A002413 -id:A002413 - cp's OEIS frontend

A211790 Rectangular array: R(k,n) = number of ordered triples (w,x,y) with all terms in {1,...,n} and w^k

1, 7, 1, 23, 7, 1, 54, 22, 7, 1, 105, 51, 22, 7, 1, 181, 97, 50, 22, 7, 1, 287, 166, 96, 50, 22, 7, 1, 428, 263, 163, 95, 50, 22, 7, 1, 609, 391, 255, 161, 95, 50, 22, 7, 1, 835, 554, 378, 253, 161, 95, 50, 22, 7, 1, 1111, 756, 534, 374, 252, 161, 95, 50, 22, 7

Offset: 1

Clark Kimberling, Apr 21 2012

...
Let R be the array in A211790 and let R' be the array in A211793.  Then R(k,n) + R'(k,n) = 3^(n-1).  Moreover, (row k of R) =(row k of A211796) for k>2, by Fermat's last theorem; likewise, (row k of R')=(row k of A211799) for k>2.
...
Generalizations:  Suppose that b,c,d are nonzero integers, and let U(k,n) be the number of ordered triples (w,x,y) with all terms in {1,...,n} and b*w*k  c*x^k+d*y^k, where the relation  is one of these: <, >=, <=, >.  What additional assumptions force the limiting row sequence to be essentially one of these: A002412, A000330, A016061, A174723, A051925?
In the following guide to related arrays and sequences, U(k,n) denotes the number of (w,x,y) as described in the preceding paragraph:
                                 first 3 rows          limiting row sequence
A211790:  w^k < x^k+y^k    A004068, A211635, A211650       A002412
A211793:  w^k >= x^k+y^k   A000292, A211636, A211651       A000330
A211796:  w^k <= x^k+y^k   A002413, A211634, A211650       A002412
A211799:  w^k > x^k+y^k    A000292, A211637, A211651       A000330
A211802:  2w^k < x^k+y^k   A182260, A211800, A211801       A016061
A211805:  2w^k >= x^k+y^k  A055232, A211803, A211804       A000330
A211808:  2w^k <= x^k+y^k  A055232, A211806, A211807       A174723
A182259:  2w^k > x^k+y^k   A182260, A211810, A211811       A051925

			Northwest corner:
  1, 7, 23, 54, 105, 181, 287, 428, 609
  1, 7, 22, 51,  97, 166, 263, 391, 554
  1, 7, 22, 50,  96, 163, 255, 378, 534
  1, 7, 22, 50,  95, 161, 253, 374, 528
  1, 7, 22, 50,  95, 161, 252, 373, 527
For n=2 and k>=1, the 7 triples (w,x,y) are (1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,2), (2,2,1), (2,2,2).

Cf. A004068 (row1), A211635 (row2), A211650 (row3), A002412.

Cf. A211793, A211796, A211799, A211802, A211805, A211808, A182259

z = 48;
t[k_, n_] := Module[{s = 0},
   (Do[If[w^k < x^k + y^k, s = s + 1],
       {w, 1, #}, {x, 1, #}, {y, 1, #}] &[n]; s)];
Table[t[1, n], {n, 1, z}]  (* A004068 *)
Table[t[2, n], {n, 1, z}]  (* A211635 *)
Table[t[3, n], {n, 1, z}]  (* A211650 *)
TableForm[Table[t[k, n], {k, 1, 12}, {n, 1, 16}]]
Flatten[Table[t[k, n - k + 1], {n, 1, 12}, {k, 1, n}]] (* A211790 *)
Table[n (n + 1) (4 n - 1)/6,
  {n, 1, z}] (* row-limit sequence, A002412 *)
(* Peter J. C. Moses, Apr 13 2012 *)

R(k,n) = n(n-1)(4n+1)/6 for 1<=k<=n, and

R(k,n) = Sum{Sum{floor[(x^k+y^k)^(1/k)] : 1<=x<=n, 1<=y<=n}} for 1<=k<=n.

A093562 (5,1) Pascal triangle.

