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.

Previous Showing 11-20 of 38 results. Next

A285900 Sum of all parts of all partitions of all positive integers <= n into consecutive parts.

Original entry on oeis.org

1, 3, 9, 13, 23, 35, 49, 57, 84, 104, 126, 150, 176, 204, 264, 280, 314, 368, 406, 446, 530, 574, 620, 668, 743, 795, 903, 959, 1017, 1137, 1199, 1231, 1363, 1431, 1571, 1679, 1753, 1829, 1985, 2065, 2147, 2315, 2401, 2489, 2759, 2851, 2945, 3041, 3188, 3338, 3542, 3646, 3752, 3968, 4188, 4300, 4528, 4644, 4762, 5002
Offset: 1

Views

Author

Omar E. Pol, May 02 2017

Keywords

Comments

a(n) is also the sum of all parts of all partitions of all positive integers <= n into an odd number of equal parts. - Omar E. Pol, Jun 05 2017

Examples

			For n = 15, there are four partitions of 15 into consecutive parts: [15], [8, 7], [6, 5, 4] and [5, 4, 3, 2, 1]. The sum of all parts of these four partitions is 60, and a(14) = 204, so a(15) = 204 + 60 = 264.
		

Crossrefs

Partial sums of A245579.

Programs

  • Mathematica
    a285900[n_] := Accumulate[Map[# DivisorSum[#, 1 &, OddQ] &, Range[n]]]
    a285900[60] (* data *) (* Hartmut F. W. Hoft, Jun 06 2017 *)
  • PARI
    a(n)=sum(i=1, n, i * sumdiv(i, d, d%2)); \\ Andrew Howroyd, Nov 06 2018
    
  • PARI
    a(n)=sum(k=1, (n+1)\2, (2*k - 1)/2 * (n\(2*k - 1)) * (1 + n\(2*k - 1))); \\ Andrew Howroyd, Nov 06 2018

Formula

a(n) = Sum_{k=1..floor((n+1)/2)} (2*k-1)/2 * floor(n/(2*k-1)) * floor(1 + n/(2*k-1)). - Daniel Suteu, Nov 06 2018

A328361 Triangle read by rows: T(n,k) is the total number of k's in all partitions of n into consecutive parts, (1 <= k <= n).

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Oct 20 2019

Keywords

Comments

Iff n is a power of 2 (A000079) then row n lists n - 1 zeros together with 1.
Iff n is an odd prime (A065091) then row n lists (n - 3)/2 zeros, 1, 1, (n - 3)/2 zeros, 1.

Examples

			Triangle begins:
1;
0, 1;
1, 1, 1;
0, 0, 0, 1;
0, 1, 1, 0, 1;
1, 1, 1, 0, 0, 1;
0, 0, 1, 1, 0, 0, 1;
0, 0, 0, 0, 0, 0, 0, 1;
0, 1, 1, 2, 1, 0, 0, 0, 1;
1, 1, 1, 1, 0, 0, 0, 0, 0, 1;
0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1;
0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1;
0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1;
0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1;
1, 1, 1, 2, 2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1;
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1;
...
For n = 9 there are three partitions of 9 into consecutive parts, they are [9], [5, 4], [4, 3, 2], so the 9th row of triangle is [0, 1, 1, 2, 1, 0, 0, 0, 1].
		

Crossrefs

Row sums give A204217.
Column 1 gives A010054, n >= 1.
Leading diagonal gives A000012.

A334466 Square array read by antidiagonals upwards: T(n,k) is the total number of parts in all partitions of n into consecutive parts that differ by k, with n >= 1, k >= 0.

Original entry on oeis.org

1, 3, 1, 4, 1, 1, 7, 3, 1, 1, 6, 1, 1, 1, 1, 12, 3, 3, 1, 1, 1, 8, 4, 1, 1, 1, 1, 1, 15, 3, 3, 3, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 18, 6, 3, 3, 3, 1, 1, 1, 1, 1, 12, 5, 4, 1, 1, 1, 1, 1, 1, 1, 1, 28, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 14, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 24, 3, 6, 3, 3, 3, 3, 1
Offset: 1

