A033455 -id:A033455 - cp's OEIS frontend

A213500 Rectangular array T(n,k): (row n) = bc, where b(h) = h, c(h) = h + n - 1, n >= 1, h >= 1, and = convolution.

1, 4, 2, 10, 7, 3, 20, 16, 10, 4, 35, 30, 22, 13, 5, 56, 50, 40, 28, 16, 6, 84, 77, 65, 50, 34, 19, 7, 120, 112, 98, 80, 60, 40, 22, 8, 165, 156, 140, 119, 95, 70, 46, 25, 9, 220, 210, 192, 168, 140, 110, 80, 52, 28, 10, 286, 275, 255, 228, 196, 161, 125, 90

Offset: 1

Clark Kimberling, Jun 14 2012

Principal diagonal: A002412.
Antidiagonal sums: A002415.
Row 1:  (1,2,3,...)**(1,2,3,...) = A000292.
Row 2:  (1,2,3,...)**(2,3,4,...) = A005581.
Row 3:  (1,2,3,...)**(3,4,5,...) = A006503.
Row 4:  (1,2,3,...)**(4,5,6,...) = A060488.
Row 5:  (1,2,3,...)**(5,6,7,...) = A096941.
Row 6:  (1,2,3,...)**(6,7,8,...) = A096957.
...
In general, the convolution of two infinite sequences is defined from the convolution of two n-tuples:  let X(n) = (x(1),...,x(n)) and Y(n)=(y(1),...,y(n)); then X(n)**Y(n) = x(1)*y(n)+x(2)*y(n-1)+...+x(n)*y(1); this sum is the n-th term in the convolution of infinite sequences:(x(1),...,x(n),...)**(y(1),...,y(n),...), for all n>=1.
...
In the following guide to related arrays and sequences, row n of each array T(n,k) is the convolution b**c of the sequences b(h) and c(h+n-1). The principal diagonal is given by T(n,n) and the n-th antidiagonal sum by S(n). In some cases, T(n,n) or S(n) differs in offset from the listed sequence.
b(h)........ c(h)........ T(n,k) .. T(n,n) .. S(n)
h .......... h .......... A213500 . A002412 . A002415
h .......... h^2 ........ A212891 . A213436 . A024166
h^2 ........ h .......... A213503 . A117066 . A033455
h^2 ........ h^2 ........ A213505 . A213546 . A213547
h .......... h*(h+1)/2 .. A213548 . A213549 . A051836
h*(h+1)/2 .. h .......... A213550 . A002418 . A005585
h*(h+1)/2 .. h*(h+1)/2 .. A213551 . A213552 . A051923
h .......... h^3 ........ A213553 . A213554 . A101089
h^3 ........ h .......... A213555 . A213556 . A213547
h^3 ........ h^3 ........ A213558 . A213559 . A213560
h^2 ........ h*(h+1)/2 .. A213561 . A213562 . A213563
h*(h+1)/2 .. h^2 ........ A213564 . A213565 . A101094
2^(h-1) .... h .......... A213568 . A213569 . A047520
2^(h-1) .... h^2 ........ A213573 . A213574 . A213575
h .......... Fibo(h) .... A213576 . A213577 . A213578
Fibo(h) .... h .......... A213579 . A213580 . A053808
Fibo(h) .... Fibo(h) .... A067418 . A027991 . A067988
Fibo(h+1) .. h .......... A213584 . A213585 . A213586
Fibo(n+1) .. Fibo(h+1) .. A213587 . A213588 . A213589
h^2 ........ Fibo(h) .... A213590 . A213504 . A213557
Fibo(h) .... h^2 ........ A213566 . A213567 . A213570
h .......... -1+2^h ..... A213571 . A213572 . A213581
-1+2^h ..... h .......... A213582 . A213583 . A156928
-1+2^h ..... -1+2^h ..... A213747 . A213748 . A213749
h .......... 2*h-1 ...... A213750 . A007585 . A002417
2*h-1 ...... h .......... A213751 . A051662 . A006325
2*h-1 ...... 2*h-1 ...... A213752 . A100157 . A071238
2*h-1 ...... -1+2^h ..... A213753 . A213754 . A213755
-1+2^h ..... 2*h-1 ...... A213756 . A213757 . A213758
2^(n-1) .... 2*h-1 ...... A213762 . A213763 . A213764
2*h-1 ...... Fibo(h) .... A213765 . A213766 . A213767
Fibo(h) .... 2*h-1 ...... A213768 . A213769 . A213770
Fibo(h+1) .. 2*h-1 ...... A213774 . A213775 . A213776
Fibo(h) .... Fibo(h+1) .. A213777 . A001870 . A152881
h .......... 1+[h/2] .... A213778 . A213779 . A213780
1+[h/2] .... h .......... A213781 . A213782 . A005712
1+[h/2] .... [(h+1)/2] .. A213783 . A213759 . A213760
h .......... 3*h-2 ...... A213761 . A172073 . A002419
3*h-2 ...... h .......... A213771 . A213772 . A132117
3*h-2 ...... 3*h-2 ...... A213773 . A214092 . A213818
h .......... 3*h-1 ...... A213819 . A213820 . A153978
3*h-1 ...... h .......... A213821 . A033431 . A176060
3*h-1 ...... 3*h-1 ...... A213822 . A213823 . A213824
3*h-1 ...... 3*h-2 ...... A213825 . A213826 . A213827
3*h-2 ...... 3*h-1 ...... A213828 . A213829 . A213830
2*h-1 ...... 3*h-2 ...... A213831 . A213832 . A212560
3*h-2 ...... 2*h-1 ...... A213833 . A130748 . A213834
h .......... 4*h-3 ...... A213835 . A172078 . A051797
4*h-3 ...... h .......... A213836 . A213837 . A071238
4*h-3 ...... 2*h-1 ...... A213838 . A213839 . A213840
2*h-1 ...... 4*h-3 ...... A213841 . A213842 . A213843
2*h-1 ...... 4*h-1 ...... A213844 . A213845 . A213846
4*h-1 ...... 2*h-1 ...... A213847 . A213848 . A180324
[(h+1)/2] .. [(h+1)/2] .. A213849 . A049778 . A213850
h .......... C(2*h-2,h-1) A213853
...
Suppose that u = (u(n)) and v = (v(n)) are sequences having generating functions U(x) and V(x), respectively. Then the convolution u**v has generating function U(x)*V(x). Accordingly, if u and v are homogeneous linear recurrence sequences, then every row of the convolution array T satisfies the same homogeneous linear recurrence equation, which can be easily obtained from the denominator of U(x)*V(x). Also, every column of T has the same homogeneous linear recurrence as v.

