A094728 -id:A094728 - cp's OEIS frontend

A269837 Irregular triangle read by rows: even terms of A094728(n+1) divided by 4.

1, 2, 4, 3, 6, 4, 9, 8, 5, 12, 10, 6, 16, 15, 12, 7, 20, 18, 14, 8, 25, 24, 21, 16, 9, 30, 28, 24, 18, 10, 36, 35, 32, 27, 20, 11, 42, 40, 36, 30, 22, 12, 49, 48, 45, 40, 33, 24, 13, 56, 54, 50, 44, 36, 26, 14, 64, 63, 60, 55, 48, 39, 28, 15

Offset: 0

Paul Curtz, Mar 06 2016

See A264798 and A261046 for the Hydrogen atom and the Janet periodic table.
a(n) odd terms are again A264798.
Decomposition by multiplication i.e. a(n) = b(n)*c(n) by irregular triangle:
1,                   1                 1,
2,                   1                 2,
4,   3,              2, 1,             2, 3,
6,   4,         =    2, 1,        *    3, 4,
9,   8, 5,           3, 2, 1,          3, 4, 5,
12, 10, 6,           3, 2, 1,          4, 5, 6,
16, 15, 12, 7,       4, 3, 2, 1,       4, 5, 6, 7,
etc.                 etc.              etc.
b(n) is duplicated A004736(n) or mirror of A122197(n+1). c(n) = A138099(n+1).
Decomposition by subtraction, a(n) = d(n) - e(n):
1,                    1                    0,
2,                    2,                   0,
4,   3,               4,  3,               0, 0,
6,   4,         =     6,  5,          -    0, 1,
9,   8, 5,            9,  8,  7,           0, 0, 2,
12, 10, 6,           12, 11, 10,           0, 1, 4,
16, 15, 12, 7,       16, 15, 14, 13,       0, 0, 2, 6,
20, 18, 14, 8,       20, 19, 18, 17,       0, 1, 4, 9,
etc.                 etc.                  etc.
d(n) is the natural numbers A000027 inverted by lines.  e(n) will be studied (see A239873).
Sum of a(n) by diagonals: 1, 5, 13, 27, 48, ... . The third differences have the period 2: repeat 2, 1. See A002717.

Cf. A000027, A000034, A002717, A004736, A094728, A122197, A138099, A167430, A239873, A261046, A264798.

Table[(n + 1)^2 - k^2, {n, 15}, {k, 0, n - 1}]/4 /. Rational -> Nothing // Flatten (* _Michael De Vlieger, Mar 07 2016 *)

A264798 Irregular triangle read by rows: odd-valued terms of A094728(n+1).

Original entry on oeis.org

1, 3, 9, 5, 15, 7, 25, 21, 9, 35, 27, 11, 49, 45, 33, 13, 63, 55, 39, 15, 81, 77, 65, 45, 17, 99, 91, 75, 51, 19, 121, 117, 105, 85, 57, 21, 143, 135, 119, 95, 63, 23, 169, 165, 153, 133, 105, 69, 25, 195, 187, 171, 147, 115, 75, 27, 225, 221, 209, 189, 161, 125, 81, 29, 255, 247

Offset: 0

Paul Curtz, Nov 25 2015

A094728(n+1) comes from A120070(n+2). a(n) approximates frequencies of the spectral lines of the hydrogen atom.
Row sums: 1, 3, 14, 22, ... = A024598(n+1).
First column: A085046(n+1).
Row sums of A261046(n) = 1, 3, 8, 12, ... = A014255(n). See the formula.

			Irregular triangle begins:
1,
3,
9,  5,
15, 7,
25, 21,  9,
35, 27, 11,
49, 45, 33, 13,
63, 55, 39, 15,
...

Cf. A014255, A024598, A085046, A094728, A120070, A167268, A249947, A261046.

Table[n^2 - k^2, {n, 14}, {k, 0, n - 1}] /. n_ /; EvenQ@ n -> Nothing // Flatten (* Michael De Vlieger, Nov 25 2015 *)

for(n=1,20,for(k=0,n-1,s=n^2-k^2;if(s%2,print1(s,", ")))) \\ Derek Orr, Dec 24 2015

a(n) = A261046(n)*A167268(n+1)/2, where A167268 is Janet's sequence.

A002412 Hexagonal pyramidal numbers, or greengrocer's numbers.