Views

Author

Omar E. Pol, May 01 2020

Keywords

Comments

The one-part partition n = n is included in the count.
The column k is related to (k+2)-gonal numbers, assuming that 2-gonals are the nonnegative numbers, 3-gonals are the triangular numbers, 4-gonals are the squares, 5-gonals are the pentagonal numbers, and so on.
Note that the number of parts for T(n,0) = A000203(n), equaling the sum of the divisors of n.
For fixed k>0, Sum_{j=1..n} T(j,k) ~ 2^(3/2) * n^(3/2) / (3*sqrt(k)). - Vaclav Kotesovec, Oct 23 2024

Examples

			Square array starts:
   n\k|   0  1  2  3  4  5  6  7  8  9 10 11 12
   ---+---------------------------------------------
   1  |   1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
   2  |   3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
   3  |   4, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
   4  |   7, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
   5  |   6, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
   6  |  12, 4, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, ...
   7  |   8, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, ...
   8  |  15, 1, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, ...
   9  |  13, 6, 4, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, ...
  10  |  18, 5. 3. 1. 3. 1, 3, 1, 3, 1, 1, 1, 1, ...
  11  |  12, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 1, 1, ...
  12  |  28, 4, 6, 4, 3, 1, 3, 1, 3, 1, 3, 1, 1, ...
  ...
For n = 9 we have that:
For k = 0 the partitions of 9 into consecutive parts that differ by 0 (or simply: the partitions of 9 into equal parts) are [9], [3,3,3], [1,1,1,1,1,1,1,1,1]. In total there are 13 parts, so T(9,0) = 13.
For k = 1 the partitions of 9 into consecutive parts that differ by 1 (or simply: the partitions of 9 into consecutive parts) are [9], [5,4], [4,3,2]. In total there are six parts, so T(9,1) = 6.
For k = 2 the partitions of 9 into consecutive parts that differ by 2 are [9], [5, 3, 1]. In total there are four parts, so T(9,2) = 4.
		

Crossrefs

Columns k: A000203 (k=0), A204217 (k=1), A066839 (k=2), A330889 (k=3), A334464 (k=4), A334732 (k=5), A334949 (k=6), A377300 (k=7), A377301 (k=8).
Triangles whose row sums give the column k: A127093 (k=0), A285914 (k=1), A330466 (k=2) (conjectured), A330888 (k=3), A334462 (k=4), A334540 (k=5), A339947 (k=6).
Sequences of number of partitions related to column k: A000005 (k=0), A001227 (k=1), A038548 (k=2), A117277 (k=3), A334461 (k=4), A334541 (k=5), A334948 (k=6).
Tables of partitions related to column k: A010766 (k=0), A286001 (k=1), A332266 (k=2), A334945 (k=3), A334618 (k=4).
Polygonal numbers related to column k: A001477 (k=0), A000217 (k=1), A000290 (k=2), A000326 (k=3), A000384 (k=4), A000566 (k=5), A000567 (k=6).

Programs

  • Mathematica
    nmax = 14;
    col[k_] := col[k] = CoefficientList[Sum[n x^(n(k n - k + 2)/2)/(1 - x^n), {n, 1, nmax}] + O[x]^(nmax+1), x];
    T[n_, k_] := col[k][[n+1]];
    Table[T[n-k, k], {n, 1, nmax}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Nov 30 2020 *)

Formula

The g.f. for column k is Sum_{n>=1} n*x^(n*(k*n-k+2)/2)/(1-x^n). (For proof, see A330889. - N. J. A. Sloane, Nov 21 2020)

A288772 a(n) is the minimum number of rows from the table described in A286000 that are required to represent the partitions of all positive integers <= n into consecutive parts.

Original entry on oeis.org

1, 2, 4, 4, 6, 8, 8, 8, 11, 13, 13, 14, 14, 17, 19, 19, 19, 21, 21, 24, 26, 26, 26, 26, 29, 29, 32, 34, 34, 34, 34, 34, 38, 38, 41, 43, 43, 43, 44, 44, 44, 48, 48, 51, 53, 53, 53, 53, 55, 55, 56, 59, 59, 62, 64, 64, 64, 64, 64, 67, 67, 67, 71, 71, 74, 76, 76, 76, 76, 76, 76, 80, 80, 80, 84, 84, 87, 89, 89, 89, 89
Offset: 1

Views

Author

Omar E. Pol, Jun 17 2017

Keywords

Comments

a(n) has the same definition related to the table A286001 which is another version of the table A286000.
First differs from A288529 at a(11), which shares infinitely many terms.

