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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A131307 (A127701 * A000012 + A000012(signed) * A127701) - A000012.

Original entry on oeis.org

1, 2, 3, 3, 2, 5, 4, 5, 2, 7, 5, 4, 7, 2, 9, 6, 7, 4, 9, 2, 11, 7, 6, 9, 4, 11, 2, 13, 8, 9, 6, 11, 4, 13, 2, 15, 9, 8, 11, 6, 13, 4, 15, 2, 17, 10, 11, 8, 13, 6, 15, 4, 17, 2, 19
Offset: 1

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Author

Gary W. Adamson, Sep 30 2007

Keywords

Comments

Row sums = A035608: (1, 5, 10, 18, 27, 39, 52, ...).

Examples

			First few rows of the triangle:
   1;
   2,  3;
   3,  2,  5;
   4,  5,  2,  7;
   5,  4,  7,  2,  9;
   6,  7,  4,  9,  2, 11;
   7,  6,  9,  4, 11,  2, 13;
   8,  9,  6, 11,  4, 13,  2, 15;
   9,  8, 11,  6, 13,  4, 15,  2, 17;
  10, 11,  8, 13,  6, 15,  4, 17,  2, 19;
  ...
		

Crossrefs

Formula

(A127701 * A000012 + A000012(signed) * A127701) - A000012; as infinite lower triangular matrices, where A000012(signed) = (1; -1,1; 1,-1,1; -1,1,-1,1; ...).

A008277 Triangle of Stirling numbers of the second kind, S2(n,k), n >= 1, 1 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 6, 1, 1, 15, 25, 10, 1, 1, 31, 90, 65, 15, 1, 1, 63, 301, 350, 140, 21, 1, 1, 127, 966, 1701, 1050, 266, 28, 1, 1, 255, 3025, 7770, 6951, 2646, 462, 36, 1, 1, 511, 9330, 34105, 42525, 22827, 5880, 750, 45, 1, 1, 1023, 28501, 145750, 246730, 179487, 63987, 11880, 1155, 55, 1
Offset: 1

Views

Author

Keywords

Comments

Also known as Stirling set numbers and written {n, k}.
S2(n,k) counts partitions of an n-set into k nonempty subsets.
From Manfred Boergens, Apr 07 2025: (Start)
With regard to the preceding comment:
For disjoint collections of subsets see A256894.
For arbitrary collections of subsets see A163353.
For arbitrary collections of nonempty subsets see A055154. (End)
Triangle S2(n,k), 1 <= k <= n, read by rows, given by [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is Deléham's operator defined in A084938.
Number of partitions of {1, ..., n+1} into k+1 nonempty subsets of nonconsecutive integers, including the partition 1|2|...|n+1 if n=k. E.g., S2(3,2)=3 since the number of partitions of {1,2,3,4} into three subsets of nonconsecutive integers is 3, i.e., 13|2|4, 14|2|3, 1|24|3. - Augustine O. Munagi, Mar 20 2005
Draw n cards (with replacement) from a deck of k cards. Let prob(n,k) be the probability that each card was drawn at least once. Then prob(n,k) = S2(n,k)*k!/k^n (see A090582). - Rainer Rosenthal, Oct 22 2005
Define f_1(x), f_2(x), ..., such that f_1(x)=e^x and for n = 2, 3, ..., f_{n+1}(x) = (d/dx)(x*f_n(x)). Then f_n(x) = e^x*Sum_{k=1..n} S2(n,k)*x^(k-1). - Milan Janjic, May 30 2008
From Peter Bala, Oct 03 2008: (Start)
For tables of restricted Stirling numbers of the second kind see A143494 - A143496.
S2(n,k) gives the number of 'patterns' of words of length n using k distinct symbols - see [Cooper & Kennedy] for an exact definition of the term 'pattern'. As an example, the words AADCBB and XXEGTT, both of length 6, have the same pattern of letters. The five patterns of words of length 3 are AAA, AAB, ABA, BAA and ABC giving row 3 of this table as (1,3,1).
Equivalently, S2(n,k) gives the number of sequences of positive integers (N_1,...,N_n) of length n, with k distinct entries, such that N_1 = 1 and N_(i+1) <= 1 + max{j = 1..i} N_j for i >= 1 (restricted growth functions). For example, Stirling(4,2) = 7 since the sequences of length 4 having 2 distinct entries that satisfy the conditions are (1,1,1,2), (1,1,2,1), (1,2,1,1), (1,1,2,2), (1,2,2,2), (1,2,2,1) and (1,2,1,2).
(End)
Number of combinations of subsets in the plane. - Mats Granvik, Jan 13 2009
S2(n+1,k+1) is the number of size k collections of pairwise disjoint, nonempty subsets of [n]. For example: S2(4,3)=6 because there are six such collections of subsets of [3] that have cardinality two: {(1)(23)},{(12)(3)}, {(13)(2)}, {(1)(2)}, {(1)(3)}, {(2)(3)}. - Geoffrey Critzer, Apr 06 2009
Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1, k+1) equals the number of ways to choose 0 or more balls of each color in such a way that exactly (n-k) colors are chosen at least once, and no two colors are chosen the same positive number of times. - Matthew Vandermast, Nov 22 2010
S2(n,k) is the number of monotonic-labeled forests on n vertices with exactly k rooted trees, each of height one or less. See link "Counting forests with Stirling and Bell numbers" below. - Dennis P. Walsh, Nov 16 2011
If D is the operator d/dx, and E the operator xd/dx, Stirling numbers are given by: E^n = Sum_{k=1..n} S2(n,k) * x^k*D^k. - Hyunwoo Jang, Dec 13 2011
The Stirling polynomials of the second kind (a.k.a. the Bell / Touchard polynomials) are the umbral compositional inverses of the falling factorials (a.k.a. the Pochhammer symbol or Stirling polynomials of the first kind), i.e., binomial(Bell(.,x),n) = x^n/n! (cf. Copeland's 2007 formulas), implying binomial(xD,n) = binomial(Bell(.,:xD:),n) = :xD:^n/n! where D = d/dx and :xD:^n = x^n*D^n. - Tom Copeland, Apr 17 2014
S2(n,k) is the number of ways to nest n matryoshkas (Russian nesting dolls) so that exactly k matryoshkas are not contained in any other matryoshka. - Carlo Sanna, Oct 17 2015
The row polynomials R(n, x) = Sum_{k=1..n} S2(n, k)*x^k appear in the numerator of the e.g.f. of n-th powers, E(n, x) = Sum_{m>=0} m^n*x^m/m!, as E(n, x) = exp(x)*x*R(n, x), for n >= 1. - Wolfdieter Lang, Apr 02 2017
With offsets 0 for n and k this is the Sheffer product matrix A007318*A048993 denoted by (exp(t), (exp(t) - 1)) with e.g.f. exp(t)*exp(x*(exp(t) - 1)). - Wolfdieter Lang, Jun 20 2017
Number of words on k+1 unlabeled letters of length n+1 with no repeated letters. - Thomas Anton, Mar 14 2019
Also coefficients of moments of Poisson distribution about the origin expressed as polynomials in lambda. [Haight] (see also A331155). - N. J. A. Sloane, Jan 14 2020
k!*S2(n,k) is the number of surjections from an n-element set to a k-element set. - Jianing Song, Jun 01 2022