Original entry on oeis.org

0, 1, 7, 22, 50, 95, 161, 252, 372, 525, 715, 946, 1222, 1547, 1925, 2360, 2856, 3417, 4047, 4750, 5530, 6391, 7337, 8372, 9500, 10725, 12051, 13482, 15022, 16675, 18445, 20336, 22352, 24497, 26775, 29190, 31746, 34447, 37297, 40300

Offset: 0

N. J. A. Sloane

Binomial transform of (1, 6, 9, 4, 0, 0, 0, ...). - Gary W. Adamson, Oct 16 2007
a(n) is the sum of the maximum(m,n) over {(m,n): m,n in positive integers, m<=n}. - Geoffrey Critzer, Oct 11 2009
We obtain these numbers for d=2 in the identity n*(n*(d*n-d+2)/2)-sum(k*(d*k-d+2)/2, k=0..n-1) = n*(n+1)*(2*d*n-2*d+3)/6 (see Klaus Strassburger in Formula lines). - Bruno Berselli, Apr 21 2010, Nov 16 2010
q^a(n) is the Hankel transform of the q-Catalan numbers. - Paul Barry, Dec 15 2010
Row 1 of the convolution array A213835. - Clark Kimberling, Jul 04 2012
From Ant King, Oct 24 2012: (Start)
For n>0, the digital roots of this sequence A010888(A002412(n)) form the purely periodic 27-cycle {1,7,4,5,5,8,9,3,3,4,1,7,8,8,2,3,6,6,7,4,1,2,2,5,6,9,9}.
For n>0, the units' digits of this sequence A010879(A002412(n)) form the purely periodic 20-cycle {1,7,2,0,5,1,2,2,5,5,6,2,7,5,0,6,7,7,0,0}.
(End)
Partial sums of A000384. - Omar E. Pol, Jan 12 2013
Row sums of A094728. - J. M. Bergot, Jun 14 2013
Number of orbits of Aut(Z^7) as function of the infinity norm (n+1) of the representative integer lattice point of the orbit, when the cardinality of the orbit is  equal to 40320. - Philippe A.J.G. Chevalier, Dec 28 2015
Coefficients in the hypergeometric series identity 1 - 7*(x - 1)/(3*x + 1) + 22*(x - 1)*(x - 2)/((3*x + 1)*(3*x + 2)) - 50*(x - 1)*(x - 2)*(x - 3)/((3*x + 1)*(3*x + 2)*(3*x + 3)) + ... = 0, valid for Re(x) > 1. Cf. A000326 and A002418. Column 3 of A103450. - Peter Bala, Mar 14 2019

			Let n=5, 2*n=10. Since 10 = 1 + 9 = 2 + 8 = 3 + 7 = 4 + 6 = 5 + 5, a(5) = 1*9 + 2*8 + 3*7 + 4*6 + 5*5 = 95. - _Vladimir Shevelev_, May 11 2012

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 194.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 2.
T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.
I. Siap, Linear codes over F_2 + u*F_2 and their complete weight enumerators, in Codes and Designs (Ohio State, May 18, 2000), pp. 259-271. De Gruyter, 2002.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe and William A. Tedeschi, Table of n, a(n) for n = 0..10000 (first 1000 terms computed by T. D. Noe)
Abdullah Atmaca and A. Yavuz Oruç, On the size of two families of unlabeled bipartite graphs, AKCE International Journal of Graphs and Combinatorics, Vo. 16, No. 2 (2019), pp. 222-229.
Bruno Berselli, A description of the transform in Comments lines: website Matem@ticamente (in Italian).
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Luis Verde-Star, A Matrix Approach to Generalized Delannoy and Schröder Arrays, J. Int. Seq., Vol. 24 (2021), Article 21.4.1.
Eric Weisstein's World of Mathematics, Hexagonal Pyramidal Number.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Bisection of A002623. Equals A000578(n) - A000330(n-1).
Cf. A000292, A016061.
a(n) = A093561(n+2, 3), (4, 1)-Pascal column.
Cf. A220084 for a list of numbers of the form n*P(k,n)-(n-1)*P(k,n-1), where P(k,n) is the n-th k-gonal pyramidal number (see Adamson's formula).
Cf. similar sequences listed in A237616.
Orbits of Aut(Z^7) as function of the infinity norm A000579, A154286, A102860, A002412, A045943, A115067, A008585, A005843, A001477, A000217.

List([0..40],n->n*(n+1)*(4*n-1)/6); # Muniru A Asiru, Mar 18 2019

[n*(n+1)*(4*n-1)/6: n in [0..40]]; // Vincenzo Librandi, Nov 28 2015

seq(sum(i*(2*k-i), i=1..k), k=0..100); # Wesley Ivan Hurt, Sep 25 2013

Figurate[ ngon_, rank_, dim_] := Binomial[rank + dim - 2, dim - 1] ((rank - 1)*(ngon - 2) + dim)/dim; Table[ Figurate[6, r, 3], {r, 0, 40}] (* Robert G. Wilson v, Aug 22 2010 *)
Table[n(n+1)(4n-1)/6, {n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1}, {0,1,7,22}, 40] (* Harvey P. Dale, Jul 16 2011 *)

