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 1-3 of 3 results.

A238005 Number of partitions of n into distinct parts such that (greatest part) - (least part) = (number of parts).

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 1, 2, 0, 2, 2, 2, 2, 2, 1, 4, 3, 2, 3, 3, 2, 4, 4, 4, 3, 4, 2, 5, 5, 3, 5, 6, 3, 5, 3, 5, 6, 6, 4, 6, 6, 4, 6, 6, 3, 7, 7, 7, 6, 6, 5, 7, 7, 5, 6, 8, 6, 8, 8, 6, 8, 8, 4, 9, 6, 7, 9, 9, 7, 7, 9, 8, 9, 9, 5, 9, 7, 8, 10, 10
Offset: 1

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Author

Clark Kimberling, Feb 17 2014

Keywords

Comments

Note that partitions into distinct parts are also called strict partitions.
a(n) is the number of strict partitions of n into nearly consecutive parts, that is, the number of ways to write n as a sum of terms i, i+1, i+2, ..., i+k (i>=1, k>=2) where one of the interior parts i+1, i+2, ..., i+k-1 is missing. Examples of nearly consecutive partitions (corresponding to the initial nonzero values of a(n)) are 13, 24, 124, 134, 35, 235, 46, ... . - Don Reble, Sep 07 2021
Let T(n) = n*(n+1)/2 = A000217(n) denote the n-th triangular number.
Theorem A. a(n) = b(n) - c(n), where b(n) is the inverse triangular number sequence A003056, that is, b(n) is the maximal i such that T_i <= n, and c(n) is the number of partitions of n into consecutive parts = number of odd divisors of n = A001227(n).
This theorem was conjectured by Omar E. Pol in February 2018, and proved independently by William J. Keith and Roland Bacher on Sep 05 2021. The elegant proof given in the link below is due to Don Reble.

Examples

			a(8) = 2 counts these partitions:  53, 431.
		

Crossrefs

a(n) is also the number of zeros in the n-th row of the triangles A196020, A211343, A231345, A236106, A237048 (simpler), A239662, A261699, A271344, A272026, A280850, A285574, A285891, A285914, A286013, A296508 (and possibly others). Omar E. Pol, Feb 17 2018
Row sums of A347579. - Omar E. Pol, Sep 07 2021

Programs

  • Mathematica
    z = 70; q[n_] := q[n] = Select[IntegerPartitions[n], Max[Length /@ Split@#] == 1 &]; p1[p_] := p1[p] = DeleteDuplicates[p]; t[p_] := t[p] = Length[p1[p]];
    Table[Count[q[n], p_ /; Max[p] - Min[p] < t[p]], {n, z}]  (* A001227 *)
    Table[Count[q[n], p_ /; Max[p] - Min[p] <= t[p]], {n, z}] (* A003056 *)
    Table[Count[q[n], p_ /; Max[p] - Min[p] == t[p]], {n, z}] (* A238005 *)
    Table[Count[q[n], p_ /; Max[p] - Min[p] > t[p]], {n, z}]  (* A238006 *)
    Table[Count[q[n], p_ /; Max[p] - Min[p] >= t[p]], {n, z}] (* A238007 *)
    {0}~Join~Array[Floor[(Sqrt[1 + 8 #] - 1)/2] - DivisorSum[#, 1 &, OddQ] &, 102] (* Michael De Vlieger, Feb 18 2018 *)
  • PARI
    a(n) = if (n, (sqrtint(8*n+1)-1)\2 - sumdiv(n, d, d%2), 0); \\ Michel Marcus, Mar 01 2018

Formula

G.f. = (x/(1-x)) * Sum_{k >= 1} x^(k*(k+1)/2) * (1 - x^(k-1)) / (1 - x^k). This follows from Theorem A and the g.f.s for A003056 and A001227. - William J. Keith, Sep 05 2021
a(n) = A238007(n) - A238006(n). - Omar E. Pol, Sep 11 2021
A001227(n) + a(n) + A238006(n) = A000009(n). - R. J. Mathar, Sep 23 2021

Extensions

Edited by N. J. A. Sloane, Sep 11 2021, mostly to add Theorem A.

A347737 Zero together with the partial sums of A238005.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 3, 5, 5, 7, 9, 11, 13, 15, 16, 20, 23, 25, 28, 31, 33, 37, 41, 45, 48, 52, 54, 59, 64, 67, 72, 78, 81, 86, 89, 94, 100, 106, 110, 116, 122, 126, 132, 138, 141, 148, 155, 162, 168, 174, 179, 186, 193, 198, 204, 212, 218, 226, 234, 240, 248, 256, 260
Offset: 0

Views

Author

Omar E. Pol, Sep 11 2021

Keywords

Comments

a(n) is also the total number of ones in the first n rows of A347579, n >= 1.

Crossrefs

a(n) is also the total number of zeros in the first n rows of the triangles A196020, A211343, A231345, A236106, A237048 (simpler), A239662, A261699, A271344, A272026, A280850, A285574, A285891, A285914, A286013, A296508 (and possibly others), n >= 1.

