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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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A000330 Square pyramidal numbers: a(n) = 0^2 + 1^2 + 2^2 + ... + n^2 = n*(n+1)*(2*n+1)/6.

Original entry on oeis.org

0, 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819, 1015, 1240, 1496, 1785, 2109, 2470, 2870, 3311, 3795, 4324, 4900, 5525, 6201, 6930, 7714, 8555, 9455, 10416, 11440, 12529, 13685, 14910, 16206, 17575, 19019, 20540, 22140, 23821, 25585, 27434, 29370
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

The sequence contains exactly one square greater than 1, namely 4900 (according to Gardner). - Jud McCranie, Mar 19 2001, Mar 22 2007 [This is a result from Watson. - Charles R Greathouse IV, Jun 21 2013] [See A351830 for further related comments and references.]
Number of rhombi in an n X n rhombus. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Number of acute triangles made from the vertices of a regular n-polygon when n is odd (cf. A007290). - Sen-Peng Eu, Apr 05 2001
Gives number of squares with sides parallel to the axes formed from an n X n square. In a 1 X 1 square, one is formed. In a 2 X 2 square, five squares are formed. In a 3 X 3 square, 14 squares are formed and so on. - Kristie Smith (kristie10spud(AT)hotmail.com), Apr 16 2002; edited by Eric W. Weisstein, Mar 05 2025
a(n-1) = B_3(n)/3, where B_3(x) = x(x-1)(x-1/2) is the third Bernoulli polynomial. - Michael Somos, Mar 13 2004
Number of permutations avoiding 13-2 that contain the pattern 32-1 exactly once.
Since 3*r = (r+1) + r + (r-1) = T(r+1) - T(r-2), where T(r) = r-th triangular number r*(r+1)/2, we have 3*r^2 = r*(T(r+1) - T(r-2)) = f(r+1) - f(r-1) ... (i), where f(r) = (r-1)*T(r) = (r+1)*T(r-1). Summing over n, the right hand side of relation (i) telescopes to f(n+1) + f(n) = T(n)*((n+2) + (n-1)), whence the result Sum_{r=1..n} r^2 = n*(n+1)*(2*n+1)/6 immediately follows. - Lekraj Beedassy, Aug 06 2004
Also as a(n) = (1/6)*(2*n^3 + 3*n^2 + n), n > 0: structured trigonal diamond numbers (vertex structure 5) (cf. A006003 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Number of triples of integers from {1, 2, ..., n} whose last component is greater than or equal to the others.
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 12 2005
Sum of the first n positive squares. - Cino Hilliard, Jun 18 2007
Maximal number of cubes of side 1 in a right pyramid with a square base of side n and height n. - Pasquale CUTOLO (p.cutolo(AT)inwind.it), Jul 09 2007
If a 2-set Y and an (n-2)-set Z are disjoint subsets of an n-set X then a(n-3) is the number of 4-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007
We also have the identity 1 + (1+4) + (1+4+9) + ... + (1+4+9+16+ ... + n^2) = n(n+1)(n+2)(n+(n+1)+(n+2))/36; ... and in general the k-fold nested sum of squares can be expressed as n(n+1)...(n+k)(n+(n+1)+...+(n+k))/((k+2)!(k+1)/2). - Alexander R. Povolotsky, Nov 21 2007
The terms of this sequence are coefficients of the Engel expansion of the following converging sum: 1/(1^2) + (1/1^2)*(1/(1^2+2^2)) + (1/1^2)*(1/(1^2+2^2))*(1/(1^2+2^2+3^2)) + ... - Alexander R. Povolotsky, Dec 10 2007
Convolution of A000290 with A000012. - Sergio Falcon, Feb 05 2008
Hankel transform of binomial(2*n-3, n-1) is -a(n). - Paul Barry, Feb 12 2008
Starting (1, 5, 14, 30, ...) = binomial transform of [1, 4, 5, 2, 0, 0, 0, ...]. - Gary W. Adamson, Jun 13 2008
Starting (1,5,14,30,...) = second partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} binomial(n+2,i+2)*b(i), where b(i)=1,2,0,0,0,... - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
Convolution of A001477 with A005408: a(n) = Sum_{k=0..n} (2*k+1)*(n-k). - Reinhard Zumkeller, Mar 07 2009
Sequence of the absolute values of the z^1 coefficients of the polynomials in the GF1 denominators of A156921. See A157702 for background information. - Johannes W. Meijer, Mar 07 2009
The sequence is related to A000217 by a(n) = n*A000217(n) - Sum_{i=0..n-1} A000217(i) and this is the case d = 1 in the identity n^2*(d*n-d+2)/2 - Sum_{i=0..n-1} i*(d*i-d+2)/2 = n*(n+1)(2*d*n-2*d+3)/6, or also the case d = 0 in n^2*(n+2*d+1)/2 - Sum_{i=0..n-1} i*(i+2*d+1)/2 = n*(n+1)*(2*n+3*d+1)/6. - Bruno Berselli, Apr 21 2010, Apr 03 2012
a(n)/n = k^2 (k = integer) for n = 337; a(337) = 12814425, a(n)/n = 38025, k = 195, i.e., the number k = 195 is the quadratic mean (root mean square) of the first 337 positive integers. There are other such numbers -- see A084231 and A084232. - Jaroslav Krizek, May 23 2010
Also the number of moves to solve the "alternate coins game": given 2n+1 coins (n+1 Black, n White) set alternately in a row (BWBW...BWB) translate (not rotate) a pair of adjacent coins at a time (1 B and 1 W) so that at the end the arrangement shall be BBBBB..BW...WWWWW (Blacks separated by Whites). Isolated coins cannot be moved. - Carmine Suriano, Sep 10 2010
From J. M. Bergot, Aug 23 2011: (Start)
Using four consecutive numbers n, n+1, n+2, and n+3 take all possible pairs (n, n+1), (n, n+2), (n, n+3), (n+1, n+2), (n+1, n+3), (n+2, n+3) to create unreduced Pythagorean triangles. The sum of all six areas is 60*a(n+1).
Using three consecutive odd numbers j, k, m, (j+k+m)^3 - (j^3 + k^3 + m^3) equals 576*a(n) = 24^2*a(n) where n = (j+1)/2. (End)
From Ant King, Oct 17 2012: (Start)
For n > 0, the digital roots of this sequence A010888(a(n)) form the purely periodic 27-cycle {1, 5, 5, 3, 1, 1, 5, 6, 6, 7, 2, 2, 9, 7, 7, 2, 3, 3, 4, 8, 8, 6, 4, 4, 8, 9, 9}.
For n > 0, the units' digits of this sequence A010879(a(n)) form the purely periodic 20-cycle {1, 5, 4, 0, 5, 1, 0, 4, 5, 5, 6, 0, 9, 5, 0, 6, 5, 9, 0, 0}. (End)
Length of the Pisano period of this sequence mod n, n>=1: 1, 4, 9, 8, 5, 36, 7, 16, 27, 20, 11, 72, 13, 28, 45, 32, 17, 108, 19, 40, ... . - R. J. Mathar, Oct 17 2012
Sum of entries of n X n square matrix with elements min(i,j). - Enrique Pérez Herrero, Jan 16 2013
The number of intersections of diagonals in the interior of regular n-gon for odd n > 1 divided by n is a square pyramidal number; that is, A006561(2*n+1)/(2*n+1) = A000330(n-1) = (1/6)*n*(n-1)*(2*n-1). - Martin Renner, Mar 06 2013
For n > 1, a(n)/(2n+1) = A024702(m), for n such that 2n+1 = prime, which results in 2n+1 = A000040(m). For example, for n = 8, 2n+1 = 17 = A000040(7), a(8) = 204, 204/17 = 12 = A024702(7). - Richard R. Forberg, Aug 20 2013
A formula for the r-th successive summation of k^2, for k = 1 to n, is (2*n+r)*(n+r)!/((r+2)!*(n-1)!) (H. W. Gould). - Gary Detlefs, Jan 02 2014
The n-th square pyramidal number = the n-th triangular dipyramidal number (Johnson 12), which is the sum of the n-th + (n-1)-st tetrahedral numbers. E.g., the 3rd tetrahedral number is 10 = 1+3+6, the 2nd is 4 = 1+3. In triangular "dipyramidal form" these numbers can be written as 1+3+6+3+1 = 14. For "square pyramidal form", rebracket as 1+(1+3)+(3+6) = 14. - John F. Richardson, Mar 27 2014
Beukers and Top prove that no square pyramidal number > 1 equals a tetrahedral number A000292. - Jonathan Sondow, Jun 21 2014
Odd numbered entries are related to dissections of polygons through A100157. - Tom Copeland, Oct 05 2014
From Bui Quang Tuan, Apr 03 2015: (Start)
We construct a number triangle from the integers 1, 2, 3, ..., n as follows. The first column contains 2*n-1 integers 1. The second column contains 2*n-3 integers 2, ... The last column contains only one integer n. The sum of all the numbers in the triangle is a(n).
Here is an example with n = 5:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
(End)
The Catalan number series A000108(n+3), offset 0, gives Hankel transform revealing the square pyramidal numbers starting at 5, A000330(n+2), offset 0 (empirical observation). - Tony Foster III, Sep 05 2016; see Dougherty et al. link p. 2. - Andrey Zabolotskiy, Oct 13 2016
Number of floating point additions in the factorization of an (n+1) X (n+1) real matrix by Gaussian elimination as e.g. implemented in LINPACK subroutines sgefa.f or dgefa.f. The number of multiplications is given by A007290. - Hugo Pfoertner, Mar 28 2018
The Jacobi polynomial P(n-1,-n+2,2,3) or equivalently the sum of dot products of vectors from the first n rows of Pascal's triangle (A007318) with the up-diagonal Chebyshev T coefficient vector (1,3,2,0,...) (A053120) or down-diagonal vector (1,-7,32,-120,400,...) (A001794). a(5) = 1 + (1,1).(1,3) + (1,2,1).(1,3,2) + (1,3,3,1).(1,3,2,0) + (1,4,6,4,1).(1,3,2,0,0) = (1 + (1,1).(1,-7) + (1,2,1).(1,-7,32) + (1,3,3,1).(1,-7,32,-120) + (1,4,6,4,1).(1,-7,32,-120,400))*(-1)^(n-1) = 55. - Richard Turk, Jul 03 2018
Coefficients in the terminating series identity 1 - 5*n/(n + 4) + 14*n*(n - 1)/((n + 4)*(n + 5)) - 30*n*(n - 1)*(n - 2)/((n + 4)*(n + 5)*(n + 6)) + ... = 0 for n = 1,2,3,.... Cf. A002415 and A108674. - Peter Bala, Feb 12 2019
n divides a(n) iff n == +- 1 (mod 6) (see A007310). (See De Koninck reference.) Examples: a(11) = 506 = 11 * 46, and a(13) = 819 = 13 * 63. - Bernard Schott, Jan 10 2020
For n > 0, a(n) is the number of ternary words of length n+2 having 3 letters equal to 2 and 0 only occurring as the last letter. For example, for n=2, the length 4 words are 2221,2212,2122,1222,2220. - Milan Janjic, Jan 28 2020
Conjecture: Every integer can be represented as a sum of three generalized square pyramidal numbers. A related conjecture is given in A336205 corresponding to pentagonal case. A stronger version of these conjectures is that every integer can be expressed as a sum of three generalized r-gonal pyramidal numbers for all r >= 3. In here "generalized" means negative indices are included. - Altug Alkan, Jul 30 2020
The natural number y is a term if and only if y = a(floor((3 * y)^(1/3))). - Robert Israel, Dec 04 2024
Also the number of directed bishop moves on an n X n chessboard, where two moves are considered the same if one can be obtained from the other by a rotation of the board. Reflections are ignored. Equivalently, number of directed bishop moves on an n X n chessboard, where two moves are considered the same if one can be obtained from the other by an axial reflection of the board (horizontal or vertical). Rotations and diagonal reflections are ignored. - Hilko Koning, Aug 22 2025

