A008606 -id:A008606 - cp's OEIS frontend

A259749 Numbers that are congruent to {1,2,5,7,10,11,13,17,19,23} mod 24.

1, 2, 5, 7, 10, 11, 13, 17, 19, 23, 25, 26, 29, 31, 34, 35, 37, 41, 43, 47, 49, 50, 53, 55, 58, 59, 61, 65, 67, 71, 73, 74, 77, 79, 82, 83, 85, 89, 91, 95, 97, 98, 101, 103, 106, 107, 109, 113, 115, 119, 121, 122, 125, 127, 130, 131, 133, 137, 139, 143, 145

Offset: 1

José María Grau Ribas, Jul 04 2015

Original name: Numbers n such that A259748(n) = 0.

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-2,2,-2,2,-2,2,-1).

Cf. A000914.

Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259750, A259751, A259752, A259754, A259755.

A[n_] := A[n] = Sum[a b, {a, 1, n}, {b, a + 1, n}] ; Select[Range[600], Mod[A[#], #]  == 0 & ]
Rest@ CoefficientList[Series[x (1 + x^2) (1 + 2 x^2 - x^3 + 2 x^4 - 2 x^5 + 3 x^6 + x^7)/((1 - x)^2*(1 - x + x^2 - x^3 + x^4) (1 + x + x^2 + x^3 + x^4)), {x, 0, 61}], x] (* Michael De Vlieger, Aug 25 2016 *)
Select[Range[150],MemberQ[{1,2,5,7,10,11,13,17,19,23},Mod[#,24]]&] (* or *) LinearRecurrence[{2,-2,2,-2,2,-2,2,-2,2,-1},{1,2,5,7,10,11,13,17,19,23},70] (* Harvey P. Dale, Jan 15 2022 *)

Vec(x*(1+x^2)*(1+2*x^2-x^3+2*x^4-2*x^5+3*x^6+x^7)/((1-x)^2*(1-x+x^2-x^3+x^4)*(1+x+x^2+x^3+x^4)) + O(x^100)) \\ Colin Barker, Aug 25 2016

A259748(a(n)) = Sum_{x*y: x,y in Z/a(n)Z, x<>y} = 0.

G.f.: x*(1+x^2)*(1+2*x^2-x^3+2*x^4-2*x^5+3*x^6+x^7) / ((1-x)^2*(1-x+x^2-x^3+x^4)*(1+x+x^2+x^3+x^4)). - Colin Barker, Aug 25 2016

Better name from Danny Rorabaugh, Oct 22 2015

A259751 Numbers that are congruent to {8, 16} mod 24.

Original entry on oeis.org

8, 16, 32, 40, 56, 64, 80, 88, 104, 112, 128, 136, 152, 160, 176, 184, 200, 208, 224, 232, 248, 256, 272, 280, 296, 304, 320, 328, 344, 352, 368, 376, 392, 400, 416, 424, 440, 448, 464, 472, 488, 496, 512, 520, 536, 544, 560, 568, 584, 592, 608, 616, 632

Offset: 1

José María Grau Ribas, Jul 06 2015

Original name: Numbers n such that n/A259748(n) = 4.

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000914, A001651.

Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259749, A259750, A259752, A259754, A259755.

A[n_] := A[n] = Sum[a b, {a, 1,  n}, {b, a + 1, n}] ; Select[Range[600], Mod[A[#], #]/# == 1/4 & ]

Vec(8*x*(1+x+x^2)/((1-x)^2*(1+x)) + O(x^100)) \\ Colin Barker, Aug 26 2016

A259748(a(n))/a(n) = 1/4.
a(n) = 8*A001651. - Danny Rorabaugh, Oct 22 2015
From Colin Barker, Aug 26 2016: (Start)
a(n) = 12*n-2*(-1)^n-6.
a(n) = a(n-1)+a(n-2)-a(n-3) for n>3.
G.f.: 8*x*(1+x+x^2) / ((1-x)^2*(1+x)).
(End)
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/72. - Amiram Eldar, Dec 31 2021
E.g.f.: 2*(4 + (6*x - 3)*exp(x) - exp(-x)). - David Lovler, Sep 05 2022

Better name from Danny Rorabaugh, Oct 22 2015

A259752 a(n) = 24*n - 18.