Examples

			The triangle S2(n, k) begins:
\ k    1       2       3        4         5         6         7         8        9
n \   10      11      12       13        14        15       ...
----------------------------------------------------------------------------------
1  |   1
2  |   1       1
3  |   1       3       1
4  |   1       7       6        1
5  |   1      15      25       10         1
6  |   1      31      90       65        15         1
7  |   1      63     301      350       140        21         1
8  |   1     127     966     1701      1050       266        28         1
9  |   1     255    3025     7770      6951      2646       462        36        1
10 |   1     511    9330    34105     42525     22827      5880       750       45
       1
11 |   1    1023   28501   145750    246730    179487     63987     11880     1155
      55       1
12 |   1    2047   86526   611501   1379400   1323652    627396    159027    22275
    1705      66       1
13 |   1    4095  261625  2532530   7508501   9321312   5715424   1899612   359502
   39325    2431      78        1
14 |   1    8191  788970 10391745  40075035  63436373  49329280  20912320  5135130
  752752   66066    3367       91         1
15 |   1   16383 2375101 42355950 210766920 420693273 408741333 216627840 67128490
12662650 1479478  106470     4550       105         1
...
----------------------------------------------------------------------------------
x^4 = 1 x_(1) + 7 x_(2) + 6 x_(3) + 1 x_(4), where x_(k) = P(x,k) = k!*C(x,k). - _Daniel Forgues_, Jan 16 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. 835.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 103ff.
  • B. A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • G. Boole, Finite Differences, 5th ed. New York, NY: Chelsea, 1970.
  • C. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, 2002, Theorem 8.11, pp. 298-299.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 310.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Springer, p. 92.
  • F. N. David, M. G. Kendall, and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
  • S.N. Elaydi, An Introduction to Difference Equations, 3rd ed. Springer, 2005.
  • H. H. Goldstine, A History of Numerical Analysis, Springer-Verlag, 1977; Section 2.7.
  • R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 244.
  • Frank Avery Haight, Handbook of the Poisson distribution, John Wiley, 1967. See pages 6,7.
  • A. D. Korshunov, Asymptotic behavior of Stirling numbers of the second kind. (Russian) Metody Diskret. Analiz. No. 39 (1983), 24-41.
  • E. Kuz'min and A. I. Shirshov: On the number e, pp. 111-119, eq.(6), in: Kvant Selecta: Algebra and Analysis, I, ed. S. Tabachnikov, Am.Math.Soc., 1999, p. 116, eq. (11).
  • J. Riordan, An Introduction to Combinatorial Analysis, p. 48.
  • J. Stirling, The Differential Method, London, 1749; see p. 7.

Crossrefs

Cf. A008275 (Stirling numbers of first kind), A048993 (another version of this triangle).
See also A331155.
Cf. A000110 (row sums), A102661 (partial row sums).

Programs

  • Haskell
    a008277 n k = a008277_tabl !! (n-1) !! (k-1)
    a008277_row n = a008277_tabl !! (n-1)
    a008277_tabl = map tail $ a048993_tabl  -- Reinhard Zumkeller, Mar 26 2012
    
  • J
    n ((] (1 % !)) * +/@((^~ * (] (1 ^ |.)) * (! {:)@]) i.@>:)) k NB. _Stephen Makdisi, Apr 06 2016
    
  • Magma
    [[StirlingSecond(n,k): k in [1..n]]: n in [1..12]]; // G. C. Greubel, May 22 2019
  • Maple
    seq(seq(combinat[stirling2](n, k), k=1..n), n=1..10); # Zerinvary Lajos, Jun 02 2007
    stirling_2 := (n,k) -> (1/k!) * add((-1)^(k-i)*binomial(k,i)*i^n, i=0..k);
  • Mathematica
    Table[StirlingS2[n, k], {n, 11}, {k, n}] // Flatten (* Robert G. Wilson v, May 23 2006 *)
    BellMatrix[f_, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
    rows = 12;
    B = BellMatrix[1&, rows];
    Table[B[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 28 2018, after Peter Luschny *)
    a[n_, n_] := 1; a[n_, 1] := 1;
    a[n_, k_] := a[n, k] = a[n-1, k-1] + k a[n-1, k]; Flatten@
    Table[a[n, k], {n, 1, 11}, {k, 1, n}] (* Oliver Seipel, Jun 12 2024 *)
    With[{m = 11},
     Flatten@MapIndexed[Take[#, #2[[1]]] &,
       Transpose@
        Table[Range[1, m]! Coefficient[(E^x-1)^k/k! + O[x]^(m+1), x,
    Range[1, m]], {k, 1, m}]]] (* Oliver Seipel, Jun 12 2024 *)
  • Maxima
    create_list(stirling2(n+1,k+1),n,0,30,k,0,n); /* Emanuele Munarini, Jun 01 2012 */
    