Programs

  • Mathematica
    Accumulate@Table[Length@Select[Select[IntegerPartitions@n,DuplicateFreeQ],Differences@MinMax@#=={Length@#}&],{n,60}] (* Giorgos Kalogeropoulos, Sep 12 2021 *)
  • Python
    from math import isqrt
    def A347737(n): return (r:=isqrt((n+1<<3)+1)-1>>1)*(6*n+4-r*(r+3))//6-((t:=isqrt(m:=n>>1))+(s:=isqrt(n)))*(t-s)-(sum(n//k for k in range(1,s+1))-sum(m//k for k in range(1,t+1))<<1) # Chai Wah Wu, Jun 07 2025

Formula

a(n) = A006463(n+1) - A060831(n).

A380231 Alternating row sums of triangle A237591.

Original entry on oeis.org

1, 2, 1, 2, 1, 4, 3, 4, 5, 4, 3, 6, 5, 4, 7, 8, 7, 8, 7, 10, 9, 8, 7, 10, 11, 10, 9, 12, 11, 14, 13, 14, 13, 12, 15, 16, 15, 14, 13, 16, 15, 18, 17, 16, 19, 18, 17, 20, 21, 22, 21, 20, 19, 22, 21, 24, 23, 22, 21, 24, 23, 22, 25, 26, 25, 28, 27, 26, 25, 28, 27, 32, 31, 30, 29, 28, 31, 30, 29
Offset: 1

Views

Author

Omar E. Pol, Jan 17 2025

Keywords

Comments

Consider the symmetric Dyck path in the first quadrant of the square grid described in the n-th row of A237593. Let C = (A240542(n), A240542(n)) be the middle point of the Dyck path.
a(n) is also the coordinate on the x axis of the point (a(n),n) and also the coordinate on the y axis of the point (n,a(n)) such that the middle point of the line segment [(a(n),n),(n,a(n))] coincides with the middle point C of the symmetric Dyck path.
The three line segments [(a(n),n),C], [(n,a(n)),C] and [(n,n),C] have the same length.
For n > 2 the points (n,n), C and (a(n),n) are the vertices of a virtual isosceles right triangle.
For n > 2 the points (n,n), C and (n,a(n)) are the vertices of a virtual isosceles right triangle.
For n > 2 the points (a(n),n), (n,n) and (n,a(n)) are the vertices of a virtual isosceles right triangle.

Examples

			For n = 14 the 14th row of A237591 is [8, 3, 1, 2] hence the alternating row sum is 8 - 3 + 1 - 2 = 4, so a(14) = 4.
On the other hand the 14th row of A237593 is the 14th row of A237591 together with the 14 th row of A237591 in reverse order as follows: [8, 3, 1, 2, 2, 1, 3, 8].
Then with the terms of the 14th row of A237593 we can draw a Dyck path in the first quadrant of the square grid as shown below:
.
         (y axis)
          .
          .
          .    (4,14)              (14,14)
          ._ _ _ . _ _ _ _            .
          .               |
          .               |
          .               |_
          .                 |
          .                 |_ _
          .                C    |_ _ _
          .                           |
          .                           |
          .                           |
          .                           |
          .                           . (14,4)
          .                           |
          .                           |
          . . . . . . . . . . . . . . | . . . (x axis)
        (0,0)
.
In the example the point C is the point (9,9).
The three line segments [(4,14),(9,9)], [(14,4),(9,9)] and [(14,14),(9,9)] have the same length.
The points (14,14), (9,9) and (4,14) are the vertices of a virtual isosceles right triangle.
The points (14,14), (9,9) and (14,4) are the vertices of a virtual isosceles right triangle.
The points (4,14), (14,14) and (14,4) are the vertices of a virtual isosceles right triangle.
		

Crossrefs

Other alternating row sums (ARS) related to the Dyck paths of A237593 and the stepped pyramid described in A245092 are as follows:
ARS of A237593 give A000004.
ARS of A196020 give A000203.
ARS of A252117 give A000203.
ARS of A271343 give A000593.
ARS of A231347 give A001065.
ARS of A236112 give A004125.
ARS of A236104 give A024916.
ARS of A249120 give A024916.
ARS of A271344 give A033879.
ARS of A231345 give A033880.
ARS of A239313 give A048050.
ARS of A237048 give A067742.
ARS of A236106 give A074400.
ARS of A235794 give A120444.
ARS of A266537 give A146076.
ARS of A236540 give A153485.
ARS of A262612 give A175254.
ARS of A353690 give A175254.
ARS of A239446 give A235796.
ARS of A239662 give A239050.
ARS of A235791 give A240542.
ARS of A272026 give A272027.
ARS of A211343 give A336305.

Programs

  • PARI
    row235791(n) = vector((sqrtint(8*n+1)-1)\2, i, 1+(n-(i*(i+1)/2))\i);
    a(n) = my(orow = concat(row235791(n), 0)); vecsum(vector(#orow-1, i, (-1)^(i+1)*(orow[i] - orow[i+1]))); \\ Michel Marcus, Apr 13 2025

Formula

a(n) = 2*A240542(n) - n.
a(n) = n - 2*A322141(n).
a(n) = A240542(n) - A322141(n).
Showing 1-3 of 3 results.