A078567 -id:A078567 - cp's OEIS frontend

A006218 a(n) = Sum_{k=1..n} floor(n/k); also Sum_{k=1..n} d(k), where d = number of divisors (A000005); also number of solutions to x*y = z with 1 <= x,y,z <= n.

0, 1, 3, 5, 8, 10, 14, 16, 20, 23, 27, 29, 35, 37, 41, 45, 50, 52, 58, 60, 66, 70, 74, 76, 84, 87, 91, 95, 101, 103, 111, 113, 119, 123, 127, 131, 140, 142, 146, 150, 158, 160, 168, 170, 176, 182, 186, 188, 198, 201, 207, 211, 217, 219, 227, 231, 239, 243, 247, 249

Offset: 0

N. J. A. Sloane

The identity Sum_{k=1..n} floor(n/k) = Sum_{k=1..n} d(k) is Equation (10), p. 58, of Apostol (1976). - N. J. A. Sloane, Dec 06 2020
The "Dirichlet divisor problem" is to find a precise asymptotic estimate for this sequence - see formula lines below, also Apostol (1976), Chap. 3.
Number of increasing arithmetic progressions where n+1 is the second or later term. - Mambetov Timur, Takenov Nurdin, Haritonova Oksana (timus(AT)post.kg; oksanka-61(AT)mail.ru), Jun 13 2002. E.g., a(3) = 5 because there are 5 such arithmetic progressions: (1, 2, 3, 4); (2, 3, 4); (1, 4); (2, 4); (3, 4).
Binomial transform of A001659.
Area covered by overlapped partitions of n, i.e., sum of maximum values of the k-th part of a partition of n into k parts. - Jon Perry, Sep 08 2005
Equals inverse Mobius transform of A116477. - Gary W. Adamson, Aug 07 2008
The Polymath project (see the Tao-Croot-Helfgott link) sketches an algorithm for computing a(n) in essentially cube root time, see section 2.1. - Charles R Greathouse IV, Oct 10 2010 [Sladkey gives another. - Charles R Greathouse IV, Oct 02 2017]
The Dirichlet inverse starts (offset 1) 1, -3, -5, 1, -10, 16, -16, 1, 2, 33, -29, -6, -37, 55, 55, -1, -52, -5, -60, ... - R. J. Mathar, Oct 17 2012
The inverse Mobius transforms yields A143356. - R. J. Mathar, Oct 17 2012
An improved approximation vs. Dirichlet is: a(n) = log(Gamma(n+1)) + 2n*gamma. Using sample ranges of {n = k^2-k to k^2 + (k-1)} the means of the new error term are < +- 0.5 up to k=150, except on two values of k. These ranges appear to give means closest to zero for such small sample sizes. It is not clear sample means remain < +- 0.5 at larger k. The standard deviations are ~(n*log(n))^(1/4)/2, with n near sample range center. - Richard R. Forberg, Jan 06 2015
The values of n for which a(n) is even are given by 4*m^2 <= n <= 4*m(m+1) for m >= 0. Example: for m=1 the values of n are 4 <= n <= 8 for which a(4) to a(8) are even. - G. C. Greubel, Sep 30 2015
For n > 0, a(n) = count(x|y), 1 <= y <= x <= n, that is, the number of pairs in the ordered list of x and y, where y divides x, up to and including n. - Torlach Rush, Jan 31 2017
a(n) is also the total number of partitions of all positive integers <= n into equal parts. - Omar E. Pol, May 29 2017
a(n) is the rank of the join of the set of elements of rank n in Young's lattice, the lattice of all integer partitions ordered by inclusion of their Ferrers diagrams. - Geoffrey Critzer, Jul 11 2018
a(n) always has the same parity as floor(sqrt(n)) = A000196(n): see A211264 (proof in Diophante link). - Bernard Schott, Feb 13 2021
From Omar E. Pol, Feb 16 2021: (Start)
Apart from initial zero this is the convolution of A341062 and A000027.
Nonzero terms convolved with A341062 gives A055507. (End)
From Bernard Schott, Apr 17 2022: (Start)
a(n-1) is the number of lattice points in the first quadrant lying under the hyperbola x*y = n, excluding the lattice points on the axes.
a(n) is the number of lattice points in the first quadrant lying on or under the hyperbola x*y = n, excluding the lattice points on the axes. (Reference Hari Kishan). (End)
Let tiles Tn (for n >= 1) be initially placed on square n on an infinite 1D board. At each step, the leftmost unblocked tile (i.e., the top tile in the leftmost stack) jumps forward exactly n squares. Tiles can stack, and only the top tile of a stack can move. This sequence gives the step number when tile n moves for the first time. - Ali Sada, May 23 2025