A002412(n):=n*(n+1)*(4*n-1)/6$ makelist(A002412(n),n,0,20); /* Martin Ettl, Dec 12 2012 */

v=vector(40,i,(i*(i+1))\2); s=0; print1(s","); forstep(i=1,40,2,s+=v[i]; print1(s","))

print([n*(n+1)*(4*n-1)//6 for n in range(40)]) # Michael S. Branicky, Mar 28 2022

a(n) = n(n + 1)(4n - 1)/6.
G.f.: x*(1+3*x)/(1-x)^4. - Simon Plouffe in his 1992 dissertation.
a(n) = n^3 - Sum_{i=1..n-1} i^2. - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de)
Partial sums of n odd-indexed triangular numbers, e.g., a(3) = t(1)+t(3)+t(5) = 1+6+15 = 22. - Jon Perry, Jul 23 2003
a(n) = Sum_{i=0..n-1} (n - i)*(n + i). - Jon Perry, Sep 26 2004
a(n) = n*A000292(n) - (n-1)*A000292(n-1) = n*binomial((n+2),3) - (n-1)*binomial((n+1),3); e.g., a(5) = 95 = 5*35 - 4*20. - Gary W. Adamson, Dec 28 2007
a(n) = Sum_{i=0..n} (2i^2 + 3i + 1), for n >= 0 (Omits the leading 0). - William A. Tedeschi, Aug 25 2010
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4), with a(0)=0, a(1)=1, a(2)=7, a(3)=22. - Harvey P. Dale, Jul 16 2011
a(n) = sum a*b, where the summing is over all unordered partitions 2*n = a+b. - Vladimir Shevelev, May 11 2012
From Ant King, Oct 24 2012: (Start)
a(n) = a(n-1) + n*(2*n-1).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 4.
a(n) = (n+1)*(2*A000384(n) + n)/6 = (4*n-1)*A000217(n)/3.
a(n) = A000292(n) + 3*A000292(n-1) = A002411(n) + A000292(n-1).
a(n) = binomial(n+2,3) + 3*binomial(n+1,3) = (4*n-1)*binomial(n+1,2)/3.
Sum_{n>=1} 1/a(n) = 6*(12*log(2)-2*Pi-1)/5 = 1.2414...
(End)
a(n) = Sum_{i=1..n} Sum_{j=1..n} max(i,j) = Sum_{i=1..n} i*(2*n-i). - Enrique Pérez Herrero, Jan 15 2013
a(n) = A005900(n+1) - A000326(n+1) = Octahedral - Pentagonal Numbers. - Richard R. Forberg, Aug 07 2013
a(n) = n*A000217(n) + Sum_{i=0..n-1} A000217(i). - Bruno Berselli, Dec 18 2013
a(n) = 2n * A000217(n) - A000330(n). - J. M. Bergot, Apr 05 2014
a(n) = A080851(4,n-1). - R. J. Mathar, Jul 28 2016
E.g.f.: x*(6 + 15*x + 4*x^2)*exp(x)/6. - Ilya Gutkovskiy, May 12 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 6*(1 + 2*sqrt(2)*Pi - 2*(3+sqrt(2))*log(2) + 4*sqrt(2)*log(2-sqrt(2)))/5. - Amiram Eldar, Jan 04 2022

A034178 Number of solutions to n = a^2 - b^2, a > b >= 0.

Original entry on oeis.org

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

Offset: 1

Erich Friedman

Also, number of ways n can be expressed as the sum of one or more consecutive odd numbers. (E.g., 45 = 45 = 13+15+17 = 5+7+9+11+13, so a(45)=3.) - Naohiro Nomoto, Feb 26 2002
a(A042965(n))>0, a(A016825(n))=0; also number of occurrences of n in A094728. - Reinhard Zumkeller, May 24 2004
It appears a(n) can be found by adding together the divisor pairs of n and finding the number of even results. For example: n=9 has the divisor pairs (1,9) and (3,3); adding the pairs: 1+9=10 is even and 3+3=6 is even, so a(9)=2. Another example: n=96 has the divisor pairs (1,96) (2,48) (3,32) (4,24) (6,16) (8,12); when each pair is added there are 4 even results, so a(96)=4. - Gregory Bryant, Dec 06 2016
It appears a(n) is the number of nonnegative integers k for which sqrt(k) + sqrt(k + n) is an integer. For example: a(2015) = 4 since there are only four nonnegative integers k for which sqrt(k) + sqrt(k + 2015) is an integer, namely k = 289, 5041, 39601, 1014049. - Joseph Barrera, Nov 29 2020

			G.f. = x + x^3 + x^4 + x^5 + x^7 + x^8 + 2*x^9 + x^11 + x^12 + x^13 + 2*x^15 + ...
From _Bernard Schott_, Apr 19 2019: (Start)
a(8) = floor((A000005(2) + 1)/2) = floor(3/2) = 1 and 8 = 3^2 - 1^2.
a(9) = floor((A000005(9) + 1)/2) = floor(4/2) = 2 and 9 = 3^2 - 0^2 = 5^2 - 4^2.
a(10) = 0 and a^2 - b^2 = 10 has no solution.
a(11) = floor(A000005(11) + 1)/2 = floor(3/2) = 1 and 11 = 6^2 - 5^2.  (End)

T. D. Noe, Table of n, a(n) for n = 1..2000
M. A. Nyblom, On the Representation of the Integers as a Difference of Squares, Fibonacci Quart., vol. 40 (2002), no. 3, 243-246.
Edward T. H. Wang, Problem 1717, Crux Mathematicorum, page 30, Vol. 19, Jan. 93.