Original entry on oeis.org

6, 30, 54, 78, 102, 126, 150, 174, 198, 222, 246, 270, 294, 318, 342, 366, 390, 414, 438, 462, 486, 510, 534, 558, 582, 606, 630, 654, 678, 702, 726, 750, 774, 798, 822, 846, 870, 894, 918, 942, 966, 990, 1014, 1038, 1062, 1086, 1110, 1134, 1158, 1182, 1206

Offset: 1

José María Grau Ribas, Jul 12 2015

Original name: Numbers n such that n/A259748(n) = 6.

Partial sums give A152746. - Leo Tavares, Jul 29 2023

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A000914, A016813, A063128, A063148, A152746.

Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259749, A259750, A259751, A259754, A259755.

A[n_] := A[n] = Sum[a b, {a, 1,  n}, {b, a + 1, n}] ; Select[Range[600], Mod[A[#], #]/# == 1/6 & ]

A259748(a(n))/a(n) = 1/6.
a(n) = 6*A016813(n-1). - Michel Marcus, Jul 18 2015
G.f.: 6*x*(3*x+1)/(x-1)^2. - Alois P. Heinz, Jul 29 2023
From Elmo R. Oliveira, Apr 04 2025: (Start)
E.g.f.: 6*(exp(x)*(4*x - 3) + 3).
a(n) = 2*a(n-1) - a(n-2) for n > 2. (End)

Better name from Danny Rorabaugh, Oct 22 2015

A259754 Numbers that are congruent to {3,9,15,18,21} mod 24.

Original entry on oeis.org

3, 9, 15, 18, 21, 27, 33, 39, 42, 45, 51, 57, 63, 66, 69, 75, 81, 87, 90, 93, 99, 105, 111, 114, 117, 123, 129, 135, 138, 141, 147, 153, 159, 162, 165, 171, 177, 183, 186, 189, 195, 201, 207, 210, 213, 219, 225, 231, 234, 237, 243, 249, 255, 258, 261, 267

Offset: 1

José María Grau Ribas, Jul 12 2015

Original name: Numbers n such that n/A259748(n) = 3/2.

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Cf. A000914.

Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259749, A259750, A259751, A259752, A259755.

A[n_] := A[n] = Sum[a b, {a, 1, n}, {b, a + 1, n}]; Select[Range[200], Mod[A[#], #]/# == 2/3 &]
Rest@ CoefficientList[Series[3 x (1 + x) (1 + x + x^2 + x^4)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4)), {x, 0, 56}], x] (* Michael De Vlieger, Aug 25 2016 *)
LinearRecurrence[{1,0,0,0,1,-1},{3,9,15,18,21,27},60] (* Harvey P. Dale, Aug 30 2016 *)

Vec(3*x*(1+x)*(1+x+x^2+x^4)/((1-x)^2*(1+x+x^2+x^3+x^4)) + O(x^100)) \\ Colin Barker, Aug 25 2016

A259748(a(n))/a(n) = 2/3.
a(n) = 3*A047584(n). - Michel Marcus, Jul 18 2015
From Colin Barker, Aug 25 2016: (Start)
a(n) = a(n-1)+a(n-5)-a(n-6) for n>6.
G.f.: 3*x*(1+x)*(1+x+x^2+x^4) / ((1-x)^2*(1+x+x^2+x^3+x^4)).
(End)

Better name from Danny Rorabaugh, Oct 22 2015

A105020 Array read by antidiagonals: row n (n >= 0) contains the numbers m^2 - n^2, m >= n+1.

Original entry on oeis.org

1, 3, 4, 5, 8, 9, 7, 12, 15, 16, 9, 16, 21, 24, 25, 11, 20, 27, 32, 35, 36, 13, 24, 33, 40, 45, 48, 49, 15, 28, 39, 48, 55, 60, 63, 64, 17, 32, 45, 56, 65, 72, 77, 80, 81, 19, 36, 51, 64, 75, 84, 91, 96, 99, 100, 21, 40, 57, 72, 85, 96, 105, 112, 117, 120, 121

Offset: 0

Andrew S. Plewe and Franklin T. Adams-Watters, Jul 11 2005

A "Goldbach Conjecture" for this sequence: when there are n terms between consecutive odd integers (2n+1) and (2n+3) for n > 0, at least one will be the product of 2 primes (not necessarily distinct). Example: n=3 for consecutive odd integers a(7)=7 and a(11)=9 and of the 3 sequence entries a(8)=12, a(9)=15 and a(10)=16 between them, one is the product of 2 primes a(9)=15=3*5. - Michael Hiebl, Jul 15 2007
A024352 gives distinct values in the array, minus the first row (1, 4, 9, 16, etc.). a(n) gives all solutions to the equation x^2 + xy = n, with y mod 2 = 0, x > 0, y >= 0. - Andrew S. Plewe, Oct 19 2007
Alternatively, triangular sequence of coefficients of Dynkin diagram weights for the Cartan groups C_n: t(n,m) = m*(2*n - m). Row sums are A002412. - Roger L. Bagula, Aug 05 2008