Original entry on oeis.org

1, 5, 1, 5, 6, 1, 5, 11, 7, 1, 5, 16, 18, 8, 1, 5, 21, 34, 26, 9, 1, 5, 26, 55, 60, 35, 10, 1, 5, 31, 81, 115, 95, 45, 11, 1, 5, 36, 112, 196, 210, 140, 56, 12, 1, 5, 41, 148, 308, 406, 350, 196, 68, 13, 1, 5, 46, 189, 456, 714, 756, 546, 264, 81, 14, 1, 5, 51, 235, 645, 1170

Offset: 0

Wolfdieter Lang, Apr 22 2004

This is the fifth member, d=5, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-1, for d=1..4.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=(1+4*z)/(1-(1+x)*z).
The SW-NE diagonals give A022095(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 4. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
The array F(5;n,m) gives in the columns m >= 1 the figurate numbers based on A016861, including the heptagonal numbers A000566 (see the W. Lang link).
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
The n-th row polynomial is (4 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018

			Triangle begins
  [1];
  [5,  1];
  [5,  6,  1];
  [5, 11,  7,  1];
  ...

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
P. Bala, A note on the diagonals of a proper Riordan Array
W. Lang, First 10 rows and array of figurate numbers .

Cf. Row sums: A007283(n-1), n>=1, 1 for n=0. A082505(n+1), alternating row sums are 1 for n=0, 4 for n=2 and 0 else.
Column sequences give for m=1..9: A016861, A000566 (heptagonal), A002413, A002418, A027800, A051946, A050484, A052255, A055844.
A007318, A093563 (d=6), A228196, A228576.

a093562 n k = a093562_tabl !! n !! k
a093562_row n = a093562_tabl !! n
a093562_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [5, 1]
-- Reinhard Zumkeller, Aug 31 2014

from math import comb, isqrt
def A093562(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+(r-a<<2))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m) = F(5;n-m, m) for 0<= m <= n, otherwise 0, with F(5;0, 0)=1, F(5;n, 0)=5 if n>=1 and F(5;n, m):=(5*n+m)*binomial(n+m-1, m-1)/m if m>=1.
G.f. column m (without leading zeros): (1+4*x)/(1-x)^(m+1), m>=0.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=5 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
T(n, k) = C(n, k) + 4*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(5 + 11*x + 7*x^2/2! + x^3/3!) = 5 + 16*x + 34*x^2/2! + 60*x^3/3! + 95*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A200785 T(n,k) is the number of arrays of n+2 elements from {0,1,...,k} with no two consecutive ascents.

Original entry on oeis.org

8, 26, 16, 60, 75, 32, 115, 225, 216, 64, 196, 530, 840, 622, 128, 308, 1071, 2425, 3136, 1791, 256, 456, 1946, 5796, 11100, 11704, 5157, 512, 645, 3270, 12152, 31395, 50775, 43681, 14849, 1024, 880, 5175, 23136, 75992, 169884, 232275, 163020, 42756, 2048

Offset: 1

R. H. Hardin Nov 22 2011

All the conjectured formulas are true, and follow from the Burstein-Mansour paper. - N. J. A. Sloane, May 21 2013

			Table starts
....8.....26......60.......115.......196........308.........456.........645
...16.....75.....225.......530......1071.......1946........3270........5175
...32....216.....840......2425......5796......12152.......23136.......40905
...64....622....3136.....11100.....31395......75992......164004......324087
..128...1791...11704.....50775....169884.....474566.....1160616.....2562633
..256...5157...43681....232275....919413....2964416.....8216484....20273247
..512..14849..163020...1062500...4975322...18514405....58154912...160338680
.1024..42756..608400...4860250..26924106..115637431...411637168..1268210421
.2048.123111.2270580..22232375.145698840..722234149..2913595712.10030582998
.4096.354484.8473921.101698250.788446400.4510869636.20622837480.79335475611
Some arrays for n=4, k=3:
..0....1....0....0....1....0....3....3....0....1....3....0....2....2....2....2
..3....0....2....2....0....2....0....0....3....1....0....0....0....3....3....3
..2....3....2....2....2....2....3....3....1....0....1....0....2....1....3....3
..1....0....2....1....0....0....2....2....2....2....1....2....2....0....0....2
..0....3....0....0....1....2....1....2....0....0....3....2....0....3....1....3
..3....3....0....3....0....2....3....2....0....3....0....0....2....2....1....3

R. H. Hardin, Table of n, a(n) for n = 1..10010
A. Burstein and T. Mansour, Words restricted by 3-letter generalized multipermutation patterns, Annals. Combin., 7 (2003), 1-14. See Th. 3.13.

