A022267 -id:A022267 - cp's OEIS frontend

A235332 a(n) = n(9n + 25)/2 + 6.

6, 23, 49, 84, 128, 181, 243, 314, 394, 483, 581, 688, 804, 929, 1063, 1206, 1358, 1519, 1689, 1868, 2056, 2253, 2459, 2674, 2898, 3131, 3373, 3624, 3884, 4153, 4431, 4718, 5014, 5319, 5633, 5956, 6288, 6629, 6979, 7338, 7706, 8083, 8469, 8864, 9268, 9681, 10103

Offset: 0

Bruno Berselli, Jan 22 2014

This is the case d=6 of n*(9*n + 4*d + 1)/2 + d. Other similar sequences are:
d=0, A022267;
d=1, A064225;
d=2, A062123;
d=3, A064226;
d=4, A022266 (with initial 0);
d=5, A178977.
First bisection of A235537.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A017257 (first differences), A022266, A022267, A062123, A064225, A064226, A081267, A178977, A235537.

Magma
```
[n*(9*n+25)/2+6: n in [0..50]];
```

Table[n (9 n + 25)/2 + 6, {n, 0, 50}]
LinearRecurrence[{3,-3,1},{6,23,49},50] (* Harvey P. Dale, Feb 12 2022 *)

a(n)=n*(9*n+25)/2+6 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (6 + 5*x - 2*x^2)/(1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
2*a(n) - a(n+1) + 12 = A081267(n).
E.g.f.: exp(x)*(12 + 34*x + 9*x^2)/2. - Elmo R. Oliveira, Nov 13 2024

A241016 Triangle read by rows: T(n, k) = sum of k-th row of n X n square filled with the numbers 1 through n^2 reading across rows left-to-right.

Original entry on oeis.org

1, 3, 7, 6, 15, 24, 10, 26, 42, 58, 15, 40, 65, 90, 115, 21, 57, 93, 129, 165, 201, 28, 77, 126, 175, 224, 273, 322, 36, 100, 164, 228, 292, 356, 420, 484, 45, 126, 207, 288, 369, 450, 531, 612, 693, 55, 155, 255, 355, 455, 555, 655, 755, 855, 955, 66, 187, 308, 429, 550

Offset: 1

Kival Ngaokrajang, Aug 08 2014

See illustration in links.

The corresponding triangle with column sums is found in A251630. - Wolfdieter Lang, Dec 09 2014

			The triangle T(n, k) begins:
n\k  1   2   3   4   5   6   7   8   9  10 ...
1:   1
2:   3   7
3:   6  15  24
4:  10  26  42  58
5:  15  40  65  90 115
6:  21  57  93 129 165 201
7:  28  77 126 175 224 273 322
8:  36 100 164 228 292 356 420 484
9:  45 126 207 288 369 450 531 612 693
10: 55 155 255 355 455 555 655 755 855 955
... reformatted - _Wolfdieter Lang_, Dec 08 2014

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Kival Ngaokrajang, Illustration of initial terms

Columns k=1..6: A000217, A005449, A005475, A022265, A022267, A022269.
Diagonals: A081436, A059270, ...
Row sums: A037270.

Table[Sum[n*(k - 1) + j, {j,1,n}], {n,1,10}, {k,1,n}] // Flatten (* G. C. Greubel, Aug 23 2017 *)

trg(nn) = {for (n=1, nn, mm = matrix(n, n, i, j, j + n*(i-1)); for (i=1, n, print1(sum(j=1, n, mm[i, j]), ", ");); print(););} \\ Michel Marcus, Sep 15 2014

T(n, k) = Sum_{j=1..n} (n*(k-1)+ j), for n >= k >= 1. See the Michel Marcus program. - Wolfdieter Lang, Dec 08 2014

T(n, k) = binomial(n+1, 2) + n^2*(k-1). - Wolfdieter Lang, Dec 09 2014

Edited. - Wolfdieter Lang, Dec 08 2014

A082286 a(n) = 18*n + 10.

Original entry on oeis.org

10, 28, 46, 64, 82, 100, 118, 136, 154, 172, 190, 208, 226, 244, 262, 280, 298, 316, 334, 352, 370, 388, 406, 424, 442, 460, 478, 496, 514, 532, 550, 568, 586, 604, 622, 640, 658, 676, 694, 712, 730, 748, 766, 784, 802, 820, 838, 856, 874, 892, 910, 928, 946

Offset: 0

Cino Hilliard, May 10 2003

Solutions to (11^x + 13^x) mod 19 = 17.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Tanya Khovanova, Recursive Sequences
Leo Tavares, Illustration: Triangles
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A006370, A008600, A016945, A161705.