			Northwest corner (the array is read by southwest falling antidiagonals):
  1,  4, 10, 20,  35,  56,  84, ...
  2,  7, 16, 30,  50,  77, 112, ...
  3, 10, 22, 40,  65,  98, 140, ...
  4, 13, 28, 50,  80, 119, 168, ...
  5, 16, 34, 60,  95, 140, 196, ...
  6, 19, 40, 70, 110, 161, 224, ...
T(6,1) = (1)**(6) = 6;
T(6,2) = (1,2)**(6,7) = 1*7+2*6 = 19;
T(6,3) = (1,2,3)**(6,7,8) = 1*8+2*7+3*6 = 40.

Clark Kimberling, Antidiagonals n = 1..60, flattened

Cf. A000027.

b[n_] := n; c[n_] := n
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213500 *)

t(n,k) = sum(i=0, k - 1, (k - i) * (n + i));
tabl(nn) = {for(n=1, nn, for(k=1, n, print1(t(k,n - k + 1),", ");); print(););};
tabl(12) \\ Indranil Ghosh, Mar 26 2017

def t(n, k): return sum((k - i) * (n + i) for i in range(k))
for n in range(1, 13):
    print([t(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Mar 26 2017

T(n,k) = 4*T(n,k-1) - 6*T(n,k-2) + 4*T(n,k-3) - T(n,k-4).
T(n,k) = 2*T(n-1,k) - T(n-2,k).
G.f. for row n:  x*(n - (n - 1)*x)/(1 - x)^4.