Cf. A000005, A058957, A016825.

nn = 100; t = Table[0, {nn}]; Do[n = a^2 - b^2; If[n <= nn, t[[n]]++], {a, nn}, {b, 0, a - 1}];t (* T. D. Noe, May 04 2011 *)
Table[Length[FindInstance[a^2-b^2==n&&a>b>=0,{a,b},Integers,10]],{n,100}] (* Harvey P. Dale, Jul 28 2021 *)

a(n)=sum(k=1, sqrtint(n), (n-k^2)%(2*k)==0) \\ Charles R Greathouse IV, Sep 27 2012

a(n)=sumdiv(n, d, n>=d^2 && (n-d^2)%(2*d)==0) \\ Charles R Greathouse IV, Sep 27 2012

from sympy import divisor_count as d
def a(n): return (d(n)+1)//2 if n%2==1 else ((d(n//4)+1)//2 if n%4==0 else 0)
# Ely Golden, Jan 26 2025

From Naohiro Nomoto, Feb 26 2002: (Start)
a(2k) = A038548(2k) - A001227(k).
a(2k+1) = A038548(2k+1). (End)
From Bernard Schott, Apr 11 2019: (Start) (see Crux link)
a(n) = 0 if n == 2 (mod 4)
a(n) = floor((A000005(n) + 1)/2) if n == 1 or n == 3 (mod 4)
a(n) = floor((A000005(n/4) + 1)/2) if n == 0 (mod 4). (End)
G.f.: Sum_{i>=1} Sum_{j>=i} Product_{k=i..j} x^(2*k-1). - Ilya Gutkovskiy, Apr 18 2019
G.f.: Sum_{n>=1} x^(n^2)/(1-x^(2*n)) (conjecture). - Joerg Arndt, Jan 04 2024

A094727 Triangle read by rows: T(n,k) = n + k, 0 <= k < n, n >= 1.

Original entry on oeis.org

1, 2, 3, 3, 4, 5, 4, 5, 6, 7, 5, 6, 7, 8, 9, 6, 7, 8, 9, 10, 11, 7, 8, 9, 10, 11, 12, 13, 8, 9, 10, 11, 12, 13, 14, 15, 9, 10, 11, 12, 13, 14, 15, 16, 17, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23

Offset: 1

Reinhard Zumkeller, May 24 2004

All numbers m occur ceiling(m/2) times, see A004526.

The LCM of the n-th row is A076100. - Michel Marcus, Mar 18 2018

			Triangle begins:
  1;
  2,  3;
  3,  4,  5;
  4,  5,  6,  7;
  5,  6,  7,  8,  9;
  6,  7,  8,  9, 10, 11;
  7,  8,  9, 10, 11, 12, 13;
  8,  9, 10, 11, 12, 13, 14, 15;
  9, 10, 11, 12, 13, 14, 15, 16, 17;
  ... - _Philippe Deléham_, Mar 30 2013

Reinhard Zumkeller, Rows n = 1..150 of triangle, flattened
Bruno Berselli, Illustration of the initial terms.
László Németh, On the Binomial Interpolated Triangles, Journal of Integer Sequences, Vol. 20 (2017), Article 17.7.8.
Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.

Cf. A002024, A002260, A003056, A004526, A004736, A005408, A005843.
Cf. A016777, A016813, A016861, A037213, A076100, A078112, A093005.
Cf. A094728, A214604, A214661, A319556.
Cf. A128076 (rows reversed).

a094727 n k = n + k
a094727_row n = a094727_tabl !! (n-1)
a094727_tabl = iterate (\row@(h:_) -> (h + 1) : map (+ 2) row) [1]
-- Reinhard Zumkeller, Jul 22 2012

z:=12; &cat[ [m+n-1: m in [1..n] ]: n in [1..z] ];

Table[n + Range[0, n-1], {n, 12}]//Flatten (* Michael De Vlieger, Dec 16 2016 *)

from math import isqrt
def A094727(n): return ((a:=(m:=isqrt(k:=n<<1))+(k>m*(m+1)))*(3-a)>>1)+n-1 # Chai Wah Wu, Jun 19 2025

flatten([[n+k for k in range(n)] for n in range(1,16)]) # G. C. Greubel, Mar 10 2024