			Array begins:
  1  4  9 16 25 36  49  64  81 100 ...
  3  8 15 24 35 48  63  80  99 120 ...
  5 12 21 32 45 60  77  96 117 140 ...
  7 16 27 40 55 72  91 112 135 160 ...
  9 20 33 48 65 84 105 128 153 180 ...
  ...
Triangle begins:
   1;
   3,  4;
   5,  8,  9;
   7, 12, 15, 16;
   9, 16, 21, 24, 25;
  11, 20, 27, 32, 35, 36;
  13, 24, 33, 40, 45, 48, 49;
  15, 28, 39, 48, 55, 60, 63, 64;
  17, 32, 45, 56, 65, 72, 77, 80, 81;
  19, 36, 51, 64, 75, 84, 91, 96, 99, 100;

R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 139.

G. C. Greubel, Antidiagonals n = 0..50, flattened

Rows give A000290, A005563, A028347, A028560, A028566, A098603, A098847, A098848, A098849, A098850.
Columns give A005408, A008586, A016945, A008590, A017329, A008594, A008598, A008602, A008606, A000567.
Diagonals give A033428, A045944, A067725.
Cf. A000384, A002412, A024352, A105020, A128076.

[(k+1)*(2*n-k+1): k in [0..n], n in [0..15]]; // G. C. Greubel, Mar 15 2023

t[n_, m_]:= (n^2 - m^2); Flatten[Table[t[i, j], {i,12}, {j,i-1,0,-1}]]
(* to view table *) Table[t[i, j], {j,0,6}, {i,j+1,10}]//TableForm (* Robert G. Wilson v, Jul 11 2005 *)
Table[(k+1)*(2*n-k+1), {n,0,15}, {k,0,n}]//Flatten (* Roger L. Bagula, Aug 05 2008 *)

def A105020(n,k): return (k+1)*(2*n-k+1)
flatten([[A105020(n,k) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Mar 15 2023

a(n) = r^2 - (r^2 + r - m)^2/4, where r = round(sqrt(m)) and m = 2*n+2. - Wesley Ivan Hurt, Sep 04 2021
a(n) = A128076(n+1) * A105020(n+1). - Wesley Ivan Hurt, Jan 07 2022
From G. C. Greubel, Mar 15 2023: (Start)
Sum_{k=0..n} T(n, k) = A002412(n+1).
Sum_{k=0..n} (-1)^k*T(n, k) = (1/2)*((1+(-1)^n)*A000384((n+2)/2) - (1- (-1)^n)*A000384((n+1)/2)). (End)

More terms from Robert G. Wilson v, Jul 11 2005

A235702 Fixed points of A001175 (Pisano periods).

Original entry on oeis.org

1, 24, 120, 600, 3000, 15000, 75000, 375000, 1875000, 9375000, 46875000, 234375000, 1171875000, 5859375000, 29296875000, 146484375000, 732421875000, 3662109375000, 18310546875000, 91552734375000, 457763671875000, 2288818359375000, 11444091796875000

Offset: 1

Reinhard Zumkeller, Jan 15 2014

Michael De Vlieger, Table of n, a(n) for n = 1..1430
Shaoshi Chen, Hanqian Fang, Sergey Kitaev, and Candice X.T. Zhang, Patterns in Multi-dimensional Permutations, arXiv:2411.02897 [math.CO], 2024. See pp. 2, 26.
J. D. Fulton and W. L. Morris, On arithmetical functions related to the Fibonacci numbers, Acta Arithmetica, 16 (1969), 105-110.
Wikipedia, Pisano period
Index entries for linear recurrences with constant coefficients, signature (5).