			a(3) = 5 because 3 + floor(3/2) + 1 = 3 + 1 + 1 = 5. Or tau(1) + tau(2) + tau(3) = 1 + 2 + 2 = 5.
a(4) = 8 because 4 + floor(4/2) + floor(4/3) + 1 = 4 + 2 + 1 + 1 = 8. Or
tau(1) + tau(2) + tau(3) + tau(4) = 1 + 2 + 2 + 3 = 8.
a(5) = 10 because 5 + floor(5/2) + floor(5/3) + floor (5/4) + 1 = 5 + 2 + 1 + 1 + 1 = 10. Or tau(1) + tau(2) + tau(3) + tau(4) + tau(5) = 1 + 2 + 2 + 3 + 2 = 10.

T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976.
K. Chandrasekharan, Introduction to Analytic Number Theory. Springer-Verlag, 1968, Chap. VI.
K. Chandrasekharan, Arithmetical Functions. Springer-Verlag, 1970, Chapter VIII, pp. 194-228. Springer-Verlag, Berlin.
P. G. L. Dirichlet, Werke, Vol. ii, pp. 49-66.
M. N. Huxley, The Distribution of Prime Numbers, Oxford Univ. Press, 1972, p. 7.
M. N. Huxley, Area, Lattice Points and Exponential Sums, Oxford, 1996; p. 239.
Hari Kishan, Number Theory, Krishna, Educational Publishers, 2014, Theorem 1, p. 133.
H. L. Montgomery, Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis, Amer. Math. Soc., 1996, p. 56.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Nurdin N. Takenov and Oksana Haritonova, Representation of positive integers by a special set of digits and sequences, in Dolmatov, S. L. et al. editors, Materials of Science, Practical seminar "Modern Mathematics".
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Exercise 3.6.13 on page 107.

N. J. A. Sloane, Table of n, a(n) for n = 0..20000 (first 1000 terms from T. D. Noe)
Dorin Andrica and Ovidiu Bagdasar, On some results concerning the polygonal polynomials, Carpathian Journal of Mathematics (2019) Vol. 35, No. 1, 1-11.
D. Andrica and E. J. Ionascu, On the number of polynomials with coefficients in [n], An. St. Univ. Ovidius Constanța, Vol. 22(1),2014, 13-23.
R. Bellman and H. N. Shapiro, On a problem in additive number theory, Annals Math., 49 (1948), 333-340. See Eq. 1.5.
D. Berkane, O. Bordellès, and O. Ramaré, Explicit upper bounds for the remainder term in the divisor problem, Math. Comp. 81:278 (2012), pp. 1025-1051.
Peter J. Cameron and Hamid Reza Dorbidi, Minimal cover groups, arXiv:2311.15652 [math.GR], 2023. See p. 13.
Diophante, A1712, La même parité (in French).
Xiaoxi Duan and M. W. Wong, The Dirichlet divisor problem, traces and determinants for complex powers of the twisted bi-Laplacian, J. of Math. Analysis and Applications, Volume 410, Issue 1, Feb 01 2014, Pages 151-157
L. Hoehn and J. Ridenhour, Summations involving computer-related functions, Math. Mag., 62 (1989), 191-196.
M. N. Huxley, Exponential Sums and Lattice Points III, Proc. London Math. Soc., 87 (2003), pp. 591-609.
Vaclav Kotesovec, Graph - The asymptotic ratio (1000000 terms)
Richard Sladkey, A successive approximation algorithm for computing the divisor summatory function, arXiv:1206.3369 [math.NT], 2012.
Terence Tao, Ernest Croot III, and Harald Helfgott, Deterministic methods to find primes, Math. Comp. 81 (2012), 1233-1246; also at arXiv:1009.3956 [math.NT], 2010-2012.

Right edge of A056535. Cf. A000005, A001659, A052511, A143236.
Row sums of triangle A003988, A010766 and A143724.
A061017 is an inverse.
It appears that the partial sums give A078567. - N. J. A. Sloane, Nov 24 2008
Cf. A116477, A051731, A055507, A161700, A004125, A212120, A341062.

a006218 n = sum $ map (div n) [1..n]
-- Reinhard Zumkeller, Jan 29 2011

[0] cat [&+[Floor(n/k):k in [1..n]]:n in [1..60]]; // Marius A. Burtea, Aug 25 2019

with(numtheory): A006218 := n->add(sigma[0](i), i=1..n);