Column 1 is A000079
Column 2 is A076264
Column 3 is A072335
Row 1 is A002413
Cf. A200781.

T(n-2,k) = \sum_{L=0}^n (-1)^L / L! * \sum_{M=0}^{min(L,[(n-L)/2])} binomial(n-L-M,M) * M! * (k+1)^(n-L-2*M) B_{L,M}(x_1,x_2,...), where B_{L,M}() are Bell polynomials, x_i = binomial(k+1,i+2) * i! * f(i), i=1,2,..., and f(i) has period of length 6: [0,1,1,0,-1,-1] (i.e., f(0)=0, f(1)=1, etc.). This formula implies that for a fixed n, T(n,k) is a polynomial in k, which is easy to compute. - Max Alekseyev, Dec 12 2011
Empirical formulas for columns:
k=1: a(n) = 2*a(n-1)
k=2: a(n) = 3*a(n-1) -a(n-3)
k=3: a(n) = 4*a(n-1) -4*a(n-3) +a(n-4)
k=4: a(n) = 5*a(n-1) -10*a(n-3) +5*a(n-4)
k=5: a(n) = 6*a(n-1) -20*a(n-3) +15*a(n-4) -a(n-6)
k=6: a(n) = 7*a(n-1) -35*a(n-3) +35*a(n-4) -7*a(n-6) +a(n-7)
k=7: a(n) = 8*a(n-1) -56*a(n-3) +70*a(n-4) -28*a(n-6) +8*a(n-7)
Empirical recurrence for general column k:
0 = sum{i=0..floor(k/3) (binomial(k+1,3*i+1)*T(n-(3*i+1),k))} - sum{i=0..floor((k+1)/3) (binomial(k+1,3*i)*T(n-3*i,k))}
Formulae for rows:
T(1,k) = (5/6)*k^3 + 3*k^2 + (19/6)*k + 1
T(2,k) = (17/24)*k^4 + (43/12)*k^3 + (151/24)*k^2 + (53/12)*k + 1
T(3,k) = (7/12)*k^5 + (47/12)*k^4 + (39/4)*k^3 + (133/12)*k^2 + (17/3)*k + 1
T(4,k) = (349/720)*k^6 + (321/80)*k^5 + (1883/144)*k^4 + (1013/48)*k^3 + (3139/180)*k^2 + (413/60)*k + 1
T(5,k) = (2017/5040)*k^7 + (1427/360)*k^6 + (5759/360)*k^5 + (607/18)*k^4 + (28459/720)*k^3 + (9113/360)*k^2 + (848/105)*k + 1
T(6,k) = (6679/20160)*k^8 + (4799/1260)*k^7 + (26449/1440)*k^6 + (2162/45)*k^5 + (212153/2880)*k^4 + (6019/90)*k^3 + (174571/5040)*k^2 + (3893/420)*k + 1
T(7,k) = (99377/362880)*k^9 + (48247/13440)*k^8 + (243673/12096)*k^7 + (60529/960)*k^6 + (2076437/17280)*k^5 + (274529/1920)*k^4 + (952027/9072)*k^3 + (152461/3360)*k^2 + (26399/2520)*k + 1

A162147 a(n) = n(n+1)(5*n + 4)/6.

Original entry on oeis.org

0, 3, 14, 38, 80, 145, 238, 364, 528, 735, 990, 1298, 1664, 2093, 2590, 3160, 3808, 4539, 5358, 6270, 7280, 8393, 9614, 10948, 12400, 13975, 15678, 17514, 19488, 21605, 23870, 26288, 28864, 31603, 34510, 37590, 40848, 44289, 47918, 51740, 55760

Offset: 0

Vladimir Joseph Stephan Orlovsky, Jun 25 2009

Partial sums of A005475.
Suppose we extend the triangle in A215631 to a symmetric array by reflection about the main diagonal. The array is defined by m(i,j) = i^2 + i*j + j^2: 3, 7, 13, ...; 7, 12, 19, ...; 13, 19, 27, .... Then a(n) is the sum of the n-th antidiagonal. Examples: 3, 7 + 7, 13 + 12 + 13, 21 + 19 + 19 + 21, etc. - J. M. Bergot, Jun 25 2013
Binomial transform of [0,3,8,5,0,0,0,...]. - Alois P. Heinz, Mar 10 2015

			For n=4, a(4) = 0*(5+0) + 1*(5+1) + 2*(5+2) + 3*(5+3) + 4*(5+4) = 80. - _Bruno Berselli_, Mar 17 2016

Jinyuan Wang, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A002412, A002413, A006331, A016061, A033994.

