    A091030     A091045
A000012        1,      1,      1,       1,         1,         1,          1, ...
A008864        3,      4,      6,       8,        12,        14,         18, ...
A060800        7,     13,     31,      57,       133,       183,        307, ...
A131991       15,     40,    156,     400,      1464,      2380,       5220, ...
A131992       31,    121,    781,    2801,     16105,     30941,      88741, ...
A131993       63,    364,   3906,   19608,    177156,    402234,    1508598, ...
.......      127,   1093,  19531,  137257,   1948717,   5229043,   25646167, ...
.......      255,   3280,  97656,  960800,  21435888,  67977560,  435984840, ...
.......      511,   9841, 488281, 6725601, 235794769, 883708281, 7411742281, ...
...
		

Crossrefs

Programs

  • PARI
    T(n, k) = sigma(prime(k)^n); \\ Michel Marcus, Sep 13 2018

Formula

T(n,k) = A000203(A000040(k)^n).
T(n,k) = Sum_{j=0..n} A000040(k)^j.
T(n,k) = Sum_{j=0..n} A319075(j,k).
T(n,k) = (A000040(k)^(n+1) - 1)/(A000040(k) - 1).
T(n,k) = (A000040(k)^(n+1) - 1)/A006093(k).
Showing 1-8 of 8 results.