Examples

			G.f. = x + 5*x^2 + 14*x^3 + 30*x^4 + 55*x^5 + 91*x^6 + 140*x^7 + 204*x^8 + ...

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. 813.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover Publications, NY, 1964, p. 194.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 215,223.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 122, see #19 (3(1)), I(n); p. 155.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 47-49.
  • H. S. M. Coxeter, Polyhedral numbers, pp. 25-35 of R. S. Cohen, J. J. Stachel and M. W. Wartofsky, eds., For Dirk Struik: Scientific, historical and political essays in honor of Dirk J. Struik, Reidel, Dordrecht, 1974.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p.165).
  • J. M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 310, pp. 46-196, Ellipses, Paris, 2004.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 2.
  • M. Gardner, Fractal Music, Hypercards and More, Freeman, NY, 1991, p. 293.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 293.
  • M. Holt, Math puzzles and games, Walker Publishing Company, 1977, p. 2 and p. 89.
  • Simon Singh, The Simpsons and Their Mathematical Secrets. London: Bloomsbury Publishing PLC (2013): 188.
  • 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).
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 126.

Links

Crossrefs

Cf. A000217, A000292, A000537, A005408, A006003, A006331, A033994, A033999, A046092, A050409, A050446, A050447, A060493, A100157, A132124, A132112, A156921, A157702, A258708, A351105, A351830.
Sums of 2 consecutive terms give A005900.
Column 0 of triangle A094414.
Column 1 of triangle A008955.
Right side of triangle A082652.
Row 2 of array A103438.
Partial sums of A000290.
Cf. similar sequences listed in A237616 and A254142.
Cf. |A084930(n, 1)|.
Cf. A253903 (characteristic function).
Cf. A034705 (differences of any two terms).

Programs

  • GAP
    List([0..30], n-> n*(n+1)*(2*n+1)/6); # G. C. Greubel, Dec 31 2019
  • Haskell
    a000330 n = n * (n + 1) * (2 * n + 1) `div` 6
a000330_list = scanl1 (+) a000290_list
-- Reinhard Zumkeller, Nov 11 2012, Feb 03 2012
  • Magma
    [n*(n+1)*(2*n+1)/6: n in [0..50]]; // Wesley Ivan Hurt, Jun 28 2014
  • Magma
    [0] cat [((2*n+3)*Binomial(n+2,2))/3: n in [0..40]]; // Vincenzo Librandi, Jul 30 2014
  • Maple
    A000330 := n -> n*(n+1)*(2*n+1)/6;
a := n->(1/6)*n*(n+1)*(2*n+1): seq(a(n),n=0..53); # Emeric Deutsch
with(combstruct): ZL:=[st, {st=Prod(left, right), left=Set(U, card=r), right=Set(U, card=r), U=Sequence(Z, card>=1)}, unlabeled]: subs(r=1, stack): seq(count(subs(r=2, ZL), size=m*2), m=1..45) ; # Zerinvary Lajos, Jan 02 2008
nmax := 44; for n from 0 to nmax do fz(n) := product( (1-(2*m-1)*z)^(n+1-m) , m=1..n); c(n) := abs(coeff(fz(n),z,1)); end do: a := n-> c(n): seq(a(n), n=0..nmax); # Johannes W. Meijer, Mar 07 2009
  • Mathematica
    Table[Binomial[w+2, 3] + Binomial[w+1, 3], {w, 0, 30}]
CoefficientList[Series[x(1+x)/(1-x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 30 2014 *)
Accumulate[Range[0,50]^2] (* Harvey P. Dale, Sep 25 2014 *)
  • Maxima
    A000330(n):=binomial(n+2,3)+binomial(n+1,3)$
makelist(A000330(n),n,0,20); /* Martin Ettl, Nov 12 2012 */
  • PARI
    {a(n) = n * (n+1) * (2*n+1) / 6};
  • PARI
    upto(n) = [x*(x+1)*(2*x+1)/6 | x<-[0..n]] \\ Cino Hilliard, Jun 18 2007, edited by M. F. Hasler, Jan 02 2024
  • Python
    a=lambda n: (n*(n+1)*(2*n+1))//6 # Indranil Ghosh, Jan 04 2017
  • Sage
    [n*(n+1)*(2*n+1)/6 for n in (0..30)] # G. C. Greubel, Dec 31 2019

Formula

G.f.: x*(1+x)/(1-x)^4. - Simon Plouffe (in his 1992 dissertation: generating function for sequence starting at a(1))
E.g.f.: (x + 3*x^2/2 + x^3/3)*exp(x).
a(n) = n*(n+1)*(2*n+1)/6 = binomial(n+2, 3) + binomial(n+1, 3).
2*a(n) = A006331(n). - N. J. A. Sloane, Dec 11 1999
Can be extended to Z with a(n) = -a(-1-n) for all n in Z.
a(n) = A002492(n)/4. - Paul Barry, Jul 19 2003
a(n) = (((n+1)^4 - n^4) - ((n+1)^2 - n^2))/12. - Xavier Acloque, Oct 16 2003
From Alexander Adamchuk, Oct 26 2004: (Start)
a(n) = sqrt(A271535(n)).
a(n) = (Sum_{k=1..n} Sum_{j=1..n} Sum_{i=1..n} (i*j*k)^2)^(1/3). (End)
a(n) = Sum_{i=1..n} i*(2*n-2*i+1); sum of squares gives 1 + (1+3) + (1+3+5) + ... - Jon Perry, Dec 08 2004
a(n+1) = A000217(n+1) + 2*A000292(n). - Creighton Dement, Mar 10 2005
Sum_{n>=1} 1/a(n) = 6*(3-4*log(2)); Sum_{n>=1} (-1)^(n+1)*1/a(n) = 6*(Pi-3). - Philippe Deléham, May 31 2005
Sum of two consecutive tetrahedral (or pyramidal) numbers a(n) = A000292(n-1) + A000292(n). - Alexander Adamchuk, May 17 2006
Euler transform of length-2 sequence [ 5, -1 ]. - Michael Somos, Sep 04 2006
a(n) = a(n-1) + n^2. - Rolf Pleisch, Jul 22 2007
a(n) = A132121(n,0). - Reinhard Zumkeller, Aug 12 2007
a(n) = binomial(n, 2) + 2*binomial(n, 3). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009, corrected by M. F. Hasler, Jan 02 2024
a(n) = A168559(n) + 1 for n > 0. - Reinhard Zumkeller, Feb 03 2012
a(n) = Sum_{i=1..n} J_2(i)*floor(n/i), where J_2 is A007434. - Enrique Pérez Herrero, Feb 26 2012
a(n) = s(n+1, n)^2 - 2*s(n+1, n-1), where s(n, k) are Stirling numbers of the first kind, A048994. - Mircea Merca, Apr 03 2012
a(n) = A001477(n) + A000217(n) + A007290(n+2) + 1. - J. M. Bergot, May 31 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 2. - Ant King, Oct 17 2012
a(n) = Sum_{i = 1..n} Sum_{j = 1..n} min(i,j). - Enrique Pérez Herrero, Jan 15 2013
a(n) = A000217(n) + A007290(n+1). - Ivan N. Ianakiev, May 10 2013
a(n) = (A047486(n+2)^3 - A047486(n+2))/24. - Richard R. Forberg, Dec 25 2013
a(n) = Sum_{i=0..n-1} (n-i)*(2*i+1), with a(0) = 0. After 0, row sums of the triangle in A101447. - Bruno Berselli, Feb 10 2014
a(n) = n + 1 + Sum_{i=1..n+1} (i^2 - 2i). - Wesley Ivan Hurt, Feb 25 2014
a(n) = A000578(n+1) - A002412(n+1). - Wesley Ivan Hurt, Jun 28 2014
a(n) = Sum_{i = 1..n} Sum_{j = i..n} max(i,j). - Enrique Pérez Herrero, Dec 03 2014
a(n) = A055112(n)/6, see Singh (2013). - Alonso del Arte, Feb 20 2015
For n >= 2, a(n) = A028347(n+1) + A101986(n-2). - Bui Quang Tuan, Apr 03 2015
For n > 0: a(n) = A258708(n+3,n-1). - Reinhard Zumkeller, Jun 23 2015
a(n) = A175254(n) + A072481(n), n >= 1. - Omar E. Pol, Aug 12 2015
a(n) = A000332(n+3) - A000332(n+1). - Antal Pinter, Dec 27 2015
Dirichlet g.f.: zeta(s-3)/3 + zeta(s-2)/2 + zeta(s-1)/6. - Ilya Gutkovskiy, Jun 26 2016
a(n) = A080851(2,n-1). - R. J. Mathar, Jul 28 2016
a(n) = (A005408(n) * A046092(n))/12 = (2*n+1)*(2*n*(n+1))/12. - Bruce J. Nicholson, May 18 2017
12*a(n) = (n+1)*A001105(n) + n*A001105(n+1). - Bruno Berselli, Jul 03 2017
a(n) = binomial(n-1, 1) + binomial(n-1, 2) + binomial(n, 3) + binomial(n+1, 2) + binomial(n+1, 3). - Tony Foster III, Aug 24 2018
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Nathan Fox, Dec 04 2019
Let T(n) = A000217(n), the n-th triangular number. Then a(n) = (T(n)+1)^2 + (T(n)+2)^2 + ... + (T(n)+n)^2 - (n+2)*T(n)^2. - Charlie Marion, Dec 31 2019
a(n) = 2*n - 1 - a(n-2) + 2*a(n-1). - Boštjan Gec, Nov 09 2023
a(n) = 2/(2*n)! * Sum_{j = 1..n} (-1)^(n+j) * j^(2*n+2) * binomial(2*n, n-j). Cf. A060493. - Peter Bala, Mar 31 2025