Examples

			Figures A..D show the evolution of the table of partitions into consecutive parts described in A286000, for n = 8..11:
.     ---------------------------------------------------------------------
Figure:      A            B                    C                  D
.     ---------------------------------------------------------------------
.    n:      8            9                   10                 11
Row   ---------------------------------------------------------------------
1     |  1;        |  1;             |   1;             |   1;            |
1     |  2;        |  2;             |   2;             |   2;            |
3     |  3,  2;    |  3,  2;         |   3,  2;         |   3,  2;        |
4     |  4,  1;    |  4,  1;         |   4,  1;         |   4,  1;        |
5     |  5,  3;    |  5,  3;         |   5,  3;         |   5,  3;        |
6     |  6,  2,  3;|  6,  2,  3;     |   6,  2,  3;     |   6,  2,  3;    |
7     |  7,  4,  2;|  7,  4,  2;     |   7,  4,  2;     |   7,  4,  2;    |
8     | [8], 3,  1;|  8,  3,  1;     |   8,  3,  1;     |   8,  3,  1;    |
9     |            | [9],[5],[4];    |   9,  5,  4;     |   9,  5,  4;    |
10    |            | 10, [4],[3],  4;| [10], 4,  3, [4];|  10,  4,  3;  4;|
11    |            | 11,  6, [2],  3;|  11,  6,  2; [3];| [11],[6], 2,  3;|
12    |            |                 |  12,  5,  5, [2];|  12, [5], 5,  2;|
13    |            |                 |  13,  7,  4, [1];|  13,  7,  4,  1;|
.     ---------------------------------------------------------------------
. a(n):      8              11                13                 13
.     ---------------------------------------------------------------------
For n = 8 we need a table with at least 8 rows, so a(8) = 8.
For n = 9 we need a table with at least 11 rows, so a(9) = 11.
For n = 10 we need a table with at least 13 rows, so a(10) = 13.
For n = 11 we need a table with at least 13 rows, so a(11) = 13.
		

Crossrefs

A285899 Total number of parts in all partitions of all positive integers <= n into consecutive parts.

Original entry on oeis.org

1, 2, 5, 6, 9, 13, 16, 17, 23, 28, 31, 35, 38, 43, 54, 55, 58, 66, 69, 75, 87, 92, 95, 99, 107, 112, 124, 132, 135, 148, 151, 152, 164, 169, 184, 196, 199, 204, 216, 222, 225, 240, 243, 252, 278, 283, 286, 290, 300, 310, 322, 331, 334, 351, 369, 377, 389, 394, 397, 414, 417, 422, 450, 451, 469, 488, 491, 500, 512, 529
Offset: 1

Views

Author

Omar E. Pol, May 02 2017

Keywords

Comments

Partial sums of A204217.
Sum of first n rows of the triangle A285914.
Where records occur in A328365. - Omar E. Pol, Oct 22 2019
Row sums of A328368. - Omar E. Pol, Nov 04 2019

Examples

			For n = 15 there are four partitions of 15 into consecutive parts: [15], [8, 7], [6, 5, 4] and [5, 4, 3, 2, 1]. The total number of parts in these four partitions is 11, and a(14) = 43, so a(15) = 43 + 11 = 54.
		

Crossrefs

A334464 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 4.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 4, 3, 1, 6, 1, 3, 4, 3, 1, 6, 1, 3, 4, 7, 1, 6, 1, 7, 4, 3, 1, 10, 1, 3, 4, 7, 1, 6, 1, 7, 9, 3, 1, 10, 1, 8, 4, 7, 1, 6, 6, 7, 4, 3, 1, 15, 1, 3, 4, 7, 6, 12, 1, 7, 4, 8, 1, 16, 1, 3, 9, 7, 1, 12, 1, 12, 4, 3, 1, 16, 6, 3, 4, 7, 1, 17, 8, 7, 4
Offset: 1

Views

Author

Omar E. Pol, May 05 2020

Keywords

Comments

The one-part partition n = n is included in the count.
For the relation to hexagonal numbers see also A334462.

Examples

			For n = 28 there are three partitions of 28 into consecutive parts that differ by 4, including 28 as a valid partition. They are [28], [16, 12] and [13, 9, 5, 1]. The number of parts of these partitions are 1, 2, 4 respectively. The total number of parts is 1 + 2 + 4 = 7, so a(28) = 7.
		