A001249 Squares of tetrahedral numbers: a(n) = binomial(n+3,n)^2.

Original entry on oeis.org

1, 16, 100, 400, 1225, 3136, 7056, 14400, 27225, 48400, 81796, 132496, 207025, 313600, 462400, 665856, 938961, 1299600, 1768900, 2371600, 3136441, 4096576, 5290000, 6760000, 8555625, 10732176, 13351716, 16483600, 20205025, 24601600, 29767936, 35808256

Offset: 0

N. J. A. Sloane

Total area of all square and rectangular regions from an n+1 X n+1 grid. E.g., n = 2, there are 9 individual squares, 4 2 X 2's and 1 3 X 3, total area 9 + 16 + 9 = 34. The rectangular regions include 6 2 X 1's, 6 1 X 2's, 3 3 X 1's, 3 1 X 3's, 2 3 X 2's and 2 2 X 3's, total area 12 + 12 + 9 + 9 + 12 + 12 = 66, hence a(2) = 34 + 66 = 100. - Jon Perry, Jul 29 2003 [Index/grid size adjusted by Rick L. Shepherd, Jun 27 2017]
Number of 3 X 3 submatrices of an n+3 X n+3 matrix. - Rick L. Shepherd, Jun 27 2017
The inverse binomial transform gives row n=2 of A087107. - R. J. Mathar, Aug 31 2022

T. D. Noe, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000290, A000292, A006542, A033455, A108674 (first diffs.), A086020 (partial sums).
Third column of triangle A008459.
Cf. A000579, A001303, A040977.

A001249 := proc(n) binomial(n+3,n)^2 end proc: seq(A001249(n),n=0..10) ; # Zerinvary Lajos, May 17 2006

Table[Binomial[n + 3, 3]^2, {n, 0, 100}] (* T. D. Noe, Jun 26 2012 *)

a(n)=binomial(n+3,3)^2 \\ Charles R Greathouse IV, Sep 24 2015

From R. J. Mathar, Aug 19 2008: (Start)
a(n) = (A000292(n+1))^2.
O.g.f.: (1+x)(x^2+8x+1)/(1-x)^7. (End)
a(n) = C(n+4, 3)*C(n+4, 4)/(n+4) + A001303(n) = C(n+4, 3)*C(n+3, 3)/4 + A001303(n) = C(n+4, 6) + 3*C(n+5, 6) + C(n+6,6) + A001303(n). - Gary Detlefs, Aug 07 2013
-n^2*a(n) + (n+3)^2*a(n-1) = 0. - R. J. Mathar, Aug 15 2013
a(n) = 9*A040977(n-1) + A000579(n+6) + A000579(n+3). - R. J. Mathar, Aug 15 2013
a(n) = (n+3)*C(n+2, 2)*C(n+3, 3)/3. - Gary Detlefs, Jan 06 2014
a(n) = A000290(n+1)*A000290(n+2)*A000290(n+3)/36. - Bruno Berselli, Nov 12 2014
G.f. 2F1(4,4;1;x). - R. J. Mathar, Aug 09 2015
E.g.f.: exp(x)*(1 + 15*x + 69*x^2/2! + 147*x^3/3! + 162*x^4/4! +  90*x^5/5! +  20*x^6/6!). Computed from the o.g.f with the formulas (23) - (25) of the W. Lang link given in A060187. - Wolfdieter Lang, Jul 27 2017
From Amiram Eldar, Jan 24 2022: (Start)
Sum_{n>=0} 1/a(n) = 9*Pi^2 - 351/4.
Sum_{n>=0} (-1)^n/a(n) = 63/4 - 3*Pi^2/2. (End)
a(n) = 7*a(n-1)-21*a(n-2)+35*a(n-3)-35*a(n-4)+21*a(n-5)-7*a(n-6)+a(n-7). - Wesley Ivan Hurt, Aug 29 2022
a(n) = a(n-1)+A000217(n+1)*A000330(n+1). - J. M. Bergot, Aug 29 2022
a(n) = A002415(n+2)^2 - 20*A006857(n-1). - Yasser Arath Chavez Reyes, Nov 08 2024

A061167 a(n) = n^5 - n.