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A007954 Product of decimal digits of n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 0, 8, 16, 24, 32, 40, 48, 56, 64, 72, 0, 9, 18, 27, 36, 45, 54, 63, 72, 81, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 0

Views

Author

R. Muller

Keywords

Comments

Moebius transform of A093811(n). a(n) = A093811(n) * A008683(n), where operation * denotes Dirichlet convolution, namely b(n) * c(n) = Sum_{d|n} b(d) * c(n/d). Simultaneously holds Dirichlet multiplication: a(n) * A000012(n) = A093811(n). - Jaroslav Krizek, Mar 22 2009
Apart from the 0's, all terms are in A002473. Further, for all m in A002473 there is some n such that a(n) = m, see A096867. - Charles R Greathouse IV, Sep 29 2013
a(n) = 0 asymptotically almost surely, namely for all n except for the set of numbers without digit '0'; this set is of density zero, since it is less and less probable to have no '0' as the number of digits of n grows. (See also A054054.) - M. F. Hasler, Oct 11 2015

Links

Crossrefs

Cf. A031347 (different from A035930), A007953, A007602, A010888, A093811, A008683, A000012, A061076 (partial sums), A230099.
Cf. A051802 (ignoring zeros).

Programs

  • Haskell
    a007954 n | n < 10 = n
          | otherwise = m * a007954 n' where (n', m) = divMod n 10
-- Reinhard Zumkeller, Oct 26 2012, Mar 14 2011
  • Magma
    [0] cat [&*Intseq(n): n in [1..110]]; // Vincenzo Librandi, Jan 03 2020
  • Maple
    A007954 := proc(n::integer)
    if n = 0 then
        0;
    else
        mul( d,d=convert(n,base,10)) ;
    end if;
end proc: # R. J. Mathar, Oct 02 2019
  • Mathematica
    Array[Times @@ IntegerDigits@ # &, 108, 0] (* Robert G. Wilson v, Mar 15 2011 *)
  • PARI
    A007954(n)= { local(resul = n % 10); n \= 10; while( n > 0, resul *= n %10; n \= 10; ); return(resul); } \\ R. J. Mathar, May 23 2006, edited by M. F. Hasler, Apr 23 2015
  • PARI
    A007954(n)=prod(i=1,#n=Vecsmall(Str(n)),n[i]-48) \\ (...eval(Vec(...)),n[i]) is about 50% slower; (...digits(n)...) about 6% slower. \\ M. F. Hasler, Dec 06 2009
  • PARI
    a(n)=if(n,factorback(digits(n)),0) \\ Charles R Greathouse IV, Apr 14 2020
  • Python
    from math import prod
def a(n): return prod(map(int, str(n)))
print([a(n) for n in range(108)]) # Michael S. Branicky, Jan 16 2022
  • Scala
    (0 to 99).map(.toString.toCharArray.map( - 48).scanRight(1)( * ).head) // Alonso del Arte, Apr 14 2020

Formula

A000035(a(A014261(n))) = 1. - Reinhard Zumkeller, Nov 30 2007
a(n) = abs(A187844(n)). - Reinhard Zumkeller, Mar 14 2011
a(n) > 0 if and only if A054054(n) > 0. a(n) = d in {1, ..., 9} if n = (10^k - 1)/9 + (d - 1)*10^m = A002275(k) + (d - 1)*A011557(m) for some k > m >= 0. The statement holds with "if and only if" for d in {1, 2, 3, 5, 7}. For d = 4, 6, 8 or 9, one has a(n) = d if n = (10^k - 1)/9 + (a - 1)*10^m + (b - 1)*10^p with integers k > m > p >= 0 and a, b > 0 such that d = a*b. - M. F. Hasler, Oct 11 2015
From Robert Israel, May 17 2016: (Start)
G.f.: Sum_{n >= 0} Product_{j = 0..n} Sum_{k = 1..9} k*x^(k*10^j).
G.f. satisfies A(x) = (x + 2*x^2 + ... + 9*x^9)*(1 + A(x^10)). (End)
a(n) <= 9^(1 + log_10(n/9)). - Lucas A. Brown, Jun 22 2023

Extensions

Error in term 25 corrected, Nov 15 1995

A007051 a(n) = (3^n + 1)/2.

Original entry on oeis.org

1, 2, 5, 14, 41, 122, 365, 1094, 3281, 9842, 29525, 88574, 265721, 797162, 2391485, 7174454, 21523361, 64570082, 193710245, 581130734, 1743392201, 5230176602, 15690529805, 47071589414, 141214768241, 423644304722, 1270932914165, 3812798742494, 11438396227481
Offset: 0

Views

Author

Colin Mallows, N. J. A. Sloane, Simon Plouffe, Robert G. Wilson v

Keywords

Comments

Number of ordered trees with n edges and height at most 4.
Number of palindromic structures using a maximum of three different symbols. - Marks R. Nester
Number of compositions of all even natural numbers into n parts <= 2 (0 is counted as a part), see example. - Adi Dani, May 14 2011
Consider the mapping f(a/b) = (a + 2*b)/(2*a + b). Taking a = 1, b = 2 to start with, and carrying out this mapping repeatedly on each new (reduced) rational number gives the sequence 1/2, 4/5, 13/14, 40/41, ... converging to 1. The sequence contains the denominators = (3^n+1)/2. The same mapping for N, i.e., f(a/b) = (a + N*b)/(a+b) gives fractions converging to N^(1/2). - Amarnath Murthy, Mar 22 2003
Second binomial transform of the expansion of cosh(x). - Paul Barry, Apr 05 2003
The sequence (1, 1, 2, 5, ...) = 3^n/6 + 1/2 + 0^n/3 has binomial transform A007581. - Paul Barry, Jul 20 2003
Number of (s(0), s(1), ..., s(2n+2)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| = 1 for i = 1, 2, ..., 2n+2, s(0) = 1, s(2n+2) = 1. - Herbert Kociemba, Jun 10 2004
Density of regular language L over {1,2,3}^* (i.e., number of strings of length n in L) described by regular expression 11*+11*2(1+2)*+11*2(1+2)*3(1+2+3)*. - Nelma Moreira, Oct 10 2004
Sums of rows of the triangle in A119258. - Reinhard Zumkeller, May 11 2006
Number of n-words from the alphabet A = {a,b,c} which contain an even number of a's. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 30 2006
Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which either 0) x and y are disjoint and for which x is not a subset of y and y is not a subset of x, or 1) x = y. - Ross La Haye, Jan 10 2008
a(n+1) gives the number of primitive periodic multiplex juggling sequences of length n with base state <2>. - Steve Butler, Jan 21 2008
a(n) is also the number of idempotent order-preserving and order-decreasing partial transformations (of an n-chain). - Abdullahi Umar, Oct 02 2008
Equals row sums of triangle A147292. - Gary W. Adamson, Nov 05 2008
Equals leftmost column of A071919^3. - Gary W. Adamson, Apr 13 2009
A010888(a(n))=5 for n >= 2, that is, the digital root of the terms >= 5 equals 5. - Parthasarathy Nambi, Jun 03 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=5, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^n*charpoly(A,2). - Milan Janjic, Jan 27 2010
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=6, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=(-1)^(n-1)*charpoly(A,3). - Milan Janjic, Feb 21 2010
It appears that if s(n) is a rational sequence of the form s(1)=2, s(n)= (2*s(n-1)+1)/(s(n-1)+2), n>1 then s(n)=a(n)/(a(n-1)-1).
Form an array with m(1,n)=1 and m(i,j) = Sum_{k=1..i-1} m(k,j) + Sum_{k=1..j-1} m(i,k), which is the sum of the terms to the left of m(i,j) plus the sum above m(i,j). The sum of the terms in antidiagonal(n-1) = a(n). - J. M. Bergot, Jul 16 2013
From Peter Bala, Oct 29 2013: (Start)
An Engel expansion of 3 to the base b := 3/2 as defined in A181565, with the associated series expansion 3 = b + b^2/2 + b^3/(2*5) + b^4/(2*5*14) + .... Cf. A034472.
More generally, for a positive integer n >= 3, the sequence [1, n - 1, n^2 - n - 1, ..., ( (n - 2)*n^k + 1 )/(n - 1), ...] is an Engel expansion of n/(n - 2) to the base n/(n - 1). Cases include A007583 (n = 4), A083065 (n = 5) and A083066 (n = 6). (End)
Diagonal elements (and one more than antidiagonal elements) of the matrix A^n where A=(2,1;1,2). - David Neil McGrath, Aug 17 2014
From M. Sinan Kul, Sep 07 2016: (Start)
a(n) is equal to the number of integer solutions to the following equation when x is equal to the product of n distinct primes: 1/x = 1/y + 1/z where 0 < x < y <= z.
If z = k*y where k is a fraction >= 1 then the solutions can be given as: y = ((k+1)/k)*x and z = (k+1)*x.
Here k can be equal to any divisor of x or to the ratio of two divisors.
For example for x = 2*3*5 = 30 (product of three distinct primes), k would have the following 14 values: 1, 6/5, 3/2, 5/3, 2, 5/2, 3, 10/3, 5, 6, 15/2, 10, 15, 30.
As an example for k = 10/3, we would have y=39, z=130 and 1/39 + 1/130 = 1/30.
Here finding the number of fractions would be equivalent to distributing n balls (distinct primes) to two bins (numerator and denominator) with no empty bins which can be found using Stirling numbers of the second kind. So another definition for a(n) is: a(n) = 2^n + Sum_{i=2..n} Stirling2(i,2)*binomial(n,i).
(End)
a(n+1) is the smallest i for which the Catalan number C(i) (see A000108) is divisible by 3^n for n > 0. This follows from the rule given by Franklin T. Adams-Watters for determining the multiplicity with which a prime divides C(n). We need to find the smallest number in base 3 to achieve a given count. Applied to prime 3, 1 is the smallest digit that counts but requires to be followed by 2 which cannot be at the end to count. Therefore the number in base 3 of the form 1{n-1 times}20 = (3^(n+1) + 1)/2 + 1 = a(n+1)+1 is the smallest number to achieve count n which implies the claim. - Peter Schorn, Mar 06 2020
Let A be a Toeplitz matrix of order n, defined by: A[i,j]=1, if ij; A[i,i]=2. Then, for n>=1, a(n) = det A. - Dmitry Efimov, Oct 28 2021
a(n) is the least number k such that A065363(k) = -(n-1), for n > 0. - Amiram Eldar, Sep 03 2022

