A056000 -id:A056000 - cp's OEIS frontend

A098832 Square array read by antidiagonals: even-numbered rows of the table are of the form n(n+m) and odd-numbered rows are of the form n(n+m)/2.

1, 3, 3, 6, 8, 2, 10, 15, 5, 5, 15, 24, 9, 12, 3, 21, 35, 14, 21, 7, 7, 28, 48, 20, 32, 12, 16, 4, 36, 63, 27, 45, 18, 27, 9, 9, 45, 80, 35, 60, 25, 40, 15, 20, 5, 55, 99, 44, 77, 33, 55, 22, 33, 11, 11, 66, 120, 54, 96, 42, 72, 30, 48, 18, 24, 6, 78, 143, 65, 117, 52, 91, 39, 65, 26, 39, 13, 13

Offset: 1

Eugene McDonnell (eemcd(AT)mac.com), Nov 02 2004

The rows of this table and that in A098737 are related. Given a function f = n/( 1 + (1+n) mod(2) ), row n of A098737 can be derived from row n of T by multiplying the latter by f(n); row n of T can be derived from row n of A098737 by dividing the latter by f(n).

			Array begins as:
  1,  3,  6, 10, 15, 21,  28,  36,  45 ... A000217;
  3,  8, 15, 24, 35, 48,  63,  80,  99 ... A005563;
  2,  5,  9, 14, 20, 27,  35,  44,  54 ... A000096;
  5, 12, 21, 32, 45, 60,  77,  96, 117 ... A028347;
  3,  7, 12, 18, 25, 33,  42,  52,  63 ... A027379;
  7, 16, 27, 40, 55, 72,  91, 112, 135 ... A028560;
  4,  9, 15, 22, 30, 39,  49,  60,  72 ... A055999;
  9, 20, 33, 48, 65, 84, 105, 128, 153 ... A028566;
  5, 11, 18, 26, 35, 45,  56,  68,  81 ... A056000;
Antidiagonals begin as:
   1;
   3,  3;
   6,  8,  2;
  10, 15,  5,  5;
  15, 24,  9, 12,  3;
  21, 35, 14, 21,  7,  7;
  28, 48, 20, 32, 12, 16,  4;
  36, 63, 27, 45, 18, 27,  9,  9;
  45, 80, 35, 60, 25, 40, 15, 20,  5;
  55, 99, 44, 77, 33, 55, 22, 33, 11, 11;

G. C. Greubel, Antidiagonals n = 1..50, flattened

Row m of array: A000217 (m=1), A005563 (m=2), A000096 (m=3), A028347 (m=4), A027379 (m=5), A028560 (m=6), A055999 (m=7), A028566 (m=8), A056000 (m=9), A098603 (m=10), A056115 (m=11), A098847 (m=12), A056119 (m=13), A098848 (m=14), A056121 (m=15), A098849 (m=16), A056126 (m=17), A098850 (m=18), A051942 (m=19).
Column m of array: A026741 (m=1), A022998 (m=2), A165351 (m=3).
Cf. A098737, A168509, A181900.

A098832:= func< n,k | (1/4)*(3+(-1)^k)*(n+1)*(n-k+1) >;
[A098832(n,k): k in [1..n], n in [1..15]]; // G. C. Greubel, Jul 31 2022

A098832[n_, k_]:= (1/4)*(3+(-1)^k)*(n+1)*(n-k+1);
Table[A098832[n,k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Jul 31 2022 *)

def A098832(n,k): return (1/4)*(3+(-1)^k)*(n+1)*(n-k+1)
flatten([[A098832(n,k) for k in (1..n)] for n in (1..15)]) # G. C. Greubel, Jul 31 2022

