cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A260706 Row sums of A260672.

Original entry on oeis.org

0, 1, 3, 6, 9, 12, 16, 20, 25, 30, 35, 40, 45, 51, 57, 63, 70, 77, 84, 91, 98, 105, 112, 120, 128, 136, 144, 153, 162, 171, 180, 189, 198, 207, 216, 225, 235, 245, 255, 265, 275, 286, 297, 308, 319, 330, 341, 352, 363, 374, 385, 396, 408, 420, 432, 444, 456
Offset: 0

Views

Author

Reinhard Zumkeller, Nov 17 2015

Keywords

Comments

a(n) = Sum(A260672(n,k): k=0..A193832(n)-1);
A193832(n+1) = a(n+1) - a(n);
a(A001318(n+1)) = A212760(n).

Crossrefs

Programs

  • Haskell
    a260706 = sum . a260672_row

A001318 Generalized pentagonal numbers: m*(3*m - 1)/2, m = 0, +-1, +-2, +-3, ....

Original entry on oeis.org

0, 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77, 92, 100, 117, 126, 145, 155, 176, 187, 210, 222, 247, 260, 287, 301, 330, 345, 376, 392, 425, 442, 477, 495, 532, 551, 590, 610, 651, 672, 715, 737, 782, 805, 852, 876, 925, 950, 1001, 1027, 1080, 1107, 1162, 1190, 1247, 1276, 1335
Offset: 0