Examples

			From _Adi Dani_, May 14 2011: (Start)
a(3)=14 because all compositions of even natural numbers into 3 parts <=2 are
for 0: (0,0,0)
for 2: (0,1,1), (1,0,1), (1,1,0), (0,0,2), (0,2,0), (2,0,0)
for 4: (0,2,2), (2,0.2), (2,2,0), (1,1,2), (1,2,1), (2,1,1)
for 6: (2,2,2).
(End)

References

  • J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 47.
  • Adi Dani, Quasicompositions of natural numbers, Proceedings of III congress of mathematicians of Macedonia, Struga Macedonia 29 IX -2 X 2005 pages 225-238.
  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section E11.
  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]
  • P. Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 60.
  • P. Ribenboim, The Little Book of Big Primes, Springer-Verlag, NY, 1991, p. 53.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A056449, A064881-A064886, A008277, A007581, A056272, A056273, A000392, A000079, A034472, A147292, A003462, A065363, A071919, A007583, A083065, A083066.
A row of the array in A278984.

Programs

Formula

a(n) = 3*a(n-1) - 1.
Binomial transform of Chebyshev coefficients A011782. - Paul Barry, Mar 16 2003
From Paul Barry, Mar 16 2003: (Start)
a(n) = 4*a(n-1) - 3*a(n-2) for n > 1, a(0)=1, a(1)=2.
G.f.: (1 - 2*x)/((1 - x)*(1 - 3*x)). (End)
E.g.f.: exp(2*x)*cosh(x). - Paul Barry, Apr 05 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*2^(n-2*k). - Paul Barry, May 08 2003
This sequence is also the partial sums of the first 3 Stirling numbers of second kind: a(n) = S(n+1, 1) + S(n+1, 2) + S(n+1, 3) for n >= 0; alternatively it is the number of partitions of [n+1] into 3 or fewer parts. - Mike Zabrocki, Jun 21 2004
For c=3, a(n) = (c^n)/c! + Sum_{k=1..c-2}((k^n)/k!*(Sum_{j=2..c-k}(((-1)^j)/j!))) or = Sum_{k=1..c} g(k, c)*k^n where g(1, 1) = 1, g(1, c) = g(1, c-1) + ((-1)^(c-1))/(c-1)! for c > 1, and g(k, c) = g(k-1, c-1)/k for c > 1 and 2 <= k <= c. - Nelma Moreira, Oct 10 2004
The i-th term of the sequence is the entry (1, 1) in the i-th power of the 2 X 2 matrix M = ((2, 1), (1, 2)). - Simone Severini, Oct 15 2005
If p[i]=fibonacci(2i-3) and if A is the Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)= det A. - Milan Janjic, May 08 2010
INVERT transform of A001519: [1, 1, 2, 5, 13, 34, ...]. - Gary W. Adamson, Jun 13 2011
a(n) = M^n*[1,1,1,0,0,0,...], leftmost column term; where M = an infinite bidiagonal matrix with all 1's in the superdiagonal and (1,2,3,...) in the main diagonal and the rest zeros. - Gary W. Adamson, Jun 23 2011
a(n) = M^n*[1,1,1,0,0,0,...], top entry term; where M is an infinite bidiagonal matrix with all 1's in the superdiagonal, (1,2,3,...) as the main diagonal, and the rest zeros. - Gary W. Adamson, Jun 24 2011
a(n) = A201730(n,0). - Philippe Deléham, Dec 05 2011
a(n) = A006342(n) + A006342(n-1). - Yuchun Ji, Sep 19 2018
From Dmitry Efimov, Oct 29 2021: (Start)
a(2*m+1) = Product_{k=-m..m} (2+i*tan(Pi*k/(2*m+1))),
a(2*m) = Product_{k=-m..m-1} (2+i*tan(Pi*(2*k+1)/(4*m))),
where i is the imaginary unit. (End)

A002283 a(n) = 10^n - 1.

Original entry on oeis.org

0, 9, 99, 999, 9999, 99999, 999999, 9999999, 99999999, 999999999, 9999999999, 99999999999, 999999999999, 9999999999999, 99999999999999, 999999999999999, 9999999999999999, 99999999999999999, 999999999999999999, 9999999999999999999, 99999999999999999999, 999999999999999999999, 9999999999999999999999
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

A friend from Germany remarks that the sequence 9, 99, 999, 9999, 99999, 999999, ... might be called the grumpy German sequence: nein!, nein! nein!, nein! nein! nein!, ...
The Regan link shows that integers of the form 10^n -1 have binary representations with exactly n trailing 1 bits. Also those integers have quinary expressions with exactly n trailing 4's. For example, 10^4 -1 = (304444)5. The first digits in quinary correspond to the number 2^n -1, in our example (30)5 = 2^4 -1. A similar pattern occurs in the binary case. Consider 9 = (1001)2. - Washington Bomfim Dec 23 2010
a(n) is the number of positive integers with less than n+1 digits. - Bui Quang Tuan, Mar 09 2015
From Peter Bala, Sep 27 2015: (Start)
For n >= 1, the simple continued fraction expansion of sqrt(a(2*n)) = [10^n - 1; 1, 2*(10^n - 1), 1, 2*(10^n - 1), ...] has period 2. The simple continued fraction expansion of sqrt(a(2*n))/a(n) = [1; 10^n - 1, 2, 10^n - 1, 2, ...] also has period 2. Note the occurrence of large partial quotients in both expansions.
A theorem of Kuzmin in the measure theory of continued fractions says that large partial quotients are the exception in continued fraction expansions.
Empirically, we also see the presence of unexpectedly large partial quotients early in the continued fraction expansions of the m-th roots of the numbers a(m*n) for m >= 3. Some typical examples are given below. (End)
For n > 0, numbers whose smallest decimal digit is 9. - Stefano Spezia, Nov 16 2023

Examples

			From _Peter Bala_, Sep 27 2015: (Start)
Continued fraction expansions showing large partial quotients:
a(12)^(1/3) = [9999; 1, 299999998, 1, 9998, 1, 449999998, 1, 7998, 1, 535714284, 1, 2, 2, 142, 2, 2, 1, 599999999, 3, 1, 1,...].
Compare with a(30)^(1/3) = [9999999999; 1, 299999999999999999998, 1, 9999999998, 1, 449999999999999999998, 1, 7999999998, 1, 535714285714285714284, 1, 2, 2, 142857142, 2, 2, 1, 599999999999999999999, 3, 1, 1,...].
a(24)^(1/4) = [999999; 1, 3999999999999999998, 1, 666665, 1, 1, 1, 799999999999999999, 3, 476190, 7, 190476190476190476, 21, 43289, 1, 229, 1, 1864801864801863, 1, 4, 6,...].
Compare with a(48)^(1/4) = [999999999999; 1, 3999999999999999999999999999999999998, 1, 666666666665, 1, 1, 1, 799999999999999999999999999999999999, 3, 476190476190, 7, 190476190476190476190476190476190476, 21, 43290043289, 1, 229, 1, 1864801864801864801864801864801863, 1, 4, 6,...].
a(25)^(1/5) = [99999, 1, 499999999999999999998, 1, 49998, 1, 999999999999999999998, 1, 33332, 3, 151515151515151515151, 5, 1, 1, 1947, 1, 1, 38, 3787878787878787878, 1, 3, 5,...].
(End)

Links

Crossrefs

Cf. A000533, A003020, A007138, A007953, A008591, A010888, A048379, A066138, A073668, A168624, A276352.
Cf. A002275, A002276, A002277, A002278, A002279, A002280, A002281, A002282.
Cf. A075412, A075415, A178630, A178631, A178632, A178633, A178634, A178635.
Cf. A010785.

Programs

Formula

From Mohammad K. Azarian, Jan 14 2009: (Start)
G.f.: 1/(1-10*x)-1/(1-x).
E.g.f.: e^(10*x)-e^x. (End)
a(n) = A075412(n)/A002275(n) = A178630(n)/A002276(n) = A178631(n)/A002277(n) = A075415(n)/A002278(n) = A178632(n)/A002279(n) = A178633(n)/A002280(n) = A178634(n)/A002281(n) = A178635(n)/A002282(n). - Reinhard Zumkeller, May 31 2010
a(n) = a(n-1) + 9*10^(n-1) with a(0)=0; Also: a(n) = 11*a(n-1) - 10*a(n-2) with a(0)=0, a(1)=9. - Vincenzo Librandi, Jul 22 2010
For n>0, A007953(a(n)) = A008591(n) and A010888(a(n)) = 9. - Reinhard Zumkeller, Aug 06 2010
A048379(a(n)) = 0. - Reinhard Zumkeller, Feb 21 2014
a(n) = Sum_{k=1..n} 9*10^k. - Carauleanu Marc, Sep 03 2016
Sum_{n>=1} 1/a(n) = A073668. - Amiram Eldar, Nov 13 2020
From Elmo R. Oliveira, Jul 19 2025: (Start)
a(n) = 9*A002275(n).
a(n) = A010785(A008591(n)). (End)

Extensions

More terms from Michael De Vlieger, Sep 27 2015

A002411 Pentagonal pyramidal numbers: a(n) = n^2*(n+1)/2.