Cf. A000217, A017221, A022267, A060544.

[18*n + 10: n in [0..60]]; // Vincenzo Librandi, May 05 2011

Range[10, 1000, 18] (* Vladimir Joseph Stephan Orlovsky, Jun 01 2011 *)

a(n)=18*n+10 \\ Charles R Greathouse IV, Jul 10 2016

a(n) = A006370(A016945(n)). - Reinhard Zumkeller, Apr 17 2008
a(n) = 2*A017221(n). - Michel Marcus, Feb 15 2014
a(n) = A060544(n+2) - 9*A000217(n-1). - Leo Tavares, Oct 15 2022
From Elmo R. Oliveira, Apr 08 2024: (Start)
G.f.: 2*(5+4*x)/(1-x)^2.
E.g.f.: 2*exp(x)*(5 + 9*x).
a(n) = 2*a(n-1) - a(n-2) for n >= 2.
a(n) = 2*(A022267(n+1) - A022267(n)). (End)

More terms from Reinhard Zumkeller, Apr 17 2008

A198392 a(n) = (6n(3n+7)+(2n+13)*(-1)^n+3)/16 + 1.

Original entry on oeis.org

2, 4, 12, 18, 31, 41, 59, 73, 96, 114, 142, 164, 197, 223, 261, 291, 334, 368, 416, 454, 507, 549, 607, 653, 716, 766, 834, 888, 961, 1019, 1097, 1159, 1242, 1308, 1396, 1466, 1559, 1633, 1731, 1809, 1912, 1994, 2102, 2188, 2301, 2391, 2509, 2603, 2726, 2824, 2952

Offset: 0

Bruno Berselli, Oct 25 2011

For an origin of this sequence, see the triangular spiral illustrated in the Links section.

First bisection gives A117625 (without the initial term).

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A152832 (by Superseeker).

Cf. sequences related to the triangular spiral: A022266, A022267, A027468, A038764, A045946, A051682, A062708, A062725, A062728, A062741, A064225, A064226, A081266-A081268, A081270-A081272, A081275 [incomplete list].

[(6*n*(3*n+7)+(2*n+13)*(-1)^n+3)/16+1: n in [0..50]];

LinearRecurrence[{1,2,-2,-1,1},{2,4,12,18,31},60] (* Harvey P. Dale, Jun 15 2022 *)

for(n=0, 50, print1((6*n*(3*n+7)+(2*n+13)*(-1)^n+3)/16+1", "));

G.f.: (2+2*x+4*x^2+2*x^3-x^4)/((1+x)^2*(1-x)^3).
a(n) = a(n-1)+2*a(n-2)-2*a(n-3)-a(n-4)+a(n-5).
a(n)-a(-n-1) = A168329(n+1).
a(n)+a(n-1) = A102214(n).
a(2n)-a(2n-1) = A016885(n).
a(2n+1)-a(2n) = A016825(n).

A262997 a(n+3) = a(n) + 24*n + 40, a(0)=0, a(1)=5, a(2)=19.

Original entry on oeis.org

0, 5, 19, 40, 69, 107, 152, 205, 267, 336, 413, 499, 592, 693, 803, 920, 1045, 1179, 1320, 1469, 1627, 1792, 1965, 2147, 2336, 2533, 2739, 2952, 3173, 3403, 3640, 3885, 4139, 4400, 4669, 4947, 5232, 5525, 5827, 6136, 6453, 6779, 7112, 7453, 7803, 8160, 8525

Offset: 0

Paul Curtz, Oct 07 2015

The hexasections of A262397(n) are
0,  1,  4,  9, 16,  25,  36, ...  = A000290(n)
0,  5, 19, 40, 69, 107, 152, ...  = a(n)
0,  1,  5, 11, 18,  28,  40, ...  = A240438(n+1)
1,  9, 25, 49, 81, 121, 169, ...  = A016754(n)
0,  2,  7, 13, 21,  32,  44, ...  = A262523(n)
3, 13, 32, 59, 93, 136, 187, ...  = e(n+1).
The five-step recurrence in FORMULA is valuable for the six sequences.
Consider a(n) extended from right to left with their first two differences:
...,  59,  32,  13,  3, 0,  5, 19, 40, 69, ...
..., -27, -19, -10, -3, 5, 14, 21, 29, 38, ...
...,   8,   9,   7,  8, 9,  7,  8,  9,  7, ... .
From 0,the first row is
1) from right to left: e(n)
2) from left to right: a(n).
a(n) and e(n) are companions.
The third row is of period 3.
The last digit of a(n) is of period 15; the same is true of e(n).