Table[Sum[DivisorSigma[0, k], {k, n}], {n, 70}]
FoldList[Plus, 0, Table[DivisorSigma[0, x], {x, 61}]] //Rest (* much faster *)
Join[{0},Accumulate[DivisorSigma[0,Range[60]]]] (* Harvey P. Dale, Jan 06 2016 *)

PARI
```
a(n)=sum(k=1,n,n\k)
```

a(n)=sum(k=1,sqrtint(n),n\k)*2-sqrtint(n)^2 \\ Charles R Greathouse IV, Oct 10 2010

from sympy import integer_nthroot
def A006218(n): return 2*sum(n//k for k in range(1,integer_nthroot(n,2)[0]+1))-integer_nthroot(n,2)[0]**2 # Chai Wah Wu, Mar 29 2021

a(n) = n * ( log(n) + 2*gamma - 1 ) + O(sqrt(n)), where gamma is the Euler-Mascheroni number ~ 0.57721... (see A001620), Dirichlet, 1849. Again, a(n) = n * ( log(n) + 2*gamma - 1 ) + O(log(n)*n^(1/3)). The determination of the precise size of the error term is an unsolved problem (the so-called Dirichlet divisor problem) - see references, especially Huxley (2003).
The bounds from Chandrasekharan lead to the explicit bounds n log(n) + (2 gamma - 1) n - 4 sqrt(n) - 1 <= a(n) <= n log(n) + (2 gamma - 1) n + 4 sqrt(n). - David Applegate, Oct 14 2008
a(n) = 2*(Sum_{i=1..floor(sqrt(n))} floor(n/i)) - floor(sqrt(n))^2. - Benoit Cloitre, May 12 2002
G.f.: (1/(1-x))*Sum_{k >= 1} x^k/(1-x^k). - Benoit Cloitre, Apr 23 2003
For n > 0: A027750(a(n-1) + k) = k-divisor of n, = k <= A000005(n). - Reinhard Zumkeller, May 10 2006
a(n) = A161886(n) - n + 1 = A161886(n-1) - A049820(n) + 2 = A161886(n-1) + A000005(n) - n + 2 = A006590(n) + A000005(n) - n = A006590(n+1) - n - 1 = A006590(n) + A000005(n) - n for n >= 2. a(n) = a(n-1) + A000005(n) for n >= 1. - Jaroslav Krizek, Nov 14 2009
D(n) = Sum_{m >= 2, r >= 1} (r/m^(r+1)) * Sum_{j = 1..m - 1} * Sum_{k = 0 .. m^(r+1) - 1} exp{ 2*k*pi i(p^n + (m - j)m^r) / m^(r+1) } where p is some fixed prime number. - A. Neves, Oct 04 2010
Let E(n) = a(n) - n(log n + 2 gamma - 1). Then Berkane-Bordellès-Ramaré show that |E(n)| <= 0.961 sqrt(n), |E(n)| <= 0.397 sqrt(n) for n > 5559, and |E(n)| <= 0.764 n^(1/3) log n for x > 9994. - Charles R Greathouse IV, Jul 02 2012
a(n) = Sum_{k = 1..floor(sqrt(n))} A005408(floor((n/k) - (k-1))). - Gregory R. Bryant, Apr 20 2013
Dirichlet g.f. for s > 2: Sum_{n>=1} a(n)/n^s = Sum_{k>=1} (Zeta(s-1) - Sum_{n=1..k-1} (HurwitzZeta(s,n/k)*n/k^s))/k. - Mats Granvik, Sep 24 2017
From Ridouane Oudra, Dec 31 2022: (Start)
a(n) = n^2 - Sum_{i=1..n} Sum_{j=1..n} floor(log(i*j)/log(n+1));
a(n) = floor(sqrt(n)) + 2*Sum_{i=1..n} floor((sqrt(i^2 + 4*n) - i)/2);
a(n) = n + Sum_{i=1..n} v_2(i)*round(n/i), where v_2(i) = A007814(i). (End)

A078651 Number of increasing geometric-progression subsequences of [1,...,n] with integral successive-term ratio and length >= 1.

Original entry on oeis.org

1, 3, 5, 9, 11, 15, 17, 23, 27, 31, 33, 40, 42, 46, 50, 59, 61, 68, 70, 77, 81, 85, 87, 97, 101, 105, 111, 118, 120, 128, 130, 141, 145, 149, 153, 165, 167, 171, 175, 185, 187, 195, 197, 204, 211, 215, 217, 231, 235, 242, 246, 253, 255, 265, 269, 279, 283, 287

Offset: 1

Robert E. Sawyer (rs.1(AT)mindspring.com), Jan 08 2003

The number of geometric-progression subsequences of [1,...,n] with integral successive-term ratio r and length k is floor(n/r^(k-1))(n > 0, r > 1, k > 0).

			a(1): [1]; a(2): [1],[2],[1,2]; a(3): [1],[2],[3],[1,2],[1,3].