Item m of row n of T is given (in infix form) by: n T m = n * (n + m) / (1 + m (mod 2)). E.g. Item 4 of row 3 of T: 3 T 4 = 14.
From G. C. Greubel, Jul 31 2022: (Start)
A(n, k) = (1/4)*(3 + (-1)^n)*k*(k+n) (array).
T(n, k) = (1/4)*(3 + (-1)^k)*(n+1)*(n-k+1) (antidiagonal triangle).
Sum_{k=1..n} T(n, k) = (1/8)*(n+1)*( (3*n-1)*(n+1) + (1+(-1)^n)/2 ).
T(2*n-1, n) = A181900(n).
T(2*n+1, n) = 2*A168509(n+1). (End)

Missing terms added by G. C. Greubel, Jul 31 2022

A219502 T(n,k)=Number of nXk arrays of the minimum value of corresponding elements and their horizontal or vertical neighbors in a random, but sorted with lexicographically nondecreasing rows and nonincreasing columns, 0..1 nXk array.

Original entry on oeis.org

2, 2, 2, 3, 3, 3, 4, 5, 5, 4, 5, 7, 11, 7, 5, 6, 9, 18, 18, 9, 6, 7, 11, 26, 35, 26, 11, 7, 8, 13, 35, 58, 58, 35, 13, 8, 9, 15, 45, 88, 107, 88, 45, 15, 9, 10, 17, 56, 126, 179, 179, 126, 56, 17, 10, 11, 19, 68, 173, 281, 325, 281, 173, 68, 19, 11, 12, 21, 81, 230, 421, 550, 550, 421

Offset: 1

R. H. Hardin Nov 20 2012

Table starts
..2..2...3...4....5....6.....7.....8.....9....10....11....12....13....14....15
..2..3...5...7....9...11....13....15....17....19....21....23....25....27....29
..3..5..11..18...26...35....45....56....68....81....95...110...126...143...161
..4..7..18..35...58...88...126...173...230...298...378...471...578...700...838
..5..9..26..58..107..179...281...421...608...852..1164..1556..2041..2633..3347
..6.11..35..88..179..325...550...885..1369..2050..2986..4246..5911..8075.10846
..7.13..45.126..281..550...995..1703..2793..4424..6804.10200.14949.21470.30277
..8.15..56.173..421..885..1703..3083..5328..8869.14306.22458.34423.51649.76017
..9.17..68.230..608.1369..2793..5328..9663.16831.28346.46382.74003
.10.19..81.298..852.2050..4424..8869.16831.30581.53601.91116
.11.21..95.378.1164.2986..6804.14306.28346.53601.97541
.12.23.110.471.1556.4246.10200.22458.46382.91116

			Some solutions for n=3 k=4
..0..0..0..0....0..0..0..0....0..0..0..0....1..0..0..0....1..0..0..0
..0..0..0..0....0..0..0..0....1..0..0..0....1..1..0..0....1..0..0..0
..1..0..0..0....1..1..1..0....1..1..1..0....1..1..1..0....1..0..0..0

R. H. Hardin, Table of n, a(n) for n = 1..239

Column 1 is A000027
Column 2 is A004273
Column 3 is A056000(n-1) for n>1

Empirical for column k:
k=1: a(n) = n for n>1
k=2: a(n) = 2*n - 1 for n>1
k=3: a(n) = (1/2)*n^2 + (7/2)*n - 4 for n>1
k=4: a(n) = (1/6)*n^3 + n^2 + (23/6)*n - 7 for n>2
k=5: a(n) = (1/24)*n^4 + (1/4)*n^3 + (35/24)*n^2 + (21/4)*n - 13 for n>2
k=6: a(n) = (1/120)*n^5 + (1/24)*n^4 + (13/24)*n^3 + (59/24)*n^2 + (59/20)*n - 17 for n>3
k=7: a(n) = (1/720)*n^6 + (1/240)*n^5 + (23/144)*n^4 + (13/16)*n^3 + (331/180)*n^2 + (311/60)*n - 27 for n>3

A096941 Fourth column of (1,5)-Pascal triangle A096940.