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

Cf. A000290, A008586, A016742, A016754, A016789, A016921, A016945, A022267, A042965, A240438, A262397, A262523.

a[0] = 0; a[1] = 5; a[2] = 19; a[n_] := a[n] = a[n - 3] + 24 (n - 3) + 40; Table[a@ n, {n, 0, 46}] (* Michael De Vlieger, Oct 09 2015 *)
LinearRecurrence[{2,-1,1,-2,1},{0,5,19,40,69},60] (* Harvey P. Dale, Dec 16 2024 *)

vector(100, n, n--; 4*n^2 + (4*(n+1)-3)\3) \\ Altug Alkan, Oct 07 2015

concat(0, Vec(-x*(x+1)*(3*x^2+4*x+5)/((x-1)^3*(x^2+x+1)) + O(x^100))) \\ Colin Barker, Oct 08 2015

a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) - a(n-5), n> 4.
a(n) = A016742(n) + A042965(n).
a(-n) = e(n).
a(-n) + a(n) = 8*n^2.
a(n+2) - 2*a(n+1) + a(n) = period 3:repeat 9, 7, 8.
a(n+3) - a(n-3) =  8*(1 + 6*n).
a(n+7) - a(n-7) = 40*(2 + 3*n).
a(2n+1) = -a(2n) + 6*n + 3.
a(2n+2) = -a(2n+1) + 4*(n+1).
a(3n) = 4*n*(9*n+1) = 8*A022267(n), a(3n+1) = 36*n^2 +28*n +5, a(3n+2) = 36*n^2 +52*n +19.
G.f.: -x*(x+1)*(3*x^2+4*x+5) / ((x-1)^3*(x^2+x+1)). - Colin Barker, Oct 08 2015

A254605 The minimum absolute difference between k*m1 and m2 (m1A075362.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 2, 1, 0, 0, 1, 1, 1, 2, 1, 0, 0, 0, 1, 0, 2, 2, 1, 0, 0, 1, 0, 1, 1, 3, 2, 1, 0, 0, 0, 1, 2, 0, 2, 3, 2, 1, 0, 0, 1, 1, 1, 1, 1, 3, 3, 2, 1, 0, 0, 0, 0, 0, 2, 0, 2, 4, 3, 2, 1, 0, 0, 1, 1, 1, 2, 1, 1, 3, 4

Offset: 1

Lei Zhou, Feb 02 2015

k is an integer that minimizes |k*m1-m2|. It is trivial that if j is the integer part of m2/m1, k is either j or j+1.

Interestingly, suppose b is the smallest n such that a(n)=c; the sequence s(c)=b is then sequence A022267.

			A075362(1)=1=1*1, 1-1=0, so a(1)=0;
A075362(5)=6=2*3, 3-2=1, 2*2-3=1, so a(5)=1;
A075362(19)=24=4*6, 6-4=2, 4*2-6=2, so a(19)=2.

Lei Zhou, Table of n, a(n) for n = 1..10000

Cf. A075362, A022267.

NumDiff[n1_, n2_] :=  Module[{c1 = n1, c2 = n2}, If[c1 < c2, c1 = c1 + c2; c2 = c1 - c2; c1 = c1 - c2];
  k = Floor[c1/c2]; a1 = c1 - k*c2; If[a1 == 0, a2 = 0, a2 = (k + 1) c2 - c1]; Return[Min[a1, a2]]];
p1 = 1; p2 = 0; Table[p2++; If[p2 > p1, p1 = p2; p2 = 1];  NumDiff[p1, p2], {n, 1, 100}]

A343053 Table read by ascending antidiagonals: T(k, n) is the maximum vertex sum in a perimeter-magic k-gon of order n.

Original entry on oeis.org

15, 24, 24, 40, 42, 33, 54, 65, 56, 42, 77, 93, 90, 74, 51, 96, 126, 126, 115, 88, 60, 126, 164, 175, 165, 140, 106, 69, 150, 207, 224, 224, 198, 165, 120, 78, 187, 255, 288, 292, 273, 237, 190, 138, 87, 216, 308, 350, 369, 352, 322, 270, 215, 152, 96, 260, 366, 429, 455, 450, 420, 371, 309, 240, 170, 105

Offset: 3

Stefano Spezia, Apr 03 2021

			The table begins:
k\n|   3    4    5    6    7 ...
---+------------------------
3  |  15   24   33   42   51 ...
4  |  24   42   56   74   88 ...
5  |  40   65   90  115  140 ...
6  |  54   93  126  165  198 ...
7  |  77  126  175  224  273 ...
...