Cf. A001175, A001178, A008606, A000351.

a235702 n = if n == 1 then 1 else 24 * 5 ^ (n - 2)
a235702_list = 1 : iterate (* 5) 24

LinearRecurrence[{5},{1,24},30] (* or *) Join[{1},NestList[5#&,24,30]] (* Harvey P. Dale, May 07 2017 *)

Vec(-x*(19*x+1)/(5*x-1) + O(x^100)) \\ Colin Barker, Jan 16 2014

A001175(a(n)) = a(n); A001178(a(n)) = 0.
From Colin Barker, Jan 16 2014: (Start)
a(n) = 24*5^(n-2) for n > 1.
a(n) = 5*a(n-1) for n > 2.
G.f.: -x*(19*x+1) / (5*x-1). (End)
E.g.f.: (24*(exp(5*x) - 1) - 95*x)/25. - Stefano Spezia, Nov 09 2024

A274824 Triangle read by rows: T(n,k) = (n-k+1)*sigma(k), n>=1, 1<=k<=n.

Original entry on oeis.org

1, 2, 3, 3, 6, 4, 4, 9, 8, 7, 5, 12, 12, 14, 6, 6, 15, 16, 21, 12, 12, 7, 18, 20, 28, 18, 24, 8, 8, 21, 24, 35, 24, 36, 16, 15, 9, 24, 28, 42, 30, 48, 24, 30, 13, 10, 27, 32, 49, 36, 60, 32, 45, 26, 18, 11, 30, 36, 56, 42, 72, 40, 60, 39, 36, 12, 12, 33, 40, 63, 48, 84, 48, 75, 52, 54, 24, 28, 13, 36, 44, 70, 54, 96, 56, 90, 65, 72, 36, 56, 14

Offset: 1

Omar E. Pol, Oct 02 2016

Theorem: for any sequence S the partial sums of the partial sums are also the antidiagonal sums of the square array in which the n-th row gives n times the sequence S. Therefore they are also the row sums of the triangular array in which the n-th diagonal gives n times the sequence S.
In this case the sequence S is A000203.
The n-th diagonal of this triangle gives n times A000203.
The row sums give A175254 which gives the partial sums of A024916 which gives the partial sums of A000203.
T(n,k) is also the total number of unit cubes that are exactly below the terraces of the k-th level (starting from the top) up the base of the stepped pyramid with n levels described in A245092. This fact is because the mentioned terraces have the same shape as the symmetric representation of sigma(k). For more information see A237593 and A237270.
In the definition of this sequence the value n-k+1 is also the height of the terraces associated to sigma(k) in the mentioned pyramid with n levels, or in other words, the distance between the mentioned terraces and the base of the pyramid.
The sum of the n-th row of triangle equals the volume (also the number of cubes) of the mentioned pyramid with n levels.
For an illustration of the pyramid, see the Links section.
The sum of the n-th row is also 1/4 of the volume of the stepped pyramid described in A244050 with n levels.
Column k lists the positive multiples of sigma(k).
The k-th term in the n-th diagonal is equal to n*sigma(k).
Note that this is also a square array read by antidiagonals upwards: T(i,j) = i*sigma(j), i>=1, j>=1. The first row of the array is A000203. So consider that the pyramid is upside down. The value of "i" is the distance between the base of the pyramid and the terraces associated to sigma(j). The antidiagonal sums give the partial sums of the partial sums of A000203.

			Triangle begins:
1;
2,  3;
3,  6,  4;
4,  9,  8,  7;
5,  12, 12, 14, 6;
6,  15, 16, 21, 12, 12;
7,  18, 20, 28, 18, 24,  8;
8,  21, 24, 35, 24, 36,  16, 15;
9,  24, 28, 42, 30, 48,  24, 30,  13;
10, 27, 32, 49, 36, 60,  32, 45,  26,  18;
11, 30, 36, 56, 42, 72,  40, 60,  39,  36,  12;
12, 33, 40, 63, 48, 84,  48, 75,  52,  54,  24, 28;
13, 36, 44, 70, 54, 96,  56, 90,  65,  72,  36, 56,  14;
14, 39, 48, 77, 60, 108, 64, 105, 78,  90,  48, 84,  28, 24;
15, 42, 52, 84, 66, 120, 72, 120, 91,  108, 60, 112, 42, 48, 24;
16, 45, 56, 91, 72, 132, 80, 135, 104, 126, 72, 140, 56, 72, 48, 31;
...
For n = 16 and k = 10 the sum of the divisors of 10 is 1 + 2 + 5 + 10 = 18, and 16 - 10 + 1 = 7, and 7*18 = 126, so T(16,10) = 126.
On the other hand, the symmetric representation of sigma(10) has two parts of 9 cells, giving a total of 18 cells. In the stepped pyramid described in A245092, with 16 levels, there are 16 - 10 + 1 = 7 cubes exactly below every cell of the symmetric representation of sigma(10) up the base of pyramid; hence the total numbers of cubes exactly below the terraces of the 10th level (starting from the top) up the base of the pyramid is equal to 7*18 = 126. So T(16,10) = 126.
The sum of the 16th row of the triangle is 16 + 45 + 56 + 91 + 72 + 132 + 80 + 135 + 104 + 126 + 72 + 140 + 56 + 72 + 48 + 31 = A175254(16) = 1276, equaling the volume (also the number of cubes) of the stepped pyramid with 16 levels described in A245092 (see Links section).