Crossrefs

Row sums of A334462.
Column k=4 of A334466.
Cf. A000384.
Sequences of the same family whose consecutive parts differs by k are: A000203 (k=0), A204217 (k=1), A066839 (k=2), A330889 (k=3), this sequence (k=4), A334732 (k=5), A334949 (k=6).
Cf. A334461.

Programs

  • Mathematica
    nmax = 100;
    CoefficientList[Sum[n x^(n(2n-1)-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* Jean-François Alcover, Nov 30 2020 *)
    Table[Sum[If[n > 2*k*(k-1), k, 0], {k, Divisors[n]}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 22 2024 *)
  • PARI
    my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(2*k-1))/(1-x^k))) \\ Seiichi Manyama, Dec 04 2020

Formula

G.f.: Sum_{n>=1} n*x^(n*(2*n-1))/(1-x^n). (For proof, see A330889. - N. J. A. Sloane, Nov 21 2020)
Sum_{k=1..n} a(k) ~ sqrt(2) * n^(3/2) / 3. - Vaclav Kotesovec, Oct 23 2024

A328362 Triangle read by rows: T(n,k) is the sum of all parts k in all partitions of n into consecutive parts, (1 <= k <= n).

Original entry on oeis.org

1, 0, 2, 1, 2, 3, 0, 0, 0, 4, 0, 2, 3, 0, 5, 1, 2, 3, 0, 0, 6, 0, 0, 3, 4, 0, 0, 7, 0, 0, 0, 0, 0, 0, 0, 8, 0, 2, 3, 8, 5, 0, 0, 0, 9, 1, 2, 3, 4, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 5, 6, 0, 0, 0, 0, 11, 0, 0, 3, 4, 5, 0, 0, 0, 0, 0, 0, 12, 0, 0, 0, 0, 0, 6, 7, 0, 0, 0, 0, 0, 13, 0, 2, 3, 4, 5, 0, 0, 0, 0, 0, 0, 0, 0, 14
Offset: 1

Views

Author

Omar E. Pol, Oct 20 2019

Keywords

Comments

Iff n is a power of 2 (A000079) then row n lists n - 1 zeros together with n.
Iff n is an odd prime (A065091) then row n lists (n - 3)/2 zeros, (n - 1)/2, (n + 1)/2, (n - 3)/2 zeros, n.

Examples

			Triangle begins:
1;
0, 2;
1, 2, 3;
0, 0, 0, 4;
0, 2, 3, 0, 5;
1, 2, 3, 0, 0, 6;
0, 0, 3, 4, 0, 0, 7;
0, 0, 0, 0, 0, 0, 0, 8;
0, 2, 3, 8, 5, 0, 0, 0, 9;
1, 2, 3, 4, 0, 0, 0, 0, 0, 10;
0, 0, 0, 0, 5, 6, 0, 0, 0,  0, 11;
0, 0, 3, 4, 5, 0, 0, 0, 0,  0,  0, 12;
0, 0, 0, 0, 0, 6, 7, 0, 0,  0,  0,  0, 13;
0, 2, 3, 4, 5, 0, 0, 0, 0,  0,  0,  0,  0, 14;
1, 2, 3, 8,10, 6, 7, 8, 0,  0,  0,  0,  0,  0, 15;
0, 0, 0, 0, 0, 0, 0, 0, 0,  0,  0,  0,  0,  0,  0, 16;
...
For n = 9 there are three partitions of 9 into consecutive parts, they are [9], [5, 4], [4, 3, 2], so the 9th row of triangle is [0, 2, 3, 8, 5, 0, 0, 0, 9].
		

Crossrefs

Row sums give A245579.
Column 1 gives A010054, n => 1.
Leading diagonal gives A000027.

Formula

T(n,k) = k*A328361(n,k).

A334732 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 5.

Original entry on oeis.org

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

Views

Author

Omar E. Pol, May 09 2020

Keywords

Comments

The one-part partition n = n is included in the count.
For the relation to heptagonal numbers see also A334540.

Examples

			For n = 27 there are three partitions of 27 into consecutive parts that differ by 5, including 27 as a valid partition. They are [27], [16, 11] and [14, 9, 4]. The number of parts of these partitions are 1, 2, 3 respectively and the total number of parts is 1 + 2 + 3 = 6, so the a(27) = 6.
		