Paolo Xausa, Table of n, a(n) for n = 1..10000

a(n) = n + A078632(n).
See A366471 for rational ratios.
See A078567 for APs.
Partial sums of A169594.

g := (n, b) -> local i; add(iquo(n, b^i), i = 1..floor(log(n, b))):
a := n -> local b; n + add(g(n, b), b = 2..n):
seq(a(n), n = 1..58);  # Peter Luschny, Apr 03 2025

Accumulate[1 + Table[Total[IntegerExponent[n, Rest[Divisors[n]]]], {n, 100}]] (* Paolo Xausa, Aug 27 2025 *)

A078651(n) = {my(s=0, k=2); while(k<=n, s+=(n - sumdigits(n, k))/(k-1); k=k+1); n + s} \\ Zhuorui He, Aug 28 2025

a(n) = n + Sum_{r > 1, j > 0} floor(n/r^j).

A051336 Number of increasing arithmetic progressions in {1,2,3,...,n}, including trivial arithmetic progressions of lengths 1 and 2.

Original entry on oeis.org

1, 3, 7, 13, 22, 33, 48, 65, 86, 110, 138, 168, 204, 242, 284, 330, 381, 434, 493, 554, 621, 692, 767, 844, 929, 1017, 1109, 1205, 1307, 1411, 1523, 1637, 1757, 1881, 2009, 2141, 2282, 2425, 2572, 2723, 2882, 3043, 3212, 3383, 3560, 3743, 3930, 4119

Offset: 1

John W. Layman, Nov 02 1999

The number of arithmetic subsequences of [1, ..., n] with successive-term increment i and length k is (n-i*(k-1))(i > 0, k > 0, n > i*(k-1)). - Robert E. Sawyer (rs.1(AT)mindspring.com)

The best algorithm known for generating a(n) from scratch has order O(sqrt(n)) (see below). If a(n-1) is known, it reduces to O(n^(1/3)). - Daniel Hoying, May 20 2020

			a(1): [1];
a(2): [1],[2],[1,2];
a(3): [1],[2],[3],[1,2],[1,3],[2,3],[1,2,3].

T. D. Noe, Table of n, a(n) for n = 1..1000
Marcel K. Goh, Jad Hamdan, and Jonah Saks, The lattice of arithmetic progressions, arXiv:2106.05949 [math.CO], 2021.
Daniel Hoying, Proof of recurrence relation, May 19 2020.

Cf. A078567.
Cf. A006218, A143127.
See A078651 and A366471 for GPs.

nmax = 48; t = Table[ DivisorSigma[0, n], {n, 1, nmax}]; Accumulate[ Accumulate[t]+1] - Accumulate[t] (* Jean-François Alcover, Nov 08 2011 *)
With[{c=Accumulate[DivisorSigma[0,Range[50]]]},Accumulate[c+1]-c] (* Harvey P. Dale, Dec 23 2015 *)
nmax = 50; RecurrenceTable[{a[n] == a[n-1]+1+p[n], p[n] == p[n-1]+DivisorSigma[0, n-1], a[1] == 1, p[1] == 0}, {a, p}, {n, 1, nmax}][[All,1]] (* Daniel Hoying, May 16 2020 *)

from math import isqrt
def A051336(n): return (((s:=isqrt(n-1))*(s+1))**2>>2)+(1-s**2)*n+sum((q:=(n-1)//k)*(2*n-k*(1+q)) for k in range(1, s+1)) # Chai Wah Wu, Oct 21 2023

Theorem: the second differences give tau(n+1), the number of divisors of n+1 (A000005).
a(n) = n + A078567(n).
a(n) = n + Sum_{ i=1..n-1, j=1..floor(n/i) } (n - i*j). - Robert E. Sawyer (rs.1(AT)mindspring.com)
From Daniel Hoying, May 15 2020: (Start)
a(n+1) = a(n) + 1 + Sum_{i=1..n} tau(i).
       = a(n) + 1 + A006218(n+1).
a(n+1) = (n + 1)*(1 + Sum_{i=1..n} floor(n/i)) - Sum_{i=1..n} i*tau(i).
       = (n + 1)*(1 + A006218(n)) - A143127(n). (End)

A341062 Sequence whose partial sums give A000005.