  • PARI
    for(n=1,22,for(k=1,n,print1(stirling(n,k,2),", "));print()); \\ Joerg Arndt, Apr 21 2013
    
  • PARI
    Stirling2(n,k)=sum(i=0,k,(-1)^i*binomial(k,i)*i^n)*(-1)^k/k!  \\ M. F. Hasler, Mar 06 2012
    
  • Sage
    stirling_number2 # Danny Rorabaugh, Oct 11 2015
    

Formula

S2(n, k) = k*S2(n-1, k) + S2(n-1, k-1), n > 1. S2(1, k) = 0, k > 1. S2(1, 1) = 1.
E.g.f.: A(x, y) = e^(y*e^x-y). E.g.f. for m-th column: (e^x-1)^m/m!.
S2(n, k) = (1/k!) * Sum_{i=0..k} (-1)^(k-i)*binomial(k, i)*i^n.
Row sums: Bell number A000110(n) = Sum_{k=1..n} S2(n, k), n>0.
S(n, k) = Sum (i_1*i_2*...*i_(n-k)) summed over all (n-k)-combinations {i_1, i_2, ..., i_k} with repetitions of the numbers {1, 2, ..., k}. Also S(n, k) = Sum (1^(r_1)*2^(r_2)*...* k^(r_k)) summed over integers r_j >= 0, for j=1..k, with Sum{j=1..k} r_j = n-k. [Charalambides]. - Wolfdieter Lang, Aug 15 2019.
A019538(n, k) = k! * S2(n, k).
A028248(n, k) = (k-1)! * S2(n, k).
For asymptotics see Hsu (1948), among other sources.
Sum_{n>=0} S2(n, k)*x^n = x^k/((1-x)(1-2x)(1-3x)...(1-kx)).
Let P(n) = the number of integer partitions of n (A000041), p(i) = the number of parts of the i-th partition of n, d(i) = the number of distinct parts of the i-th partition of n, p(j, i) = the j-th part of the i-th partition of n, m(i, j) = multiplicity of the j-th part of the i-th partition of n, and Sum_{i=1..P(n), p(i)=m} = sum running from i=1 to i=P(n) but taking only partitions with p(i)=m parts into account. Then S2(n, m) = Sum_{i=1..P(n), p(i)=m} n!/(Product_{j=1..p(i)} p(i, j)!) * 1/(Product_{j=1..d(i)} m(i, j)!). For example, S2(6, 3) = 90 because n=6 has the following partitions with m=3 parts: (114), (123), (222). Their complexions are: (114): 6!/(1!*1!*4!) * 1/(2!*1!) = 15, (123): 6!/(1!*2!*3!) * 1/(1!*1!*1!) = 60, (222): 6!/(2!*2!*2!) * 1/(3!) = 15. The sum of the complexions is 15+60+15 = 90 = S2(6, 3). - Thomas Wieder, Jun 02 2005
Sum_{k=1..n} k*S2(n,k) = B(n+1)-B(n), where B(q) are the Bell numbers (A000110). - Emeric Deutsch, Nov 01 2006
Recurrence: S2(n+1,k) = Sum_{i=0..n} binomial(n,i)*S2(i,k-1). With the starting conditions S2(n,k) = 1 for n = 0 or k = 1 and S2(n,k) = 0 for k = 0 we have the closely related recurrence S2(n,k) = Sum_{i=k..n} binomial(n-1,i-1)*S2(i-1,k-1). - Thomas Wieder, Jan 27 2007
Representation of Stirling numbers of the second kind S2(n,k), n=1,2,..., k=1,2,...,n, as special values of hypergeometric function of type (n)F(n-1): S2(n,k)= (-1)^(k-1)*hypergeom([ -k+1,2,2,...,2],[1,1,...,1],1)/(k-1)!, i.e., having n parameters in the numerator: one equal to -k+1 and n-1 parameters all equal to 2; and having n-1 parameters in the denominator all equal to 1 and the value of the argument equal to 1. Example: S2(6,k)= seq(evalf((-1)^(k-1)*hypergeom([ -k+1,2,2,2,2,2],[1,1,1,1,1],1)/(k-1)!),k=1..6)=1,31,90,65,15,1. - Karol A. Penson, Mar 28 2007
From Tom Copeland, Oct 10 2007: (Start)
Bell_n(x) = Sum_{j=0..n} S2(n,j) * x^j = Sum_{j=0..n} E(n,j) * Lag(n,-x, j-n) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,-x, j-n)) = Sum_{j=0..n} E(n,j) * binomial(Bell.(x)+j, n) umbrally where Bell_n(x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m.
For x = 0, the equation gives Sum_{j=0..n} E(n,j) * binomial(j,n) = 1 for n=0 and 0 for all other n. By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*binomial(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = Sum_{j=0..n} E(n,j) * binomial(y+j,n).
Note that E(n,j)/n! = E(n,j)/(Sum_{k=0..n}E(n,k)). Also (n!*Lag(n, -1, j-n)) is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to the equation for x=1; n!*Bell_n(1) = n!*Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * (n!*Lag(n, -1, j-n)).
(Appended Sep 16 2020) For connections to the Bernoulli numbers, extensions, proofs, and a clear presentation of the number arrays involved in the identities above, see my post Reciprocity and Umbral Witchcraft. (End)