Original entry on oeis.org

0, 1, 6, 18, 40, 75, 126, 196, 288, 405, 550, 726, 936, 1183, 1470, 1800, 2176, 2601, 3078, 3610, 4200, 4851, 5566, 6348, 7200, 8125, 9126, 10206, 11368, 12615, 13950, 15376, 16896, 18513, 20230, 22050, 23976, 26011, 28158, 30420, 32800, 35301, 37926, 40678
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n) = n^2(n+1)/2 is half the number of colorings of three points on a line with n+1 colors. - R. H. Hardin, Feb 23 2002
Sum of n smallest multiples of n. - Amarnath Murthy, Sep 20 2002
a(n) = number of (n+6)-bit binary sequences with exactly 7 1's none of which is isolated. A 1 is isolated if its immediate neighbor(s) are 0. - David Callan, Jul 15 2004
Also as a(n) = (1/6)*(3*n^3+3*n^2), n > 0: structured trigonal prism numbers (cf. A100177 - structured prisms; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Nov 18 2005
If Y is a 3-subset of an n-set X then, for n >= 5, a(n-4) is the number of 5-subsets of X having at least two elements in common with Y. - Milan Janjic, Nov 23 2007
a(n-1), n >= 2, is the number of ways to have n identical objects in m=2 of altogether n distinguishable boxes (n-2 boxes stay empty). - Wolfdieter Lang, Nov 13 2007
a(n+1) is the convolution of (n+1) and (3n+1). - Paul Barry, Sep 18 2008
The number of 3-character strings from an alphabet of n symbols, if a string and its reversal are considered to be the same.
Partial sums give A001296. - Jonathan Vos Post, Mar 26 2011
a(n-1):=N_1(n), n >= 1, is the number of edges of n planes in generic position in three-dimensional space. See a comment under A000125 for general arrangement. Comment to Arnold's problem 1990-11, see the Arnold reference, p.506. - Wolfdieter Lang, May 27 2011
Partial sums of pentagonal numbers A000326. - Reinhard Zumkeller, Jul 07 2012
From Ant King, Oct 23 2012: (Start)
For n > 0, the digital roots of this sequence A010888(A002411(n)) form the purely periodic 9-cycle {1,6,9,4,3,9,7,9,9}.
For n > 0, the units' digits of this sequence A010879(A002411(n)) form the purely periodic 20-cycle {1,6,8,0,5,6,6,8,5,0,6,6,3,0,0,6,1,8,0,0}.
(End)
a(n) is the number of inequivalent ways to color a path graph having 3 nodes using at most n colors. Note, here there is no restriction on the color of adjacent nodes as in the above comment by R. H. Hardin (Feb 23 2002). Also, here the structures are counted up to graph isomorphism, where as in the above comment the "three points on a line" are considered to be embedded in the plane. - Geoffrey Critzer, Mar 20 2013
After 0, row sums of the triangle in A101468. - Bruno Berselli, Feb 10 2014
Latin Square Towers: Take a Latin square of order n, with symbols from 1 to n, and replace each symbol x with a tower of height x. Then the total number of unit cubes used is a(n). - Arun Giridhar, Mar 29 2015
This is the case k = n+4 of b(n,k) = n*((k-2)*n-(k-4))/2, which is the n-th k-gonal number. Therefore, this is the 3rd upper diagonal of the array in A139600. - Luciano Ancora, Apr 11 2015
For n > 0, a(n) is the number of compositions of n+7 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016
Also the Wiener index of the n-antiprism graph. - Eric W. Weisstein, Sep 07 2017
For n > 0, a(2n+1) is the number of non-isomorphic 5C_m-snakes, where m = 2n+1 or m = 2n (for n >= 2). A kC_n-snake is a connected graph in which the k >= 2 blocks are isomorphic to the cycle C_n and the block-cutpoint graph is a path. - Christian Barrientos, May 15 2019
For n >= 1, a(n-1) is the number of 0°- and 45°-tilted squares that can be drawn by joining points in an n X n lattice. - Paolo Xausa, Apr 13 2021
a(n) is the number of all possible products of n rolls of a six-sided die. This can be easily seen by the recursive formula a(n) = a(n - 1) + 2 * binomial(n, 2) + binomial(n + 1, 2). - Rafal Walczak, Jun 15 2024
a(n) is the number of all triples consisting of nonnegative integers smaller than n such that the sum of the first two integers is less than n. - Ruediger Jehn, Aug 17 2025

Examples

			a(3)=18 because 4 identical balls can be put into m=2 of n=4 distinguishable boxes in binomial(4,2)*(2!/(1!*1!) + 2!/2!) = 6*(2+1) = 18 ways. The m=2 part partitions of 4, namely (1,3) and (2,2), specify the filling of each of the 6 possible two-box choices. - _Wolfdieter Lang_, Nov 13 2007

References

  • V. I. Arnold (ed.), Arnold's Problems, Springer, 2004, comments on Problem 1990-11 (p. 75), pp. 503-510. Numbers N_1.
  • Christian Barrientos, Graceful labelings of cyclic snakes, Ars Combin., Vol. 60 (2001), pp. 85-96.
  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 194.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 166, Table 10.4/I/5).
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see Vol. 2, p. 2.
  • 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).

Links

Crossrefs

Cf. A001296, A011379, A015223, A015224, A014799, A014800, A127739, A132118, A139600.
A006002(n) = -a(-1-n).
a(n) = A093560(n+2, 3), (3, 1)-Pascal column.
A row or column of A132191.
Second column of triangle A103371.
Cf. similar sequences listed in A237616.

Programs

  • GAP
    List([0..45], n->n^2*(n+1)/2); # Muniru A Asiru, Feb 19 2018
  • Haskell
    a002411 n = n * a000217 n  -- Reinhard Zumkeller, Jul 07 2012
  • Magma
    [n^2*(n+1)/2: n in [0..40]]; // Wesley Ivan Hurt, May 25 2014
  • Maple
    seq(n^2*(n+1)/2, n=0..40);
  • Mathematica
    Table[n^2 (n + 1)/2, {n, 0, 40}]
LinearRecurrence[{4, -6, 4, -1}, {0, 1, 6, 18}, 50] (* Harvey P. Dale, Oct 20 2011 *)
Nest[Accumulate, Range[1, 140, 3], 2] (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *)
CoefficientList[Series[x (1 + 2 x) / (1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Jan 08 2016 *)
  • PARI
    a(n)=n^2*(n+1)/2
  • PARI
    concat(0, Vec(x*(1+2*x)/(1-x)^4 + O(x^100))) \\ Altug Alkan, Jan 07 2016

Formula

Average of n^2 and n^3.
G.f.: x*(1+2*x)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
a(n) = n*Sum_{k=0..n} (n-k) = n*Sum_{k=0..n} k. - Paul Barry, Jul 21 2003
a(n) = n*A000217(n). - Xavier Acloque, Oct 27 2003
a(n) = (1/2)*Sum_{j=1..n} Sum_{i=1..n} (i+j) = (1/2)*(n^2+n^3) = (1/2)*A011379(n). - Alexander Adamchuk, Apr 13 2006
Row sums of triangle A127739, triangle A132118; and binomial transform of [1, 5, 7, 3, 0, 0, 0, ...] = (1, 6, 18, 40, 75, ...). - Gary W. Adamson, Aug 10 2007
G.f.: x*F(2,3;1;x). - Paul Barry, Sep 18 2008
Sum_{j>=1} 1/a(j) = hypergeom([1, 1, 1], [2, 3], 1) = -2 + 2*zeta(2) = A195055 - 2. - Stephen Crowley, Jun 28 2009
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(0)=0, a(1)=1, a(2)=6, a(3)=18. - Harvey P. Dale, Oct 20 2011
From Ant King, Oct 23 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 3.
a(n) = (n+1)*(2*A000326(n)+n)/6 = A000292(n) + 2*A000292(n-1).
a(n) = A000330(n)+A000292(n-1) = A000217(n) + 3*A000292(n-1).
a(n) = binomial(n+2,3) + 2*binomial(n+1,3).
(End)
a(n) = (A000330(n) + A002412(n))/2 = (A000292(n) + A002413(n))/2. - Omar E. Pol, Jan 11 2013
a(n) = (24/(n+3)!)*Sum_{j=0..n} (-1)^(n-j)*binomial(n,j)*j^(n+3). - Vladimir Kruchinin, Jun 04 2013
Sum_{n>=1} a(n)/n! = (7/2)*exp(1). - Richard R. Forberg, Jul 15 2013
E.g.f.: x*(2 + 4*x + x^2)*exp(x)/2. - Ilya Gutkovskiy, May 31 2016
From R. J. Mathar, Jul 28 2016: (Start)
a(n) = A057145(n+4,n).
a(n) = A080851(3,n-1). (End)
For n >= 1, a(n) = (Sum_{i=1..n} i^2) + Sum_{i=0..n-1} i^2*((i+n) mod 2). - Paolo Xausa, Apr 13 2021
a(n) = Sum_{k=1..n} GCD(k,n) * LCM(k,n). - Vaclav Kotesovec, May 22 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 2 + Pi^2/6 - 4*log(2). - Amiram Eldar, Jan 03 2022

A000034 Period 2: repeat [1, 2]; a(n) = 1 + (n mod 2).

Original entry on oeis.org

1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Also continued fraction for (sqrt(3)+1)/2 (cf. A040001) and base-3 digital root of n+1 (cf. A007089, A010888). - Henry Bottomley, Jul 05 2001
The sequence 1,-2,-1,2,1,-2,-1,2,... with g.f. (1-2x)/(1+x^2) has a(n) = cos(Pi*n/2)-2*sin(Pi*n/2). - Paul Barry, Oct 18 2004
Hankel transform is [1,-3,0,0,0,0,0,0,0,...]. - Philippe Deléham, Mar 29 2007
4/33 = 0.121212... - Eric Desbiaux, Nov 03 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=A[i,i]:=1, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1) = charpoly(A,2). - Milan Janjic, Jan 24 2010
First differences of A032766. - Tom Edgar, Jul 17 2014
Denominator of the harmonic mean of the first n triangular numbers. - Colin Barker, Nov 13 2014
This is the lexicographically earliest sequence of positive integers such that no polynomial of degree d can be fitted to d+2 consecutive terms (equivalently, such that no iterated difference is zero). - Pontus von Brömssen, Dec 26 2021 [See A300002 for the case where not only consecutive terms are considered. - Pontus von Brömssen, Jan 03 2023]
Number of maximum antichains in the power set of {1,2,...,n} partially ordered by set inclusion. For even n, there is a unique maximum antichain formed by all subsets of size n/2; for odd n, there are two maximum antichains, one formed by all subsets of size (n-1)/2 and the other formed by all subsets of size (n+1)/2. See the David Guichard link below for a proof. - Jianing Song, Jun 19 2022

References

  • Jozsef Beck, Combinatorial Games, Cambridge University Press, 2008.
  • J.-M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 545 pages 73 and 260, Ellipses, Paris 2004.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A000035, A003945 (binomial transf), A007089, A010693, A010704, A010888, A032766, A040001, A123344, A134451, A300002.
Cf. sequences listed in Comments section of A283393.