Original entry on oeis.org

0, 0, 30, 240, 1020, 3120, 7770, 16800, 32760, 59040, 99990, 161040, 248820, 371280, 537810, 759360, 1048560, 1419840, 1889550, 2476080, 3199980, 4084080, 5153610, 6436320, 7962600, 9765600, 11881350, 14348880, 17210340, 20511120, 24299970, 28629120

Offset: 0

Henry Bottomley, Apr 18 2001

(b^2+c^2)/(bc+1) is an integer if {b,c} are of the form {0,n}, {n,n^3}, {n^3,n^5-n}, {n^5-n,n^7-2n^3}, {n^7-2n^3,n^9-3n^5+n}, etc. for some n, in which case the division results in n^2. Cf. A052530.

Convolution of A033429 by A033581. - R. J. Mathar, Aug 19 2008

			a(2) = 32 - 2 = 30.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
D. Zagier, Problems posed at the St Andrews Colloquium, 1996
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Subsequence of A249674.

Cf. A000584, A024002, A033429, A033455, A033581, A052530.

[n^5-n: n in [0..40]]; // Vincenzo Librandi, May 02 2011

Table[n^5 - n, {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2012 *)

a(n) = 30*A033455(n-1). [Corrected by Bernard Schott, Mar 16 2021]
a(n) = -n*A024002(n).
a(n) = A000584(n) - n.
O.g.f.: 30x^2(1+x)^2/(1-x)^6. - R. J. Mathar, Aug 19 2008
a(n) = n * (n-1) * (n+1) * (n^2+1). - Bernard Schott, Mar 16 2021
E.g.f.: exp(x)*x^2*(15 + 25*x + 10*x^2 + x^3). - Stefano Spezia, Dec 27 2021

A098359 Multiplication table of the square numbers read by antidiagonals.

Original entry on oeis.org

1, 4, 4, 9, 16, 9, 16, 36, 36, 16, 25, 64, 81, 64, 25, 36, 100, 144, 144, 100, 36, 49, 144, 225, 256, 225, 144, 49, 64, 196, 324, 400, 400, 324, 196, 64, 81, 256, 441, 576, 625, 576, 441, 256, 81, 100, 324, 576, 784, 900, 900, 784, 576, 324, 100, 121, 400, 729, 1024, 1225, 1296, 1225, 1024, 729, 400, 121

Offset: 1

Douglas Stones (dssto1(AT)student.monash.edu.au), Sep 04 2004

sum_{k=0..2n-2} (-1)^k*a(A000124(2n-2)+k-1) = n. See A003991. - Charlie Marion, Apr 22 2013

			Square array A(n,k) begins:
   1,   4,   9,  16,   25,   36,   49, ...
   4,  16,  36,  64,  100,  144,  196, ...
   9,  36,  81, 144,  225,  324,  441, ...
  16,  64, 144, 256,  400,  576,  784, ...
  25, 100, 225, 400,  625,  900, 1225, ...
  36, 144, 324, 576,  900, 1296, 1764, ...
  49, 196, 441, 784, 1225, 1764, 2401, ...

Cf. A000290, A003991, A098358, A098360, A098361, A213547.
Antidiagonal sums give A033455.
Main diagonal gives A000583.

A:= (n,k)-> (n*k)^2:
seq(seq(A(n, 1+d-n), n=1..d), d=1..12);  # Alois P. Heinz, May 19 2025

A(n,k) = n^2*k^2.
G.f.: [xy(1+x)(1+y)] / [(1-x)^3 * (1-y)^3 ]. - Ralf Stephan, Oct 27 2004
Sum_{j=1..n} A(j,1+n-j)*j = A213547(n). - Alois P. Heinz, May 19 2025

Offset corrected by Alois P. Heinz, May 19 2025

A130713 a(0)=a(2)=1, a(1)=2, a(n)=0 for n > 2.

