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.

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A134442 Last two digits of primes of form 4n+3 (A002145). Leading 0's omitted.

Original entry on oeis.org

3, 7, 11, 19, 23, 27, 31, 39, 43, 47, 51, 59, 63, 67, 71, 79, 83, 87, 91, 99
Offset: 1

Views

Author

Lekraj Beedassy, Jan 19 2008

Keywords

Crossrefs

Cf. A002145.

Programs

  • Mathematica
    Union[FromDigits[Take[IntegerDigits[#],-2]]&/@Select[4Range[2,130]+3, PrimeQ]] (* Harvey P. Dale, May 15 2011 *)

A136403 Absolute values of negative fundamental discriminants (A003657) that are not 3 mod 4 (A002145).

Original entry on oeis.org

4, 8, 15, 20, 24, 35, 39, 40, 51, 52, 55, 56, 68
Offset: 1

Views

Author

Paul Muljadi, Mar 30 2008

Keywords

Comments

All terms are composite numbers (A002808).

Crossrefs

A141624 A008846 (5, 13, 17, 25, ...) - A002145 (3, 7, 11, 19, ...).

Original entry on oeis.org

2, 6, 6, 6, 6, 6, -2, 6, 2, -2, 2, 6, 6, -6, -6, -18, -18, -14, -14, -18, -18, -22
Offset: 0

Views

Author

Paul Curtz, Aug 24 2008

Keywords

A347942 Number of jellyfish for AGM(F(q)) where q are primes == 3 (mod 4) (A002145).

Original entry on oeis.org

0, 1, 3, 8, 5, 10, 7, 4, 7, 30, 25, 18, 6, 41, 9, 54, 46, 33, 45, 38, 11, 14, 14, 101, 120, 18, 12, 40, 31, 17, 34, 35
Offset: 1

Views

Author

Michel Marcus, Oct 26 2021

Keywords

Crossrefs

Cf. A002145.

A000984 Central binomial coefficients: binomial(2*n,n) = (2*n)!/(n!)^2.

Original entry on oeis.org

1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520, 601080390, 2333606220, 9075135300, 35345263800, 137846528820, 538257874440, 2104098963720, 8233430727600, 32247603683100, 126410606437752, 495918532948104, 1946939425648112
Offset: 0