Crossrefs

Row sums of A334540.
Column k=5 of A334466.
Sequences of the same family whose consecutive parts differs by k are: A000203 (k=0), A204217 (k=1), A066839 (k=2), A330889 (k=3), A334464 (k=4), this sequence (k=5), A334949 (k=6).

Programs

  • Maple
    A334732 := proc(n)
        local a,k;
        a := 0 ;
        for k from 1 do
            if n>= A000566(k) then
                a := a+A334540(n,k);
            else
                return a;
            end if;
        end do:
    end proc:
    seq(A334732(n),n=1..120) ; # R. J. Mathar, Oct 02 2020
  • Mathematica
    nmax = 100;
    CoefficientList[Sum[n x^(n(5n-3)/2-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* Jean-François Alcover, Nov 30 2020 *)
    Table[Sum[If[n > 5*k*(k-1)/2 && IntegerQ[n/k - 5*(k-1)/2], k, 0], {k, Divisors[2*n]}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 23 2024 *)
  • PARI
    my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(5*k-3)/2)/(1-x^k))) \\ Seiichi Manyama, Dec 04 2020

Formula

G.f.: Sum_{n>=1} n*x^(n*(5*n-3)/2)/(1-x^n) (for proof see A330889).
Sum_{k=1..n} a(k) ~ 2^(3/2) * n^(3/2) / (3*sqrt(5)). - Vaclav Kotesovec, Oct 23 2024

Extensions

More terms from R. J. Mathar, Oct 02 2020

A286014 Sum of smallest parts of all partitions of n into consecutive parts.

Original entry on oeis.org

1, 2, 4, 4, 7, 7, 10, 8, 15, 11, 16, 15, 19, 16, 27, 16, 25, 26, 28, 22, 38, 26, 34, 31, 40, 31, 50, 29, 43, 49, 46, 32, 62, 41, 59, 48, 55, 46, 74, 46, 61, 67, 64, 46, 94, 56, 70, 63, 77, 69, 98, 55, 79, 85, 92, 61, 110, 71, 88, 93, 91, 76, 131, 64, 110, 103
Offset: 1

Views

Author

Omar E. Pol, Apr 30 2017

Keywords

Comments

If n is a power of 2 then a(n) = n, the same as A286015(n).
Conjecture: this is also the row sums of A211343.

Examples

			For n = 15 there are four partitions of 15 into consecutive parts: [15], [8, 7], [6, 5, 4] and [5, 4, 3, 2, 1]. The sum of the smallest parts is 15 + 7 + 4 + 1 = 27, so a(15) = 27.
		

Crossrefs

Programs

  • Mathematica
    Table[Total[Select[IntegerPartitions@ n, Or[Length@ # == 1, Union@ Differences@ # == {-1}] &][[All, -1]]], {n, 66}] (* Michael De Vlieger, Jul 21 2017 *)

Extensions

More terms from Alois P. Heinz, May 01 2017

A286015 Sum of largest parts of all partitions of n into consecutive parts.

Original entry on oeis.org

1, 2, 5, 4, 8, 9, 11, 8, 18, 14, 17, 17, 20, 19, 34, 16, 26, 31, 29, 26, 46, 29, 35, 33, 45, 34, 58, 35, 44, 58, 47, 32, 70, 44, 70, 57, 56, 49, 82, 50, 62, 78, 65, 53, 114, 59, 71, 65, 84, 76, 106, 62, 80, 98, 106, 67, 118, 74, 89, 106, 92, 79, 153, 64, 124
Offset: 1

Views

Author

Omar E. Pol, Apr 30 2017

Keywords

Comments

If n is a power of 2 then a(n) = n, the same as A286014(n).
Conjecture: this is also the row sums of A286013.

Examples

			For n = 15 there are four partitions of 15 into consecutive parts: [15], [8, 7], [6, 5, 4] and [5, 4, 3, 2, 1]. The sum of the largest parts is 15 + 8 + 6 + 5 = 34, so a(15) = 34.
		

Crossrefs

Programs

  • Mathematica
    Table[Total[Select[IntegerPartitions@ n, Or[Length@ # == 1, Union@ Differences@ # == {-1}] &][[All, 1]]], {n, 65}] (* Michael De Vlieger, Jul 21 2017 *)

Extensions

More terms from Alois P. Heinz, May 01 2017
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