Programs

Formula

G.f.: (1+2*x)/(1-x^2).
a(n) = 2^((1-(-1)^n)/2) = 2^(ceiling(n/2) - floor(n/2)). - Paul Barry, Jun 03 2003
a(n) = (3-(-1)^n)/2; a(n) = 1 + (n mod 2) = 3-a(n-1) = a(n-2) = a(-n).
a(n) = gcd(n-1, n+1). - Paul Barry, Sep 16 2004
Binomial transform of A123344, inverse binomial transform of A003945. - Philippe Deléham, Jun 04 2007
a(n) = A134451(n+1). - Reinhard Zumkeller, Oct 27 2007
a(n) = if(n=0,1,if(mod(a(n-1),2)=0,a(n-1)/2,(3*a(n-1)+1)/2)). See Collatz conjecture. - Paul Barry, Mar 31 2008
a(n) = 2^n (mod 3). - Vincenzo Librandi, Feb 05 2011
a(n) = A000035(n) + 1. - M. F. Hasler, Jan 13 2012
a(n) = abs(sin(n*Pi/2) - 2*cos(n*Pi/2)). - Mohammad K. Azarian, Mar 12 2012
a(n) = A010704(n) / 3. - Reinhard Zumkeller, Jul 03 2012
a(n) = floor((4/33)*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 03 2013
a(n) = floor((5/8)*3^(n+1)) mod 3. - Hieronymus Fischer, Jan 03 2013
a(n) = floor((n+1)*3/2) - floor((n)*3/2). - Hailey R. Olafson, Jul 23 2014
a(n) = denominator(n/2). - Wesley Ivan Hurt, Sep 11 2014
Dirichlet g.f.: zeta(s)*(1 + 1/2^s). - Mats Granvik, Jul 18 2016
E.g.f.: 2*sinh(x) + cosh(x). - Ilya Gutkovskiy, Jul 18 2016
a(n) = A010693(n) - 1. - Filip Zaludek, Oct 29 2016
a(n) = n + 1 - 2*floor(n/2). - Lorenzo Sauras Altuzarra, Jun 28 2019
Limit_{n->oo} (1/n)*Sum_{k=1..n} a(k) = 3/2 (De Koninck reference). - Bernard Schott, Nov 09 2021

Extensions

Better definition from M. F. Hasler, Jan 13 2012

A002412 Hexagonal pyramidal numbers, or greengrocer's numbers.

Original entry on oeis.org

0, 1, 7, 22, 50, 95, 161, 252, 372, 525, 715, 946, 1222, 1547, 1925, 2360, 2856, 3417, 4047, 4750, 5530, 6391, 7337, 8372, 9500, 10725, 12051, 13482, 15022, 16675, 18445, 20336, 22352, 24497, 26775, 29190, 31746, 34447, 37297, 40300
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Binomial transform of (1, 6, 9, 4, 0, 0, 0, ...). - Gary W. Adamson, Oct 16 2007
a(n) is the sum of the maximum(m,n) over {(m,n): m,n in positive integers, m<=n}. - Geoffrey Critzer, Oct 11 2009
We obtain these numbers for d=2 in the identity n*(n*(d*n-d+2)/2)-sum(k*(d*k-d+2)/2, k=0..n-1) = n*(n+1)*(2*d*n-2*d+3)/6 (see Klaus Strassburger in Formula lines). - Bruno Berselli, Apr 21 2010, Nov 16 2010
q^a(n) is the Hankel transform of the q-Catalan numbers. - Paul Barry, Dec 15 2010
Row 1 of the convolution array A213835. - Clark Kimberling, Jul 04 2012
From Ant King, Oct 24 2012: (Start)
For n>0, the digital roots of this sequence A010888(A002412(n)) form the purely periodic 27-cycle {1,7,4,5,5,8,9,3,3,4,1,7,8,8,2,3,6,6,7,4,1,2,2,5,6,9,9}.
For n>0, the units' digits of this sequence A010879(A002412(n)) form the purely periodic 20-cycle {1,7,2,0,5,1,2,2,5,5,6,2,7,5,0,6,7,7,0,0}.
(End)
Partial sums of A000384. - Omar E. Pol, Jan 12 2013
Row sums of A094728. - J. M. Bergot, Jun 14 2013
Number of orbits of Aut(Z^7) as function of the infinity norm (n+1) of the representative integer lattice point of the orbit, when the cardinality of the orbit is equal to 40320. - Philippe A.J.G. Chevalier, Dec 28 2015
Coefficients in the hypergeometric series identity 1 - 7*(x - 1)/(3*x + 1) + 22*(x - 1)*(x - 2)/((3*x + 1)*(3*x + 2)) - 50*(x - 1)*(x - 2)*(x - 3)/((3*x + 1)*(3*x + 2)*(3*x + 3)) + ... = 0, valid for Re(x) > 1. Cf. A000326 and A002418. Column 3 of A103450. - Peter Bala, Mar 14 2019

Examples

			Let n=5, 2*n=10. Since 10 = 1 + 9 = 2 + 8 = 3 + 7 = 4 + 6 = 5 + 5, a(5) = 1*9 + 2*8 + 3*7 + 4*6 + 5*5 = 95. - _Vladimir Shevelev_, May 11 2012

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 194.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 2.
  • T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.
  • I. Siap, Linear codes over F_2 + u*F_2 and their complete weight enumerators, in Codes and Designs (Ohio State, May 18, 2000), pp. 259-271. De Gruyter, 2002.
  • 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).

Links

Crossrefs

Bisection of A002623. Equals A000578(n) - A000330(n-1).
Cf. A000292, A016061.
a(n) = A093561(n+2, 3), (4, 1)-Pascal column.
Cf. A220084 for a list of numbers of the form n*P(k,n)-(n-1)*P(k,n-1), where P(k,n) is the n-th k-gonal pyramidal number (see Adamson's formula).
Cf. similar sequences listed in A237616.
Orbits of Aut(Z^7) as function of the infinity norm A000579, A154286, A102860, A002412, A045943, A115067, A008585, A005843, A001477, A000217.

Programs

  • GAP
    List([0..40],n->n*(n+1)*(4*n-1)/6); # Muniru A Asiru, Mar 18 2019
  • Magma
    [n*(n+1)*(4*n-1)/6: n in [0..40]]; // Vincenzo Librandi, Nov 28 2015
  • Maple
    seq(sum(i*(2*k-i), i=1..k), k=0..100); # Wesley Ivan Hurt, Sep 25 2013
  • Mathematica
    Figurate[ ngon_, rank_, dim_] := Binomial[rank + dim - 2, dim - 1] ((rank - 1)*(ngon - 2) + dim)/dim; Table[ Figurate[6, r, 3], {r, 0, 40}] (* Robert G. Wilson v, Aug 22 2010 *)
Table[n(n+1)(4n-1)/6, {n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1}, {0,1,7,22}, 40] (* Harvey P. Dale, Jul 16 2011 *)
  • Maxima
    A002412(n):=n*(n+1)*(4*n-1)/6$ makelist(A002412(n),n,0,20); /* Martin Ettl, Dec 12 2012 */
  • PARI
    v=vector(40,i,(i*(i+1))\2); s=0; print1(s","); forstep(i=1,40,2,s+=v[i]; print1(s","))
  • Python
    print([n*(n+1)*(4*n-1)//6 for n in range(40)]) # Michael S. Branicky, Mar 28 2022

Formula

a(n) = n(n + 1)(4n - 1)/6.
G.f.: x*(1+3*x)/(1-x)^4. - Simon Plouffe in his 1992 dissertation.
a(n) = n^3 - Sum_{i=1..n-1} i^2. - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de)
Partial sums of n odd-indexed triangular numbers, e.g., a(3) = t(1)+t(3)+t(5) = 1+6+15 = 22. - Jon Perry, Jul 23 2003
a(n) = Sum_{i=0..n-1} (n - i)*(n + i). - Jon Perry, Sep 26 2004
a(n) = n*A000292(n) - (n-1)*A000292(n-1) = n*binomial((n+2),3) - (n-1)*binomial((n+1),3); e.g., a(5) = 95 = 5*35 - 4*20. - Gary W. Adamson, Dec 28 2007
a(n) = Sum_{i=0..n} (2i^2 + 3i + 1), for n >= 0 (Omits the leading 0). - William A. Tedeschi, Aug 25 2010
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4), with a(0)=0, a(1)=1, a(2)=7, a(3)=22. - Harvey P. Dale, Jul 16 2011
a(n) = sum a*b, where the summing is over all unordered partitions 2*n = a+b. - Vladimir Shevelev, May 11 2012
From Ant King, Oct 24 2012: (Start)
a(n) = a(n-1) + n*(2*n-1).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 4.
a(n) = (n+1)*(2*A000384(n) + n)/6 = (4*n-1)*A000217(n)/3.
a(n) = A000292(n) + 3*A000292(n-1) = A002411(n) + A000292(n-1).
a(n) = binomial(n+2,3) + 3*binomial(n+1,3) = (4*n-1)*binomial(n+1,2)/3.
Sum_{n>=1} 1/a(n) = 6*(12*log(2)-2*Pi-1)/5 = 1.2414...
(End)
a(n) = Sum_{i=1..n} Sum_{j=1..n} max(i,j) = Sum_{i=1..n} i*(2*n-i). - Enrique Pérez Herrero, Jan 15 2013
a(n) = A005900(n+1) - A000326(n+1) = Octahedral - Pentagonal Numbers. - Richard R. Forberg, Aug 07 2013
a(n) = n*A000217(n) + Sum_{i=0..n-1} A000217(i). - Bruno Berselli, Dec 18 2013
a(n) = 2n * A000217(n) - A000330(n). - J. M. Bergot, Apr 05 2014
a(n) = A080851(4,n-1). - R. J. Mathar, Jul 28 2016
E.g.f.: x*(6 + 15*x + 4*x^2)*exp(x)/6. - Ilya Gutkovskiy, May 12 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 6*(1 + 2*sqrt(2)*Pi - 2*(3+sqrt(2))*log(2) + 4*sqrt(2)*log(2-sqrt(2)))/5. - Amiram Eldar, Jan 04 2022

A001110 Square triangular numbers: numbers that are both triangular and square.