Views

Author

Keywords

Comments

Devadoss refers to these numbers as type B Catalan numbers (cf. A000108).
Equal to the binomial coefficient sum Sum_{k=0..n} binomial(n,k)^2.
Number of possible interleavings of a program with n atomic instructions when executed by two processes. - Manuel Carro (mcarro(AT)fi.upm.es), Sep 22 2001
Convolving a(n) with itself yields A000302, the powers of 4. - T. D. Noe, Jun 11 2002
Number of ordered trees with 2n+1 edges, having root of odd degree and nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Also number of directed, convex polyominoes having semiperimeter n+2.
Also number of diagonally symmetric, directed, convex polyominoes having semiperimeter 2n+2. - Emeric Deutsch, Aug 03 2002
The second inverse binomial transform of this sequence is this sequence with interpolated zeros. Its g.f. is (1 - 4*x^2)^(-1/2), with n-th term C(n,n/2)(1+(-1)^n)/2. - Paul Barry, Jul 01 2003
Number of possible values of a 2n-bit binary number for which half the bits are on and half are off. - Gavin Scott (gavin(AT)allegro.com), Aug 09 2003
Ordered partitions of n with zeros to n+1, e.g., for n=4 we consider the ordered partitions of 11110 (5), 11200 (30), 13000 (20), 40000 (5) and 22000 (10), total 70 and a(4)=70. See A001700 (esp. Mambetov Bektur's comment). - Jon Perry, Aug 10 2003
Number of nondecreasing sequences of n integers from 0 to n: a(n) = Sum_{i_1=0..n} Sum_{i_2=i_1..n}...Sum_{i_n=i_{n-1}..n}(1). - J. N. Bearden (jnb(AT)eller.arizona.edu), Sep 16 2003
Number of peaks at odd level in all Dyck paths of semilength n+1. Example: a(2)=6 because we have U*DU*DU*D, U*DUUDD, UUDDU*D, UUDUDD, UUU*DDD, where U=(1,1), D=(1,-1) and * indicates a peak at odd level. Number of ascents of length 1 in all Dyck paths of semilength n+1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(2)=6 because we have uDuDuD, uDUUDD, UUDDuD, UUDuDD, UUUDDD, where an ascent of length 1 is indicated by a lower case letter. - Emeric Deutsch, Dec 05 2003
a(n-1) = number of subsets of 2n-1 distinct elements taken n at a time that contain a given element. E.g., n=4 -> a(3)=20 and if we consider the subsets of 7 taken 4 at a time with a 1 we get (1234, 1235, 1236, 1237, 1245, 1246, 1247, 1256, 1257, 1267, 1345, 1346, 1347, 1356, 1357, 1367, 1456, 1457, 1467, 1567) and there are 20 of them. - Jon Perry, Jan 20 2004
The dimension of a particular (necessarily existent) absolutely universal embedding of the unitary dual polar space DSU(2n,q^2) where q>2. - J. Taylor (jt_cpp(AT)yahoo.com), Apr 02 2004.
Number of standard tableaux of shape (n+1, 1^n). - Emeric Deutsch, May 13 2004
Erdős, Graham et al. conjectured that a(n) is never squarefree for sufficiently large n (cf. Graham, Knuth, Patashnik, Concrete Math., 2nd ed., Exercise 112). Sárközy showed that if s(n) is the square part of a(n), then s(n) is asymptotically (sqrt(2)-2) * (sqrt(n)) * zeta(1/2). Granville and Ramare proved that the only squarefree values are a(1)=2, a(2)=6 and a(4)=70. - Jonathan Vos Post, Dec 04 2004 [For more about this conjecture, see A261009. - N. J. A. Sloane, Oct 25 2015]
The MathOverflow link contains the following comment (slightly edited): The Erdős squarefree conjecture (that a(n) is never squarefree for n>4) was proved in 1980 by Sárközy, A. (On divisors of binomial coefficients. I. J. Number Theory 20 (1985), no. 1, 70-80.) who showed that the conjecture holds for all sufficiently large values of n, and by A. Granville and O. Ramaré (Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients. Mathematika 43 (1996), no. 1, 73-107) who showed that it holds for all n>4. - Fedor Petrov, Nov 13 2010. [From N. J. A. Sloane, Oct 29 2015]
p divides a((p-1)/2)-1=A030662(n) for prime p=5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, ... = A002144(n) Pythagorean primes: primes of form 4n+1. - Alexander Adamchuk, Jul 04 2006
The number of direct routes from my home to Granny's when Granny lives n blocks south and n blocks east of my home in Grid City. To obtain a direct route, from the 2n blocks, choose n blocks on which one travels south. For example, a(2)=6 because there are 6 direct routes: SSEE, SESE, SEES, EESS, ESES and ESSE. - Dennis P. Walsh, Oct 27 2006
Inverse: With q = -log(log(16)/(pi a(n)^2)), ceiling((q + log(q))/log(16)) = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 26 2007
Number of partitions with Ferrers diagrams that fit in an n X n box (including the empty partition of 0). Example: a(2) = 6 because we have: empty, 1, 2, 11, 21 and 22. - Emeric Deutsch, Oct 02 2007
So this is the 2-dimensional analog of A008793. - William Entriken, Aug 06 2013
The number of walks of length 2n on an infinite linear lattice that begins and ends at the origin. - Stefan Hollos (stefan(AT)exstrom.com), Dec 10 2007
The number of lattice paths from (0,0) to (n,n) using steps (1,0) and (0,1). - Joerg Arndt, Jul 01 2011
Integral representation: C(2n,n)=1/Pi Integral [(2x)^(2n)/sqrt(1 - x^2),{x,-1, 1}], i.e., C(2n,n)/4^n is the moment of order 2n of the arcsin distribution on the interval (-1,1). - N-E. Fahssi, Jan 02 2008
Also the Catalan transform of A000079. - R. J. Mathar, Nov 06 2008
Straub, Amdeberhan and Moll: "... it is conjectured that there are only finitely many indices n such that C_n is not divisible by any of 3, 5, 7 and 11." - Jonathan Vos Post, Nov 14 2008
Equals INVERT transform of A081696: (1, 1, 3, 9, 29, 97, 333, ...). - Gary W. Adamson, May 15 2009
Also, in sports, the number of ordered ways for a "Best of 2n-1 Series" to progress. For example, a(2) = 6 means there are six ordered ways for a "best of 3" series to progress. If we write A for a win by "team A" and B for a win by "team B" and if we list the played games chronologically from left to right then the six ways are AA, ABA, BAA, BB, BAB, and ABB. (Proof: To generate the a(n) ordered ways: Write down all a(n) ways to designate n of 2n games as won by team A. Remove the maximal suffix of identical letters from each of these.) - Lee A. Newberg, Jun 02 2009
Number of n X n binary arrays with rows, considered as binary numbers, in nondecreasing order, and columns, considered as binary numbers, in nonincreasing order. - R. H. Hardin, Jun 27 2009
Hankel transform is 2^n. - Paul Barry, Aug 05 2009
It appears that a(n) is also the number of quivers in the mutation class of twisted type BC_n for n>=2.
Central terms of Pascal's triangle: a(n) = A007318(2*n,n). - Reinhard Zumkeller, Nov 09 2011
Number of words on {a,b} of length 2n such that no prefix of the word contains more b's than a's. - Jonathan Nilsson, Apr 18 2012
From Pascal's triangle take row(n) with terms in order a1,a2,..a(n) and row(n+1) with terms b1,b2,..b(n), then 2*(a1*b1 + a2*b2 + ... + a(n)*b(n)) to get the terms in this sequence. - J. M. Bergot, Oct 07 2012. For example using rows 4 and 5: 2*(1*(1) + 4*(5) + 6*(10) + 4*(10) + 1*(5)) = 252, the sixth term in this sequence.
Take from Pascal's triangle row(n) with terms b1, b2, ..., b(n+1) and row(n+2) with terms c1, c2, ..., c(n+3) and find the sum b1*c2 + b2*c3 + ... + b(n+1)*c(n+2) to get A000984(n+1). Example using row(3) and row(5) gives sum 1*(5)+3*(10)+3*(10)+1*(5) = 70 = A000984(4). - J. M. Bergot, Oct 31 2012
a(n) == 2 mod n^3 iff n is a prime > 3. (See Mestrovic link, p. 4.) - Gary Detlefs, Feb 16 2013
Conjecture: For any positive integer n, the polynomial sum_{k=0}^n a(k)x^k is irreducible over the field of rational numbers. In general, for any integer m>1 and n>0, the polynomial f_{m,n}(x) = Sum_{k=0..n} (m*k)!/(k!)^m*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 23 2013
This comment generalizes the comment dated Oct 31 2012 and the second of the sequence's original comments. For j = 1 to n, a(n) = Sum_{k=0..j} C(j,k)* C(2n-j, n-k) = 2*Sum_{k=0..j-1} C(j-1,k)*C(2n-j, n-k). - Charlie Marion, Jun 07 2013
The differences between consecutive terms of the sequence of the quotients between consecutive terms of this sequence form a sequence containing the reciprocals of the triangular numbers. In other words, a(n+1)/a(n)-a(n)/a(n-1) = 2/(n*(n+1)). - Christian Schulz, Jun 08 2013
Number of distinct strings of length 2n using n letters A and n letters B. - Hans Havermann, May 07 2014
From Fung Lam, May 19 2014: (Start)
Expansion of G.f. A(x) = 1/(1+q*x*c(x)), where parameter q is positive or negative (except q=-1), and c(x) is the g.f. of A000108 for Catalan numbers. The case of q=-1 recovers the g.f. of A000108 as xA^2-A+1=0. The present sequence A000984 refers to q=-2. Recurrence: (1+q)*(n+2)*a(n+2) + ((q*q-4*q-4)*n + 2*(q*q-q-1))*a(n+1) - 2*q*q*(2*n+1)*a(n) = 0, a(0)=1, a(1)=-q. Asymptotics: a(n) ~ ((q+2)/(q+1))*(q^2/(-q-1))^n, q<=-3, a(n) ~ (-1)^n*((q+2)/(q+1))*(q^2/(q+1))^n, q>=5, and a(n) ~ -Kq*2^(2*n)/sqrt(Pi*n^3), where the multiplicative constant Kq is given by K1=1/9 (q=1), K2=1/8 (q=2), K3=3/25 (q=3), K4=1/9 (q=4). These formulas apply to existing sequences A126983 (q=1), A126984 (q=2), A126982 (q=3), A126986 (q=4), A126987 (q=5), A127017 (q=6), A127016 (q=7), A126985 (q=8), A127053 (q=9), and to A007854 (q=-3), A076035 (q=-4), A076036 (q=-5), A127628 (q=-6), A126694 (q=-7), A115970 (q=-8). (End)
a(n)*(2^n)^(j-2) equals S(n), where S(n) is the n-th number in the self-convolved sequence which yields the powers of 2^j for all integers j, n>=0. For example, when n=5 and j=4, a(5)=252; 252*(2^5)^(4-2) = 252*1024 = 258048. The self-convolved sequence which yields powers of 16 is {1, 8, 96, 1280, 17920, 258048, ...}; i.e., S(5) = 258048. Note that the convolved sequences will be composed of numbers decreasing from 1 to 0, when j<2 (exception being j=1, where the first two numbers in the sequence are 1 and all others decreasing). - Bob Selcoe, Jul 16 2014
The variance of the n-th difference of a sequence of pairwise uncorrelated random variables each with variance 1. - Liam Patrick Roche, Jun 04 2015
Number of ordered trees with n edges where vertices at level 1 can be of 2 colors. Indeed, the standard decomposition of ordered trees leading to the equation C = 1 + zC^2 (C is the Catalan function), yields this time G = 1 + 2zCG, from where G = 1/sqrt(1-4z). - Emeric Deutsch, Jun 17 2015
Number of monomials of degree at most n in n variables. - Ran Pan, Sep 26 2015
Let V(n, r) denote the volume of an n-dimensional sphere with radius r, then V(n, 2^n) / Pi = V(n-1, 2^n) * a(n/2) for all even n. - Peter Luschny, Oct 12 2015
a(n) is the number of sets {i1,...,in} of length n such that n >= i1 >= i2 >= ... >= in >= 0. For instance, a(2) = 6 as there are only 6 such sets: (2,2) (2,1) (2,0) (1,1) (1,0) (0,0). - Anton Zakharov, Jul 04 2016
From Ralf Steiner, Apr 07 2017: (Start)
By analytic continuation to the entire complex plane there exist regularized values for divergent sums such as:
Sum_{k>=0} a(k)/(-2)^k = 1/sqrt(3).
Sum_{k>=0} a(k)/(-1)^k = 1/sqrt(5).
Sum_{k>=0} a(k)/(-1/2)^k = 1/3.
Sum_{k>=0} a(k)/(1/2)^k = -1/sqrt(7)i.
Sum_{k>=0} a(k)/(1)^k = -1/sqrt(3)i.
Sum_{k>=0} a(k)/2^k = -i. (End)
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) > e(j). [Martinez and Savage, 2.18] - Eric M. Schmidt, Jul 17 2017
The o.g.f. for the sequence equals the diagonal of any of the following the rational functions: 1/(1 - (x + y)), 1/(1 - (x + y*z)), 1/(1 - (x + x*y + y*z)) or 1/(1 - (x + y + y*z)). - Peter Bala, Jan 30 2018
From Colin Defant, Sep 16 2018: (Start)
Let s denote West's stack-sorting map. a(n) is the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 231, and 321. a(n) is also the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 312, and 321.
a(n) is the number of permutations of [n+1] that avoid the patterns 1342, 3142, 3412, and 3421. (End)
All binary self-dual codes of length 4n, for n>0, must contain at least a(n) codewords of weight 2n. More to the point, there will always be at least one, perhaps unique, binary self-dual code of length 4n that will contain exactly a(n) codewords that have a hamming weight equal to half the length of the code (2n). This code can be constructed by direct summing the unique binary self-dual code of length 2 (up to permutation equivalence) to itself an even number of times. A permutation equivalent code can be constructed by augmenting two identity matrices of length 2n together. - Nathan J. Russell, Nov 25 2018
From Isaac Saffold, Dec 28 2018: (Start)
Let [b/p] denote the Legendre symbol and 1/b denote the inverse of b mod p. Then, for m and n, where n is not divisible by p,
[(m+n)/p] == [n/p]*Sum_{k=0..(p-1)/2} (-m/(4*n))^k * a(k) (mod p).
Evaluating this identity for m = -1 and n = 1 demonstrates that, for all odd primes p, Sum_{k=0..(p-1)/2} (1/4)^k * a(k) is divisible by p. (End)
Number of vertices of the subgraph of the (2n-1)-dimensional hypercube induced by all bitstrings with n-1 or n many 1s. The middle levels conjecture asserts that this graph has a Hamilton cycle. - Torsten Muetze, Feb 11 2019
a(n) is the number of walks of length 2n from the origin with steps (1,1) and (1,-1) that stay on or above the x-axis. Equivalently, a(n) is the number of walks of length 2n from the origin with steps (1,0) and (0,1) that stay in the first octant. - Alexander Burstein, Dec 24 2019
Number of permutations of length n>0 avoiding the partially ordered pattern (POP) {3>1, 1>2} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the second element but smaller than the third elements. - Sergey Kitaev, Dec 08 2020
From Gus Wiseman, Jul 21 2021: (Start)
Also the number of integer compositions of 2n+1 with alternating sum 1, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(0) = 1 through a(2) = 6 compositions are:
(1) (2,1) (3,2)
(1,1,1) (1,2,2)
(2,2,1)
(1,1,2,1)
(2,1,1,1)
(1,1,1,1,1)
The following relate to these compositions:
- The unordered version is A000070.
- The alternating sum -1 version is counted by A001791, ranked by A345910/A345912.
- The alternating sum 0 version is counted by A088218, ranked by A344619.
- Including even indices gives A126869.
- The complement is counted by A202736.
- Ranked by A345909 (reverse: A345911).
Equivalently, a(n) counts binary numbers with 2n+1 digits and one more 1 than 0's. For example, the a(2) = 6 binary numbers are: 10011, 10101, 10110, 11001, 11010, 11100.
(End)
From Michael Wallner, Jan 25 2022: (Start)
a(n) is the number of nx2 Young tableaux with a single horizontal wall between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021].
Example for a(2)=6:
3 4 2 4 3 4 3|4 4|3 2|4
1|2, 1|3, 2|1, 1 2, 1 2, 1 3
a(n) is also the number of nx2 Young tableaux with n "walls" between the first and second column.
Example for a(2)=6:
3|4 2|4 4|3 3|4 4|3 4|2
1|2, 1|3, 1|2, 2|1, 2|1, 3|1 (End)
From Shel Kaphan, Jan 12 2023: (Start)
a(n)/4^n is the probability that a fair coin tossed 2n times will come up heads exactly n times and tails exactly n times, or that a random walk with steps of +-1 will return to the starting point after 2n steps (not necessarily for the first time). As n becomes large, this number asymptotically approaches 1/sqrt(n*Pi), using Stirling's approximation for n!.
a(n)/(4^n*(2n-1)) is the probability that a random walk with steps of +-1 will return to the starting point for the first time after 2n steps. The absolute value of the n-th term of A144704 is denominator of this fraction.
Considering all possible random walks of exactly 2n steps with steps of +-1, a(n)/(2n-1) is the number of such walks that return to the starting point for the first time after 2n steps. See the absolute values of A002420 or A284016 for these numbers. For comparison, as mentioned by Stefan Hollos, Dec 10 2007, a(n) is the number of such walks that return to the starting point after 2n steps, but not necessarily for the first time. (End)
p divides a((p-1)/2) + 1 for primes p of the form 4*k+3 (A002145). - Jules Beauchamp, Feb 11 2023
Also the size of the shuffle product of two words of length n, such that the union of the two words consist of 2n distinct elements. - Robert C. Lyons, Mar 15 2023
a(n) is the number of vertices of the n-dimensional cyclohedron W_{n+1}. - Jose Bastidas, Mar 25 2025
Consider a stack of pancakes of height n, where the only allowed operation is reversing the top portion of the stack. First, perform a series of reversals of increasing sizes, followed by a series of reversals of decreasing sizes. The number of distinct permutations of the initial stack that can be reached through these operations is a(n). - Thomas Baruchel, May 12 2025