Indranil Ghosh, Rows 1..100 of triangle, flattened
Omar E. Pol, Illustration of the stepped pyramid with 16 levels

Row sums of triangle give A175254.
Diagonals 1-6: A000203, A074400, A272027, A239050, A274535, A274536.
Column 1 is A000027.
Initial zeros should be omitted in the following sequences related to the columns of triangle:
Columns 2-5: A008585, A008586, A008589, A008588.
Columns 6 and 11: A008594.
Columns 7-9: A008590, A008597, A008595.
Columns 10 and 17: A008600.
Columns 12-13: A135628, A008596.
Columns 14, 15 and 23: A008606.
Columns 16 and 25: A135631.
(There are many other OEIS sequences that are also columns of this triangle.)
Cf. A004736, A024916, A054973, A196020, A236104, A235791, A237591, A237593, A237270, A237271, A244050, A245092, A245093, A262612.

T(n,k) = (n-k+1) * A000203(k).

T(n,k) = A004736(n,k) * A245093(n,k).

A317324 Multiples of 24 and odd numbers interleaved.

Original entry on oeis.org

0, 1, 24, 3, 48, 5, 72, 7, 96, 9, 120, 11, 144, 13, 168, 15, 192, 17, 216, 19, 240, 21, 264, 23, 288, 25, 312, 27, 336, 29, 360, 31, 384, 33, 408, 35, 432, 37, 456, 39, 480, 41, 504, 43, 528, 45, 552, 47, 576, 49, 600, 51, 624, 53, 648, 55, 672, 57, 696, 59, 720, 61, 744, 63, 768, 65, 792, 67, 816, 69

Offset: 0

Omar E. Pol, Jul 25 2018

Partial sums give the generalized 28-gonal numbers (A303812).

a(n) is also the length of the n-th line segment of the rectangular spiral whose vertices are the generalized 28-gonal numbers.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A008606 and A005408 interleaved.
Column 24 of A195151.
Sequences whose partial sums give the generalized k-gonal numbers: A026741 (k=5), A001477 (k=6), zero together with A080512 (k=7), A022998 (k=8), A195140 (k=9), zero together with A165998 (k=10), A195159 (k=11), A195161 (k=12), A195312 (k=13), A195817 (k=14).
Cf. A303812.

&cat[[24*n, 2*n + 1]: n in [0..30]]; // Vincenzo Librandi, Jul 28 2018

Table[If[EvenQ[n], 24 (n/2), n], {n, 0, 70}] (* Vincenzo Librandi, Jul 28 2018 *)
With[{nn=40},Riffle[24*Range[0,nn],2*Range[0,nn]+1]] (* or *) LinearRecurrence[ {0,2,0,-1},{0,1,24,3},80] (* Harvey P. Dale, Jul 06 2019 *)

concat(0, Vec(x*(1 + 24*x + x^2) / ((1 - x)^2*(1 + x)^2) + O(x^60))) \\ Colin Barker, Jul 29 2018

a(2n) = 24*n, a(2n+1) = 2*n + 1.
G.f.: x*(1 + 24*x + x^2)/((1-x)^2*(1+x)^2). - Vincenzo Librandi, Jul 28 2018
a(n) = 2*a(n-2) - a(n-4) for n>3. - Colin Barker, Jul 29 2018
Multiplicative with a(2^e) = 3*2^(e+2), and a(p^e) = p^e for an odd prime p. - Amiram Eldar, Oct 14 2023
Dirichlet g.f.: zeta(s-1) * (1 + 11*2^(1-s)). - Amiram Eldar, Oct 26 2023

A319073 Square array read by antidiagonals upwards: T(n,k) = k*sigma(n), n >= 1, k >= 1.