Original entry on oeis.org

5, 16, 34, 60, 95, 140, 196, 264, 345, 440, 550, 676, 819, 980, 1160, 1360, 1581, 1824, 2090, 2380, 2695, 3036, 3404, 3800, 4225, 4680, 5166, 5684, 6235, 6820, 7440, 8096, 8789, 9520, 10290, 11100, 11951, 12844, 13780, 14760, 15785, 16856, 17974, 19140

Offset: 0

Wolfdieter Lang, Jul 16 2004

If Y is a 5-subset of an n-set X then, for n>=7, a(n-7) is the number of 3-subsets of X having at most one element in common with Y. - Milan Janjic, Dec 08 2007

Third column: A056000; fifth column: A096942.

Table[(n^3 + 15 n^2 + 14 n)/6, {n, 100}] (* Vladimir Joseph Stephan Orlovsky, Jul 06 2011 *)

a(n)= (n+15)*(n+2)*(n+1)/6 = 5*b(n)-4*b(n-1), with b(n):=A000292(n)=binomial(n+3, 3).

G.f.: (5-4*x)/(1-x)^4.

A337940 Triangle read by rows: T(n, k) = T(n+2) - T(n-k), with the triangular numbers T = A000217, for n >= 1, k = 1, 2, ..., n.

Original entry on oeis.org

6, 9, 10, 12, 14, 15, 15, 18, 20, 21, 18, 22, 25, 27, 28, 21, 26, 30, 33, 35, 36, 24, 30, 35, 39, 42, 44, 45, 27, 34, 40, 45, 49, 52, 54, 55, 30, 38, 45, 51, 56, 60, 63, 65, 66, 33, 42, 50, 57, 63, 68, 72, 75, 77, 78, 36, 46, 55, 63, 70, 76, 81, 85, 88, 90, 91

Offset: 1

Wolfdieter Lang, Nov 23 2020

This number triangle results from the array A(n, m) = T(n+m+1) - T(n-1), with T = A000217, for n, m >= 1. For this array see the example by Bob Selcoe, in A111774 (but with rows continued). The present triangle is obtained by reading the array by upwards antidiagonals: T(n, k) = A(n+1-k, k). See also the Jul 09 2019 comment by Ralf Steiner with the formula c_k(n) (rows k >= 1, columns n >= 3), rewritten for A(n, m) = (m+2)*(2*n+m+1)/2, leading to T(n, k) = (k+2)*(2*n-k+3)/2.
Therefore this triangle is related to the problem of giving the numbers which are sums of at least three consecutive positive integers given as sequence A111774. It allows us to find the multiplicities for the numbers of A111774. They are given in A338428(n).
To obtain the multiplicity for number N (>= 6) from A111774 one has to consider only the triangle rows n = 1, 2, ..., floor((N-3)/3).
The row reversed triangle, considered by Bob Selcoe in A111774, is T(n, n-k+1) = T(n+2) - T(k-1), for n >= 1, and k=1, 2, ..., n.
This triangle contains no odd prime numbers and no exact powers 2^m, for m >= 0. This can be seen by considering the diagonal sequences D(d, k), for d >= 1, k >= 1 or the row sequences of the array A(n, m), for n >= 1 and m >= 1. The result is A(r+1, s-2) = s*(s + 2*r + 1)/2, for r >= 0 and s >= 3 (from the g.f. of the diagonals of T given below). This is also given in the Jul 09 2019 comment by Ralf Steiner in A111774. Therefore A(r+1, s-2) is a product of two numbers >= 2, hence not a prime. And in both cases (i) s/2 integer or (ii) (s + 2*r + 1)/2 integer not both numbers can be powers of 2 by simple parity arguments.
The previous comment means that each T(n, k) has at least one odd prime as a proper divisor.
A number N appears in this triangle, or in A111774, if and only if floor(N/2) - delta(N) >= 1, where delta(N) = A055034(N). For the sequence b(n) := floor(n/2) - delta(n), for n >= 2, see A219839(n), b(1) = -1. See a W. Lang comment in A111774 for the proof.