[n*(n+1)*(5*n+4)/6: n in [0..40]]; // G. C. Greubel, Apr 01 2021

A162147:= n-> n*(n+1)*(5*n+4)/6; seq(A162147(n), n=0..40); # G. C. Greubel, Apr 01 2021

Table[(n(n+1)(5n+4))/6,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,3,14,38},50] (* Harvey P. Dale, May 04 2013 *)

a(n)=n*(n+1)*(5*n+4)/6 \\ Charles R Greathouse IV, Oct 07 2015

[n*(n+1)*(5*n+4)/6 for n in (0..40)] # G. C. Greubel, Apr 01 2021

From R. J. Mathar, Jun 27 2009: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4)
a(n) = A033994(n) + A000217(n).
G.f.: x*(3+2*x)/(1-x)^4. (End)
a(n) = A035005(n+1)/4. - Johannes W. Meijer, Feb 04 2010
a(n) = Sum_{i=0..n} i*(n + 1 + i). - Bruno Berselli, Mar 17 2016
E.g.f.: x*(18 + 24*x + 5*x^2)*exp(x)/6. - G. C. Greubel, Apr 01 2021

Definition rephrased by R. J. Mathar, Jun 27 2009

A162148 a(n) = n(n+1)(5*n+7)/6.

Original entry on oeis.org

0, 4, 17, 44, 90, 160, 259, 392, 564, 780, 1045, 1364, 1742, 2184, 2695, 3280, 3944, 4692, 5529, 6460, 7490, 8624, 9867, 11224, 12700, 14300, 16029, 17892, 19894, 22040, 24335, 26784, 29392, 32164, 35105, 38220, 41514, 44992, 48659, 52520, 56580

Offset: 0

Vladimir Joseph Stephan Orlovsky, Jun 25 2009

Partial sums of A147875.
Equals the fourth right hand column of A175136 for n>=1. - Johannes W. Meijer, May 06 2011
a(n) is the number of triples (w,x,y) havingt all terms in {0,...,n} and x+y>w. - Clark Kimberling, Jun 14 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A002412, A002413, A006331, A016061, A033994.
Cf. A162147, A175136, A190048, A190049, A254407.
Related to A000292 and A091894.

[n*(n+1)*(5*n+7)/6: n in [0..50]]; // Vincenzo Librandi, May 07 2011

A162148:= n-> n*(n+1)*(5*n+7)/6; seq(A162148(n), n=0..50); # G. C. Greubel, Mar 31 2021

Table[(n(n+1)(5n+7))/6,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,4,17,44}, 50] (* Harvey P. Dale, May 20 2014 *)

a(n)=n*(n+1)*(5*n+7)/6 \\ Charles R Greathouse IV, Oct 07 2015

[n*(n+1)*(5*n+7)/6 for n in (0..50)] # G. C. Greubel, Mar 31 2021

a(n) = A162147(n) + A000217(n).
From Johannes W. Meijer, May 06 2011: (Start)
G.f.: x*(4+x)/(1-x)^4.
a(n) = 4*binomial(n+2,3) + binomial(n+1,3).
a(n) = A091894(3,0)*binomial(n+2,3) + A091894(3,1)*binomial(n+1,3). (End)
a(n) = (n+1)*A000290(n+1) - Sum_{i=1..n+1} A000217(i).
a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4), a(0)=0, a(1)=4, a(2)=17, a(3)=44. - Harvey P. Dale, May 20 2014
E.g.f.: x*(24 +27*x +5*x^2)*exp(x)/6. - G. C. Greubel, Mar 31 2021

Definition rephrased by R. J. Mathar, Jun 27 2009

A220084 a(n) = (n + 1)(20n^2 + 19*n + 6)/6.