Original entry on oeis.org

0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025, 63955431761796, 2172602007770041, 73804512832419600, 2507180834294496361, 85170343853180456676, 2893284510173841030625, 98286503002057414584576, 3338847817559778254844961, 113422539294030403250144100
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Satisfies a recurrence of S_r type for r=36: 0, 1, 36 and a(n-1)*a(n+1)=(a(n)-1)^2. First observed by Colin Dickson in alt.math.recreational, Mar 07 2004. - Rainer Rosenthal, Mar 14 2004
For every n, a(n) is the first of three triangular numbers in geometric progression. The third number in the progression is a(n+1). The middle triangular number is sqrt(a(n)*a(n+1)). Chen and Fang prove that four distinct triangular numbers are never in geometric progression. - T. D. Noe, Apr 30 2007
The sum of any two terms is never equal to a Fermat number. - Arkadiusz Wesolowski, Feb 14 2012
Conjecture: No a(2^k), where k is a nonnegative integer, can be expressed as a sum of a positive square number and a positive triangular number. - Ivan N. Ianakiev, Sep 19 2012
For n=2k+1, A010888(a(n))=1 and for n=2k, k > 0, A010888(a(n))=9. - Ivan N. Ianakiev, Oct 12 2013
For n > 0, these are the triangular numbers which are the sum of two consecutive triangular numbers, for instance 36 = 15 + 21 and 1225 = 595 + 630. - Michel Marcus, Feb 18 2014
The sequence is the case P1 = 36, P2 = 68, Q = 1 of the 3-parameter family of 4th order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014
For n=2k, k > 0, a(n) is divisible by 12 and is therefore abundant. I conjecture that for n=2k+1 a(n) is deficient [true for k up to 43 incl.]. - Ivan N. Ianakiev, Sep 30 2014
The conjecture is true for all k > 0 because: For n=2k+1, k > 0, a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. - Muniru A Asiru, Apr 13 2016
Numbers k for which A139275(k) is a perfect square. - Bruno Berselli, Jan 16 2018

Examples

			a(2) = ((17 + 12*sqrt(2))^2 + (17 - 12*sqrt(2))^2 - 2)/32 = (289 + 24*sqrt(2) + 288 + 289 - 24*sqrt(2) + 288 - 2)/32 = (578 + 576 - 2)/32 = 1152/32 = 36 and 6^2 = 36 = 8*9/2 => a(2) is both the 6th square and the 8th triangular number.

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 38, 204.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923; see Vol. 2, p. 10.
  • Martin Gardner, Time Travel and other Mathematical Bewilderments, Freeman & Co., 1988, pp. 16-17.
  • Miodrag S. Petković, Famous Puzzles of Great Mathematicians, Amer. Math. Soc. (AMS), 2009, p. 64.
  • J. H. Silverman, A Friendly Introduction to Number Theory, Prentice Hall, 2001, p. 196.
  • 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 257-259.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 93.

Links

Crossrefs

Cf. A001108, A001109.
Other S_r type sequences are S_4=A000290, S_5=A004146, S_7=A054493, S_8=A001108, S_9=A049684, S_20=A049683, S_36=this sequence, S_49=A049682, S_144=A004191^2.
Cf. A001014; intersection of A000217 and A000290; A010052(a(n))*A010054(a(n)) = 1.
Cf. A005214, A054686, A232847 and also A233267 (reveals an interesting divisibility pattern for this sequence).
Cf. A240129 (triangular numbers that are squares of triangular numbers), A100047.
See A229131, A182334, A299921 for near-misses.

Programs

  • Haskell
    a001110 n = a001110_list !! n
a001110_list = 0 : 1 : (map (+ 2) $
   zipWith (-) (map (* 34) (tail a001110_list)) a001110_list)
-- Reinhard Zumkeller, Oct 12 2011
  • Magma
    [n le 2 select n-1 else Floor((6*Sqrt(Self(n-1)) - Sqrt(Self(n-2)))^2): n in [1..20]]; // Vincenzo Librandi, Jul 22 2015
  • Maple
    a:=17+12*sqrt(2); b:=17-12*sqrt(2); A001110:=n -> expand((a^n + b^n - 2)/32); seq(A001110(n), n=0..20); # Jaap Spies, Dec 12 2004
A001110:=-(1+z)/((z-1)*(z**2-34*z+1)); # Simon Plouffe in his 1992 dissertation
  • Mathematica
    f[n_]:=n*(n+1)/2; lst={}; Do[If[IntegerQ[Sqrt[f[n]]],AppendTo[lst,f[n]]],{n,0,10!}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)
Table[(1/8) Round[N[Sinh[2 n ArcSinh[1]]^2, 100]], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *)
Transpose[NestList[Flatten[{Rest[#],34Last[#]-First[#]+2}]&, {0,1},20]][[1]]  (* Harvey P. Dale, Mar 25 2011 *)
LinearRecurrence[{35, -35, 1}, {0, 1, 36}, 20] (* T. D. Noe, Mar 25 2011 *)
LinearRecurrence[{6,-1},{0,1},20]^2 (* Harvey P. Dale, Oct 22 2012 *)
(* Square = Triangular = Triangular = A001110 *)
ChebyshevU[#-1,3]^2==Binomial[ChebyshevT[#/2,3]^2,2]==Binomial[(1+ChebyshevT[#,3])/2,2]=={1,36,1225,41616,1413721}[[#]]&@Range[5]
True (* Bill Gosper, Jul 20 2015 *)
L=0;r={};Do[AppendTo[r,L];L=1+17*L+6*Sqrt[L+8*L^2],{i,1,19}];r (* Kebbaj Mohamed Reda, Aug 02 2023 *)
  • PARI
    a=vector(100);a[1]=1;a[2]=36;for(n=3,#a,a[n]=34*a[n-1]-a[n-2]+2);a \\ Charles R Greathouse IV, Jul 25 2011
  • Scheme
    ;; With memoizing definec-macro from Antti Karttunen's IntSeq-library.
(definec (A001110 n) (if (< n 2) n (+ 2 (- (* 34 (A001110 (- n 1))) (A001110 (- n 2))))))
;; Antti Karttunen, Dec 06 2013
  • Scheme
    ;; For testing whether n is in this sequence:
(define (inA001110? n) (and (zero? (A068527 n)) (inA001109? (floor->exact (sqrt n)))))
(define (inA001109? n) (= (* 8 n n) (floor->exact (* (sqrt 8) n (ceiling->exact (* (sqrt 8) n))))))
;; Antti Karttunen, Dec 06 2013

Formula

a(0) = 0, a(1) = 1; for n >= 2, a(n) = 34 * a(n-1) - a(n-2) + 2.
G.f.: x*(1 + x) / (( 1 - x )*( 1 - 34*x + x^2 )).
a(n-1) * a(n+1) = (a(n)-1)^2. - Colin Dickson, posting to alt.math.recreational, Mar 07 2004
If L is a square-triangular number, then the next one is 1 + 17*L + 6*sqrt(L + 8*L^2). - Lekraj Beedassy, Jun 27 2001
a(n) - a(n-1) = A046176(n). - Sophie Kuo (ejiqj_6(AT)yahoo.com.tw), May 27 2006
a(n) = A001109(n)^2 = A001108(n)*(A001108(n)+1)/2 = (A000129(n)*A001333(n))^2 = (A000129(n)*(A000129(n) + A000129(n-1)))^2. - Henry Bottomley, Apr 19 2000
a(n) = (((17+12*sqrt(2))^n) + ((17-12*sqrt(2))^n)-2)/32. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002
Limit_{n->oo} a(n+1)/a(n) = 17 + 12*sqrt(2). See UWC problem link and solution. - Jaap Spies, Dec 12 2004
From Antonio Alberto Olivares, Nov 07 2003: (Start)
a(n) = 35*(a(n-1) - a(n-2)) + a(n-3);
a(n) = -1/16 + ((-24 + 17*sqrt(2))/2^(11/2))*(17 - 12*sqrt(2))^(n-1) + ((24 + 17*sqrt(2))/2^(11/2))*(17 + 12*sqrt(2))^(n-1). (End)
a(n+1) = (17*A029547(n) - A091761(n) - 1)/16. - R. J. Mathar, Nov 16 2007
a(n) = A001333^2 * A000129^2 = A000129(2*n)^2/4 = binomial(A001108,2). - Bill Gosper, Jul 28 2008
Closed form (as square = triangular): ( (sqrt(2)+1)^(2*n)/(4*sqrt(2)) - (1-sqrt(2))^(2*n)/(4*sqrt(2)) )^2 = (1/2) * ( ( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2 + 1 )*( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2. - Bill Gosper, Jul 25 2008
a(n) = (1/8)*(sinh(2*n*arcsinh(1)))^2. - Artur Jasinski, Feb 10 2010
a(n) = floor((17 + 12*sqrt(2))*a(n-1)) + 3 = floor(3*sqrt(2)/4 + (17 + 12*sqrt(2))*a(n-1) + 1). - Manuel Valdivia, Aug 15 2011
a(n) = (A011900(n) + A001652(n))^2; see the link about the generalized proof of square triangular numbers. - Kenneth J Ramsey, Oct 10 2011
a(2*n+1) = A002315(n)^2*(A002315(n)^2 + 1)/2. - Ivan N. Ianakiev, Oct 10 2012
a(2*n+1) = ((sqrt(t^2 + (t+1)^2))*(2*t+1))^2, where t = (A002315(n) - 1)/2. - Ivan N. Ianakiev, Nov 01 2012
a(2*n) = A001333(2*n)^2 * (A001333(2*n)^2 - 1)/2, and a(2*n+1) = A001333(2*n+1)^2 * (A001333(2*n+1)^2 + 1)/2. The latter is equivalent to the comment above from Ivan using A002315, which is a bisection of A001333. Using A001333 shows symmetry and helps show that a(n) are both "squares of triangular" and "triangular of squares". - Richard R. Forberg, Aug 30 2013
a(n) = (A001542(n)/2)^2.
From Peter Bala, Apr 03 2014: (Start)
a(n) = (T(n,17) - 1)/16, where T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = U(n-1,3)^2, for n >= 1, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -17; 1, 18].
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
a(n) = A096979(2*n-1) for n > 0. - Ivan N. Ianakiev, Jun 21 2014
a(n) = (6*sqrt(a(n-1)) - sqrt(a(n-2)))^2. - Arkadiusz Wesolowski, Apr 06 2015
From Daniel Poveda Parrilla, Jul 16 2016 and Sep 21 2016: (Start)
a(n) = A000290(A002965(2*n)*A002965(2*n + 1)) (after Hugh Darwen).
a(n) = A000217(2*(A000129(n))^2 - (A000129(n) mod 2)).
a(n) = A000129(n)^4 + Sum_{k=0..(A000129(n)^2 - (A000129(n) mod 2))} 2*k. This formula can be proved graphically by taking the corresponding triangle of a square triangular number and cutting both acute angles, one level at a time (sum of consecutive even numbers), resulting in a square of squares (4th powers).
a(n) = A002965(2*n)^4 + Sum_{k=A002965(2*n)^2..A002965(2*n)*A002965(2*n + 1) - 1} 2*k + 1. This formula takes an equivalent sum of consecutives, but odd numbers. (End)
E.g.f.: (exp((17-12*sqrt(2))*x) + exp((17+12*sqrt(2))*x) - 2*exp(x))/32. - Ilya Gutkovskiy, Jul 16 2016