Views

Author

Keywords

Comments

Partial sums of A026741. - Jud McCranie; corrected by Omar E. Pol, Jul 05 2012
From R. K. Guy, Dec 28 2005: (Start)
"Conway's relation twixt the triangular and pentagonal numbers: Divide the triangular numbers by 3 (when you can exactly):
0 1 3 6 10 15 21 28 36 45 55 66 78 91 105 120 136 153 ...
0 - 1 2 .- .5 .7 .- 12 15 .- 22 26 .- .35 .40 .- ..51 ...
.....-.-.....+..+.....-..-.....+..+......-...-.......+....
"and you get the pentagonal numbers in pairs, one of positive rank and the other negative.
"Append signs according as the pair have the same (+) or opposite (-) parity.
"Then Euler's pentagonal number theorem is easy to remember:
"p(n-0) - p(n-1) - p(n-2) + p(n-5) + p(n-7) - p(n-12) - p(n-15) ++-- = 0^n
where p(n) is the partition function, the left side terminates before the argument becomes negative and 0^n = 1 if n = 0 and = 0 if n > 0.
"E.g. p(0) = 1, p(7) = p(7-1) + p(7-2) - p(7-5) - p(7-7) + 0^7 = 11 + 7 - 2 - 1 + 0 = 15."
(End)
The sequence may be used in order to compute sigma(n), as described in Euler's article. - Thomas Baruchel, Nov 19 2003
Number of levels in the partitions of n + 1 with parts in {1,2}.
a(n) is the number of 3 X 3 matrices (symmetrical about each diagonal) M = {{a, b, c}, {b, d, b}, {c, b, a}} such that a + b + c = b + d + b = n + 2, a,b,c,d natural numbers; example: a(3) = 5 because (a,b,c,d) = (2,2,1,1), (1,2,2,1), (1,1,3,3), (3,1,1,3), (2,1,2,3). - Philippe Deléham, Apr 11 2007
Also numbers a(n) such that 24*a(n) + 1 = (6*m - 1)^2 are odd squares: 1, 25, 49, 121, 169, 289, 361, ..., m = 0, +-1, +-2, ... . - Zak Seidov, Mar 08 2008
From Matthew Vandermast, Oct 28 2008: (Start)
Numbers n for which A000326(n) is a member of A000332. Cf. A145920.
This sequence contains all members of A000332 and all nonnegative members of A145919. For values of n such that n*(3*n - 1)/2 belongs to A000332, see A145919. (End)
Starting with offset 1 = row sums of triangle A168258. - Gary W. Adamson, Nov 21 2009
Starting with offset 1 = Triangle A101688 * [1, 2, 3, ...]. - Gary W. Adamson, Nov 27 2009
Starting with offset 1 can be considered the first in an infinite set generated from A026741. Refer to the array in A175005. - Gary W. Adamson, Apr 03 2010
Vertex number of a square spiral whose edges have length A026741. The two axes of the spiral forming an "X" are A000326 and A005449. The four semi-axes forming an "X" are A049452, A049453, A033570 and the numbers >= 2 of A033568. - Omar E. Pol, Sep 08 2011
A general formula for the generalized k-gonal numbers is given by n*((k - 2)*n - k + 4)/2, n=0, +-1, +-2, ..., k >= 5. - Omar E. Pol, Sep 15 2011
a(n) is the number of 3-tuples (w,x,y) having all terms in {0,...,n} and 2*w = 2*x + y. - Clark Kimberling, Jun 04 2012
Generalized k-gonal numbers are second k-gonal numbers and positive terms of k-gonal numbers interleaved, k >= 5. - Omar E. Pol, Aug 04 2012
a(n) is the sum of the largest parts of the partitions of n+1 into exactly 2 parts. - Wesley Ivan Hurt, Jan 26 2013
Conway's relation mentioned by R. K. Guy is a relation between triangular numbers and generalized pentagonal numbers, two sequences from different families, but as triangular numbers are also generalized hexagonal numbers in this case we have a relation between two sequences from the same family. - Omar E. Pol, Feb 01 2013
Start with the sequence of all 0's. Add n to each value of a(n) and the next n - 1 terms. The result is the generalized pentagonal numbers. - Wesley Ivan Hurt, Nov 03 2014
(6k + 1) | a(4k). (3k + 1) | a(4k+1). (3k + 2) | a(4k+2). (6k + 5) | a(4k+3). - Jon Perry, Nov 04 2014
Enge, Hart and Johansson proved: "Every generalised pentagonal number c >= 5 is the sum of a smaller one and twice a smaller one, that is, there are generalised pentagonal numbers a, b < c such that c = 2a + b." (see link theorem 5). - Peter Luschny, Aug 26 2016
The Enge, et al. result for c >= 5 also holds for c >= 2 if 0 is included as a generalized pentagonal number. That is, 2 = 2*1 + 0. - Michael Somos, Jun 02 2018
Suggestion for title, where n actually matches the list and b-file: "Generalized pentagonal numbers: k(n)*(3*k(n) - 1)/2, where k(n) = A001057(n) = [0, 1, -1, 2, -2, 3, -3, ...], n >= 0" - Daniel Forgues, Jun 09 2018 & Jun 12 2018
Generalized k-gonal numbers are the partial sums of the sequence formed by the multiples of (k - 4) and the odd numbers (A005408) interleaved, with k >= 5. - Omar E. Pol, Jul 25 2018
The last digits form a symmetric cycle of length 40 [0, 1, 2, 5, ..., 5, 2, 1, 0], i.e., a(n) == a(n + 40) (mod 10) and a(n) == a(40*k - n - 1) (mod 10), 40*k > n. - Alejandro J. Becerra Jr., Aug 14 2018
Only 2, 5, and 7 are prime. All terms are of the form k*(k+1)/6, where 3 | k or 3 | k+1. For k > 6, the value divisible by 3 must have another factor d > 2, which will remain after the division by 6. - Eric Snyder, Jun 03 2022
8*a(n) is the product of two even numbers one of which is n + n mod 2. - Peter Luschny, Jul 15 2022
a(n) is the dot product of [1, 2, 3, ..., n] and repeat[1, 1/2]. a(5) = 12 = [1, 2, 3, 4, 5] dot [1, 1/2, 1, 1/2, 1] = [1 + 1 + 3 + 2 + 5]. - Gary W. Adamson, Dec 10 2022
Every nonnegative number is the sum of four terms of this sequence [S. Realis]. - N. J. A. Sloane, May 07 2023
From Peter Bala, Jan 06 2025: (Start)
The sequence terms are the exponents in the expansions of the following infinite products:
1) Product_{n >= 1} (1 - s(n)*q^n) = 1 + q + q^2 + q^5 + q^7 + q^12 + q^15 + ..., where s(n) = (-1)^(1 + mod(n+1,3)).
2) Product_{n >= 1} (1 - q^(2*n))*(1 - q^(3*n))^2/((1 - q^n)*(1 - q^(6*n))) = 1 + q + q^2 + q^5 + q^7 + q^12 + q^15 + ....
3) Product_{n >= 1} (1 - q^n)*(1 - q^(4*n))*(1 - q^(6*n))^5/((1 - q^(2*n))*(1 - q^(3*n))*(1 - q^(12*n)))^2 = 1 - q + q^2 - q^5 - q^7 + q^12 - q^15 + q^22 + q^26 - q^35 + ....
4) Product_{n >= 1} (1 - q^(2*n))^13/((1 - (-1)^n*q^n)*(1 - q^(4*n)))^5 = 1 - 5*q + 7*q^2 - 11*q^5 + 13*q^7 - 17*q^12 + 19*q^15 - + .... See Oliver, Theorem 1.1. (End)