Original entry on oeis.org

1, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Paul Curtz and Tanya Khovanova, Jul 01 2007

Self-convolution of A019590. Up to a sign the convolutional inverse of the natural numbers sequence. - Tanya Khovanova, Jul 14 2007

Iterated partial sums give the chain A130713 -> A113311 -> A008574 -> A001844 -> A005900 -> A006325 -> A033455 -> A259181, up to index. The k-th term of the n-th partial sums is (n^2-7n+14 + 4k(k+n-4))(k+n-4)!/(k-1)!/(n-1)!, for k > 3-n. Iterating partial sums in reverse (n-th differences with n zeros prepended) gives row (n+3) of A182533, modulo signs and trailing zeros. - Travis Scott, Feb 19 2023

Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
Dominika Závacká, Cristina Dalfó, and Miquel Angel Fiol, Integer sequences from k-iterated line digraphs, CEUR: Proc. 24th Conf. Info. Tech. - Appl. and Theory (ITAT 2024) Vol 3792, 156-161. See p. 161, Table 2.

A130713:=n->binomial(2*n, n^2); seq(A130713(n), n=0..100); # Wesley Ivan Hurt, Mar 08 2014

Table[Binomial[2 n, n^2], {n, 0, 100}] (* Wesley Ivan Hurt, Mar 08 2014 *)

G.f.: 1 + 2*x + x^2.

a(n) = binomial(2n, n^2). - Wesley Ivan Hurt, Mar 08 2014

A213505 Rectangular array: (row n) = bc, where b(h) = h^2, c(h) = (n-1+h)^2, n>=1, h>=1, and = convolution.

Original entry on oeis.org

1, 8, 4, 34, 25, 9, 104, 88, 52, 16, 259, 234, 170, 89, 25, 560, 524, 424, 280, 136, 36, 1092, 1043, 899, 674, 418, 193, 49, 1968, 1904, 1708, 1384, 984, 584, 260, 64, 3333, 3252, 2996, 2555, 1979, 1354, 778, 337, 81, 5368, 5268, 4944, 4368, 3584

Offset: 1

Clark Kimberling, Jun 16 2012

Principal diagonal: A213546.
Antidiagonal sums: A213547.
Row 1, (1,4,9,...)**(1,4,9,...): A033455.
Row 2, (1,4,9,...)**(4,9,16,...): (k^5 + 10*k^4 + 40*k^3 + 50*k^2 +19*k)/30.
Row 3, (1,4,9,...)**(9,16,25,...):  (k^5 + 15*k^4 + 90*k^3 + 120*k^2+44*k)/30.
For a guide to related arrays, see A213500.

			Northwest corner (the array is read by falling antidiagonals):
1....8.....34....104...259....560
4....25....88....234...524....1043
9....52....170...424...899....1708
16...89....280...674...1384...2555
25...136...418...984...1979...3584
...
T(5,1) = (1)**(25) = 25
T(5,2) = (1,4)**(25,36) = 1*36+4*25 = 136
T(5,3) = (1,4,9)**(25,36,49) = 1*49+4*36+9*25 = 418

Clark Kimberling, Antidiagonals n = 1..60, flattened
Henri Muehle, Proper Mergings of Stars and Chains are Counted by Sums of Antidiagonals in Certain Convolution Arrays -- The Details, arXiv preprint arXiv:1301.1654, 2013.

Cf. A213500.

b[n_] := n^2; c[n_] := n^2
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213505 *)
d = Table[t[n, n], {n, 1, 40}] (* A213546 *)
s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A213547 *)

T(n,k) = 6*T(n,k-1) - 15*T(n,k-2) + 20*T(n,k-3) - 15*T(n,k-4) + 6*T(n,k-5) - T(n,k-6).

G.f. for row n: f(x)/g(x), where f(x) = n^2 - (n^2 - 2*n - 1)*x - (n^2 - 2)*x^2 - ((n - 1)^2)*x^3 and g(x) = (1 - x)^6.

A271662 Convolution of nonzero pentagonal numbers (A000326) with themselves.