A037074 Numbers that are the product of a pair of twin primes.

Original entry on oeis.org

15, 35, 143, 323, 899, 1763, 3599, 5183, 10403, 11663, 19043, 22499, 32399, 36863, 39203, 51983, 57599, 72899, 79523, 97343, 121103, 176399, 186623, 213443, 272483, 324899, 359999, 381923, 412163, 435599, 656099, 675683, 685583, 736163
Offset: 1

Views

Author

Felice Russo

Keywords

Comments

Each entry is the product of p and p+2 where both p and p+2 are prime, i.e., the product of the lesser and greater of a twin prime pair.
Except for the first term, all entries have digital root 8. - Lekraj Beedassy, Jun 11 2004
The above statement follows from p > 3 => (p,p+2) = (6k-1,6k+1) => p*(p+2) = 36k^2 - 1 == 8 (mod 9), and A010888 === A010878 (mod 9). - M. F. Hasler, Jan 11 2013
Albert A. Mullin states that m is a product of twin primes iff phi(m)*sigma(m) = (m-3)*(m+1), where phi(m) = A000010(m) and sigma(m) = A000203(m). Of course, for a product of distinct primes p*q we know sigma(p*q) = (p+1)*(q+1) and if p, q, are twin primes, say q = p + 2, then sigma(p*q) = (p+1)*(q+1) = (p+1)*(p+3). - Jonathan Vos Post, Feb 21 2006
Also the area of twin prime rectangles. A twin prime rectangle is a rectangle whose sides are components of twin prime pairs. E.g., the twin prime pair (3,5) produces a 3 X 5 unit rectangle which has area 15 square units. - Cino Hilliard, Jul 28 2006
Except for 15, a product of twin primes is of the form 36k^2 - 1 (cf. A136017, A002822). - Artur Jasinski, Dec 12 2007
A072965(a(n)) = 1; A072965(m) mod A037074(n) > 0 for all m. - Reinhard Zumkeller, Jan 29 2008
The number of terms less than 10^(2n) is A007508(n). - Robert G. Wilson v, Feb 08 2012
If m is the product of twin primes, then sigma(m) = m + 1 + 2*sqrt(m + 1), phi(m) = m + 1 - 2*sqrt(m + 1). pmin(m) = sqrt(m + 1) - 1, pmax(m) = sqrt(m + 1) + 1. - Wesley Ivan Hurt, Jan 06 2013
Semiprimes of the form 4*k^2 - 1. - Vincenzo Librandi, Apr 13 2013

Examples

			a(2)=35 because 5*7=35, that is (5,7) is the 2nd pair of twin primes.

References

  • Albert A. Mullin, "Bicomposites, twin primes and arithmetic progression", Abstract 04T-11-48, Abstracts of AMS, Vol. 25, No. 4, 2004, p. 795.

Links

Crossrefs

Cf. A000010, A000203, A001359, A006512, A014574, A136017, A074480 (multiplicative closure), A209328.
Cf. A071700 (subsequence).
Cf. A075369.

Programs

  • Haskell
    a037074 = subtract 1 . a075369  -- Reinhard Zumkeller, Feb 10 2015
-- Reinhard Zumkeller, Feb 10 2015, Aug 14 2011
  • Magma
    [p*(p+2): p in PrimesUpTo(1000) | IsPrime(p+2)];  // Bruno Berselli, Jul 08 2011
  • Magma
    IsSemiprime:=func; [s: n in [1..500] | IsSemiprime(s) where s is 4*n^2-1]; // Vincenzo Librandi, Apr 13 2013
  • Maple
    ZL:=[]: for p from 1 to 863 do if (isprime(p) and isprime(p+2) ) then ZL:=[op(ZL),(p*(p+2))]; fi; od; print(ZL); # Zerinvary Lajos, Mar 07 2007
for i from 1 to 150 do if ithprime(i+1) = ithprime(i) + 2 then print({ithprime(i)*ithprime(i+1)}); fi; od; # Zerinvary Lajos, Mar 19 2007
  • Mathematica
    s = Select[ Prime@ Range@170, PrimeQ[ # + 2] &]; s(s + 2) (* Robert G. Wilson v, Feb 21 2006 *)
(* For checking large numbers, the following code is better. For instance, we could use the fQ function to determine that 229031718473564142083 is in this sequence. *) fQ[n_] := Block[{fi = FactorInteger[n]}, Last@# & /@ fi == {1, 1} && Differences[ First@# & /@ fi] == {2}]; Select[ Range[750000], fQ] (* Robert G. Wilson v, Feb 08 2012 *)
Times@@@Select[Partition[Prime[Range[500]],2,1],Last[#]-First[#]==2&] (* Harvey P. Dale, Oct 16 2012 *)
  • PARI
    g(n) = for(x=1,n,if(prime(x+1)-prime(x)==2,print1(prime(x)*prime(x+1)","))) \\ Cino Hilliard, Jul 28 2006

Formula

a(n) = A001359(n)*A006512(n). A000010(a(n))*A000203(a(n)) = (a(n)-3)*(a(n)+1). - Jonathan Vos Post, Feb 21 2006
a(n) = (A014574(n))^2 - 1. a(n+1) = (6*A002822(n))^2 - 1. - Lekraj Beedassy, Sep 02 2006
a(n) = A075369(n) - 1. - Reinhard Zumkeller, Feb 10 2015
Sum_{n>=1} 1/a(n) = A209328. - Amiram Eldar, Nov 20 2020
A000010(a(n)) == 0 (mod 8). - Darío Clavijo, Oct 26 2022

Extensions

More terms from Erich Friedman

A031347 Multiplicative digital root of n (keep multiplying digits of n until reaching a single digit).

Original entry on oeis.org

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

Views

Author

Eric W. Weisstein

Keywords

Comments

a(n) = 0 for almost all n. - Charles R Greathouse IV, Oct 02 2013
More precisely, a(n) = 0 asymptotically almost surely, namely, among others, for all numbers n which have a digit '0', and as n has more and more digits, it becomes increasingly less probable that no digit is equal to zero. (The set A011540 has density 1.) Thus the density of numbers for which a(n) > 0 is zero, although this happens for infinitely many numbers, for example all repunits n = (10^k - 1)/9 = A002275(k). - M. F. Hasler, Oct 11 2015

Links

Crossrefs

Cf. A007954, A007953, A003001, A010888 (additive digital root of n), A031286 (additive persistence of n), A031346 (multiplicative persistence of n).
Numbers having multiplicative digital roots 0-9: A034048, A002275, A034049, A034050, A034051, A034052, A034053, A034054, A034055, A034056.

Programs

  • Haskell
    a031347 = until (< 10) a007954
-- Reinhard Zumkeller, Oct 17 2011, Sep 22 2011
  • Maple
    A007954 := proc(n) return mul(d, d=convert(n,base,10)): end: A031347 := proc(n) local m: m:=n: while(length(m)>1)do m:=A007954(m): od: return m: end: seq(A031347(n),n=0..100); # Nathaniel Johnston, May 04 2011
  • Mathematica
    mdr[n_] := NestWhile[Times @@ IntegerDigits@# &, n, UnsameQ, All]; Table[ mdr[n], {n, 0, 104}] (* Robert G. Wilson v, Aug 04 2006 *)
Table[NestWhile[Times@@IntegerDigits[#] &, n, # > 9 &], {n, 0, 90}] (* Harvey P. Dale, Mar 10 2019 *)
  • PARI
    A031347(n)=local(resul); if(n<10, return(n) ); resul = n % 10; n = (n - n%10)/10; while( n > 0, resul *= n %10; n = (n - n%10)/10; ); return(A031347(resul))
for(n=1,80, print1(A031347(n),",")) \\ R. J. Mathar, May 23 2006
  • PARI
    A031347(n)={while(n>9,n=prod(i=1,#n=digits(n),n[i]));n} \\ M. F. Hasler, Dec 07 2014
  • Python
    from operator import mul
from functools import reduce
def A031347(n):
    while n > 9:
       n = reduce(mul, (int(d) for d in str(n)))
    return n
# Chai Wah Wu, Aug 23 2014
  • Python
    from math import prod
def A031347(n):
    while n > 9: n = prod(map(int, str(n)))
    return n
print([A031347(n) for n in range(100)]) # Michael S. Branicky, Apr 17 2024
  • Scala
    def iterDigitProd(n: Int): Int = n.toString.length match {
  case 1 => n
  case  => iterDigitProd(n.toString.toCharArray.map( - 48).scanRight(1)( * ).head)
}
(0 to 99).map(iterDigitProd) // Alonso del Arte, Apr 11 2020

Formula

a(n) = d in {1, ..., 9} if (but not only if) n = (10^k - 1)/9 + (d - 1)*10^m = A002275(k) + (d - 1)*A011557(m) for some k > m >= 0. - M. F. Hasler, Oct 11 2015
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