Examples

			G.f. = x + 2*x^2 + 5*x^3 + 7*x^4 + 12*x^5 + 15*x^6 + 22*x^7 + 26*x^8 + 35*x^9 + ...
		

References

  • Enoch Haga, A strange sequence and a brilliant discovery, chapter 5 of Exploring prime numbers on your PC and the Internet, first revised ed., 2007 (and earlier ed.), pp. 53-70.
  • Ross Honsberger, Ingenuity in Mathematics, Random House, 1970, p. 117.
  • Donald E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, (to appear), section 7.2.1.4, equation (18).
  • Ivan Niven and Herbert S. Zuckerman, An Introduction to the Theory of Numbers, 2nd ed., Wiley, NY, 1966, p. 231.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A080995 (characteristic function), A026741 (first differences), A034828 (partial sums), A165211 (mod 2).
Cf. A000326 (pentagonal numbers), A005449 (second pentagonal numbers), A000217 (triangular numbers).
Indices of nonzero terms of A010815, i.e., the (zero-based) indices of 1-bits of the infinite binary word to which the terms of A068052 converge.
Union of A036498 and A036499.
Sequences of generalized k-gonal numbers: this sequence (k=5), A000217 (k=6), A085787 (k=7), A001082 (k=8), A118277 (k=9), A074377 (k=10), A195160 (k=11), A195162 (k=12), A195313 (k=13), A195818 (k=14), A277082 (k=15), A274978 (k=16), A303305 (k=17), A274979 (k=18), A303813 (k=19), A218864 (k=20), A303298 (k=21), A303299 (k=22), A303303 (k=23), A303814 (k=24), A303304 (k=25), A316724 (k=26), A316725 (k=27), A303812 (k=28), A303815 (k=29), A316729 (k=30).
Column 1 of A195152.
Squares in APs: A221671, A221672.
Quadrisection: A049453(k), A033570(k), A033568(k+1), A049452(k+1), k >= 0.
Cf. A002620.

Programs

  • GAP
    a:=[0,1,2,5];; for n in [5..60] do a[n]:=2*a[n-2]-a[n-4]+3; od; a; # Muniru A Asiru, Aug 16 2018
    
  • Haskell
    a001318 n = a001318_list !! n
    a001318_list = scanl1 (+) a026741_list -- Reinhard Zumkeller, Nov 15 2015
    
  • Magma
    [(6*n^2 + 6*n + 1 - (2*n + 1)*(-1)^n)/16 : n in [0..50]]; // Wesley Ivan Hurt, Nov 03 2014
    