T(n+1, k) = T(n, k) + 1 = T(n, k+1); T(n+1, k+1) = T(n, k) + 2.
T(n, n - A005843(k)) = A005843(n-k) for 0 <= k <= n/2.
T(n, n - A005408(k)) = A005408(n-k) for 0 <= k < n/2.
T(A005408(n), n) = A016777(n), n >= 0.
Sum_{k=1..n} T(n, k) = A000326(n) (row sums).
T(n, k) = A002024(n,k) + A002260(n,k) - 1. - Reinhard Zumkeller, Apr 27 2006
As a sequence rather than as a table: If m = floor((sqrt(8n-7)+1)/2), a(n) = n - m*(m-3)/2 - 1. - Carl R. White, Jul 30 2009
T(n, k) = n+k-1, n >= k >= 1. - Vincenzo Librandi, Nov 23 2009 [corrected by Klaus Brockhaus, Nov 23 2009]
T(n,k) = A037213((A214604(n,k) + A214661(n,k)) / 2). - Reinhard Zumkeller, Jul 25 2012
From Boris Putievskiy, Jan 16 2013: (Start)
a(n) = A002260(n) + A003056(n).
a(n) = i+t, where i=n-t*(t+1)/2, t=floor((-1+sqrt(8*n-7))/2). (End)
From G. C. Greubel, Mar 10 2024: (Start)
T(3*n-3, n) = A016813(n-1).
T(4*n-4, n) = A016861(n-1).
Sum_{k=0..n-1} (-1)^k*T(n, k) = A319556(n).
Sum_{k=0..floor((n-1)/2)} T(n-k, k) = A093005(n).
Sum_{k=0..floor((n-1)/2)} (-1)^k*T(n-k, k) = A078112(n-1).
Sum_{j=1..n} (Sum_{k=0..n-1} T(j, k)) = A002411(n) (sum of n rows). (End)

A049777 Triangular array read by rows: T(m,n) = n + n+1 + ... + m = (m+n)(m-n+1)/2.

Original entry on oeis.org

1, 3, 2, 6, 5, 3, 10, 9, 7, 4, 15, 14, 12, 9, 5, 21, 20, 18, 15, 11, 6, 28, 27, 25, 22, 18, 13, 7, 36, 35, 33, 30, 26, 21, 15, 8, 45, 44, 42, 39, 35, 30, 24, 17, 9, 55, 54, 52, 49, 45, 40, 34, 27, 19, 10, 66, 65, 63, 60, 56, 51, 45, 38, 30, 21, 11, 78, 77, 75, 72, 68, 63, 57, 50

Offset: 1

Clark Kimberling

Triangle read by rows, T(n,k) = A000217(n) - A000217(k), 0 <= k < n. - Philippe Deléham, Mar 07 2013
Subtriangle of triangle in A049780. - Philippe Deléham, Mar 07 2013
No primes and all composite numbers (except 2^x) are generated after the first two columns of the square array for this sequence. In other words, no primes and all composites except 2^x are generated when m-n >= 2. - Bob Selcoe, Jun 18 2013
Diagonal sums in the square array equal partial sums of squares (A000330). - Bob Selcoe, Feb 14 2014
From Bob Selcoe, Oct 27 2014: (Start)
The following apply to the triangle as a square array read by rows unless otherwise specified (see Table link);
Conjecture: There is at least one prime in interval [T(n,k), T(n,k+1)]. Since T(n,k+1)/T(n,k) decreases to (k+1)/k as n increases, this is true for k=1 ("Bertrand's Postulate", first proved by P. Chebyshev), k=2 (proved by El Bachraoui) and k=3 (proved by Loo).
Starting with T(1,1), The falling diagonal of the first 2 numbers in each column (read by column) are the generalized pentagonal numbers (A001318). That is, the coefficients of T(1,1), T(2,1), T(2,2), T(3,2), T(3,3), T(4,3), T(4,4) etc. are the generalized pentagonal numbers. These are A000326 and A005449 (Pentagonal and Second pentagonal numbers: n*(3*n+1)/2, respectively), interweaved.
Let D(n,k) denote falling diagonals starting with T(n,k):
Treating n as constant: pentagonal numbers of the form n*k + 3*k*(k-1)/2 are D(n,1); sequences A000326, 005449, A045943, A115067, A140090, A140091, A059845, A140672, A140673, A140674, A140675, A151542 are formed by n = 1 through 12, respectively.
Treating k as constant: D(1,k) are (3*n^2 + (4k-5)*n + (k-1)*(k-2))/2. When k = 2(mod3), D(1,k), is same as D(k+1,1) omitting the first (k-2)/3 numbers in the sequences. So D(1,2) is same as D(3,1); D(1,5) is same as D(6,1) omitting the 6; D(1,8) is same as D(9,1) omitting the 9 and 21; etc.
D(1,3) and D(1,4) are sequences A095794 and A140229, respectively.
(End)

			Rows: {1}; {3,2}; {6,5,3}; ...
Triangle begins:
   1;
   3,  2;
   6,  5,  3;
  10,  9,  7,  4;
  15, 14, 12,  9,  5;
  21, 20, 18, 15, 11,  6;
  28, 27, 25, 22, 18, 13,  7;
  36, 35, 33, 30, 26, 21, 15,  8;
  45, 44, 42, 39, 35, 30, 24, 17,  9;
  55, 54, 52, 49, 45, 40, 34, 27, 19, 10; ...

Vincenzo Librandi, Table of n, a(n) for n = 1..5000
M. El Bachraoui, Primes in the interval [2n,3n], Int. J. Contemp. Math. Sciences 1:13 (2006), pp. 617-621.
A. Loo, On the primes in the interval [3n,4n], Int. J. Contemp. Math. Sciences 6 (2011), no. 38, 1871-1882.
S. Ramanujan, A proof of Bertrand's postulate, J. Indian Math. Soc., 11 (1919), 181-182.
Vladimir Shevelev, Charles R. Greathouse IV, Peter J. C. Moses, On intervals (kn, (k+1)n) containing a prime for all n>1, Journal of Integer Sequences, Vol. 16 (2013), Article 13.7.3. arXiv, arXiv:1212.2785 [math.NT], 2012.