Examples

			G.f.: 1 + 2*x + 6*x^2 + 20*x^3 + 70*x^4 + 252*x^5 + 924*x^6 + ...
For n=2, a(2) = 4!/(2!)^2 = 24/4 = 6, and this is the middle coefficient of the binomial expansion (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4. - _Michael B. Porter_, Jul 06 2016
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, id. 160.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 575, line -3, with a=b=n.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 101.
  • Emeric Deutsch and Louis W. Shapiro, Seventeen Catalan identities, Bulletin of the Institute of Combinatorics and its Applications, 31 (2001), 31-38.
  • Henry W. Gould, Combinatorial Identities, Morgantown, 1972, (3.66), page 30.
  • Ronald. L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, Second Ed., see Exercise 112.
  • Martin Griffiths, The Backbone of Pascal's Triangle, United Kingdom Mathematics Trust (2008), 3-124.
  • Leonard Lipshitz and A. van der Poorten, "Rational functions, diagonals, automata and arithmetic", in Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990), 339-358.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • 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).

Crossrefs

Cf. A000108, A002420, A002457, A030662, A002144, A135091, A081696, A182400. Differs from A071976 at 10th term.
Bisection of A001405 and of A226302. See also A025565, the same ordered partitions but without all in which are two successive zeros: 11110 (5), 11200 (18), 13000 (2), 40000 (0) and 22000 (1), total 26 and A025565(4)=26.
Cf. A226078, A051924 (first differences).
Cf. A258290 (arithmetic derivative). Cf. A098616, A214377.
See A261009 for a conjecture about this sequence.
Cf. A046521 (first column).
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.

Programs

  • GAP
    List([1..1000], n -> Binomial(2*n,n)); # Muniru A Asiru, Jan 30 2018
  • Haskell
    a000984 n = a007318_row (2*n) !! n  -- Reinhard Zumkeller, Nov 09 2011
    
  • Magma
    a:= func< n | Binomial(2*n,n) >; [ a(n) : n in [0..10]];
    
  • Maple
    A000984 := n-> binomial(2*n,n); seq(A000984(n), n=0..30);
    with(combstruct); [seq(count([S,{S=Prod(Set(Z,card=i),Set(Z,card=i))}, labeled], size=(2*i)), i=0..20)];
    with(combstruct); [seq(count([S,{S=Sequence(Union(Arch,Arch)), Arch=Prod(Epsilon, Sequence(Arch),Z)},unlabeled],size=i), i=0..25)];
    with(combstruct):bin := {B=Union(Z,Prod(B,B))}: seq (count([B,bin,unlabeled],size=n)*n, n=1..25); # Zerinvary Lajos, Dec 05 2007
    A000984List := proc(m) local A, P, n; A := [1,2]; P := [1];
    for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), 2*P[-1]]);
    A := [op(A), 2*P[-1]] od; A end: A000984List(28); # Peter Luschny, Mar 24 2022
  • Mathematica
    Table[Binomial[2n, n], {n, 0, 24}] (* Alonso del Arte, Nov 10 2005 *)
    CoefficientList[Series[1/Sqrt[1-4x],{x,0,25}],x]  (* Harvey P. Dale, Mar 14 2011 *)
  • Maxima
    A000984(n):=(2*n)!/(n!)^2$ makelist(A000984(n),n,0,30); /* Martin Ettl, Oct 22 2012 */
    
  • PARI
    A000984(n)=binomial(2*n,n) \\ much more efficient than (2n)!/n!^2. \\ M. F. Hasler, Feb 26 2014
    
  • PARI
    fv(n,p)=my(s);while(n\=p,s+=n);s
    a(n)=prodeuler(p=2,2*n,p^(fv(2*n,p)-2*fv(n,p))) \\ Charles R Greathouse IV, Aug 21 2013
    
  • PARI
    fv(n,p)=my(s);while(n\=p,s+=n);s
    a(n)=my(s=1);forprime(p=2,2*n,s*=p^(fv(2*n,p)-2*fv(n,p)));s \\ Charles R Greathouse IV, Aug 21 2013
    
  • Python
    from _future_ import division
    A000984_list, b = [1], 1
    for n in range(10**3):
        b = b*(4*n+2)//(n+1)
        A000984_list.append(b) # Chai Wah Wu, Mar 04 2016
    