			The triangle T(n, k) begins:
n \ k  1  2  3  4  5   6   7   8   9  10  11  12  13  14  15 ...
1:     6
2:     9 10
3:    12 14 15
4:    15 18 20 21
5:    18 22 25 27 28
6:    21 26 30 33 35  36
7:    24 30 35 39 42  44  45
8:    27 34 40 45 49  52  54  55
9:    30 38 45 51 56  60  63  65  66
10:   33 42 50 57 63  68  72  75  77  78
11:   36 46 55 63 70  76  81  85  88  90  91
12:   39 50 60 69 77  84  90  95  99 102 104 105
13:   42 54 65 75 84  92  99 105 110 114 117 119 120
14:   45 58 70 81 91 100 108 115 121 126 130 133 135 136
15:   48 62 75 87 98 108 117 125 132 138 143 147 150 152 153
...
N = 15 appears precisely twice from the sums 4+5+6 = A(4, 1) = T(4, 1), and (1+2+3)+4+5 = A(1, 3) = T(3, 3), i.e., with a sum of 3 and 5 consecutive positive integers.
N = 42 appears three times from the sums 13+14+15 = A(13, 1) = T(13, 1), 9+10+11 +12 = A(9, 2) = T(10, 2), 3+4+5+6+7+8+9 = A(3, 5) = T(7, 5); i.e., 42 can be written as a sum of 3, 4 and 7 consecutive positive integers.

Cf. A055034, A111774, A338428 (multiplicities), A219839.
For columns k = 1, 2, ..., 10 see A008585, A016825, A008587, A016945, A008589, A017113, A008591, A017329, A008593, A017593.
For diagonals d = 1, 2, ..., 10 see A000217, A000096, A055998, A055999, A056000, A056115, A056119 , A056121, A056126, A051942.

Flatten[Table[((n+2)*(n+3)-(n-k)*(n-k+1))/2,{n,11},{k,n}]] (* Stefano Spezia, Nov 24 2020 *)

T(n, k) = ((n+2)*(n+3) - (n-k)*(n-k+1))/2, for n >= 1 and k = 1, 2, ..., n (see the name).
T(n, k) = (k+2)*(2*n-k+3)/2 (factorized).
G.f. columns k = 2*j+1, for j >= 0: Go(j, x) = x^(2*j+1)*(2*j+3)*(j+2 - (j+1)*x)/(1-x)^2,
G.f. columns k = 2*j, for j >= 1: Ge(j, x) = x^(2*j)*(j+1)*(2*j+3 - (2*j+1)*x)/(1-x)^2.
G.f. row polynomials: G(z,x) = z*x*(1 + z*x)^3*{3*(2-z) - (8-3*z)*(z*x) + (3-z)*(z*x)^2}/((1 - z)^2*(1 - (z*x)^2)^3).
G.f. diagonals d >= 1: GD(d, x) = ((d+1)*3 - (5*d+3)*x + (2*d+1)*x^2)/(1-x)^3.
G.f. of GD(d, x): GGD(z,x) = (6-8*x+3*x^2 - (3-3*x+x^2)*z)/((1-x)^3*(1-z)^2).

A212427 a(n) = 17*n + A000217(n-1).

Original entry on oeis.org

0, 17, 35, 54, 74, 95, 117, 140, 164, 189, 215, 242, 270, 299, 329, 360, 392, 425, 459, 494, 530, 567, 605, 644, 684, 725, 767, 810, 854, 899, 945, 992, 1040, 1089, 1139, 1190, 1242, 1295, 1349, 1404, 1460, 1517, 1575, 1634, 1694, 1755, 1817, 1880, 1944, 2009

Offset: 0

Jesse Han, May 16 2012

Generalization: T(n,i) = A000217(i-1+n) - A000217(i-1) = i*n + A000217(n-1); in this case is i=17. See also the comment in A212428.