Original entry on oeis.org

1, 1, 0, 1, -1, 2, -2, 2, -1, 1, -2, 4, -4, 2, 0, 1, -3, 4, -4, 4, -2, 0, -2, 6, -5, 1, 0, 2, -4, 6, -6, 4, -2, 0, 0, 5, -7, 2, 0, 4, -6, 6, -6, 4, 0, -2, -2, 8, -7, 3, -2, 2, -4, 6, -4, 4, -4, 0, -2, 10, -10, 2, 2, 1, -3, 4, -6, 4, -2, 4, -6, 10, -10, 2, 2, 0, -2, 4, -6, 8, -5, -1, -2, 10, -8, 0, 0, 4, -6, 10

Offset: 1

Omar E. Pol, Feb 04 2021

Essentially a duplicate of A051950.
Convolved with A000041 gives A138137.
Convolved with A000027 gives the nonzero terms of A006218.
Convolved with A000070 gives the nonzero terms of A006128.
Convolved with A014153 gives the nonzero terms of A284870.
Convolved with A036469 gives the nonzero terms of A305082.
Convolved with the nonzero terms of A006218 gives A055507.
Convolved with the nonzero terms of A000217 gives the nonzero terms of A078567.

1 together with A051950.
Cf. A000005 (partial sums).
Cf. A000027, A000041, A000070, A000217, A006128, A006218, A014153, A036469, A055507, A078567, A138137, A284870, A305082, A340793.

Join[{1}, Differences[Table[DivisorSigma[0, n], {n, 1, 90}]]] (* Amiram Eldar, Feb 06 2021 *)

a(n) = A051950(n) for n > 1.

A106847 a(n) = Sum {k + jm <= n} (k + jm), with 0 < k,j,m <= n.

Original entry on oeis.org

0, 0, 2, 11, 31, 71, 131, 229, 357, 537, 767, 1064, 1412, 1867, 2385, 3000, 3720, 4570, 5506, 6608, 7808, 9194, 10734, 12436, 14260, 16360, 18622, 21079, 23739, 26668, 29758, 33199, 36815, 40742, 44924, 49369, 54085, 59265, 64661, 70355

Offset: 0

Ralf Stephan, May 06 2005

nonn

			We have 1+1*1=2<=3, 1+2*1=3, 1+1*2=3, 2+1*1=3, thus a(3)=2+3+3+3=11.

M. F. Hasler, Table of n, a(n) for n = 0..9999

Cf. A106633, A106634, A106846, A078567 (number of terms).

Cf. A006218, A143272, A143274, A143127, A319085.

A106847 := proc(n)
    local a,k,l,m ;
    a := 0 ;
    for k from 1 to n do
        for l from 1 to n-k do
            m := floor((n-k)/l) ;
            if m >=1 then
                m := min(m,n) ;
                a := a+m*k+l*m*(m+1)/2 ;
            end if;
        end do:
    end do:
    a ;
end proc: # R. J. Mathar, Oct 17 2012

A106847[n_] := Module[{a, k, l, m}, a = 0; For[k = 1, k <= n, k++, For[l = 1, l <= n - k, l++, If[l == 0, m = n, m = Floor[(n - k)/l]]; If[m >= 1, m = Min[m, n]; a = a + m*k + l*m*(m + 1)/2]]]; a];
Table[A106847[n], {n, 0, 40}] (* Jean-François Alcover, Apr 04 2024, after R. J. Mathar *)

A106847(n)=sum(m=1,n-1,sum(k=1,(n-1)\m,(n-m*k)*(n+m*k+1)))/2  \\ M. F. Hasler, Oct 17 2012

From Ridouane Oudra, Jun 02 2024: (Start)
a(n) = (1/2)*Sum_{k=1..n} (n^2 + n - k^2 - k)*tau(k);
a(n) = (1/2)*(n^2 + n)*A006218(n) - Sum_{k=1..n} A143272(k);
a(n) = (1/2)*((n + 1)*A143274(n) - A143127(n) - A319085(n)). (End)
a(n) ~ n^3 * (log(n) + 2*gamma - 4/3)/3, where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Jun 15 2024

A134546 Triangle read by rows: T(n, k) = Sum_{j=0..n} floor(j / k).