Row sums = A000330.
Cf. A001318 (generalized pentagonal numbers).
Cf. A000217, A002260, A049780, A094728, A095794, A140229.
Cf. A000326, 005449, A045943, A115067, A140090, A140091, A059845, A140672, A140673, A140674, A140675, A151542 (pentagonal numbers of form n*k + 3*k*(k-1)/2).

/* As triangle */ [[(m+n)*(m-n+1) div 2: n in [1..m]]: m in [1.. 15]]; // Vincenzo Librandi, Oct 27 2014

Flatten[Table[(n+k) (n-k+1)/2,{n,15},{k,n}]] (* Harvey P. Dale, Feb 27 2012 *)

{T(n,k) = if( k<1 || nMichael Somos, Oct 06 2007 */

Partial sums of A002260 row terms, starting from the right; e.g., row 3 of A002260 = (1, 2, 3), giving (6, 5, 3). - Gary W. Adamson, Oct 23 2007
Sum_{k=0..n-1} (-1)^k*(2*k+1)*A000203(T(n,k)) = (-1)^(n-1)*A000330(n). - Philippe Deléham, Mar 07 2013
Read as a square array: T(n,k) = k*(k+2n-1)/2. - Bob Selcoe, Oct 27 2014

A128624 Row sums of A128623.

Original entry on oeis.org

1, 4, 12, 24, 45, 72, 112, 160, 225, 300, 396, 504, 637, 784, 960, 1152, 1377, 1620, 1900, 2200, 2541, 2904, 3312, 3744, 4225, 4732, 5292, 5880, 6525, 7200, 7936, 8704, 9537, 10404, 11340, 12312, 13357, 14440, 15600, 16800, 18081, 19404, 20812, 22264, 23805

Offset: 1

Gary W. Adamson, Mar 14 2007

Also the number of (w,x,y) with all terms in {0,...,n-1} and w <= R <= x, where R = max(w,x,y)-min(w,x,y), see A212959. - Clark Kimberling, Jun 10 2012

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Cf. A128621, A128623, A212959.

Cf. A094728 (diagonal row sums).

[n*((n+1)^2-1+(n mod 2))/4: n in [1..50]]; // G. C. Greubel, Mar 12 2024

Table[n*(n^2 +2*n +Mod[n,2])/4, {n,50}] (* G. C. Greubel, Mar 12 2024 *)

Vec(x*(1+2*x+3*x^2)/((1-x)^4*(1+x)^2) + O(x^100)) \\ Colin Barker, Jan 31 2016

[n*((n+1)^2-1+(n%2))//4 for n in range(1,51)] # G. C. Greubel, Mar 12 2024

G.f.: x*(1+2*x+3*x^2) / ((1+x)^2*(1-x)^4). - R. J. Mathar, Jun 27 2012
From Colin Barker, Jan 31 2016: (Start)
a(n) = n*(2*n^2 + 4*n + 1 - (-1)^n)/8.
a(n) = n^2*(n + 2)/4 for n even.
a(n) = n*(n^2 + 2*n + 1)/4 for n odd. (End)
From G. C. Greubel, Mar 12 2024: (Start)
a(n) = Sum_{k=0..floor((n-1)/2)} A094728(n, k).
E.g.f.: (1/8)*x*(exp(-x) + (7 + 10*x + 2*x^2)*exp(x)). (End)

Incorrect formula removed by R. J. Mathar, Jun 27 2012

A370707 Triangle read by rows: T(n, k) = (-1)^kProduct_{j=0..k-1} (j - n)(j + n), for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 4, 12, 1, 9, 72, 360, 1, 16, 240, 2880, 20160, 1, 25, 600, 12600, 201600, 1814400, 1, 36, 1260, 40320, 1088640, 21772800, 239500800, 1, 49, 2352, 105840, 4233600, 139708800, 3353011200, 43589145600, 1, 64, 4032, 241920, 13305600, 638668800, 24908083200, 697426329600, 10461394944000

Offset: 0

Peter Luschny, Feb 27 2024

The definition, and also the representation T(n, k) = ff(n, k) * rf(n, k) (see the first formula), makes it natural to call this triangle the central factorial numbers.

			Triangle starts:
  [0] 1;
  [1] 1,  1;
  [2] 1,  4,   12;
  [3] 1,  9,   72,    360;
  [4] 1, 16,  240,   2880,   20160;
  [5] 1, 25,  600,  12600,  201600,   1814400;
  [6] 1, 36, 1260,  40320, 1088640,  21772800,  239500800;
  [7] 1, 49, 2352, 105840, 4233600, 139708800, 3353011200, 43589145600;
.
T(n, k) is a product where 'n' is the 'center' and 'k' is the 'half-length' of the product. For instance, T(5, 4) = (5-3)*(5-2)*(5-1)*5 * 5*(5+1)*(5+2)*(5+3) = 201600. Now consider the polynomial P(4, x) = -36*x^2 + 49*x^4 - 14*x^6 + x^8. Evaluating this polynomial at x = 5 shows P(4, 5) = 201600 = T(5, 4). The coefficients of the polynomial are row 4 of A269944.