  • Magma
    [(3*n^2 + 2*n + (n mod 2) * (2*n + 1)) div 8: n in [0..70]]; // Vincenzo Librandi, Nov 04 2014
    
  • Maple
    A001318 := -(1+z+z**2)/(z+1)**2/(z-1)**3; # Simon Plouffe in his 1992 dissertation; gives sequence without initial zero
    A001318 := proc(n) (6*n^2+6*n+1)/16-(2*n+1)*(-1)^n/16 ; end proc: # R. J. Mathar, Mar 27 2011
  • Mathematica
    Table[n*(n+1)/6, {n, Select[Range[0, 100], Mod[#, 3] != 1 &]}]
    Select[Accumulate[Range[0,200]]/3,IntegerQ] (* Harvey P. Dale, Oct 12 2014 *)
    CoefficientList[Series[x (1 + x + x^2) / ((1 + x)^2 (1 - x)^3), {x, 0, 70}], x] (* Vincenzo Librandi, Nov 04 2014 *)
    LinearRecurrence[{1,2,-2,-1,1},{0,1,2,5,7},70] (* Harvey P. Dale, Jun 05 2017 *)
    a[ n_] := With[{m = Quotient[n + 1, 2]}, m (3 m + (-1)^n) / 2]; (* Michael Somos, Jun 02 2018 *)
  • PARI
    {a(n) = (3*n^2 + 2*n + (n%2) * (2*n + 1)) / 8}; /* Michael Somos, Mar 24 2011 */
    
  • PARI
    {a(n) = if( n<0, n = -1-n); polcoeff( x * (1 - x^3) / ((1 - x) * (1-x^2))^2 + x * O(x^n), n)}; /* Michael Somos, Mar 24 2011 */
    
  • PARI
    {a(n) = my(m = (n+1) \ 2); m * (3*m + (-1)^n) / 2}; /* Michael Somos, Jun 02 2018 */
    
  • Python
    def a(n):
        p = n % 2
        return (n + p)*(3*n + 2 - p) >> 3
    print([a(n) for n in range(60)])  # Peter Luschny, Jul 15 2022
    
  • Python
    def A001318(n): return n*(n+1)-(m:=n>>1)*(m+1)>>1 # Chai Wah Wu, Nov 23 2024
  • Sage
    @CachedFunction
    def A001318(n):
        if n == 0 : return 0
        inc = n//2 if is_even(n) else n
        return inc + A001318(n-1)
    [A001318(n) for n in (0..59)] # Peter Luschny, Oct 13 2012
    