Original entry on oeis.org

1, 3, 1, 6, 2, 1, 10, 4, 2, 1, 15, 6, 3, 2, 1, 21, 9, 5, 3, 2, 1, 28, 12, 7, 4, 3, 2, 1, 36, 16, 9, 6, 4, 3, 2, 1, 45, 20, 12, 8, 5, 4, 3, 2, 1, 55, 25, 15, 10, 7, 5, 4, 3, 2, 1, 66, 30, 18, 12, 9, 6, 5, 4, 3, 2, 1, 78, 36, 22, 15, 11, 8, 6, 5, 4, 3, 2, 1, 91, 42, 26, 18, 13, 10, 7, 6, 5, 4, 3, 2, 1

Offset: 1

Gary W. Adamson, Oct 31 2007

From Bob Selcoe, Aug 08 2016: (Start)
Columns are partial sums of k-repeating increasing positive integers:
  Column 1 is {1+2+3+4+5+...}         = A000217 (triangular numbers);
  Column 2 is {1+1+2+2+3+3+4+4+...}   = A002620 (quarter-squares);
  Column 3 is {1+1+1+2+2+2+3+3+3+...} = A130518.
  Columns k = 4..7 are A130519, A130520, A174709 and A174738, respectively.
T(n, k) is the number of positive multiples of k which can be expressed as i-j, {i=1..n; j=0..n-1}. So for example, T(5, 2) = 6 because there are 6 ways to express i-j {i<=5} as a multiple of 2: {5-3, 4-2, 3-1, 2-0, 5-1 and 4-0}. (End)
Conjecture: For T(n, k) n >= k^(3/2), there is at least one prime in the interval [T(n-1, k+1), T(n, k)]. - Bob Selcoe, Aug 21 2016
Theorem: For n >= 3*k, T(n, k) is composite. - Daniel Hoying, Jul 08 2020

			The triangle T(n, k) begins:
   n\k  1   2   3   4  5  6  7  8  9  10 ...
   1:   1
   2:   3   1
   3:   6   2   1
   4:  10   4   2   1
   5:  15   6   3   2  1
   6:  21   9   5   3  2  1
   7:  28  12   7   4  3  2  1
   8:  36  16   9   6  4  3  2  1
   9:  45  20  12   8  5  4  3  2  1
  10:  55  25  15  10  7  5  4  3  2   1
... Reformatted. - _Wolfdieter Lang_, Feb 04 2015
T(10,3) = 15: 3*floor(10/3)*floor(13/3)/2 - floor(10/3)*(3-1 - 13 mod 3) = 3*3*4/2 - 3*(3-1-1) = 18 - 3 = 15. - _Bob Selcoe_, Aug 21 2016

Michael De Vlieger, Table of n, a(n) for n = 1..11325 (1 <= n <= 150)

Cf. A078567 (row sums), A000217 (column 1).

Cf. A004736, A051731, A002620, A130518, A130519, A130520, A174709, A174738.

T := proc(n, k) option remember: `if`(n = k, 1, T(n-1, k) + iquo(n,k)) end:
seq(seq(T(n,k), k=1..n),n=1..16); # Peter Luschny, May 26 2020

nn = 12; f[w_] := Map[PadRight[#, nn] &, w]; MapIndexed[Take[#1, First@ #2] &, f@ Table[Reverse@ Range@ n, {n, nn}].f@ Table[Boole@ Divisible[n, #] & /@ Range@ n, {n, nn}]] // Flatten (* Michael De Vlieger, Aug 10 2016 *)

t(n, k) = if (k>n, 0, if (n==1, 1, t(n-1, k) + n\k));
tabl(nn) = {m = matrix(nn, nn, n , k, t(n,k)); for (n=1, nn, for (k=1, n, print1(m[n, k], ", ");); print(););} \\ Michel Marcus, Jan 18 2015

trg(nn) = {ma = matrix(nn, nn, n, k, if (k<=n, n-k+1, 0)); mb = matrix(nn, nn, n, k, if (k<=n, !(n%k), 0)); ma*mb;} \\ Michel Marcus, Jan 20 2015

Original definition: T = A004736 * A051731 as infinite lower triangular matrices.
In other words: T(n, k) = Sum_{m=k..n} A004736(n, m)*A051731(m, k).
T(n, k) = 0 if n < k, T(1, 1) = 1, and T(n, k) = T(n-1, k) + floor(n/k), for n >= 2. - Richard R. Forberg, Jan 17 2015
T(n, k) = k*floor(n/k)*floor((n+k)/k)/2 - floor(n/k)*(k-1-(n mod k)). - Bob Selcoe, Aug 21 2016
T(n, k) = k*A000217(b) + (b+1)*[(n +1)-(b + 1)*k] for 1 <= k <= floor[(n + 1) / 2] where b = floor[(n - k + 1) / k], T(n, k) = n-k+1 for floor[(n + 1) / 2] < k <= n and T(n, k) = 0 for k > n. - Henri Gonin, May 12 2020
T(n, k) = (-k/2)*floor(n/k)^2+(n-k/2+1)*floor(n/k). - Daniel Hoying, May 25 2020
From Daniel Hoying, Jul 06 2020: (Start)
T(m + 2*n - 1, m + n) = n for n > 0, m >= 0.
T(3*m + 3*ceiling((n-3)/6) + (n+1)/2, 2*m + 2*ceiling((n-3)/6) + 1) = n for n > 0, n odd, 0 <= m <= floor(n/3).
T(3*m + 3*ceiling(n/6) + n/2 - 1, 2*m + 2*ceiling(n/6)) = n for n > 0, n even, 0 <= m <= floor(n/3). (End)