Diagonals: A002674, A327882.
Columns: A000290, A047928.
Cf. A370704 (row sums), A370706, A094728, A048994 (Stirling1), A130595 (order 0), A269947 (order 3)

T := (n, k) -> local j; (-1)^k * mul((j - n)*(j + n), j = 0..k-1):
seq(seq(T(n, k), k = 0..n), n = 0..8);
# The central factorial numbers:
cf := (n, k) -> ifelse(k = 0, 1, n*(n + k - 1)! / (n - k)! ):
for n from 0 to 6 do seq(cf(n, k), k = 0..n) od;
# Alternative (recurrence):
T := proc(n, k) option remember;
if k = 0 then 1 else T(n, k - 1)*(n^2 - (k - 1)^2) fi end:
for n from 0 to 7 do seq(T(n, k), k = 0..n) od;
# Illustrating the connection with the cf-polynomials and their coefficients:
cfpoly := (n,x) -> local k; mul(x^2 - k^2, k = 0..n-1):
A370707row := n -> local k; [seq(cfpoly(k, n), k = 0..n)]:
A204579row := n -> local k; [seq(coeff(cfpoly(n, x), x, 2*k), k = 0..n)]:
for n from 0 to 5 do lprint([n], A370707row(n), A204579row(n)) od;

T[n_, k_] := If[n == 0, 1, -n Pochhammer[1 - n - k, 2 k - 1]];
Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten

from math import prod
def T(n, k): return (-1)**k * prod((j - n)*(j + n) for j in range(k))
print([T(n, k) for n in range(8) for k in range(n + 1)])

def T(n, k): return falling_factorial(n, k) * rising_factorial(n, k)
for n in range(9): print([T(n, k) for k in range(n + 1)])

T(n, k) = FallingFactorial(n, k) * RisingFactorial(n, k).
T(n, k) = (n*(n + k - 1)!)/(n - k)! if k > 0, and T(n, 0) = 1.
Calling the numbers in the second formula cf leads to the memorable form cf(n, k) = ff(n, k) * rf(n, k). This identity generalizes to the function
  cf(x, n) = x*Gamma(x + n)/Gamma(x - n + 1) for n > 0 and cf(x, 0) = 1.
The last equation shows that the variable 'n' does not have to be an integer but can be any complex number if only the quotient remains defined (which one often can achieve by taking the limit). Indeed, in the classical Steffensen-Riordan case, n/2 is used instead of n, which leads to the complex situation Sloane discusses in A008955.
T(n, k) = -n*Pochhammer(1 - n - k, 2*k - 1) for n > 0.
T(n, k) = k!*binomial(n, k)*Pochhammer(n, k) = k!*A370706(n, k).
T(n, n) = n!*Pochhammer(n, n) (valid for n >= 0, whereas T(n, n) = (2*n)!/2 = A002674(n) is valid for n >= 1 only).
T(n, k) = T(n, k - 1)*(n^2 - (k - 1)^2) if k > 0, otherwise 1. (Recurrence)
The cf(n, k) are values of the polynomials Pcf(n, x) = Product_{k=0..n-1} (x^2 - k^2), whose coefficients vanish for odd powers and for even powers are A269944.
T(n, k) = Pcf(k, n) where Pcf(k,x) = Sum_{j=0..k} (-1)^(k-j)*A269944(k,j)*x^(2*j).
The central factorials can be described in three different ways: By the product T(n, k) = f(n, k) * rf(n, k), by the complex function cf(x, n), and through the polynomials Pcf(n, x). Although these relations are self-contained, they are regarded as only one-half of a more general notion, namely as central factorials of the first kind.
There is a fundamental connection with the Stirling numbers of first kind (A048994). The easiest way to see this is to generalize the definition: Let CF(z, s) = Product_{j=0..n-1} (z - s(j)), where s(j) is some complex sequence. Then the coefficients of CF(z, s) are equal to the Stirling_1 numbers if s = 0, 1, 2, ..., n, ..., and they are equal to the coefficients of our Pcf(n, z) polynomials if s = 0, 1, 4, ..., n^2, .... (This is also why A269944 is called the 'Stirling cycle numbers of order 2'. For completeness, if s = 1, 1, 1, ..., then the coefficients of CF(z, s), the 'Stirling cycle numbers of order 0', are the signed Pascal triangle A130595. See A269947 for order 3.)

A101447 Triangle read by rows: T(n,k) = (2k+1)(n+1-k), 0 <= k < n.