Formula

Euler: Product_{n>=1} (1 - x^n) = Sum_{n=-oo..oo} (-1)^n*x^(n*(3*n - 1)/2).
A080995(a(n)) = 1: complement of A090864; A000009(a(n)) = A051044(n). - Reinhard Zumkeller, Apr 22 2006
Euler transform of length-3 sequence [2, 2, -1]. - Michael Somos, Mar 24 2011
a(-1 - n) = a(n) for all n in Z. a(2*n) = A005449(n). a(2*n - 1) = A000326(n). - Michael Somos, Mar 24 2011. [The extension of the recurrence to negative indices satisfies the signature (1,2,-2,-1,1), but not the definition of the sequence m*(3*m -1)/2, because there is no m such that a(-1) = 0. - Klaus Purath, Jul 07 2021]
a(n) = 3 + 2*a(n-2) - a(n-4). - Ant King, Aug 23 2011
Product_{k>0} (1 - x^k) = Sum_{k>=0} (-1)^k * x^a(k). - Michael Somos, Mar 24 2011
G.f.: x*(1 + x + x^2)/((1 + x)^2*(1 - x)^3).
a(n) = n*(n + 1)/6 when n runs through numbers == 0 or 2 mod 3. - Barry E. Williams
a(n) = A008805(n-1) + A008805(n-2) + A008805(n-3), n > 2. - Ralf Stephan, Apr 26 2003
Sequence consists of the pentagonal numbers (A000326), followed by A000326(n) + n and then the next pentagonal number. - Jon Perry, Sep 11 2003
a(n) = (6*n^2 + 6*n + 1)/16 - (2*n + 1)*(-1)^n/16; a(n) = A034828(n+1) - A034828(n). - Paul Barry, May 13 2005
a(n) = Sum_{k=1..floor((n+1)/2)} (n - k + 1). - Paul Barry, Sep 07 2005
a(n) = A000217(n) - A000217(floor(n/2)). - Pierre CAMI, Dec 09 2007
If n even a(n) = a(n-1) + n/2 and if n odd a(n) = a(n-1) + n, n >= 2. - Pierre CAMI, Dec 09 2007
a(n)-a(n-1) = A026741(n) and it follows that the difference between consecutive terms is equal to n if n is odd and to n/2 if n is even. Hence this is a self-generating sequence that can be simply constructed from knowledge of the first term alone. - Ant King, Sep 26 2011
a(n) = (1/2)*ceiling(n/2)*ceiling((3*n + 1)/2). - Mircea Merca, Jul 13 2012
a(n) = (A008794(n+1) + A000217(n))/2 = A002378(n) - A085787(n). - Omar E. Pol, Jan 12 2013
a(n) = floor((n + 1)/2)*((n + 1) - (1/2)*floor((n + 1)/2) - 1/2). - Wesley Ivan Hurt, Jan 26 2013
From Oskar Wieland, Apr 10 2013: (Start)
a(n) = a(n+1) - A026741(n),
a(n) = a(n+2) - A001651(n),
a(n) = a(n+3) - A184418(n),
a(n) = a(n+4) - A007310(n),
a(n) = a(n+6) - A001651(n)*3 = a(n+6) - A016051(n),
a(n) = a(n+8) - A007310(n)*2 = a(n+8) - A091999(n),
a(n) = a(n+10)- A001651(n)*5 = a(n+10)- A072703(n),
a(n) = a(n+12)- A007310(n)*3,
a(n) = a(n+14)- A001651(n)*7. (End)
a(n) = (A007310(n+1)^2 - 1)/24. - Richard R. Forberg, May 27 2013; corrected by Zak Seidov, Mar 14 2015; further corrected by Jianing Song, Oct 24 2018
a(n) = Sum_{i = ceiling((n+1)/2)..n} i. - Wesley Ivan Hurt, Jun 08 2013
G.f.: x*G(0), where G(k) = 1 + x*(3*k + 4)/(3*k + 2 - x*(3*k + 2)*(3*k^2 + 11*k + 10)/(x*(3*k^2 + 11*k + 10) + (k + 1)*(3*k + 4)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 16 2013
Sum_{n>=1} 1/a(n) = 6 - 2*Pi/sqrt(3). - Vaclav Kotesovec, Oct 05 2016
a(n) = Sum_{i=1..n} numerator(i/2) = Sum_{i=1..n} denominator(2/i). - Wesley Ivan Hurt, Feb 26 2017
a(n) = A000292(A001651(n))/A001651(n), for n>0. - Ivan N. Ianakiev, May 08 2018
a(n) = ((-5 + (-1)^n - 6n)*(-1 + (-1)^n - 6n))/96. - José de Jesús Camacho Medina, Jun 12 2018
a(n) = Sum_{k=1..n} k/gcd(k,2). - Pedro Caceres, Apr 23 2019
Quadrisection. For r = 0,1,2,3: a(r + 4*k) = 6*k^2 + sqrt(24*a(r) + 1)*k + a(r), for k >= 1, with inputs (k = 0) {0,1,2,5}. These are the sequences A049453(k), A033570(k), A033568(k+1), A049452(k+1), for k >= 0, respectively. - Wolfdieter Lang, Feb 12 2021
a(n) = a(n-4) + sqrt(24*a(n-2) + 1), n >= 4. - Klaus Purath, Jul 07 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 6*(log(3)-1). - Amiram Eldar, Feb 28 2022
a(n) = A002620(n) + A008805(n-1). Gary W. Adamson, Dec 10 2022
E.g.f.: (x*(7 + 3*x)*cosh(x) + (1 + 5*x + 3*x^2)*sinh(x))/8. - Stefano Spezia, Aug 01 2024

A054440 Number of ordered pairs of partitions of n with no common parts.