Original entry on oeis.org

1, 15, 62, 162, 335, 601, 980, 1492, 2157, 2995, 4026, 5270, 6747, 8477, 10480, 12776, 15385, 18327, 21622, 25290, 29351, 33825, 38732, 44092, 49925, 56251, 63090, 70462, 78387, 86885, 95976, 105680, 116017, 127007, 138670, 151026, 164095, 177897, 192452

Offset: 0

Bruno Berselli, Dec 11 2012

Sequence related to heptagonal pyramidal numbers (A002413) by a(n) = n*A002413(n) - (n-1)*A002413(n-1).
Other sequences of numbers of the form m*P(k,m)-(m-1)*P(k,m-1), where P(k,m) is the m-th k-gonal pyramidal number:
k=3, A002412(m) = m*A000292(m)-(m-1)*A000292(m-1);
k=4, A051662(m) = (m+1)*A000330(m+1)-m*A000330(m);
k=5, A213772(m) = m*A002411(m)-(m-1)*A002411(m-1);
k=6, A213837(m) = m*A002412(m)-(m-1)*A002412(m-1);
k=7, this sequence;
k=8, A130748(m) = m*A002414(m)-(m-1)*A002414(m-1).
Also, first bisection of A212983.
Binomial transform of (1, 14, 33, 20, 0, 0, 0, ...). - Gary W. Adamson, Aug 26 2015

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000292, A000330, A000566, A002411, A002412, A002413, A002414, A051662, A130748, A212983, A213772, A213837.

[(n+1)*(20*n^2+19*n+6)/6: n in [0..40]]; // Bruno Berselli, Jun 28 2016

/* By first comment: */  A002413:=func; [n*A002413(n)-(n-1)*A002413(n-1): n in [1..40]];

I:=[1,15,62,162]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013

Table[(n + 1) (20 n^2 + 19 n + 6)/6, {n, 0, 40}]
LinearRecurrence[{4,-6,4,-1},{1,15,62,162},40] (* Harvey P. Dale, Dec 23 2012 *)
CoefficientList[Series[(1 + 11 x + 8 x^2) / (1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Aug 18 2013 *)

makelist((n+1)*(20*n^2+19*n+6)/6, n, 0, 20); /* Martin Ettl, Dec 12 2012 */

a(n)=(n+1)*(20*n^2+19*n+6)/6 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (1+11*x+8*x^2)/(1-x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), for n>3, a(0)=1, a(1)=15, a(2)=62, a(3)=162. - Harvey P. Dale, Dec 23 2012
a(n) = (n+1)*A000566(n+1) + Sum_{i=0..n} A000566(i). - Bruno Berselli, Dec 18 2013
E.g.f.: exp(x)*(6 + 84*x + 99*x^2 + 20*x^3)/6. - Elmo R. Oliveira, Aug 06 2025

A279218 Expansion of Product_{k>=1} 1/(1 - x^k)^(k(k+1)(5*k-2)/6).

Original entry on oeis.org

1, 1, 9, 35, 131, 454, 1601, 5325, 17467, 55588, 173858, 532809, 1607056, 4769263, 13957660, 40302923, 114962909, 324157109, 904247056, 2496917319, 6829241131, 18510038697, 49741367504, 132582175873, 350655140642, 920568519505, 2399692063845, 6213105691838, 15982216140168, 40855658807127, 103814659491641

Offset: 0

Ilya Gutkovskiy, Dec 08 2016

nonn

Euler transform of the heptagonal pyramidal numbers (A002413).

Vaclav Kotesovec, Table of n, a(n) for n = 0..1000
M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to arXiv version]
M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to Lin. Alg. Applic. version together with omitted figures]
N. J. A. Sloane, Transforms
Eric Weisstein's World of Mathematics, Heptagonal Pyramidal Number
Index to sequences related to pyramidal numbers

Cf. A000335, A002413, A279215, A279216, A279217, A279219.

nmax=30; CoefficientList[Series[Product[1/(1 - x^k)^(k (k + 1) (5 k - 2)/6), {k, 1, nmax}], {x, 0, nmax}], x]

G.f.: Product_{k>=1} 1/(1 - x^k)^(k*(k+1)*(5*k-2)/6).