Original entry on oeis.org

1, 10, 49, 164, 434, 980, 1974, 3648, 6303, 10318, 16159, 24388, 35672, 50792, 70652, 96288, 128877, 169746, 220381, 282436, 357742, 448316, 556370, 684320, 834795, 1010646, 1214955, 1451044, 1722484, 2033104, 2387000, 2788544, 3242393, 3753498, 4327113, 4968804, 5684458

Offset: 0

Ilya Gutkovskiy, Apr 12 2016

More generally, the ordinary generating function for the convolution of nonzero k-gonal numbers with themselves is (1 + (k - 3)*x)^2/(1 - x)^6.

G. C. Greubel, Table of n, a(n) for n = 0..1000
OEIS Wiki, Figurate numbers
Eric Weisstein's World of Mathematics, Pentagonal Number
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1)

Cf. A000326.

Cf. similar sequences of the convolution of k-gonal numbers with themselves: A000389 (k=3, without zeros), A033455 (k=4), this sequence (k=5), A271870 (k=6).

/* From definition: */ P:=func; /*, where P(n,k) is the n-th k-gonal number, */ [&+[P(n+1-i,5)*P(i,5): i in [1..n]]: n in [1..40]]; // Bruno Berselli, Apr 13 2016

[(n+1)*(n+2)*(n+3)*(9*n^2+21*n+20)/120: n in [0..40]]; // Bruno Berselli, Apr 13 2016

LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 10, 49, 164, 434, 980}, 40]
Table[(n + 1) (n + 2) (n + 3) (9 n^2 + 21 n + 20)/120, {n, 0, 40}]
With[{nmax = 50}, CoefficientList[Series[(120 + 1080*x + 1800*x^2 + 920*x^3 + 165*x^4 + 9*x^5)*Exp[x]/120, {x, 0, nmax}], x]*Range[0, nmax]!] (* G. C. Greubel, Jun 07 2017 *)

vector(40, n, n--; (n+1)*(n+2)*(n+3)*(9*n^2+21*n+20)/120) \\ Altug Alkan, Apr 12 2016

O.g.f.: (1 + 2*x)^2/(1 - x)^6.
E.g.f.: (120 + 1080*x + 1800*x^2 + 920*x^3 + 165*x^4 + 9*x^5)*exp(x)/120.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
a(n) = (n + 1)*(n + 2)*(n + 3)*(9*n^2 + 21*n + 20)/120.
Sum_{n>=0} 1/a(n) = 1.13108002...

Edited by Bruno Berselli, Apr 13 2016

A349966 a(n) = Sum_{k=0..n} (k * (n-k))^n.

Original entry on oeis.org

1, 0, 1, 16, 418, 17600, 1086979, 92223488, 10292241540, 1462309109760, 257739952352133, 55188518041440256, 14111052911099343782, 4246668467339066589184, 1485904567816768099571207, 598145009954138900489830400

Offset: 0

Seiichi Manyama, Dec 07 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..240

Cf. A033455, A100262, A145216, A165817, A306548, A349964, A349965.

a[0] = 1; a[n_] := Sum[(k*(n - k))^n, {k, 0, n}]; Array[a, 16, 0] (* Amiram Eldar, Dec 07 2021 *)

PARI
```
a(n) = sum(k=0, n, (k*(n-k))^n);
```

a(n) = [x^n] (Sum_{k=0..n} k^n * x^k)^2.

a(n) ~ sqrt(Pi) * n^(2*n + 1/2) / 2^(2*n + 1). - Vaclav Kotesovec, Dec 07 2021

A129717 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k 101's (n >= 0, 0 <= k <= floor((n-1)/2)). A Fibonacci binary word is a binary word having no 00 subword.