Original entry on oeis.org

1, 2, 3, 3, 6, 5, 4, 9, 10, 7, 5, 12, 15, 14, 9, 6, 15, 20, 21, 18, 11, 7, 18, 25, 28, 27, 22, 13, 8, 21, 30, 35, 36, 33, 26, 15, 9, 24, 35, 42, 45, 44, 39, 30, 17, 10, 27, 40, 49, 54, 55, 52, 45, 34, 19, 11, 30, 45, 56, 63, 66, 65, 60, 51, 38, 21, 12, 33, 50, 63, 72, 77, 78, 75, 68, 57, 42, 23

Offset: 0

Lambert Klasen (lambert.klasen(AT)gmx.de) and Gary W. Adamson, Jan 19 2005

The triangle is generated from the product of matrix A and matrix B, i.e., A * B where A = the infinite lower triangular matrix:
  1 0 0 0 0 ...
  1 1 0 0 0 ...
  1 1 1 0 0 ...
  1 1 1 1 0 ...
  1 1 1 1 1 ...
... and B = the infinite lower triangular matrix:
  1 0 0 0 0 ...
  1 3 0 0 0 ...
  1 3 5 0 0 ...
  1 3 5 7 0 ...
  1 3 5 7 9 ...
  ...
Row sums give the square pyramidal numbers A000330.
T(n+0,0)=1*n=A000027(n+1); T(n+1,1)=3*n=A008585(n); T(n+2,2)=5*n=A008587(n); T(n+3,3)=7*n=A008589(n); etc. So T(n,0)*T(n,1)=3*n*(n+1)=A028896(n) (6 times triangular numbers). T(n,1)*T(n,2)/10=3*n*(n+1)/2=A045943(n) for n>0 T(n,2)*T(n,3)/10=7/2*n*(n+1)=A024966(n) for n>1 (7 times triangular numbers), etc.
From Gary W. Adamson, Apr 25 2010: (Start)
Consider the following array, signed as shown:
  ...
  1,   3,  5,   7,  9,  11, ...
  2,  -6, 10, -14, 18, -22, ...
  3,   9, 15,  21, 27,  33, ...
  4, -12, 20, -28, 36, -44, ...
  5,  15, 25,  35, 45,  55, ...
  6, -18, 30, -42, 54, -66, ...
  7,  21, 35,  49, 63,  77, ...
  ...
Let each term (+, -)k = (+, -) phi^(-k).
Consider the inverse terms of the Lucas series (1/1, 1/3, 1/4, 1/7, ...).
By way of example, let q = phi = 1.6180339...; then
...
1/1  = q^(-1) + q^(-3) + q^(-5) + q^(-7) + q^(-9) + ...
1/3  = q^(-2) - q^(-6) + q^(-10) - q^(-14) + q^(-18) + ...
1/4  = q^(-3) + q^(-9) + q^(-15) + q^(-21) + q^(-27) +...
1/7  = q^(-4) - q^(-12) + q^(-20) - q^(-28) + q^(-36) + ...
1/11 = q^(-5) + q^(-15) + q^(-25) + q^(-35) + q^(-45) + ...
...
Relating to the Pell series, the corresponding "Lucas"-like series is (2, 6, 14, 34, 82, 198, ...) such that herein, q = 2.414213... = (1 + sqrt(2)).
Then analogous to the previous set,
...
1/2 = q^(-1) + q^(-3) + q^(-5) + q^(-7) + ...
1/6 = q^(-2) - q^(-6) + q^(-10) - q^(-14) + q^(-18) + ...
... (End)

			From _Bruno Berselli_, Feb 10 2014: (Start)
Triangle begins:
   1;
   2,  3;
   3,  6,  5;
   4,  9, 10,  7;
   5, 12, 15, 14,  9;
   6, 15, 20, 21, 18, 11;
   7, 18, 25, 28, 27, 22, 13;
   8, 21, 30, 35, 36, 33, 26, 15;
   9, 24, 35, 42, 45, 44, 39, 30, 17;
  10, 27, 40, 49, 54, 55, 52, 45, 34, 19;
  11, 30, 45, 56, 63, 66, 65, 60, 51, 38, 21;
  etc.
(End)

Cf. A094728 (triangle generated by B*A), A000330.

t[n_, k_] := If[n < k, 0, (2*k + 1)*(n - k + 1)]; Flatten[ Table[ t[n, k], {n, 0, 11}, {k, 0, n}]] (* Robert G. Wilson v, Jan 20 2005 *)

PARI
```
T(n,k)=if(n
				
```

cp's OEIS Frontend

A269837 Irregular triangle read by rows: even terms of A094728(n+1) divided by 4.

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A264798 Irregular triangle read by rows: odd-valued terms of A094728(n+1).

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A002412 Hexagonal pyramidal numbers, or greengrocer's numbers.

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A034178 Number of solutions to n = a^2 - b^2, a > b >= 0.

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A094727 Triangle read by rows: T(n,k) = n + k, 0 <= k < n, n >= 1.

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A049777 Triangular array read by rows: T(m,n) = n + n+1 + ... + m = (m+n)(m-n+1)/2.

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A370707 Triangle read by rows: T(n, k) = (-1)^kProduct_{j=0..k-1} (j - n)(j + n), for 0 <= k <= n.

A101447 Triangle read by rows: T(n,k) = (2k+1)(n+1-k), 0 <= k < n.