Original entry on oeis.org

1, 0, 2, 4, 12, 16, 48, 60, 148, 220, 438, 618, 1302, 1740, 3216, 4788, 8170, 11512, 19862, 27570, 45448, 64600, 100808, 141724, 223080, 307512, 465736, 652518, 968180, 1334030, 1972164, 2691132, 3902432, 5347176, 7611484, 10358426, 14697028, 19790508, 27691500
Offset: 0

Views

Author

Herbert S. Wilf, May 13 2000

Keywords

Examples

			a(3)=4 because of the 4 pairs of partitions of 3: (3,21),(3,111),(21,3),(111,3).
		

Crossrefs

Programs

  • Haskell
    a054440 = sum . zipWith (*) a087960_list . map a001255 . a260672_row
    -- Reinhard Zumkeller, Nov 15 2015
  • Maple
    with(combinat): p1 := sum(numbpart(n)^2*x^n, n=0..500): it := p1*product((1-x^i), i=1..500): s := series(it, x, 500): for i from 0 to 100 do printf(`%d,`,coeff(s,x,i)) od:
  • Mathematica
    nmax = 50; CoefficientList[Series[Sum[PartitionsP[k]^2*x^k, {k, 0, nmax}]/Sum[PartitionsP[k]*x^k, {k, 0, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jul 04 2016 *)

Formula

G.f.: Sum[p(n)^2*x^n]/Sum[p(n)*x^n], with p(n)=number of partitions of n.
a(n) ~ sqrt(3) * exp(Pi*sqrt(2*n)) / (64 * 2^(1/4) * n^(7/4)). - Vaclav Kotesovec, May 20 2018
a(n) = [(x*y)^n] Product_{k>=1} (1 + x^k / (1 - x^k) + y^k / (1 - y^k)). - Ilya Gutkovskiy, Apr 24 2025

Extensions

Corrected and extended by James Sellers, May 23 2000

A260664 Number of ordered triples of partitions of n with no common parts.

Original entry on oeis.org

1, 0, 6, 18, 90, 192, 864, 1710, 5970, 13110, 36810, 75984, 210546, 410130, 1003908, 2045808, 4616730, 8950176, 19746720, 37297710, 78247344, 147410640, 294299424, 543058032, 1067679540, 1925323308, 3653769792, 6555529158, 12129597486, 21348640230
Offset: 0

Views

Author

Reinhard Zumkeller, Nov 15 2015

Keywords

Examples

			a(3) = 18 because of the 18 triples of partitions of 3: (3,3,21), (3,3,111), (3,21,3), (3,21,21), (3,21,111), (3,111,3), (3,111,21), (3,111,111), (21,3,3), (21,3,21), (21,3,111), (21,21,3), (21,111,3), (111,3,3), (111,3,21), (111,3,111), (111,21,3) and (111,111,3);
a(3) = A000041(3-A001318(0))^3 - A000041(3-A001318(1))^3 - A000041(3-A001318(2))^3 = 3^3 - 2^3 - 1^3 = 27 - 8 - 1 = 18.
		