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000096, A001008, A002805, A055998-A056000, A056115, A056119, A056121, A056126, A051942, A101859, A132754-A132758, A212428.

For n > 22, T(n,17) matches A074170 but with opposite sign.

[n*(n+33)/2: n in [0..49]]; // Bruno Berselli, Jun 22 2012

Table[-17 (17 - 1)/2 + (17 + n) (16 + n)/2, {n, 0, 100}]

a(n)=n*(n+33)/2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = (16+n)*(17+n)/2 - 16*17/2 = 17*n + (n-1)*n/2 = n*(n+33)/2.
G.f.: x*(17-16*x)/(1-x)^3. - Bruno Berselli, Jun 22 2012
a(n) = 17*n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
From Amiram Eldar, Jan 11 2021: (Start)
Sum_{n>=1} 1/a(n) = 2*A001008(33)/(33*A002805(33)) = 53676090078349/216605329340400.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/33 - 14606816124167/340379803249200. (End)
From Elmo R. Oliveira, Dec 12 2024: (Start)
E.g.f.: exp(x)*x*(34 + x)/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A212428 a(n) = 18*n + A000217(n-1).

Original entry on oeis.org

0, 18, 37, 57, 78, 100, 123, 147, 172, 198, 225, 253, 282, 312, 343, 375, 408, 442, 477, 513, 550, 588, 627, 667, 708, 750, 793, 837, 882, 928, 975, 1023, 1072, 1122, 1173, 1225, 1278, 1332, 1387, 1443, 1500, 1558, 1617, 1677, 1738, 1800, 1863, 1927, 1992, 2058

Offset: 0

Jesse Han, May 16 2012

Generalization: T(n,i) = A000217(i-1+n) - A000217(i-1) = i*n + A000217(n-1) (corrected by Zak Seidov, Jun 21 2012); in this case is i=18.

For i = 11..16, Milan Janjic observed that if we define f(n,b,i) = Sum_{k=0..n-b} binomial(n,k)*Stirling1(n-k,b)*Product_{j=0..k-1} (-i - j), then T(n-1,i) = -f(n,n-1,i) for n >= 1.

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

Cf. A000217, A000096, A001008, A002805, A055998-A056000, A056115, A056119, A056121, A056126, A051942, A101859, A132754-A132758, A212427.

[n*(n+35)/2: n in [0..48]]; // Bruno Berselli, Jun 21 2012

Table[-18 (18 - 1)/2 + (18 + n) (17 + n)/2, {n, 0, 100}]
LinearRecurrence[{3,-3,1},{0,18,37},60] (* Harvey P. Dale, Jun 09 2024 *)

a(n)=n*(n+35)/2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = (17+n)*(18+n)/2 - 17*18/2 = 18*n + (n-1)*n/2 = n*(n+35)/2.
G.f.: x*(18-17*x)/(1-x)^3. - Bruno Berselli, Jun 21 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jul 10 2012
a(n) = 18*n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
From Amiram Eldar, Jan 11 2021: (Start)
Sum_{n>=1} 1/a(n) = 2*A001008(35)/(35*A002805(35)) = 54437269998109/229732925058000.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/35 - 102126365345729/2527062175638000. (End)
E.g.f.: exp(x)*x*(36 + x)/2. - Elmo R. Oliveira, Dec 12 2024

A174183 a(n) is the period k such that binomial(m, n) (mod 10) = binomial(m + k, n) (mod 10).

Original entry on oeis.org

1, 10, 20, 60, 240, 1200, 7200, 50400, 403200, 3628800, 36288000, 399168000, 4790016000, 62270208000, 871782912000, 13076743680000, 209227898880000, 3556874280960000, 64023737057280000, 1216451004088320000

Offset: 0

Michel Lagneau, Mar 11 2010

a(n) is the period (mod 10) of the numbers in each column n of Pascal's triangle.

			x(0)= 0.C(1,0)C(2,0)C(3,0) ... = 0.11111111111... and p(0)=1 ;
x(1)= 0.C(1,1)C(2,1)C(3,1) ... = 0.12345678901234... and p(1) = 10 ;
x(2)= 0.C(2,2)C(3,2)C(4,2) ... = 0.13605186556815063100 13605186556815063100... and p(2)=20.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.