n-th row = leftmost column of nonzero terms of A127701^(n-1). Also, (n+1)-th row of the triangle = A127701 * n-th row; deleting the zeros. Example: A127701 * [1, 3, 1, 0, 0, 0, ...] = [1, 7, 6, 1, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007
The row polynomials are given by D^n(e^(x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A147315 and A094198. See also A185422. - Peter Bala, Nov 25 2011
Let f(x) = e^(e^x). Then for n >= 1, 1/f(x)*(d/dx)^n(f(x)) = 1/f(x)*(d/dx)^(n-1)(e^x*f(x)) = Sum_{k=1..n} S2(n,k)*e^(k*x). Similar formulas hold for A039755, A105794, A111577, A143494 and A154537. - Peter Bala, Mar 01 2012
S2(n,k) = A048993(n,k), 1 <= k <= n. - Reinhard Zumkeller, Mar 26 2012
O.g.f. for the n-th diagonal is D^n(x), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012
n*i!*S2(n-1,i) = Sum_{j=(i+1)..n} (-1)^(j-i+1)*j!/(j-i)*S2(n,j). - Leonid Bedratyuk, Aug 19 2012
G.f.: (1/Q(0)-1)/(x*y), where Q(k) = 1 - (y+k)*x - (k+1)*y*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
From Tom Copeland, Apr 17 2014: (Start)
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result as A007318(x) = P(x).
With Bell(n,x)=B(n,x) defined above, D = d/dx, and :xD:^n = x^n*D^n, a Dobinski formula gives umbrally f(y)^B(.,x) = e^(-x)*e^(f(y)*x). Then f(y)^B(.,:xD:)g(x) = [f(y)^(xD)]g(x) = e^[-(1-f(y)):xD:]g(x) = g[f(y)x].
In particular, for f(y) = (1+y),
A) (1+y)^B(.,x) = e^(-x)*e^((1+y)*x) = e^(x*y) = e^[log(1+y)B(.,x)],
B) (I+dP)^B(.,x) = e^(x*dP) = P(x) = e^[x*(e^M-I)]= e^[M*B(.,x)] with dP = A132440, M = A238385-I = log(I+dP), and I = identity matrix, and
C) (1+dP)^(xD) = e^(dP:xD:) = P(:xD:) = e^[(e^M-I):xD:] = e^[M*xD] with action e^(dP:xD:)g(x) = g[(I+dP)*x].
D) P(x)^m = P(m*x), which implies (Sum_{k=1..m} a_k)^j = B(j,m*x) where the sum is umbrally evaluated only after exponentiation with (a_k)^q = B(.,x)^q = B(q,x). E.g., (a1+a2+a3)^2=a1^2+a2^2+a3^2+2(a1*a2+a1*a3+a2*a3) = 3*B(2,x)+6*B(1,x)^2 = 9x^2+3x = B(2,3x).
E) P(x)^2 = P(2x) = e^[M*B(.,2x)] = A038207(x), the face vectors of the n-Dim hypercubes.
(End)
As a matrix equivalent of some inversions mentioned above, A008277*A008275 = I, the identity matrix, regarded as lower triangular matrices. - Tom Copeland, Apr 26 2014
O.g.f. for the n-th diagonal of the triangle (n = 0,1,2,...): Sum_{k>=0} k^(k+n)*(x*e^(-x))^k/k!. Cf. the generating functions of the diagonals of A039755. Also cf. A112492. - Peter Bala, Jun 22 2014
Floor(1/(-1 + Sum_{n>=k} 1/S2(n,k))) = A034856(k-1), for k>=2. The fractional portion goes to zero at large k. - Richard R. Forberg, Jan 17 2015
From Daniel Forgues, Jan 16 2016: (Start)
Let x_(n), called a factorial term (Boole, 1970) or a factorial polynomial (Elaydi, 2005: p. 60), denote the falling factorial Product_{k=0..n-1} (x-k). Then, for n >= 1, x_(n) = Sum_{k=1..n} A008275(n,k) * x^k, x^n = Sum_{k=1..n} T(n,k) * x_(k), where A008275(n,k) are Stirling numbers of the first kind.
For n >= 1, the row sums yield the exponential numbers (or Bell numbers): Sum_{k=1..n} T(n,k) = A000110(n), and Sum_{k=1..n} (-1)^(n+k) * T(n,k) = (-1)^n * Sum_{k=1..n} (-1)^k * T(n,k) = (-1)^n * A000587(n), where A000587 are the complementary Bell numbers. (End)
Sum_{k=1..n} k*S2(n,k) = A138378(n). - Alois P. Heinz, Jan 07 2022
O.g.f. for the m-th column: x^m/(Product_{j=1..m} 1-j*x). - Daniel Checa, Aug 25 2022
S2(n,k) ~ (k^n)/k!, for fixed k as n->oo. - Daniel Checa, Nov 08 2022
S2(2n+k, n) ~ (2^(2n+k-1/2) * n^(n+k-1/2)) / (sqrt(Pi*(1-c)) * exp(n) * c^n * (2-c)^(n+k)), where c = -LambertW(-2 * exp(-2)). - Miko Labalan, Dec 21 2024
From Mikhail Kurkov, Mar 05 2025: (Start)
For a general proof of the formulas below via generating functions, see Mathematics Stack Exchange link.
Recursion for the n-th row (independently of other rows): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} (j-2)!*binomial(-k,j)*T(n,k+j-1) for 1 <= k < n with T(n,n) = 1 (see Fedor Petrov link).
Recursion for the k-th column (independently of other columns): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} binomial(n,j)*T(n-j+1,k)*(-1)^j for 1 <= k < n with T(n,n) = 1. (End)