Formula

a(n)/(n+1) = A000108(n), the Catalan numbers.
G.f.: A(x) = (1 - 4*x)^(-1/2) = 1F0(1/2;;4x).
a(n+1) = 2*A001700(n) = A030662(n) + 1. a(2*n) = A001448(n), a(2*n+1) = 2*A002458(n) =A099976.
D-finite with recurrence: n*a(n) + 2*(1-2*n)*a(n-1)=0.
a(n) = 2^n/n! * Product_{k=0..n-1} (2*k+1).
a(n) = a(n-1)*(4-2/n) = Product_{k=1..n} (4-2/k) = 4*a(n-1) + A002420(n) = A000142(2*n)/(A000142(n)^2) = A001813(n)/A000142(n) = sqrt(A002894(n)) = A010050(n)/A001044(n) = (n+1)*A000108(n) = -A005408(n-1)*A002420(n). - Henry Bottomley, Nov 10 2000
Using Stirling's formula in A000142 it is easy to get the asymptotic expression a(n) ~ 4^n / sqrt(Pi * n). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
Integral representation as n-th moment of a positive function on the interval [0, 4]: a(n) = Integral_{x=0..4}(x^n*((x*(4-x))^(-1/2))/Pi), n=0, 1, ... This representation is unique. - Karol A. Penson, Sep 17 2001
Sum_{n>=1} 1/a(n) = (2*Pi*sqrt(3) + 9)/27. [Lehmer 1985, eq. (15)] - Benoit Cloitre, May 01 2002 (= A073016. - Bernard Schott, Jul 20 2022)
a(n) = Max_{ (i+j)!/(i!j!) | 0<=i,j<=n }. - Benoit Cloitre, May 30 2002
a(n) = Sum_{k=0..n} binomial(n+k-1,k), row sums of A059481. - Vladeta Jovovic, Aug 28 2002
E.g.f.: exp(2*x)*I_0(2x), where I_0 is Bessel function. - Michael Somos, Sep 08 2002
E.g.f.: I_0(2*x) = Sum a(n)*x^(2*n)/(2*n)!, where I_0 is Bessel function. - Michael Somos, Sep 09 2002
a(n) = Sum_{k=0..n} binomial(n, k)^2. - Benoit Cloitre, Jan 31 2003
Determinant of n X n matrix M(i, j) = binomial(n+i, j). - Benoit Cloitre, Aug 28 2003
Given m = C(2*n, n), let f be the inverse function, so that f(m) = n. Letting q denote -log(log(16)/(m^2*Pi)), we have f(m) = ceiling( (q + log(q)) / log(16) ). - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Oct 30 2003
a(n) = 2*Sum_{k=0..(n-1)} a(k)*a(n-k+1)/(k+1). - Philippe Deléham, Jan 01 2004
a(n+1) = Sum_{j=n..n*2+1} binomial(j, n). E.g., a(4) = C(7,3) + C(6,3) + C(5,3) + C(4,3) + C(3,3) = 35 + 20 + 10 + 4 + 1 = 70. - Jon Perry, Jan 20 2004
a(n) = (-1)^(n)*Sum_{j=0..(2*n)} (-1)^j*binomial(2*n, j)^2. - Helena Verrill (verrill(AT)math.lsu.edu), Jul 12 2004
a(n) = Sum_{k=0..n} binomial(2n+1, k)*sin((2n-2k+1)*Pi/2). - Paul Barry, Nov 02 2004
a(n-1) = (1/2)*(-1)^n*Sum_{0<=i, j<=n}(-1)^(i+j)*binomial(2n, i+j). - Benoit Cloitre, Jun 18 2005
a(n) = C(2n, n-1) + C(n) = A001791(n) + A000108(n). - Lekraj Beedassy, Aug 02 2005
G.f.: c(x)^2/(2*c(x)-c(x)^2) where c(x) is the g.f. of A000108. - Paul Barry, Feb 03 2006
a(n) = A006480(n) / A005809(n). - Zerinvary Lajos, Jun 28 2007
a(n) = Sum_{k=0..n} A106566(n,k)*2^k. - Philippe Deléham, Aug 25 2007
a(n) = Sum_{k>=0} A039599(n, k). a(n) = Sum_{k>=0} A050165(n, k). a(n) = Sum_{k>=0} A059365(n, k)*2^k, n>0. a(n+1) = Sum_{k>=0} A009766(n, k)*2^(n-k+1). - Philippe Deléham, Jan 01 2004
a(n) = 4^n*Sum_{k=0..n} C(n,k)(-4)^(-k)*A000108(n+k). - Paul Barry, Oct 18 2007
a(n) = Sum_{k=0..n} A039598(n,k)*A059841(k). - Philippe Deléham, Nov 12 2008
A007814(a(n)) = A000120(n). - Vladimir Shevelev, Jul 20 2009
From Paul Barry, Aug 05 2009: (Start)
G.f.: 1/(1-2x-2x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction);
G.f.: 1/(1-2x/(1-x/(1-x/(1-x/(1-... (continued fraction). (End)
If n>=3 is prime, then a(n) == 2 (mod 2*n). - Vladimir Shevelev, Sep 05 2010
Let A(x) be the g.f. and B(x) = A(-x), then B(x) = sqrt(1-4*x*B(x)^2). - Vladimir Kruchinin, Jan 16 2011
a(n) = (-4)^n*sqrt(Pi)/(gamma((1/2-n))*gamma(1+n)). - Gerry Martens, May 03 2011
a(n) = upper left term in M^n, M = the infinite square production matrix:
2, 2, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, ...
1, 1, 1, 1, 0, 0, ...
1, 1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, 1, .... - Gary W. Adamson, Jul 14 2011
a(n) = Hypergeometric([-n,-n],[1],1). - Peter Luschny, Nov 01 2011
E.g.f.: hypergeometric([1/2],[1],4*x). - Wolfdieter Lang, Jan 13 2012
a(n) = 2*Sum_{k=0..n-1} a(k)*A000108(n-k-1). - Alzhekeyev Ascar M, Mar 09 2012
G.f.: 1 + 2*x/(U(0)-2*x) where U(k) = 2*(2*k+1)*x + (k+1) - 2*(k+1)*(2*k+3)*x/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jun 28 2012
a(n) = Sum_{k=0..n} binomial(n,k)^2*H(k)/(2*H(n)-H(2*n)), n>0, where H(n) is the n-th harmonic number. - Gary Detlefs, Mar 19 2013
G.f.: Q(0)*(1-4*x), where Q(k) = 1 + 4*(2*k+1)*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 2*x*(2*k+1)/(2*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - 2*x/(2*x + (k+1)^2/(2*k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
Special values of Jacobi polynomials, in Maple notation: a(n) = 4^n*JacobiP(n,0,-1/2-n,-1). - Karol A. Penson, Jul 27 2013
a(n) = 2^(4*n)/((2*n+1)*Sum_{k=0..n} (-1)^k*C(2*n+1,n-k)/(2*k+1)). - Mircea Merca, Nov 12 2013
a(n) = C(2*n-1,n-1)*C(4*n^2,2)/(3*n*C(2*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
Sum_{n>=0} a(n)/n! = A234846. - Richard R. Forberg, Feb 10 2014
0 = a(n)*(16*a(n+1) - 6*a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 17 2014
a(n+1) = 4*a(n) - 2*A000108(n). Also a(n) = 4^n*Product_{k=1..n}(1-1/(2*k)). - Stanislav Sykora, Aug 09 2014
G.f.: Sum_{n>=0} x^n/(1-x)^(2*n+1) * Sum_{k=0..n} C(n,k)^2 * x^k. - Paul D. Hanna, Nov 08 2014
a(n) = (-4)^n*binomial(-1/2,n). - Jean-François Alcover, Feb 10 2015
a(n) = 4^n*hypergeom([-n,1/2],[1],1). - Peter Luschny, May 19 2015
a(n) = Sum_{k=0..floor(n/2)} C(n,k)*C(n-k,k)*2^(n-2*k). - Robert FERREOL, Aug 29 2015
a(n) ~ 4^n*(2-2/(8*n+2)^2+21/(8*n+2)^4-671/(8*n+2)^6+45081/(8*n+2)^8)/sqrt((4*n+1) *Pi). - Peter Luschny, Oct 14 2015
A(-x) = 1/x * series reversion( x*(2*x + sqrt(1 + 4*x^2)) ). Compare with the o.g.f. B(x) of A098616, which satisfies B(-x) = 1/x * series reversion( x*(2*x + sqrt(1 - 4*x^2)) ). See also A214377. - Peter Bala, Oct 19 2015
a(n) = GegenbauerC(n,-n,-1). - Peter Luschny, May 07 2016
a(n) = gamma(1+2*n)/gamma(1+n)^2. - Andres Cicuttin, May 30 2016
Sum_{n>=0} (-1)^n/a(n) = 4*(5 - sqrt(5)*log(phi))/25 = 0.6278364236143983844442267..., where phi is the golden ratio. - Ilya Gutkovskiy, Jul 04 2016
From Peter Bala, Jul 22 2016: (Start)
This sequence occurs as the closed-form expression for several binomial sums:
a(n) = Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)*binomial(2*n + 1,k).
a(n) = 2*Sum_{k = 0..2*n-1} (-1)^(n+k)*binomial(2*n - 1,k)*binomial(2*n,k) for n >= 1.
a(n) = 2*Sum_{k = 0..n-1} binomial(n - 1,k)*binomial(n,k) for n >= 1.
a(n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x - k,n)*binomial(y - k,n) for arbitrary x and y.
For m = 3,4,5,... both Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y + k,n) and Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x - k,n)*binomial(y - k,n) appear to equal Kronecker's delta(n,0).
a(n) = (-1)^n*Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y - k,n) for arbitrary x and y.
For m = 3,4,5,... Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y - k,n) appears to equal Kronecker's delta(n,0).
a(n) = Sum_{k = 0..2n} (-1)^k*binomial(2*n,k)*binomial(3*n - k,n)^2 = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)* binomial(n + k,n)^2. (Gould, Vol. 7, 5.23).
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n,n + k)*binomial(n + k,n)^2. (End)
From Ralf Steiner, Apr 07 2017: (Start)
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p-4q)) for q in N, p in Z/{-4q< (some p) <-2}.
...
Sum_{k>=0} a(k)/(-4)^k = 1/sqrt(2).
Sum_{k>=0} a(k)/(17/4)^k = sqrt(17).
Sum_{k>=0} a(k)/(18/4)^k = 3.
Sum_{k>=0} a(k)/5^k = sqrt(5).
Sum_{k>=0} a(k)/6^k = sqrt(3).
Sum_{k>=0} a(k)/8^k = sqrt(2).
...
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p-4q)) for p>4q.(End)
Boas-Buck recurrence: a(n) = (2/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n, 0). See a comment there. - Wolfdieter Lang, Aug 10 2017
a(n) = Sum_{k = 0..n} (-1)^(n-k) * binomial(2*n+1, k) for n in N. - Rene Adad, Sep 30 2017
a(n) = A034870(n,n). - Franck Maminirina Ramaharo, Nov 26 2018
From Jianing Song, Apr 10 2022: (Start)
G.f. for {1/a(n)}: 4*(sqrt(4-x) + sqrt(x)*arcsin(sqrt(x)/2)) / (4-x)^(3/2).
E.g.f. for {1/a(n)}: 1 + exp(x/4)*sqrt(Pi*x)*erf(sqrt(x)/2)/2.
Sum_{n>=0} (-1)^n/a(n) = 4*(1/5 - arcsinh(1/2)/(5*sqrt(5))). (End)
From Peter Luschny, Sep 08 2022: (Start)
a(n) = 2^(2*n)*Product_{k=1..2*n} k^((-1)^(k+1)) = A056040(2*n).
a(n) = A001316(n) * A356637(n) * A261130(n) for n >= 2. (End)
a(n) = 4^n*binomial(n-1/2,-1/2) = 4^n*GegenbauerC(n,1/4,1). - Gerry Martens, Oct 19 2022
Occurs on the right-hand side of the binomial sum identities Sum_{k = -n..n} (-1)^k * (n + x - k) * binomial(2*n, n+k)^2 = (x + n)*a(n) and Sum_{k = -n..n} (-1)^k * (n + x - k)^2 * binomial(2*n, n+k)^3 = x*(x + 2*n)*a(n) (x arbitrary). Compare with the identity: Sum_{k = -n..n} (-1)^k * binomial(2*n, n+k)^2 = a(n). - Peter Bala, Jul 31 2023
From Peter Bala, Mar 31 2024: (Start)
4^n*a(n) = Sum_{k = 0..2*n} (-1)^k*a(k)*a(2*n-k).
16^n = Sum_{k = 0..2*n} a(k)*a(2*n-k). (End)
From Gary Detlefs, May 28 2024: (Start)
a(n) = Sum_{k=0..floor(n/2)} binomial(n,2k)*binomial(2*k,k)*2^(n-2*k). (H. W. Gould) - Gary Detlefs, May 28 2024
a(n) = Sum_{k=0..2*n} (-1)^k*binomial(2n,k)*binomial(2*n+2*k,n+k)*3^(2*n-k). (H. W. Gould) (End)
a(n) = Product_{k>=n+1} k^2/(k^2 - n^2). - Antonio Graciá Llorente, Sep 08 2024
a(n) = Product_{k=1..n} A003418(floor(2*n/k))^((-1)^(k+1)) (Golomb, 2003). - Amiram Eldar, Aug 08 2025