Harvey P. Dale, Table of n, a(n) for n = 0..449
Michel Lagneau, Proof
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081, 2014

Cf. A000142, A002415, A007318, A002024, A000096, A000124, A002378, A000292, A000330, A055998, A055999, A056000, A056115, A056119, A056121, A056126, A051942, A101859, A001477.

Join[{1},Array[10#!&,20]] (* Harvey P. Dale, Feb 18 2018 *)

from math import factorial
def A174183(n): return 10*factorial(n) if n else 1 # Chai Wah Wu, Aug 07 2025

a(0)=1, and a(n) = 10 * n! for n >= 1.

Additional comments, and errors in examples corrected by Michel Lagneau, May 07 2010

A270048 a(1) = 0; a(n+1) = a(n) + n * the number of digits of a(n).

Original entry on oeis.org

0, 1, 3, 6, 10, 20, 32, 46, 62, 80, 100, 133, 169, 208, 250, 295, 343, 394, 448, 505, 565, 628, 694, 763, 835, 910, 988, 1069, 1181, 1297, 1417, 1541, 1669, 1801, 1937, 2077, 2221, 2369, 2521, 2677, 2837, 3001, 3169, 3341, 3517, 3697, 3881, 4069, 4261, 4457, 4657, 4861, 5069, 5281, 5497, 5717, 5941

Offset: 1

Francesco Di Matteo, Mar 09 2016

In this sequence each a(n) term is the sum of k-terms, where k is the number of digits of a(n-1).
This is easy to verify by observing the following table:
+----+---------+---------+---------+--+-----+
|  n | A000217 | A056000 | A101859 |..| a(n)|
+----+---------+---------+---------+--+-----+
|  1 |       0 |       . |       . | .|   0 |
|  2 |       1 |       . |       . | .|   1 |
|  3 |       3 |       . |       . | .|   3 |
|  4 |       6 |       . |       . | .|   6 |
|  5 |      10 |       0 |       . | .|  10 |
|  6 |      15 |       5 |       . | .|  20 |
|  7 |      21 |      11 |       . | .|  32 |
|  8 |      28 |      18 |       . | .|  46 |
|  9 |      36 |      26 |       . | .|  62 |
| 10 |      45 |      35 |       . | .|  80 |
| 11 |      55 |      45 |       0 | .| 100 |
| 12 |      66 |      56 |      11 | .| 133 |
| 13 |      78 |      68 |      23 | .| 169 |
| 14 |      91 |      81 |      36 | .| 208 |
| 15 |     105 |      95 |      50 | .| 250 |
| 16 |     120 |     110 |      65 | .| 295 |
| 17 |     136 |     126 |      81 | .| 343 |
.
As we can see each of those terms is a term of a different subsequence, that is generated with the same construction rule, that is: a(n) = n + a(n-1) + Z.
In fact:
A000217 --> a(n) = n + a(n-1) + 0;
A056000 --> a(n) = n + a(n-1) + 4;
A101859 --> a(n) = n + a(n-1) + 10.
And so on, where the Z value is the n value of this sequence when the number of digits of a(n) is greater than that of a(n-1), or Z = Sum_{j=1..i} k(j) where k(j) is A270270(j).

			a(4) = 3 + 3*1 = 6;
a(5) = 6 + 4*1 = 10;
a(6) = 10 + 5*2 = 20.