Terrel Trotter, Perimeter-Magic Polygons, Journal of Recreational Mathematics Vol. 7, No. 1, 1974, pp. 14-20 (see equations 6 and 8).

Cf. A005475 (n = 4), A022267 (n = 6), A059270, A179805 (k = 3), A343052 (minimum).

T[k_,n_]:=k(1+k(2n-3)-Mod[n,2](1-Mod[k,2]))/2; Table[T[k+3-n,n],{k,3,14},{n,3,k}]//Flatten

T(k, n) = k*(1 + k*(2n - 3) - (n mod 2)*(1 - (k mod 2)))/2.

T(n, n) = A059270(n-1).

A360665 Square array T(n, k) = k((2n-1)*k+1)/2 read by rising antidiagonals.

Original entry on oeis.org

0, 0, 0, 0, 1, -1, 0, 2, 3, -3, 0, 3, 7, 6, -6, 0, 4, 11, 15, 10, -10, 0, 5, 15, 24, 26, 15, -15, 0, 6, 19, 33, 42, 40, 21, -21, 0, 7, 23, 42, 58, 65, 57, 28, -28, 0, 8, 27, 51, 74, 90, 93, 77, 36, -36, 0, 9, 31, 60, 90, 115, 129, 126, 100, 45, -45

Offset: 0

Paul Curtz, Mar 17 2023

			By rows:
   0,   0,  -1,  -3,  -6,  -10,  -15,  -21,  -28, ...   = -A161680
   0,   1,   3,   6,  10,   15,   21,   28,   36, ...   =  A000217
   0,   2,   7,  15,  26,   40,   57,   77,  100, ...   =  A005449
   0,   3,  11,  24,  42,   65,   93,  126,  164, ...   =  A005475
   0,   4,  15,  33,  58,   90,  129,  175,  228, ...   =  A022265
   0,   5,  19,  42,  74,  115,  165,  224,  292, ...   =  A022267
   0,   6,  23,  51,  90,  140,  201,  273,  356, ...   =  A022269
   ... .

Cf. Antidiagonal sums: A034827(n+1).
Cf. A000217, A000290, A005449, A005475, A022265.
Cf. A022267, A022269, A059270, A081436, A101986.
Cf. A161680, A162254, A179297, A360962, A361226.

T[n_, k_] := ((2*n - 1)*k^2 + k)/2; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Mar 31 2023 *)

T(n, k) = ((2*n-1)*k^2+k)/2 \\ Thomas Scheuerle, Mar 17 2023

T(n,k) = T(n,k-1)+k^2.
T(n,n) = A081436(n-1).
T(n,n+1) = A059270(n).
T(n,n+4) = -3*A179297(n+4).
T(n+3,n) = A162254(n).
T(n+5,n) = 3*A101986(n).
From Stefano Spezia, Mar 31 2023: (Start)
O.g.f.: (x*y - y^2 + 2*x*y^2)/((1 - x)^2*(1 - y)^3).
E.g.f.: exp(x+y)*y*(2*x - y + 2*x*y)/2. (End)

cp's OEIS Frontend

A235332 a(n) = n*(9*n + 25)/2 + 6.

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A241016 Triangle read by rows: T(n, k) = sum of k-th row of n X n square filled with the numbers 1 through n^2 reading across rows left-to-right.

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A082286 a(n) = 18*n + 10.

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A198392 a(n) = (6*n*(3*n+7)+(2*n+13)*(-1)^n+3)/16 + 1.

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A262997 a(n+3) = a(n) + 24*n + 40, a(0)=0, a(1)=5, a(2)=19.

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A254605 The minimum absolute difference between k*m1 and m2 (m1A075362.

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A343053 Table read by ascending antidiagonals: T(k, n) is the maximum vertex sum in a perimeter-magic k-gon of order n.

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A235332 a(n) = n(9n + 25)/2 + 6.

A198392 a(n) = (6n(3n+7)+(2n+13)*(-1)^n+3)/16 + 1.

A360665 Square array T(n, k) = k((2n-1)*k+1)/2 read by rising antidiagonals.