a(n) ~ exp(-Zeta'(-1)/3 - Zeta(3)/(8*Pi^2) - Pi^16/(388800000000*Zeta(5)^3) - Pi^8*Zeta(3)/(5400000*Zeta(5)^2) - Zeta(3)^2/(450*Zeta(5)) + 5*Zeta'(-3)/6 + (Pi^12/(270000000*2^(2/5)*5^(1/5)*Zeta(5)^(11/5)) + Pi^4*Zeta(3)/(4500*2^(2/5) * 5^(1/5)*Zeta(5)^(6/5))) * n^(1/5) + (-Pi^8/(180000*2^(4/5)*5^(2/5)*Zeta(5)^(7/5)) - Zeta(3)/(3*2^(4/5)*(5*Zeta(5))^(2/5))) * n^(2/5) + (Pi^4/(180*2^(1/5)*(5*Zeta(5))^(3/5))) * n^(3/5) + ((5*(5*Zeta(5))^(1/5))/(2^(8/5))) * n^(4/5)) * Zeta(5)^(67/720) / (2^(113/360) * 5^(293/720) * sqrt(Pi) * n^(427/720)). - Vaclav Kotesovec, Dec 08 2016

A322636 Numbers that are sums of consecutive heptagonal numbers (A000566).

Original entry on oeis.org

0, 1, 7, 8, 18, 25, 26, 34, 52, 55, 59, 60, 81, 89, 107, 112, 114, 115, 136, 148, 170, 188, 189, 193, 195, 196, 235, 248, 260, 282, 286, 300, 307, 308, 337, 341, 342, 396, 403, 424, 430, 448, 449, 455, 456, 469, 521, 530, 540, 572, 585, 616, 619, 628, 637, 644, 645, 684, 697

Offset: 1

Ilya Gutkovskiy, Dec 21 2018

nonn

Robert Israel, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Heptagonal Number

Cf. A000566, A002413, A034705, A034706, A319184, A319185, A322637.

N:= 1000: # for terms up to N
Hepta:= [seq(n*(5*n-3)/2,n=0..floor((3+sqrt(9+40*N))/10))]:
PS:= ListTools:-PartialSums(Hepta):
S:= select(`<=`,{0,seq(seq(PS[i]-PS[j],j=1..i-1),i=1..nops(PS))},N):
sort(convert(S,list)); # Robert Israel, May 22 2025

terms = 59;
nmax = 17; kmax = 9; (* empirical *)
T = Table[n(5n-3)/2, {n, 0, nmax}];
Union[T, Table[k MovingAverage[T, k], {k, 2, kmax}]//Flatten][[1 ;; terms]] (* Jean-François Alcover, Dec 26 2018 *)

A261720 Array of pyramidal (3-dimensional figurate numbers) read by antidiagonals.

Original entry on oeis.org

1, 1, 4, 1, 5, 10, 1, 6, 14, 20, 1, 7, 18, 30, 35, 1, 8, 22, 40, 55, 56, 1, 9, 26, 50, 75, 91, 84, 1, 10, 30, 60, 95, 126, 140, 120, 1, 11, 34, 70, 115, 161, 196, 204, 165, 1, 12, 38, 80, 135, 196, 252, 288, 285, 220, 1, 13, 42, 90, 155, 231, 308, 372, 405, 385, 286

Offset: 1

Gary W. Adamson, Aug 29 2015

First few sequences in the array:
  1,  4, 10, 20,  35,  56,  84, 120, 165,  220, ... A000292
  1,  5, 14, 30,  55,  91, 140, 204, 285,  385, ... A000330
  1,  6, 18, 40,  75, 126, 196, 288, 405,  550, ... A002411
  1,  7, 22, 50,  95, 161, 252, 372, 525,  715, ... A002412
  1,  8, 26, 60, 115, 196, 308, 456, 645,  880, ... A002413
  1,  9, 30, 70, 135, 231, 364, 540, 765, 1045, ... A002414
  1, 10, 34, 80, 155, 266, 420, 624, 885, 1210, ... A007584
  ...
The corresponding bases to rows are: Triangle, Square, Pentagon, Hexagon, Heptagon, Octagon, ...

			Row 2: (1, 5, 14, 30, 55, ...) = (1, 4, 10, 20, 35, ...) + (0, 1, 4, 10, 20, 35, ...).
(1, 7, 22, 50, ...) is the binomial transform of (1, 6, 9, 4, 0, 0, 0, ...) 3rd row in Pascal's triangle (1,4) followed by zeros. (1, 7, 22, 50, ...) is the third partial sum of (1, 4, 4, 4, ...).