A028387 a(n) = n + (n+1)^2.

Original entry on oeis.org

1, 5, 11, 19, 29, 41, 55, 71, 89, 109, 131, 155, 181, 209, 239, 271, 305, 341, 379, 419, 461, 505, 551, 599, 649, 701, 755, 811, 869, 929, 991, 1055, 1121, 1189, 1259, 1331, 1405, 1481, 1559, 1639, 1721, 1805, 1891, 1979, 2069, 2161, 2255, 2351, 2449, 2549, 2651
Offset: 0

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Author

Keywords

Comments

a(n+1) is the least k > a(n) + 1 such that A000217(a(n)) + A000217(k) is a square. - David Wasserman, Jun 30 2005
Values of Fibonacci polynomial n^2 - n - 1 for n = 2, 3, 4, 5, ... - Artur Jasinski, Nov 19 2006
A127701 * [1, 2, 3, ...]. - Gary W. Adamson, Jan 24 2007
Row sums of triangle A135223. - Gary W. Adamson, Nov 23 2007
Equals row sums of triangle A143596. - Gary W. Adamson, Aug 26 2008
a(n-1) gives the number of n X k rectangles on an n X n chessboard (for k = 1, 2, 3, ..., n). - Aaron Dunigan AtLee, Feb 13 2009
sqrt(a(0) + sqrt(a(1) + sqrt(a(2) + sqrt(a(3) + ...)))) = sqrt(1 + sqrt(5 + sqrt(11 + sqrt(19 + ...)))) = 2. - Miklos Kristof, Dec 24 2009
When n + 1 is prime, a(n) gives the number of irreducible representations of any nonabelian group of order (n+1)^3. - Andrew Rupinski, Mar 17 2010
a(n) = A176271(n+1, n+1). - Reinhard Zumkeller, Apr 13 2010
The product of any 4 consecutive integers plus 1 is a square (see A062938); the terms of this sequence are the square roots. - Harvey P. Dale, Oct 19 2011
Or numbers not expressed in the form m + floor(sqrt(m)) with integer m. - Vladimir Shevelev, Apr 09 2012
Left edge of the triangle in A214604: a(n) = A214604(n+1,1). - Reinhard Zumkeller, Jul 25 2012
Another expression involving phi = (1 + sqrt(5))/2 is a(n) = (n + phi)(n + 1 - phi). Therefore the numbers in this sequence, even if they are prime in Z, are not prime in Z[phi]. - Alonso del Arte, Aug 03 2013
a(n-1) = n*(n+1) - 1, n>=0, with a(-1) = -1, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 5 for b = 2*n+1. In general D = b^2 - 4ac > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013
a(n) has prime factors given by A038872. - Richard R. Forberg, Dec 10 2014
A253607(a(n)) = -1. - Reinhard Zumkeller, Jan 05 2015
An example of a quadratic sequence for which the continued square root map (see A257574) produces the number 2. There are infinitely many sequences with this property - another example is A028387. See Popular Computing link. - N. J. A. Sloane, May 03 2015
Left edge of the triangle in A260910: a(n) = A260910(n+2,1). - Reinhard Zumkeller, Aug 04 2015
Numbers m such that 4m+5 is a square. - Bruce J. Nicholson, Jul 19 2017
The numbers represented as 131 in base n: 131_4 = 29, 131_5 = 41, ... . If 'digits' larger than the base are allowed then 131_2 = 11 and 131_1 = 5 also. - Ron Knott, Nov 14 2017
From Klaus Purath, Mar 18 2019: (Start)
Let m be a(n) or a prime factor of a(n). Then, except for 1 and 5, there are, if m is a prime, exactly two squares y^2 such that the difference y^2 - m contains exactly one pair of factors {x,z} such that the following applies: x*z = y^2 - m, x + y = z with
x < y, where {x,y,z} are relatively prime numbers. {x,y,z} are the initial values of a sequence of the Fibonacci type. Thus each a(n) > 5, if it is a prime, and each prime factor p > 5 of an a(n) can be assigned to exactly two sequences of the Fibonacci type. a(0) = 1 belongs to the original Fibonacci sequence and a(1) = 5 to the Lucas sequence.
But also the reverse assignment applies. From any sequence (f(i)) of the Fibonacci type we get from its 3 initial values by f(i)^2 - f(i-1)*f(i+1) with f(i-1) < f(i) a term a(n) or a prime factor p of a(n). This relation is also valid for any i. In this case we get the absolute value |a(n)| or |p|. (End)
a(n-1) = 2*T(n) - 1, for n>=1, with T = A000217, is a proper subsequence of A089270, and the terms are 0,-1,+1 (mod 5). - Wolfdieter Lang, Jul 05 2019
a(n+1) is the number of wedged n-dimensional spheres in the homotopy of the neighborhood complex of Kneser graph KG_{2,n}. Here, KG_{2,n} is a graph whose vertex set is the collection of subsets of cardinality 2 of set {1,2,...,n+3,n+4} and two vertices are adjacent if and only if they are disjoint. - Anurag Singh, Mar 22 2021
Also the number of squares between (n+2)^2 and (n+2)^4. - Karl-Heinz Hofmann, Dec 07 2021
(x, y, z) = (A001105(n+1), -a(n-1), -a(n)) are solutions of the Diophantine equation x^3 + 4*y^3 + 4*z^3 = 8. - XU Pingya, Apr 25 2022
The least significant digit of terms of this sequence cycles through 1, 5, 1, 9, 9. - Torlach Rush, Jun 05 2024

Examples

			From _Ilya Gutkovskiy_, Apr 13 2016: (Start)
Illustration of initial terms:
                                        o               o
                        o           o   o o           o o
            o       o   o o       o o   o o o       o o o
    o   o   o o   o o   o o o   o o o   o o o o   o o o o
o   o o o   o o o o o   o o o o o o o   o o o o o o o o o
n=0  n=1       n=2           n=3               n=4
(End)
From _Klaus Purath_, Mar 18 2019: (Start)
Examples:
a(0) = 1: 1^1-0*1 = 1, 0+1 = 1 (Fibonacci A000045).
a(1) = 5: 3^2-1*4 = 5, 1+3 = 4 (Lucas A000032).
a(2) = 11: 4^2-1*5 = 11, 1+4 = 5 (A000285); 5^2-2*7 = 11, 2+5 = 7 (A001060).
a(3) = 19: 5^2-1*6 = 19, 1+5 = 6 (A022095); 7^2-3*10 = 19, 3+7 = 10 (A022120).
a(4) = 29: 6^2-1*7 = 29, 1+6 = 7 (A022096); 9^2-4*13 = 29, 4+9 = 13 (A022130).
a(11)/5 = 31: 7^2-2*9 = 31, 2+7 = 9 (A022113); 8^2-3*11 = 31, 3+8 = 11 (A022121).
a(24)/11 = 59: 9^2-2*11 = 59, 2+9 = 11 (A022114); 12^2-5*17 = 59, 5+12 = 17 (A022137).
(End)
		

Crossrefs

Complement of A028392. Third column of array A094954.
Cf. A000217, A002522, A062392, A062786, A127701, A135223, A143596, A052905, A162997, A062938 (squares of this sequence).
A110331 and A165900 are signed versions.
Cf. A002327 (primes), A094210.
Frobenius number for k successive numbers: this sequence (k=2), A079326 (k=3), A138984 (k=4), A138985 (k=5), A138986 (k=6), A138987 (k=7), A138988 (k=8).