Crossrefs

Programs

  • Haskell
    a260664 = sum . zipWith (*) a087960_list . map a133042 . a260672_row
  • Mathematica
    Table[Sum[(Cos[Pi*j/2] - Sin[Pi*j/2]) * PartitionsP[n - ((6*j^2 + 6*j + 1)/16 - (2*j + 1)*(-1)^j/16)]^3, {j, 0, Floor[Sqrt[8*n/3]]}], {n, 0, 30}] (* Vaclav Kotesovec, Jul 04 2016 *)
    nmax = 50; CoefficientList[Series[Sum[PartitionsP[k]^3*x^k, {k, 0, nmax}] / Sum[PartitionsP[k]*x^k, {k, 0, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jul 04 2016 *)

Formula

a(n) = p(n)^3 - p(n-k(1))^3 - p(n-k(2))^3 + p(n-k(3))^3 + p(n-k(4))^3 - p(n-k(5))^3 - ..., with p=A000041 and k=A001318, see Wilf link: p. 2, (3).
G.f.: Sum[p(n)^3*x^n]/Sum[p(n)*x^n], with p(n)=number of partitions of n. - Vaclav Kotesovec, Jul 04 2016
a(n) ~ 2^(3/2) * exp(4*Pi*sqrt(n/3)) / (729 * 3^(1/4) * n^(11/4)). - Vaclav Kotesovec, May 20 2018

A193832 Irregular triangle read by rows in which row n lists 2n-1 copies of 2n-1 and n copies of 2n, for n >= 1.

Original entry on oeis.org

1, 2, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14
Offset: 1

Views

Author

Omar E. Pol, Aug 22 2011

Keywords

Comments

Sequence of successive positive integers k in which if k is odd then k appears k times, otherwise if k is even then k appears k/2 times.
Note that an arrangement of the blocks of this sequence shows the growth of the generalized pentagonal numbers A001318 (see example).
The sums of each block give the positive integers of A129194: 1, 2, 9, 8, 25, 18, 49,...
Partial sums of A080995. - Paolo P. Lava, Aug 23 2011.
Concatenations of rows of triangles A001650 and A111650; also, seen as a flat list, the row lengths of triangle A260672 and the first differences of its row sums (cf. A260706). - Reinhard Zumkeller, Nov 17 2015
Also a(n) = number of squares in the arithmetic progression {24k + 1: 0 <= k <= n-1} [Granville]. - N. J. A. Sloane, Dec 13 2017

Examples

			a) If written as a triangle the initial rows are
  1, 2,
  3, 3, 3, 4, 4,
  5, 5, 5, 5, 5, 6, 6, 6,
  7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8,
  9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10,
  ...
Row sums give A126587.
b) An application using the blocks of this sequence: the illustration of the growth of an arrangement which represents the generalized pentagonal numbers A001318. For example; the first 9 positive initial terms: 1, 2, 5, 7, 12, 15, 22, 26, 35.
.
.         9
.       8 9
.     8 7 9
.   8 6 7 9
. 8 6 5 7 9
. 6 4 5 7 9
. 4 3 5 7 9
. 2 3 5 7 9
. 1 3 5 7 9
...
		

Crossrefs

Programs

  • Haskell
    a193832 n k = a193832_tabf !! (n-1) !! (k-1)
    a193832_row n = a193832_tabf !! (n-1)
    a193832_tabf = zipWith (++) a001650_tabf a111650_tabl
    a193832' n = a193832_list !! (n - 1)
    a193832_list = concat a193832_tabf
    -- Reinhard Zumkeller, Nov 15 2015
  • Mathematica
    Array[Join @@ MapIndexed[ConstantArray[#, #/(1 + Boole[First@ #2 == 2])] &, {2 # - 1, 2 #}] &, 7] // Flatten (* or *)
    Table[If[k <= 2 n - 1, 2 n - 1, 2 n], {n, 7}, {k, 3 n - 1}] // Flatten (* Michael De Vlieger, Dec 14 2017 *)

Formula

a(n) = sqrt(8n/3) plus or minus 1 [Granville] - N. J. A. Sloane, Dec 13 2017
If 8 <= n <= 52, then a(n-1) < a(n) if and only if n is in A221672. - Jonathan Sondow, Dec 14 2017

Extensions

Edited by N. J. A. Sloane, Dec 13 2017
Showing 1-5 of 5 results.