A002144 Pythagorean primes: primes of the form 4*k + 1.

Original entry on oeis.org

5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, 101, 109, 113, 137, 149, 157, 173, 181, 193, 197, 229, 233, 241, 257, 269, 277, 281, 293, 313, 317, 337, 349, 353, 373, 389, 397, 401, 409, 421, 433, 449, 457, 461, 509, 521, 541, 557, 569, 577, 593, 601, 613, 617
Offset: 1

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Author

Keywords

Comments

Rational primes that decompose in the field Q(sqrt(-1)). - N. J. A. Sloane, Dec 25 2017
These are the prime terms of A009003.
-1 is a quadratic residue mod a prime p if and only if p is in this sequence.
Sin(a(n)*Pi/2) = 1 with Pi = 3.1415..., see A070750. - Reinhard Zumkeller, May 04 2002
If at least one of the odd primes p, q belongs to the sequence, then either both or neither of the congruences x^2 = p (mod q), x^2 = q (mod p) are solvable, according to Gauss reciprocity law. - Lekraj Beedassy, Jul 17 2003
Odd primes such that binomial(p-1, (p-1)/2) == 1 (mod p). - Benoit Cloitre, Feb 07 2004
Primes that are the hypotenuse of a right triangle with integer sides. The Pythagorean triple is {A002365(n), A002366(n), a(n)}.
Also, primes of the form a^k + b^k, k > 1. - Amarnath Murthy, Nov 17 2003
The square of a(n) is the average of two other squares. This fact gives rise to a class of monic polynomials x^2 + bx + c with b = a(n) that will factor over the integers regardless of the sign of c. See A114200. - Owen Mertens (owenmertens(AT)missouristate.edu), Nov 16 2005
Also such primes p that the last digit is always 1 for the Nexus numbers of form n^p - (n-1)^p. - Alexander Adamchuk, Aug 10 2006
The set of Pythagorean primes is a proper subset of the set of positive fundamental discriminants (A003658). - Paul Muljadi, Mar 28 2008
A079260(a(n)) = 1; complement of A137409. - Reinhard Zumkeller, Oct 11 2008
From Artur Jasinski, Dec 10 2008: (Start)
If we take 4 numbers: 1, A002314(n), A152676(n), A152680(n) then multiplication table modulo a(n) is isomorphic to the Latin square:
1 2 3 4
2 4 1 3
3 1 4 2
4 3 2 1
and isomorphic to the multiplication table of {1, i, -i, -1} where i is sqrt(-1), A152680(n) is isomorphic to -1, A002314(n) with i or -i and A152676(n) vice versa -i or i. 1, A002314(n), A152676(n), A152680(n) are subfield of Galois field [a(n)]. (End)
Primes p such that the arithmetic mean of divisors of p^3 is an integer. There are 2 sequences of such primes: this one and A002145. - Ctibor O. Zizka, Oct 20 2009
Equivalently, the primes p for which the smallest extension of F_p containing the square roots of unity (necessarily F_p) contains the 4th roots of unity. In this respect, the n = 2 case of a family of sequences: see n=3 (A129805) and n=5 (A172469). - Katherine E. Stange, Feb 03 2010
Subsequence of A007969. - Reinhard Zumkeller, Jun 18 2011
A151763(a(n)) = 1.
k^k - 1 is divisible by 4*k + 1 if 4*k + 1 is a prime (see Dickson reference). - Gary Detlefs, May 22 2013
Not only are the squares of these primes the sum of two nonzero squares, but the primes themselves are also. 2 is the only prime equal to the sum of two nonzero squares and whose square is not. 2 is therefore not a Pythagorean prime. - Jean-Christophe Hervé, Nov 10 2013
The statement that these primes are the sum of two nonzero squares follows from Fermat's theorem on the sum of two squares. - Jerzy R Borysowicz, Jan 02 2019
The decompositions of the prime and its square into two nonzero squares are unique. - Jean-Christophe Hervé, Nov 11 2013. See the Dickson reference, Vol. II, (B) on p. 227. - Wolfdieter Lang, Jan 13 2015
p^e for p prime of the form 4*k+1 and e >= 1 is the sum of 2 nonzero squares. - Jon Perry, Nov 23 2014
Primes p such that the area of the isosceles triangle of sides (p, p, q) for some integer q is an integer. - Michel Lagneau, Dec 31 2014
This is the set of all primes that are the average of two squares. - Richard R. Forberg, Mar 01 2015
Numbers k such that ((k-3)!!)^2 == -1 (mod k). - Thomas Ordowski, Jul 28 2016
This is a subsequence of primes of A004431 and also of A016813. - Bernard Schott, Apr 30 2022
In addition to the comment from Jean-Christophe Hervé, Nov 10 2013: All powers as well as the products of any of these primes are the sum of two nonzero squares. They are terms of A001481, which is closed under multiplication. - Klaus Purath, Nov 19 2023