Original entry on oeis.org

1, 3, 2, 4, 6, 3, 7, 8, 9, 4, 6, 14, 12, 12, 5, 12, 12, 21, 16, 15, 6, 8, 24, 18, 28, 20, 18, 7, 15, 16, 36, 24, 35, 24, 21, 8, 13, 30, 24, 48, 30, 42, 28, 24, 9, 18, 26, 45, 32, 60, 36, 49, 32, 27, 10, 12, 36, 39, 60, 40, 72, 42, 56, 36, 30, 11, 28, 24, 54, 52, 75, 48, 84, 48, 63, 40, 33, 12

Offset: 1

Omar E. Pol, Sep 22 2018

			The corner of the square array begins:
         A000203 A074400 A272027 A239050 A274535 A274536 A319527 A319528
A000027:       1,      2,      3,      4,      5,      6,      7,      8, ...
A008585:       3,      6,      9,     12,     15,     18,     21,     24, ...
A008586:       4,      8,     12,     16,     20,     24,     28,     32, ...
A008589:       7,     14,     21,     28,     35,     42,     49,     56, ...
A008588:       6,     12,     18,     24,     30,     36,     42,     48, ...
A008594:      12,     24,     36,     48,     60,     72,     84,     96, ...
A008590:       8,     16,     24,     32,     40,     48,     56,     64, ...
A008597:      15,     30,     45,     60,     75,     90,    105,    120, ...
A008595:      13,     26,     39,     52,     65,     78,     91,    104, ...
A008600:      18,     36,     54,     72,     90,    108,    126,    144, ...
...

Index entries for sequences related to sigma(n)

Another version of A274824.
Antidiagonal sums give A175254.
Main diagonal gives A064987.
Row n lists the multiples of A000203(n).
Row 1 is A000027.
Initial zeros should be omitted in the following sequences related to the rows of the array:
Row 2-5: A008585, A008586, A008589, A008588.
Rows 6 and 11: A008594.
Rows 7-9: A008590, A008597, A008595.
Rows 10 and 17: A008600.
Rows 12-13: A135628, A008596.
Rows 14, 15 and 23: A008606.
Rows 16 and 25: A135631.
(Note that in the OEIS there are many other sequences that are also rows of this square array.)
Cf. A000203, A237593, A319526.

T:=Flat(List([1..12],n->List([1..n],k->k*Sigma(n-k+1))));; Print(T); # Muniru A Asiru, Jan 01 2019

with(numtheory): T:=(n,k)->k*sigma(n-k+1): seq(seq(T(n,k),k=1..n),n=1..12); # Muniru A Asiru, Jan 01 2019

Table[k DivisorSigma[1, #] &[m - k + 1], {m, 12}, {k, m}] // Flatten (* Michael De Vlieger, Dec 31 2018 *)

A280097 Sum of the divisors of 24*n - 1.

Original entry on oeis.org

24, 48, 72, 120, 144, 168, 168, 192, 264, 240, 264, 336, 312, 408, 360, 384, 456, 432, 672, 480, 504, 576, 600, 744, 600, 720, 648, 744, 840, 720, 744, 840, 912, 984, 840, 864, 888, 912, 1296, 1104, 984, 1080, 1032, 1272, 1176, 1104, 1368, 1152, 1488, 1320, 1224, 1320, 1344, 1824, 1320

Offset: 1

Omar E. Pol, Dec 25 2016

All terms are multiples of 24 [Gupta, Sierpinski]. - Vincenzo Librandi, Apr 07 2011

Note that 24n - 1 is also the denominator of the Bruinier-Ono finite algebraic formula for the number of partitions of n (Cf. A183010).

			For n = 5 we have that 24*5 - 1 = 119, and the sum of the divisors of 119 is 1 + 7 + 17 + 119 = 144, so a(5) = 144.

Hansraj Gupta, Congruent properties of sigma(n), Math. Student 13 (1945), 25-29; entire issue.
Wacław Sierpiński, Elementary Theory of numbers, Monografie Mathematyczne, Vol. 42 (1964), chapter 4, p. 168.

Cf. A000203, A008606, A183010, A280098.

DivisorSigma[1,24*Range[60]-1] (* Harvey P. Dale, Jan 25 2024 *)

a(n) = sigma(24*n - 1); \\ Amiram Eldar, Jan 09 2025

a(n) = A000203(A183010(n)).

Sum_{k=1..n} a(k) = c * n^2 + O(n*log(n)), where c = 4*Pi^2/3 = 13.159472... . - Amiram Eldar, Mar 28 2024

cp's OEIS Frontend

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