Edited. Name clarified. Formulas rewritten. - Wolfdieter Lang, Feb 04 2015
Corrected and extended by Michael De Vlieger, Aug 10 2016
Edited and new name from Peter Luschny, Apr 02 2025

A309099 Number of partitions of n avoiding the partition (4,3,1).

Original entry on oeis.org

1, 1, 2, 3, 5, 7, 11, 15, 21, 28, 37, 46, 59, 72, 87, 104, 124, 144, 168, 192, 220, 250, 282, 314, 352, 391, 432, 475, 522, 569, 622, 675, 732, 791, 852, 915, 985, 1055, 1127, 1201, 1281, 1361, 1447, 1533, 1623, 1717, 1813, 1909, 2013, 2118, 2227, 2338, 2453

Offset: 0

Jonathan S. Bloom, Jul 12 2019

nonn

We say a partition alpha contains mu provided that one can delete rows and columns from (the Ferrers board of) alpha and then top/right justify to obtain mu. If this is not possible then we say alpha avoids mu. For example, the only partitions avoiding (2,1) are those whose Ferrers boards are rectangles.

Alois P. Heinz, Table of n, a(n) for n = 0..10000
Jonathan Bloom and Nathan McNew, Counting pattern-avoiding integer partitions, arXiv:1908.03953 [math.CO], 2019.
J. Bloom and D. Saracino, On Criteria for rook equivalence of Ferrers boards, arXiv:1808.04221 [math.CO], 2018.
J. Bloom and D. Saracino, Rook and Wilf equivalence of integer partitions, arXiv:1808.04238 [math.CO], 2018.
J. Bloom and D. Saracino, Rook and Wilf equivalence of integer partitions, European J. Combin., 71 (2018), 246-267.
J. Bloom and D. Saracino, On Criteria for rook equivalence of Ferrers boards, European J. Combin., 76 (2018), 199-207.
Joachim König, Closed-form for the number of partitions of n avoiding the partition (4,3,1), answer to question on MathOverflow (2023).

Cf. A002378, A078567, A309097, A309098, A309058.

b:= proc(n) option remember; `if`(n<1, [0$2],
      (p-> p+[numtheory[tau](n), p[1]])(b(n-1)))
    end:
a:= n-> b(n+1)[2]+`if`(n=0, 1, n*(1-n)):
seq(a(n), n=0..55);  # Alois P. Heinz, Dec 20 2023

b[n_] := b[n] = If[n < 1, {0, 0}, With[{p = b[n-1]},
   p + {DivisorSigma[0, n], p[[1]]}]];
a[n_] := b[n+1][[2]] + If[n == 0, 1, n*(1-n)];
Table[a[n], {n, 0, 55}] (* Jean-François Alcover, Jan 29 2025, after Alois P. Heinz *)

a(n) = if(n == 0, 1, sum(i = 1, n, (n - i + 1) * numdiv(i)) - n * (n - 1)) \\ Mikhail Kurkov, Dec 20 2023 [verification needed]

a(n) = A078567(n+1) - A002378(n-1) for n > 0 with a(0) = 1. - Mikhail Kurkov, Dec 20 2023 [verification needed]

More terms from Alois P. Heinz, Jul 12 2019

A381886 Triangle read by rows: T(n, k) = Sum_{j=1..floor(log[k](n))} floor(n / k^j) if k >= 2, T(n, 1) = n, T(n, 0) = 0^n.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 3, 1, 1, 0, 4, 3, 1, 1, 0, 5, 3, 1, 1, 1, 0, 6, 4, 2, 1, 1, 1, 0, 7, 4, 2, 1, 1, 1, 1, 0, 8, 7, 2, 2, 1, 1, 1, 1, 0, 9, 7, 4, 2, 1, 1, 1, 1, 1, 0, 10, 8, 4, 2, 2, 1, 1, 1, 1, 1, 0, 11, 8, 4, 2, 2, 1, 1, 1, 1, 1, 1, 0, 12, 10, 5, 3, 2, 2, 1, 1, 1, 1, 1, 1