Examples

			The following table shows the relationship between several closely related sequences:
Here p = A002144 = primes == 1 (mod 4), p = a^2+b^2 with a < b;
a = A002331, b = A002330, t_1 = ab/2 = A070151;
p^2 = c^2 + d^2 with c < d; c = A002366, d = A002365,
t_2 = 2ab = A145046, t_3 = b^2 - a^2 = A070079,
with {c,d} = {t_2, t_3}, t_4 = cd/2 = ab(b^2-a^2).
  ---------------------------------
   p  a  b  t_1  c   d t_2 t_3  t_4
  ---------------------------------
   5  1  2   1   3   4   4   3    6
  13  2  3   3   5  12  12   5   30
  17  1  4   2   8  15   8  15   60
  29  2  5   5  20  21  20  21  210
  37  1  6   3  12  35  12  35  210
  41  4  5  10   9  40  40   9  180
  53  2  7   7  28  45  28  45  630
  ...
a(7) = 53 = A002972(7)^2 + (2*A002973(7))^2 = 7^2 + (2*1)^2 = 49 + 4, and this is the only way. - _Wolfdieter Lang_, Jan 13 2015
		

References

  • David A. Cox, "Primes of the Form x^2 + n y^2", Wiley, 1989.
  • L. E. Dickson, "History of the Theory of Numbers", Chelsea Publishing Company, 1919, Vol I, page 386
  • L. E. Dickson, History of the Theory of Numbers, Carnegie Institution, Publ. No. 256, Vol. II, Washington D.C., 1920, p. 227.
  • G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, Cambridge, University Press, 1940, p. 132.
  • M. du Sautoy, The Music of the Primes, Fourth Estate / HarperCollins, 2003; see p. 76.
  • 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, pages 241, 243.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 90.

Crossrefs

Cf. A004613 (multiplicative closure).
Apart from initial term, same as A002313.
For values of n see A005098.
Primes in A020668.

Programs

  • Haskell
    a002144 n = a002144_list !! (n-1)
    a002144_list = filter ((== 1) . a010051) [1,5..]
    -- Reinhard Zumkeller, Mar 06 2012, Feb 22 2011
    
  • Magma
    [a: n in [0..200] | IsPrime(a) where a is 4*n + 1 ]; // Vincenzo Librandi, Nov 23 2014
    
  • Maple
    a := []; for n from 1 to 500 do if isprime(4*n+1) then a := [op(a),4*n+1]; fi; od: A002144 := n->a[n];
    # alternative
    A002144 := proc(n)
        option remember ;
        local a;
        if n = 1 then
            5;
        else
            for a from procname(n-1)+4 by 4 do
                if isprime(a) then
                    return a;
                end if;
            end do:
        end if;
    end proc:
    seq(A002144(n),n=1..100) ; # R. J. Mathar, Jan 31 2024
  • Mathematica
    Select[4*Range[140] + 1, PrimeQ[ # ] &] (* Stefan Steinerberger, Apr 16 2006 *)
    Select[Prime[Range[150]],Mod[#,4]==1&] (* Harvey P. Dale, Jan 28 2021 *)
  • PARI
    select(p->p%4==1,primes(1000))
    
  • PARI
    A002144_next(p=A2144[#A2144])={until(isprime(p+=4),);p} /* NB: p must be of the form 4k+1. Beyond primelimit, this is *much* faster than forprime(p=...,, p%4==1 && return(p)). */
    A2144=List(5); A002144(n)={while(#A2144A002144_next())); A2144[n]}
    \\ M. F. Hasler, Jul 06 2024
    
  • Python
    from sympy import prime
    A002144 = [n for n in (prime(x) for x in range(1,10**3)) if not (n-1) % 4]
    # Chai Wah Wu, Sep 01 2014
    
  • Python
    from sympy import isprime
    print(list(filter(isprime, range(1, 618, 4)))) # Michael S. Branicky, May 13 2021
    
  • SageMath
    def A002144_list(n): # returns all Pythagorean primes <= n
        return [x for x in prime_range(5,n+1) if x % 4 == 1]
    A002144_list(617) # Peter Luschny, Sep 12 2012

Formula

Odd primes of form x^2 + y^2, (x=A002331, y=A002330, with x < y) or of form u^2 + 4*v^2, (u = A002972, v = A002973, with u odd). - Lekraj Beedassy, Jul 16 2004
p^2 - 1 = 12*Sum_{i = 0..floor(p/4)} floor(sqrt(i*p)) where p = a(n) = 4*n + 1. [Shirali]
a(n) = A000290(A002972(n)) + A000290(2*A002973(n)) = A000290(A002331(n+1)) + A000290(A002330(n+1)). - Reinhard Zumkeller, Feb 16 2010
a(n) = A002972(n)^2 + (2*A002973(n))^2, n >= 1. See the Jean-Christophe Hervé Nov 11 2013 comment. - Wolfdieter Lang, Jan 13 2015
a(n) = 4*A005098(n) + 1. - Zak Seidov, Sep 16 2018
From Vaclav Kotesovec, Apr 30 2020: (Start)
Product_{k>=1} (1 - 1/a(k)^2) = A088539.
Product_{k>=1} (1 + 1/a(k)^2) = A243380.
Product_{k>=1} (1 - 1/a(k)^3) = A334425.
Product_{k>=1} (1 + 1/a(k)^3) = A334424.
Product_{k>=1} (1 - 1/a(k)^4) = A334446.
Product_{k>=1} (1 + 1/a(k)^4) = A334445.
Product_{k>=1} (1 - 1/a(k)^5) = A334450.
Product_{k>=1} (1 + 1/a(k)^5) = A334449. (End)
From Vaclav Kotesovec, May 05 2020: (Start)
Product_{k>=1} (1 + 1/A002145(k)) / (1 + 1/a(k)) = Pi/(4*A064533^2) = 1.3447728438248695625516649942427635670667319092323632111110962...
Product_{k>=1} (1 - 1/A002145(k)) / (1 - 1/a(k)) = Pi/(8*A064533^2) = 0.6723864219124347812758324971213817835333659546161816055555481... (End)
Sum_{k >= 1} 1/a(k)^s = (1/2) * Sum_{n >= 1 odd numbers} moebius(n) * log((2*n*s)! * zeta(n*s) * abs(EulerE(n*s - 1)) / (Pi^(n*s) * 2^(2*n*s) * BernoulliB(2*n*s) * (2^(n*s) + 1) * (n*s - 1)!))/n, s >= 3 odd number. - Dimitris Valianatos, May 21 2020
Legendre symbol (-1, a(n)) = +1, for n >= 1. - Wolfdieter Lang, Mar 03 2021

A065091 Odd primes.

Original entry on oeis.org

3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277
Offset: 1

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Author

Labos Elemer, Nov 12 2001

Keywords

Comments

Rayes et al. prove that the a(n)-th Chebyshev-T polynomial, divided by x, is irreducible over the integers.
Odd primes can be written as a sum of no more than two consecutive positive integers. Powers of 2 do not have a representation as a sum of k consecutive positive integers (other than the trivial n=n, for k=1). See A111774. - Jaap Spies, Jan 04 2007
Intersection of A005408 and A000040. - Reinhard Zumkeller, Oct 14 2008
Primes which are the sum of two consecutive numbers. - Juri-Stepan Gerasimov, Nov 07 2009
The arithmetic mean of divisors of p^3, (1+p)(1+p^2)/4, for odd primes p is an integer. - Ctibor O. Zizka, Oct 20 2009
Primes == -+ 1 (mod 4). - Juri-Stepan Gerasimov, Apr 27 2010
a(n) = A053670(A179675(n)) and a(n) <> A053670(m) for m < A179675(n). - Reinhard Zumkeller, Jul 23 2010
Triads of the form <2*a(n+1), a(n+1), 3*a(n+1)> like <6,3,9>, <10,5,15>, <14,7,21> appear in the EKG sequence A064413, see Theorem (3) there. - Paul Curtz, Feb 13 2011.
Complement of A065090; abs(A151763(a(n))) = 1. - Reinhard Zumkeller, Oct 06 2011
Right edge of the triangle in A065305. - Reinhard Zumkeller, Jan 30 2012
Numbers with two odd divisors. - Omar E. Pol, Mar 24 2012
Odd prime p divides some (2^k + 1) or (2^k - 1), (k>0, minimal, cf. A003558) depending on the parity of A179480((p+1)/2) = r. This is a consequence of the Quasi-order theorem and corollaries, [Hilton and Pederson, pp. 260-264]: 2^k == (-1)^r mod b, b odd; and b divides 2^k - (-1)^r, where p is a subset of b. - Gary W. Adamson, Aug 26 2012
Subset of the arithmetic numbers (A003601). - Wesley Ivan Hurt, Sep 27 2013
Odd primes p satisfy the identity: p = (product(2*cos((2*k+1)*Pi/(2*p)), k=0..(p-3)/2))^2. This follows from C(2*p, 0) = (-1)^((p-1)/2)*p, n>=2, with the minimal polynomial C(k,x) of rho(k) := 2*cos(Pi/k). See A187360 for C and the W. Lang link on the field Q(rho(n)), eqs. (20) and (37). - Wolfdieter Lang, Oct 23 2013
Numbers m > 1 such that m^2 divides (2m-1)!! + m. - Thomas Ordowski, Nov 28 2014
Numbers m such that m divides 2*(m-3)! + 1. - Thomas Ordowski, Jun 20 2015
Numbers m such that (2m-3)!! == m (mod m^2). - Thomas Ordowski, Jul 24 2016
Odd numbers m such that ((m-3)!!)^2 == +-1 (mod m). - Thomas Ordowski, Jul 27 2016
Primes of the form x^2 - y^2. - Thomas Ordowski, Feb 27 2017
Conjecture: a(n) is the smallest odd number m > prime(n) such that Sum_{k=1..prime(n)-1} k^(m-1) == prime(n)-1 (mod m). This is an extension of the Agoh-Giuga conjecture. - Thomas Ordowski, Aug 01 2018
Numbers k > 1 such that either Phi(k,x) == 1 (mod k) or Phi(k,x) == k (mod k^2) holds, where Phi(k,x) is the k-th cyclotomic polynomial. - Jianing Song, Aug 02 2018