Programs

Formula

a(n) = sqrt(A062938(n)). - Floor van Lamoen, Oct 08 2001
a(0) = 1, a(1) = 5, a(n) = (n+1)*a(n-1) - (n+2)*a(n-2) for n > 1. - Gerald McGarvey, Sep 24 2004
a(n) = A105728(n+2, n+1). - Reinhard Zumkeller, Apr 18 2005
a(n) = A109128(n+2, 2). - Reinhard Zumkeller, Jun 20 2005
a(n) = 2*T(n+1) - 1, where T(n) = A000217(n). - Gary W. Adamson, Aug 15 2007
a(n) = A005408(n) + A002378(n); A084990(n+1) = Sum_{k=0..n} a(k). - Reinhard Zumkeller, Aug 20 2007
Binomial transform of [1, 4, 2, 0, 0, 0, ...] = (1, 5, 11, 19, ...). - Gary W. Adamson, Sep 20 2007
G.f.: (1+2*x-x^2)/(1-x)^3. a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - R. J. Mathar, Jul 11 2009
a(n) = (n + 2 + 1/phi) * (n + 2 - phi); where phi = 1.618033989... Example: a(3) = 19 = (5 + .6180339...) * (3.381966...). Cf. next to leftmost column in A162997 array. - Gary W. Adamson, Jul 23 2009
a(n) = a(n-1) + 2*(n+1), with n > 0, a(0) = 1. - Vincenzo Librandi, Nov 18 2010
For k < n, a(n) = (k+1)*a(n-k) - k*a(n-k-1) + k*(k+1); e.g., a(5) = 41 = 4*11 - 3*5 + 3*4. - Charlie Marion, Jan 13 2011
a(n) = lower right term in M^2, M = the 2 X 2 matrix [1, n; 1, (n+1)]. - Gary W. Adamson, Jun 29 2011
G.f.: (x^2-2*x-1)/(x-1)^3 = G(0) where G(k) = 1 + x*(k+1)*(k+4)/(1 - 1/(1 + (k+1)*(k+4)/G(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 16 2012
Sum_{n>0} 1/a(n) = 1 + Pi*tan(sqrt(5)*Pi/2)/sqrt(5). - Enrique Pérez Herrero, Oct 11 2013
E.g.f.: exp(x) (1+4*x+x^2). - Tom Copeland, Dec 02 2013
a(n) = A005408(A000217(n)). - Tony Foster III, May 31 2016
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2).
Product_{n>=1} (1 - 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2)/6. (End)
a(5*n+1)/5 = A062786(n+1). - Torlach Rush, Jun 05 2024

Extensions

Minor edits by N. J. A. Sloane, Jul 04 2010, following suggestions from the Sequence Fans Mailing List

A001040 a(n+1) = n*a(n) + a(n-1) with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 1, 3, 10, 43, 225, 1393, 9976, 81201, 740785, 7489051, 83120346, 1004933203, 13147251985, 185066460993, 2789144166880, 44811373131073, 764582487395121, 13807296146243251, 263103209266016890, 5275871481466581051, 111056404320064218961, 2448516766522879398193
Offset: 0

Views

Author

Keywords

Comments

If the initial 0 and 1 are omitted, CONTINUANT transform of 1, 2, 3, 4, 5, ...
a(n+1) is the numerator of the continued fraction given by C(n) = [n, n-1,...,3,2,1], e.g., [1] = 1, [2,1]=3, [3,2,1] = 10/3, [4,3,2,1] = 43/10 etc. Cf. A001053. - Amarnath Murthy, May 02 2001
Along those lines, a(n) is the denominator of the continued fraction [n,n-1,...3,2,1] and is the numerator of the continued fraction [1,2,3,...,n-1]. - Greg Dresden, Feb 20 2020
Starting (1, 3, 10, 43, ...) = eigensequence of triangle A127701. - Gary W. Adamson, Dec 29 2008
For n >=2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 1's along the superdiagonal and the subdiagonal, and consecutive integers from 1 to n along the main diagonal (see Mathematica program below). - John M. Campbell, Jul 08 2011
Generally, solution of the recurrence a(n+1) = n*a(n) + a(n-1) is a(n) = BesselI(n,-2)*(2*a(0)*BesselK(1,2)-2*a(1)*BesselK(0,2)) + (2*a(0)*BesselI(1,2)+2*a(1)*BesselI(0,2))*BesselK(n,2), and asymptotic is a(n) ~ (a(0)*BesselI(1,2)+a(1)*BesselI(0,2)) * (n-1)!. - Vaclav Kotesovec, Jan 05 2013
For n > 0: a(n) = A058294(n,n) = A102473(n,n) = A102472(n,1). - Reinhard Zumkeller, Sep 14 2014
Conjecture: 2*n!*a(n) is the number of open tours by a rook on an (n X 2) chessboard which ends at the opposite line of length n. - Mikhail Kurkov, Nov 19 2019

Examples

			G.f. = x + x^2 + 3*x^3 + 10*x^4 + 43*x^5 + 225*x^6 + 1393*x^7 + 9976*x^8 + ...
		