Francesco Di Matteo, Table of n, a(n) for n = 1..1000

Cf. A000217, A056000, A101859, A264847, A270270.

a[1] = 0; a[n_] := a[n] = # + (n - 1) If[# == 0, 1, IntegerLength@ #] &@ a[n - 1]; Table[a@ n, {n, 57}] (* Michael De Vlieger, Mar 09 2016 *)

a(n) = if (n==1, 0, prec = a(n-1); prec + (n-1)*#Str(prec)); \\ Michel Marcus, Apr 03 2016

b = 0
print(b, end=',')
for g in range(1, 100):
   b += g*len(str(b))
   print(b, end=',')

A356754 Triangle read by rows: T(n,k) = ((n-1)(n+2))/2 + 2k.

Original entry on oeis.org

2, 4, 6, 7, 9, 11, 11, 13, 15, 17, 16, 18, 20, 22, 24, 22, 24, 26, 28, 30, 32, 29, 31, 33, 35, 37, 39, 41, 37, 39, 41, 43, 45, 47, 49, 51, 46, 48, 50, 52, 54, 56, 58, 60, 62, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87

Offset: 1

Torlach Rush, Aug 25 2022

The first column of the triangle is the Lazy Caterer's sequence A000124.
Each subsequent column starts with A000124(n) + (2 * (n-1)).
The first downward diagonal is A046691(n).
Columns and downward diagonals of the triangle identify many sequences (possibly shifted) in the database. Examples can be found in crossrefs below.
The sum of the n-th upward diagonal of the triangle is A356288(n).

			Triangle T(n,k) begins:
  n\k   1   2   3   4   5   6   7   8   9  10  11  ...
   1:   2
   2:   4   6
   3:   7   9  11
   4:  11  13  15  17
   5:  16  18  20  22  24
   6:  22  24  26  28  30  32
   7:  29  31  33  35  37  39  41
   8:  37  39  41  43  45  47  49  51
   9:  46  48  50  52  54  56  58  60  62
  10:  56  58  60  62  64  66  68  70  72  74
  11:  67  69  71  73  75  77  79  81  83  85  87
  ...

Cf. A000124, A004120, A046691, A051938, A055999, A056000, A155212, A167487, A167499, A167614, A246172, A334563, A356288.

Table[((n-1)(n+2))/2+2k,{n,20},{k,n}]//Flatten (* Harvey P. Dale, May 26 2023 *)

def T(n, k): return ((n-1) * (n+2))//2 + 2*k
for n in range(1, 12):
    for k in range(1,(n+1)): print(T(n,k), end = ', ')

# Indexed as a linear sequence.
def a000124(n): return n*(n+1)//2 + 1
def a(n):
    l = m = 0
    for k in range(1,n):
        lc = a000124(k - 1)
        if n >= lc:
            l = lc
            m = k
        else: break
    return n + m + (n - l)

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

T(n,k+1) = T(n,k) + 2, k < n.

cp's OEIS Frontend

A098832 Square array read by antidiagonals: even-numbered rows of the table are of the form n*(n+m) and odd-numbered rows are of the form n*(n+m)/2.

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A219502 T(n,k)=Number of nXk arrays of the minimum value of corresponding elements and their horizontal or vertical neighbors in a random, but sorted with lexicographically nondecreasing rows and nonincreasing columns, 0..1 nXk array.

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A096941 Fourth column of (1,5)-Pascal triangle A096940.

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A337940 Triangle read by rows: T(n, k) = T(n+2) - T(n-k), with the triangular numbers T = A000217, for n >= 1, k = 1, 2, ..., n.

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A212427 a(n) = 17*n + A000217(n-1).

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A212428 a(n) = 18*n + A000217(n-1).

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A174183 a(n) is the period k such that binomial(m, n) (mod 10) = binomial(m + k, n) (mod 10).

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A098832 Square array read by antidiagonals: even-numbered rows of the table are of the form n(n+m) and odd-numbered rows are of the form n(n+m)/2.

A356754 Triangle read by rows: T(n,k) = ((n-1)(n+2))/2 + 2k.