References

  • Paulo Ribenboim, The little book of big primes, Springer 1991, p. 106.

Crossrefs

Cf. A000040, A033270, union of A002144 and A002145.
Cf. A230953 (boustrophedon transform).

Programs

  • Haskell
    a065091 n = a065091_list !! (n-1)
    a065091_list = tail a000040_list  -- Reinhard Zumkeller, Jan 30 2012
    
  • Magma
    [NthPrime(n): n in [2..100]]; // Vincenzo Librandi, Jun 21 2015
    
  • Maple
    A065091 := proc(n) RETURN(ithprime(n+1)) end:
  • Mathematica
    Prime[Range[2, 33]] (* Vladimir Joseph Stephan Orlovsky, Aug 22 2008 *)
  • PARI
    forprime(p=3, 200, print1(p, ", ")) \\ Felix Fröhlich, Jun 30 2014
    
  • Python
    from sympy import prime
    def A065091(n): return prime(n+1) # Chai Wah Wu, Jul 13 2024
  • Sage
    def A065091_list(limit):  # after Minác's formula
        f = 3; P = [f]
        for n in range(3, limit, 2):
            if (f+1)>n*(f//n)+1: P.append(n)
            f = f*n
        return P
    A065091_list(100)  # Peter Luschny, Oct 17 2013
    

Formula

a(n) = A000040(n+1). - M. F. Hasler, Oct 26 2013

Extensions

More terms from Francisco Salinas (franciscodesalinas(AT)hotmail.com), Jan 05 2002
Edited (moved contributions from A000040 to here) by M. F. Hasler, Oct 26 2013

A013663 Decimal expansion of zeta(5).

Original entry on oeis.org

1, 0, 3, 6, 9, 2, 7, 7, 5, 5, 1, 4, 3, 3, 6, 9, 9, 2, 6, 3, 3, 1, 3, 6, 5, 4, 8, 6, 4, 5, 7, 0, 3, 4, 1, 6, 8, 0, 5, 7, 0, 8, 0, 9, 1, 9, 5, 0, 1, 9, 1, 2, 8, 1, 1, 9, 7, 4, 1, 9, 2, 6, 7, 7, 9, 0, 3, 8, 0, 3, 5, 8, 9, 7, 8, 6, 2, 8, 1, 4, 8, 4, 5, 6, 0, 0, 4, 3, 1, 0, 6, 5, 5, 7, 1, 3, 3, 3, 3
Offset: 1

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Author

Keywords

Comments

In a widely distributed May 2011 email, Wadim Zudilin gave a rebuttal to v1 of Kim's 2011 preprint: "The mistake (unfixable) is on p. 6, line after eq. (3.3). 'Without loss of generality' can be shown to work only for a finite set of n_k's; as the n_k are sufficiently large (and N is fixed), the inequality for epsilon is false." In a May 2013 email, Zudilin extended his rebuttal to cover v2, concluding that Kim's argument "implies that at least one of zeta(2), zeta(3), zeta(4) and zeta(5) is irrational, which is trivial." - Jonathan Sondow, May 06 2013
General: zeta(2*s + 1) = (A000364(s)/A331839(s)) * Pi^(2*s + 1) * Product_{k >= 1} (A002145(k)^(2*s + 1) + 1)/(A002145(k)^(2*s + 1) - 1), for s >= 1. - Dimitris Valianatos, Apr 27 2020

Examples

			1/1^5 + 1/2^5 + 1/3^5 + 1/4^5 + 1/5^5 + 1/6^5 + 1/7^5 + ... =
1 + 1/32 + 1/243 + 1/1024 + 1/3125 + 1/7776 + 1/16807 + ... = 1.036927755143369926331365486457...
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 262.
  • Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 811.

Crossrefs

Programs

Formula

From Peter Bala, Dec 04 2013: (Start)
Definition: zeta(5) = Sum_{n >= 1} 1/n^5.
zeta(5) = 2^5/(2^5 - 1)*(Sum_{n even} n^5*p(n)*p(1/n)/(n^2 - 1)^6 ), where p(n) = n^2 + 3. See A013667, A013671 and A013675. (End)
zeta(5) = Sum_{n >= 1} (A010052(n)/n^(5/2)) = Sum_{n >= 1} ((floor(sqrt(n)) - floor(sqrt(n-1)))/n^(5/2)). - Mikael Aaltonen, Feb 22 2015
zeta(5) = Product_{k>=1} 1/(1 - 1/prime(k)^5). - Vaclav Kotesovec, Apr 30 2020
From Artur Jasinski, Jun 27 2020: (Start)
zeta(5) = (-1/30)*Integral_{x=0..1} log(1-x^4)^5/x^5.
zeta(5) = (1/24)*Integral_{x=0..infinity} x^4/(exp(x)-1).
zeta(5) = (2/45)*Integral_{x=0..infinity} x^4/(exp(x)+1).
zeta(5) = (1/(1488*zeta(1/2)^5))*(-5*Pi^5*zeta(1/2)^5 + 96*zeta'(1/2)^5 - 240*zeta(1/2)*zeta'(1/2)^3*zeta''(1/2) + 120*zeta(1/2)^2*zeta'(1/2)*zeta''(1/2)^2 + 80*zeta(1/2)^2*zeta'(1/2)^2*zeta'''(1/2)- 40*zeta(1/2)^3*zeta''(1/2)*zeta'''(1/2) - 20*zeta(1/2)^3*zeta'(1/2)*zeta''''(1/2)+4*zeta(1/2)^4*zeta'''''(1/2)). (End).
From Peter Bala, Oct 29 2023: (Start)
zeta(3) = (8/45)*Integral_{x >= 1} x^3*log(x)^3*(1 + log(x))*log(1 + 1/x^x) dx = (2/45)*Integral_{x >= 1} x^4*log(x)^4*(1 + log(x))/(1 + x^x) dx.
zeta(5) = 131/128 + 26*Sum_{n >= 1} (n^2 + 2*n + 40/39)/(n*(n + 1)*(n + 2))^5.
zeta(5) = 5162893/4976640 - 1323520*Sum_{n >= 1} (n^2 + 4*n + 56288/12925)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4))^5. Taking 10 terms of the series gives a value for zeta(5) correct to 20 decimal places.
Conjecture: for k >= 1, there exist rational numbers A(k), B(k) and c(k) such that zeta(5) = A(k) + B(k)*Sum_{n >= 1} (n^2 + 2*k*n + c(k))/(n*(n + 1)*...*(n + 2*k))^5. A similar conjecture can be made for the constant zeta(3). (End)
zeta(5) = (694/204813)*Pi^5 - Sum_{n >= 1} (6280/3251)*(1/(n^5*(exp(4*Pi*n)-1))) + Sum_{n >= 1} (296/3251)*(1/(n^5*(exp(5*Pi*n)-1))) - Sum_{n >= 1} (1073/6502)*(1/(n^5*(exp(10*Pi*n)-1))) + Sum_{n >= 1} (37/6502)*(1/(n^5*(exp(20*Pi*n)-1))). - Simon Plouffe, Jan 06 2024
From Peter Bala, Apr 27 2025: (Start)
zeta(5) = 1/5! * Integral_{x >= 0} x^5 * exp(x)/(exp(x) - 1)^2 dx = (16/15) * 1/5! * Integral_{x >= 0} x^5 * exp(x)/(exp(x) + 1)^2 dx.
zeta(5) = 1/6! * Integral_{x >= 0} x^6 * exp(x)*(exp(x) + 1)/(exp(x) - 1)^3 dx = 1/(3^3 * 5^2) * Integral_{x >= 0} x^6 * exp(x)*(exp(x) - 1)/(exp(x) + 1)^3 dx. (End)
zeta(5) = Sum_{i, j >= 1} 1/((i^4)*j*binomial(i+j, i)). More generally, zeta(n+1) = Sum_{i, j >= 1} 1/((i^n)*j*binomial(i+j, i)) for n >= 1. - Peter Bala, Aug 07 2025

A007528 Primes of the form 6k-1.