References

  • Archimedeans Problems Drive, Eureka, 22 (1959), 15.
  • 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

A column of A058294. Cf. A001053.
Cf. A127701. - Gary W. Adamson, Dec 29 2008
Similar recurrences: A001053, A058279, A058307. - Wolfdieter Lang, May 19 2010

Programs

  • Haskell
    a001040 n = a001040_list !! n
    a001040_list = 0 : 1 : zipWith (+)
       a001040_list (zipWith (*) [1..] $ tail a001040_list)
    -- Reinhard Zumkeller, Mar 05 2013
    
  • Magma
    a:=[1,1]; [0] cat [n le 2 select a[n] else (n-1)*Self(n-1) + Self(n-2): n in [1..23]]; // Marius A. Burtea, Nov 19 2019
  • Maple
    A001040 := proc(n)
        if n <= 1 then
            n;
        else
            (n-1)*procname(n-1)+procname(n-2) ;
        end if;
    end proc: # R. J. Mathar, Mar 13 2015
  • Mathematica
    Table[Permanent[Array[KroneckerDelta[#1, #2]*(#1) + KroneckerDelta[#1, #2 - 1] + KroneckerDelta[#1, #2 + 1] &, {n - 1, n - 1}]], {n, 2, 30}] (* John M. Campbell, Jul 08 2011 *)
    Join[{0},RecurrenceTable[{a[0]==1,a[1]==1,a[n]==n a[n-1]+a[n-2]}, a[n], {n,30}]] (* Harvey P. Dale, Aug 14 2011 *)
    FullSimplify[Table[2(-BesselI[n,-2]BesselK[0,2]+BesselI[0,2]BesselK[n,2]),{n,0,20}]] (* Vaclav Kotesovec, Jan 05 2013 *)
  • PARI
    {a(n) = contfracpnqn( vector(abs(n), i, i))[1, 2]}; /* Michael Somos, Sep 25 2005 */
    
  • Sage
    def A001040(n):
        if n < 2: return n
        return factorial(n-1)*hypergeometric([1-n/2,-n/2+1/2], [1,1-n,1-n], 4)
    [round(A001040(n).n(100)) for n in (0..23)] # Peter Luschny, Sep 10 2014
    

Formula

Generalized Fibonacci sequence for (unsigned) Laguerre triangle A021009. a(n+1) = sum{k=0..floor(n/2), C(n-k, k)(n-k)!/k!}. - Paul Barry, May 10 2004
a(-n) = a(n) for all n in Z. - Michael Somos, Sep 25 2005
E.g.f.: -I*Pi*(BesselY(1, 2*I)*BesselI(0, 2*sqrt(1-x)) - I*BesselI(1, 2)*BesselY(0, 2*I*sqrt(1-x))). Such e.g.f. computations were the result of an e-mail exchange with Gary Detlefs. After differentiation and putting x=0 one has to use simplifications. See the Abramowitz-Stegun handbook, p. 360, 9.1.16 and p. 375, 9.63. - Wolfdieter Lang, May 19 2010
Limit_{n->infinity} a(n)/(n-1)! = BesselI(0,2) = 2.279585302336... (see A070910). - Vaclav Kotesovec, Jan 05 2013
a(n) = 2*(BesselI(0,2)*BesselK(n,2) - BesselI(n,-2)*BesselK(0,2)). - Vaclav Kotesovec, Jan 05 2013
a(n) = (n-1)!*hypergeometric([1-n/2,1/2-n/2],[1,1-n,1-n], 4) for n >= 2. - Peter Luschny, Sep 10 2014
0 = a(n)*(-a(n+2)) + a(n+1)*(+a(n+1) + a(n+2) - a(n+3)) + a(n+2)*(+a(n+2)) for all n in Z. - Michael Somos, Sep 13 2014
Observed: a(n) = A070910*(n-1)!*(1 + 1/(n-1) + 1/(2*(n-1)^2) + O((n-1)^-3)). - A.H.M. Smeets, Aug 19 2018
a(n) mod 2 = A166486(n). - Alois P. Heinz, Jul 03 2023

Extensions

Definition clarified by A.H.M. Smeets, Aug 19 2018

A127736 a(n) = n*(n^2 + 2*n - 1)/2.

Original entry on oeis.org

1, 7, 21, 46, 85, 141, 217, 316, 441, 595, 781, 1002, 1261, 1561, 1905, 2296, 2737, 3231, 3781, 4390, 5061, 5797, 6601, 7476, 8425, 9451, 10557, 11746, 13021, 14385, 15841, 17392, 19041, 20791, 22645, 24606, 26677, 28861, 31161, 33580, 36121, 38787, 41581
Offset: 1

Views

Author

Gary W. Adamson, Jan 26 2007

Keywords

Comments

Row sums of A127735.
Row sums of A162610. - Reinhard Zumkeller, Jan 19 2013
For n > 0, a(n) is the number of compositions of n+10 into n parts avoiding parts 2 and 3. - Milan Janjic, Jan 07 2016
Sum of the numbers in the top row and last column of an n X n square array whose elements are the numbers from 1..n^2, listed in increasing order by rows (see example). - Wesley Ivan Hurt, May 18 2021

Examples

			From _Wesley Ivan Hurt_, May 18 2021: (Start)
Add all the numbers in the top row and last column.
                                                      [1   2  3  4  5]
                                      [1   2  3  4]   [6   7  8  9 10]
                            [1 2 3]   [5   6  7  8]   [11 12 13 14 15]
                   [1 2]    [4 5 6]   [9  10 11 12]   [16 17 18 19 20]
           [1]     [3 4]    [7 8 9]   [13 14 15 16]   [21 22 23 24 25]
------------------------------------------------------------------------
  n         1        2         3            4                 5
------------------------------------------------------------------------
  a(n)      1        7        21           46                85
------------------------------------------------------------------------
(End)
		

Crossrefs

Programs

Formula

Row sums of triangle A131416. Also, binomial transform of [1, 6, 8, 3, 0, 0, 0, ...). - Gary W. Adamson, Oct 23 2007
a(n) = (n+1)*A000217(n) - n = A006002(n) - n. - R. J. Mathar, Jul 21 2009
From Colin Barker, Mar 12 2014: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
G.f.: -x*(x^2-3*x-1) / (x-1)^4. (End)
a(n) = A057145(n+5,n). - R. J. Mathar, Jul 28 2016

Extensions

More terms and new name from R. J. Mathar, Jul 21 2009

A128223 a(n) = if n mod 2 = 0 then n*(n+1)/2 otherwise (n+1)^2/2-1.