Offset: 0

Peter Luschny, Apr 03 2025

			Triangle starts:
  [ 0] 1;
  [ 1] 0,  1;
  [ 2] 0,  2,  1;
  [ 3] 0,  3,  1, 1;
  [ 4] 0,  4,  3, 1, 1;
  [ 5] 0,  5,  3, 1, 1, 1;
  [ 6] 0,  6,  4, 2, 1, 1, 1;
  [ 7] 0,  7,  4, 2, 1, 1, 1, 1;
  [ 8] 0,  8,  7, 2, 2, 1, 1, 1, 1;
  [ 9] 0,  9,  7, 4, 2, 1, 1, 1, 1, 1;
  [10] 0, 10,  8, 4, 2, 2, 1, 1, 1, 1, 1;
  [11] 0, 11,  8, 4, 2, 2, 1, 1, 1, 1, 1, 1;
  [12] 0, 12, 10, 5, 3, 2, 2, 1, 1, 1, 1, 1, 1;

Jeffrey C. Lagarias and Wijit Yangjit, The factorial function and generalizations, extended, arXiv:2310.12949 [math.NT], 2023.
A. M. Oller-Marcen and J. Maria Grau, On the Base-b Expansion of the Number of Trailing Zeros of b^k!, J. Int. Seq. 14 (2011) 11.6.8.

Cf. A011371 (column 2), A054861 (column 3), A054893 (column 4), A027868 (column 5), A054895 (column 6), A054896 (column 7), A054897 (column 8), A054898 (column 9), A078651 (row sums).

Cf. A078632, A078567, A153216, A366471.

T := (n, b) -> local i; ifelse(b = 0, b^n, ifelse(b = 1, n, add(iquo(n, b^i), i = 1..floor(log(n, b))))): seq(seq(T(n, b), b = 0..n), n = 0..12);
# Alternative:
T := (n, k) -> local j; ifelse(k = 0, k^n, ifelse(k = 1, n, add(padic:-ordp(j, k), j = 1..n))): for n from 0 to 12 do seq(T(n, k), k = 0..n) od;

T[n_, 0] := If[n == 0, 1, 0]; T[n_, 1] := n;
T[n_, k_] := Last@Accumulate[IntegerExponent[Range[n], k]];
Table[T[n, k], {n, 0, 12}, {k, 0, n}] // MatrixForm
(* Alternative: *)
T[n_, k_] := Sum[Floor[n/k^j], {j, Floor[Log[k, n]]}]; T[n_, 1] := n; T[n_, 0] := 0^n; T[0, 0] = 1; Flatten@ Table[T[n, k], {n, 0, 12}, {k, 0, n}] (* Michael De Vlieger, Apr 03 2025 *)

T(n,k) = if (n==0, 1, if (n==1, k, if (k==0, 0, if (k==1, n, sum(j=1, n, valuation(j, k))))));
row(n) = vector(n+1, k, T(n,k-1)); \\ Michel Marcus, Apr 04 2025

from math import log
def T(n: int, b: int) -> int:
    return (b**n if b == 0 else n if b == 1 else
        sum(n // (b**i) for i in range(1, 1 + int(log(n, b)))))
print([[T(n, b) for b in range(n+1)] for n in range(12)])

def T(n, b): return (b^n if b == 0 else n if b == 1 else sum(valuation(j, b) for j in (1..n)))
print(flatten([[T(n, b) for b in range(n+1)] for n in srange(13)]))

T(n, k) = Sum_{j=1..n} valuation(j, k) for n >= 2.

cp's OEIS Frontend

A006218 a(n) = Sum_{k=1..n} floor(n/k); also Sum_{k=1..n} d(k), where d = number of divisors (A000005); also number of solutions to x*y = z with 1 <= x,y,z <= n.

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A078651 Number of increasing geometric-progression subsequences of [1,...,n] with integral successive-term ratio and length >= 1.

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A051336 Number of increasing arithmetic progressions in {1,2,3,...,n}, including trivial arithmetic progressions of lengths 1 and 2.

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A341062 Sequence whose partial sums give A000005.

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A106847 a(n) = Sum {k + j*m <= n} (k + j*m), with 0 < k,j,m <= n.

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A134546 Triangle read by rows: T(n, k) = Sum_{j=0..n} floor(j / k).

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A106847 a(n) = Sum {k + jm <= n} (k + jm), with 0 < k,j,m <= n.