Albert H. Beiler, "Recreations in the Theory of Numbers"; Dover, 1966, p. 194.

Cf. A000292, A000330, A002411, A002412, A002413, A002414, A007584, A220084.

Similar to A080851 but without row n=0.

T[n_,k_]:=k(k+1)((k-1)n+3)/6; Flatten[Table[T[n-k+1,k],{n,11},{k,n}]] (* Stefano Spezia, Aug 15 2024 *)

T(n,k) = A080851(n,k).
Given: first sequence in the array is A000292: (1, 4, 10, 20, 35, ...) Subsequent rows are generated by adding (0, 1, 4, 10, 20, 35, ...) to the current row.
n-th row is the binomial transform of row 3 in Pascal's triangle (1,n) followed by zeros.  Alternatively, begin with (1, 4, 10, 20, ...) being the binomial transform of (1, 3, 3, 1, 0, 0, 0, ...). Add (0, 1, 2, 1, 0, 0, 0, ...) to the latter to obtain the inverse binomial transform of the next row: (1, 5, 14, 30, 55,..); then repeat the operation.
The row starting (1, N, ...) is the 3rd partial sum of (1, (N-3), (N-3), (N-3), ...).
From Stefano Spezia, Aug 15 2024: (Start)
T(n,k) = k*(k + 1)*((k - 1)*n + 3)/6.
G.f. as array: x*y*(1 + x*(y - 1))/((1 - x)^2*(1 - y)^4).
E.g.f. as array: exp(y)*y*(exp(x)*(6 + 3*(1 + x)*y + x*y^2) - 3*(2 + y))/6. (End)

A366016 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^4 / (1 - 4 * A(x)).

Original entry on oeis.org

0, 1, 8, 102, 1580, 27193, 499828, 9609372, 190869948, 3886281300, 80681111940, 1701418017390, 36345240847188, 784821812522062, 17103169093916120, 375670490644949624, 8308349385885678684, 184856293637482503660, 4134886240989315235840, 92928784113832360511800, 2097399158679611824619120

Offset: 0

Ilya Gutkovskiy, Sep 26 2023

nonn

Reversion of g.f. for heptagonal pyramidal numbers (with signs).

Eric Weisstein's World of Mathematics, Heptagonal Pyramidal Number
Eric Weisstein's World of Mathematics, Series Reversion

Cf. A002293, A002413, A064088, A365754, A365817, A366014, A366015, A366017.

nmax = 20; A[] = 0; Do[A[x] = x (1 + A[x])^4/(1 - 4 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
CoefficientList[InverseSeries[Series[x (1 - 4 x)/(1 + x)^4, {x, 0, 20}], x], x]	
Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[4 n, n - k - 1] 4^k, {k, 0, n - 1}], {n, 1, 20}]]

a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(4*n,n-k-1) * 4^k for n > 0.

a(n) ~ sqrt(163 - 1521/sqrt(89)) * (4933 + 801*sqrt(89))^n / (sqrt(Pi) * n^(3/2) * 2^(9*n + 9/2)). - Vaclav Kotesovec, Sep 27 2023

cp's OEIS Frontend

A211790 Rectangular array: R(k,n) = number of ordered triples (w,x,y) with all terms in {1,...,n} and w^k

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A093562 (5,1) Pascal triangle.

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A200785 T(n,k) is the number of arrays of n+2 elements from {0,1,...,k} with no two consecutive ascents.

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A162147 a(n) = n*(n+1)*(5*n + 4)/6.

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A162148 a(n) = n*(n+1)*(5*n+7)/6.

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A220084 a(n) = (n + 1)*(20*n^2 + 19*n + 6)/6.

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A279218 Expansion of Product_{k>=1} 1/(1 - x^k)^(k*(k+1)*(5*k-2)/6).

A162147 a(n) = n(n+1)(5*n + 4)/6.

A162148 a(n) = n(n+1)(5*n+7)/6.

A220084 a(n) = (n + 1)(20n^2 + 19*n + 6)/6.

A279218 Expansion of Product_{k>=1} 1/(1 - x^k)^(k(k+1)(5*k-2)/6).