Original entry on oeis.org

5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89, 101, 107, 113, 131, 137, 149, 167, 173, 179, 191, 197, 227, 233, 239, 251, 257, 263, 269, 281, 293, 311, 317, 347, 353, 359, 383, 389, 401, 419, 431, 443, 449, 461, 467, 479, 491, 503, 509, 521, 557, 563, 569, 587
Offset: 1

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Author

Keywords

Comments

For values of k see A024898.
Also primes p such that p^q - 2 is not prime where q is an odd prime. These numbers cannot be prime because the binomial p^q = (6k-1)^q expands to 6h-1 some h. Then p^q-2 = 6h-1-2 is divisible by 3 thus not prime. - Cino Hilliard, Nov 12 2008
a(n) = A211890(3,n-1) for n <= 4. - Reinhard Zumkeller, Jul 13 2012
There exists a polygonal number P_s(3) = 3s - 3 = a(n) + 1. These are the only primes p with P_s(k) = p + 1, s >= 3, k >= 3, since P_s(k) - 1 is composite for k > 3. - Ralf Steiner, May 17 2018
From Bernard Schott, Feb 14 2019: (Start)
A theorem due to Andrzej Mąkowski: every integer greater than 161 is the sum of distinct primes of the form 6k-1. Examples: 162 = 5 + 11 + 17 + 23 + 47 + 59; 163 = 17 + 23 + 29 + 41 + 53. (See Sierpiński and David Wells.)
{2,3} Union A002476 Union {this sequence} = A000040.
Except for 2 and 3, all Sophie Germain primes are of the form 6k-1.
Except for 3, all the lesser of twin primes are also of the form 6k-1.
Dirichlet's theorem on arithmetic progressions states that this sequence is infinite. (End)
For all elements of this sequence p=6*k-1, there are no (x,y) positive integers such that k=6*x*y-x+y. - Pedro Caceres, Apr 06 2019

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 870.
  • A. Mąkowski, Partitions into unequal primes, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astr. Phys. 8 (1960), 125-126.
  • Wacław Sierpiński, Elementary Theory of Numbers, p. 144, Warsaw, 1964.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, Revised edition, 1997, p. 127.

Crossrefs

Intersection of A016969 and A000040.
Prime sequences A# (k,r) of the form k*n+r with 0 <= r <= k-1 (i.e., primes == r (mod k), or primes p with p mod k = r) and gcd(r,k)=1: A000040 (1,0), A065091 (2,1), A002476 (3,1), A003627 (3,2), A002144 (4,1), A002145 (4,3), A030430 (5,1), A045380 (5,2), A030431 (5,3), A030433 (5,4), A002476 (6,1), this sequence (6,5), A140444 (7,1), A045392 (7,2), A045437 (7,3), A045471 (7,4), A045458 (7,5), A045473 (7,6), A007519 (8,1), A007520 (8,3), A007521 (8,5), A007522 (8,7), A061237 (9,1), A061238 (9,2), A061239 (9,4), A061240 (9,5), A061241 (9,7), A061242 (9,8), A030430 (10,1), A030431 (10,3), A030432 (10,7), A030433 (10,9), A141849 (11,1), A090187 (11,2), A141850 (11,3), A141851 (11,4), A141852 (11,5), A141853 (11,6), A141854 (11,7), A141855 (11,8), A141856 (11,9), A141857 (11,10), A068228 (12,1), A040117 (12,5), A068229 (12,7), A068231 (12,11).
Cf. A034694 (smallest prime == 1 (mod n)).
Cf. A038700 (smallest prime == n-1 (mod n)).
Cf. A038026 (largest possible value of smallest prime == r (mod n)).
Cf. A001359 (lesser of twin primes), A005384 (Sophie Germain primes).

Programs

  • GAP
    Filtered(List([1..100],n->6*n-1),IsPrime); # Muniru A Asiru, May 19 2018
  • Haskell
    a007528 n = a007528_list !! (n-1)
    a007528_list = [x | k <- [0..], let x = 6 * k + 5, a010051' x == 1]
    -- Reinhard Zumkeller, Jul 13 2012
    
  • Maple
    select(isprime,[seq(6*n-1,n=1..100)]); # Muniru A Asiru, May 19 2018
  • Mathematica
    Select[6 Range[100]-1,PrimeQ]  (* Harvey P. Dale, Feb 14 2011 *)
  • PARI
    forprime(p=2, 1e3, if(p%6==5, print1(p, ", "))) \\ Charles R Greathouse IV, Jul 15 2011
    
  • PARI
    forprimestep(p=5,1000,6, print1(p", ")) \\ Charles R Greathouse IV, Mar 03 2025
    

Formula

A003627 \ {2}. - R. J. Mathar, Oct 28 2008
Conjecture: Product_{n >= 1} ((a(n) - 1) / (a(n) + 1)) * ((A002476(n) + 1) / (A002476(n) - 1)) = 3/4. - Dimitris Valianatos, Feb 11 2020
From Vaclav Kotesovec, May 02 2020: (Start)
Product_{k>=1} (1 - 1/a(k)^2) = 9*A175646/Pi^2 = 1/1.060548293.... =4/(3*A333240).
Product_{k>=1} (1 + 1/a(k)^2) = A334482.
Product_{k>=1} (1 - 1/a(k)^3) = A334480.
Product_{k>=1} (1 + 1/a(k)^3) = A334479. (End)
Legendre symbol (-3, a(n)) = -1 and (-3, A002476(n)) = +1, for n >= 1. For prime 3 one sets (-3, 3) = 0. - Wolfdieter Lang, Mar 03 2021

A001348 Mersenne numbers: 2^p - 1, where p is prime.

Original entry on oeis.org

3, 7, 31, 127, 2047, 8191, 131071, 524287, 8388607, 536870911, 2147483647, 137438953471, 2199023255551, 8796093022207, 140737488355327, 9007199254740991, 576460752303423487, 2305843009213693951, 147573952589676412927, 2361183241434822606847
Offset: 1

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Author

Keywords

Comments

Mersenne numbers A000225 whose indices are primes. - Omar E. Pol, Aug 31 2008
All terms are of the form 4k-1. - Paul Muljadi, Jan 31 2011
Smallest number with Hamming weight A000120 = prime(n). - M. F. Hasler, Oct 16 2018
The 5th, 9th, 10th, ... terms are not prime. See A000668 and A065341 for the primes and for the composites in this sequence. - M. F. Hasler, Nov 14 2018 [corrected by Jerzy R Borysowicz, Apr 08 2025]
Except for the first term 3: all prime factors of 2^p-1 must be 1 or -1 (mod 8), and 1 (mod 2p). - William Hu, Mar 10 2024

References

  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 16.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 47.
  • 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).

Crossrefs

Programs

  • Magma
    [2^NthPrime(n)-1: n in [1..30]]; // Vincenzo Librandi, Feb 04 2016
    
  • Maple
    A001348 := n -> 2^(ithprime(n))-1: seq (A001348(n), n=1..18);
  • Mathematica
    Table[2^Prime[n]-1, {n, 20}] (* Vladimir Joseph Stephan Orlovsky, Aug 26 2008 *)
  • PARI
    a(n)=1<Charles R Greathouse IV, Jun 10 2011
    
  • Python
    from sympy import prime
    def a(n): return 2**prime(n)-1
    print([a(n) for n in range(1, 21)]) # Michael S. Branicky, Mar 28 2022

Formula

a(n) = 2^A000040(n) - 1, n >= 1. - Wolfdieter Lang, Oct 26 2014
a(n) = A000225(A000040(n)). - Omar E. Pol, Aug 31 2008
A000668(n) = a(A016027(n)). - Omar E. Pol, Jun 29 2012
Sum_{n>=1} 1/a(n) = A262153. - Amiram Eldar, Nov 20 2020
Product_{n>=1} (1 - 1/a(n)) = A184085. - Amiram Eldar, Nov 22 2022
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