Original entry on oeis.org

0, 1, 3, 7, 10, 17, 21, 31, 36, 49, 55, 71, 78, 97, 105, 127, 136, 161, 171, 199, 210, 241, 253, 287, 300, 337, 351, 391, 406, 449, 465, 511, 528, 577, 595, 647, 666, 721, 741, 799, 820, 881, 903, 967, 990, 1057, 1081, 1151, 1176, 1249, 1275, 1351, 1378, 1457, 1485
Offset: 0

Views

Author

Gary W. Adamson, Feb 19 2007

Keywords

Comments

a(n-1) is the length of the shortest path along the edges of the complete graph with n vertices. - Martin Fuller, Dec 06 2007
From Peter Kagey, Jan 25 2015: (Start)
For an irreflexive, non-transitive, symmetric relation, a(n) is the length of a relation chain required to demonstrate that a != b for all distinct elements a and b in S, where S contains n+1 elements.
For example, for the set {1,2,3} the chain requires a(2) = 3 relations (e.g., 1 != 2 != 3 != 1). For the set {1,2,3,4}, the chain requires a(3) = 7 relations (e.g., 1 != 2 != 3 != 4 != 1 != 3 != 2 != 4 -- noting the redundancy of 2!=3 and 3!=2). (End)
Given a set of n lots of n distinct items, it is possible to sort the items from fully collated (ABCABCABC) to fully sorted (AAABBBCCC), or vice versa, using a sorting algorithm whereby at each step a portion of the overall string is selected and its contents reversed. The minimum number of steps such an algorithm will take is a(n-1). For example, when n=3, a(n-1)=3: ABCABCABC -> ABBACBACC -> ABBAABCCC -> AAABBBCCC. - Elliott Line, Aug 02 2019

Examples

			a(5) = 17 = (5 + 1 + 5 + 1 + 5), row 5 of A128222.
		

Crossrefs

Row sums of A128222.
Cf. A024206, row sums of A128221 = A128174 * A127701.

Programs

  • Haskell
    a128223 n = if even n then n*(n + 1) `div` 2 else (n+1)^2 `div` 2 - 1 -- Peter Kagey, Jul 14 2015
    
  • Magma
    [(-1+(-1)^n-(-3+(-1)^n)*n+2*n^2)/4: n in [0..60]]; // Vincenzo Librandi, Mar 18 2015
    
  • Maple
    f:=n-> if n mod 2 = 0 then n*(n+1)/2 else (n+1)^2/2-1; fi;
  • Mathematica
    f[n_] := If[EvenQ@ n, n (n + 1)/2, (n + 1)^2/2 - 1]; Array[f, 54] (* Michael De Vlieger, Mar 17 2015 *)
    Table[(- 1 + (-1)^n - (- 3 + (-1)^n) n + 2 n^2) / 4, {n, 0, 60}] (* Vincenzo Librandi, Mar 18 2015 *)
    CoefficientList[ Series[(-x - 2x^2 - 2x^3 + x^4)/((-1 + x)^3 (1 + x)^2), {x, 0, 54}], x] (* Robert G. Wilson v, Nov 16 2016 *)
    LinearRecurrence[{1,2,-2,-1,1},{0,1,3,7,10},60] (* Harvey P. Dale, Mar 17 2020 *)
  • PARI
    main(size)={my(n,m,v=vector(size),i);for(i=0,size-1,v[i+1]=if(i%2==0,i*(i+1)/2,(i+1)^2/2-1));return(v);} /* Anders Hellström, Jul 14 2015 */

Formula

a(n) = (-1+(-1)^n-(-3+(-1)^n)*n+2*n^2)/4. a(n) = a(n-1)+2*a(n-2)-2*a(n-3)-a(n-4)+a(n-5). G.f.: x*(x^3-2*x^2-2*x-1) / ((x-1)^3*(x+1)^2). - Colin Barker, Oct 16 2013
a(n) = A053439(n) - 1. - Peter Kagey, Nov 16 2016

Extensions

Edited by N. J. A. Sloane, Dec 06 2007

A127705 a(n) = mu(n) + Sum_{k|n, k>1} (k+1)*mu(n/k), where mu = A008683.

Original entry on oeis.org

1, 2, 3, 2, 5, 1, 7, 4, 6, 3, 11, 4, 13, 5, 7, 8, 17, 6, 19, 8, 11, 9, 23, 8, 20, 11, 18, 12, 29, 9, 31, 16, 19, 15, 23, 12, 37, 17, 23, 16, 41, 13, 43, 20, 24, 21, 47, 16, 42, 20, 31, 24, 53, 18, 39, 24, 35, 27, 59, 16, 61, 29, 36, 32, 47, 21, 67, 32, 43, 25, 71, 24, 73
Offset: 1

Views

Author

Gary W. Adamson, Jan 24 2007

Keywords

Comments

Previous name: Row sums of A127704.
From Robert Israel, Dec 29 2016: (Start)
a(n) = n if and only if n is in A008578.
a(p^j) = p^j - p^(j-1) if p is prime and j >= 2.
a(Product_{i=1..k} p_i) = Product_{i=1..k} (p_i-1) - (-1)^k if p_1, ..., p_k are distinct primes. (End)

Crossrefs

Programs

  • Maple
    N := 100: # to get a(1)..a(N)
    A:= Vector(N, numtheory:-mobius):
    for k from 2 to N do
      for j from 1 to floor(N/k) do
        A[j*k]:= A[j*k] + (k+1)*numtheory:-mobius(j)
    od od:
    convert(A, list); # Robert Israel, Dec 29 2016
  • Python
    def A127705_list(n):
        upto = n + 1
        p = [i for i in range(upto)]
        for i in range(2, upto):
            for j in range(i + i, upto , i):
                p[j] -= p[i]
        return p[1::]
    print(A127705_list(1000))  # Peter Luschny, Sep 02 2023

Formula

a(n) = mu(n) + Sum_{k|n,k>1} (k+1)*mu(n/k), where mu = A008683. - Robert Israel, Dec 29 2016

Extensions

Terms a(11) and beyond from Robert Israel, Dec 29 2016
New name using a formula of Robert Israel by Peter Luschny, Sep 02 2023
Previous Showing 11-17 of 17 results.