Original entry on oeis.org

1, 2, 3, 4, 1, 4, 4, 4, 8, 1, 4, 12, 5, 4, 16, 13, 1, 4, 20, 25, 6, 4, 24, 41, 19, 1, 4, 28, 61, 44, 7, 4, 32, 85, 85, 26, 1, 4, 36, 113, 146, 70, 8, 4, 40, 145, 231, 155, 34, 1, 4, 44, 181, 344, 301, 104, 9, 4, 48, 221, 489, 532, 259, 43, 1, 4, 52, 265, 670, 876, 560, 147, 10, 4, 56

Offset: 0

Emeric Deutsch, May 12 2007

Row n has 1+floor((n-1)/2) terms for n >= 1.
Row sums are the Fibonacci numbers (A000045).
T(n,1) = A008574(n-3).
T(n,2) = A001844(n-5).
T(n,3) = A005900(n-6).
T(n,4) = A006325(n-7).
T(n,5) = A033455(n-10).
T(n,k) = A129718(n,k+1) (since in each word: 1 + the number of 101's = number of runs of 1's).
Sum_{k>=0} k*T(n,k) = A004798(n-2).

			T(6,2)=5 because we have 110101, 101101, 101010, 101011 and 010101.
Triangle starts:
  1;
  2;
  3;
  4,  1;
  4,  4;
  4,  8,  1;
  4, 12,  5;

Michael De Vlieger, Table of n, a(n) for n = 0..10100 (rows 0 <= n <= 200, flattened.)

Cf. A000045, A008574, A001844, A005900, A006325, A033455, A129718, A004798.

T:=proc(n,k) if n=0 and k=0 then 1 elif n=1 and k=0 then 2 elif n=2 and k=0 then 3 elif n=3 and k=1 then 1 elif k

MapAt[{0, 1} + # &, #, 4] /. {} -> {1} &@ Table[If[n < 3, n + 1, Binomial[n - k - 1, k] + 2 Binomial[n - k - 2, k] + Binomial[n - k - 3, k]], {n, 0, 17}, {k, 0, Floor[(n - 1)/2]}] // Flatten (* Michael De Vlieger, Nov 15 2019 *)

G.f.: G(t,z) = (1+z)*(1 + z^2 - t*z^2)/(1 - z - t*z^2).
G.f. of col 0: (1+z)(1+z^2)/(1-z), leading to the partial sums of 1,1,1,1,0,0,0,...
G.f. of col k: z^(2k+1)*(1+z)^2/(1-z)^(k+1) (k >= 1).
T(n,k) = binomial(n-k-1, k) + 2*binomial(n-k-2, k) + binomial(n-k-3, k) for n >= 4 and 0 <= k < n/2.

A202241 Array F(n,m) read by antidiagonals: F(0,m)=1, F(n,0) = A130713(n), and column m+1 is recursively defined as the partial sums of column m.

Original entry on oeis.org

1, 2, 1, 1, 3, 1, 0, 4, 4, 1, 0, 4, 8, 5, 1, 0, 4, 12, 13, 6, 1, 0, 4, 16, 25, 19, 7, 1, 0, 4, 20, 41, 44, 26, 8, 1, 0, 4, 24, 61, 85, 70, 34, 9, 1, 0, 4, 28, 85, 146, 155, 104, 43, 10, 1, 0, 4, 32, 113, 231, 301, 259, 147, 53, 11, 1, 0, 4, 36, 145, 344, 532, 560, 406, 200, 64, 12, 1

Offset: 0

Paul Curtz, Dec 16 2011

The array F(n,m), beginning with row n=0, is:
  1, 1,  1,  1,   1,   1,    1,
  2, 3,  4,  5,   6,   7,    8,
  1, 4,  8, 13,  19,  26,   34,
  0, 4, 12, 25,  44,  70,  104,
  0, 4, 16, 41,  85, 155,  259,
  0, 4, 20, 61, 146, 301,  560,
  0, 4, 24, 85, 231, 532, 1092.
Columns after A130713, A113311, A008574 have signatures (3,-3,1), (4,-6,4,-1), (5,-10,10,-5,1), (6,-15,20,-15,6,-1) (from A135278(n+3)).
Inserting columns of zeros and pushing the columns down, plus alternating sign switches defines the following triangle T(n,2m) = (-1)^(m/2)*F(n-2m,m):
  1,
  2 0,
  1 0 -1,
  0 0 -3 0,
  0 0 -4 0 1,
  0 0 -4 0 4 0,
  0 0 -4 0 8 0 -1
The row sums in the triangle are (-1)^n*A099838(n).
The companion to A201863 is
  1
  1 0
  1 0   0
  1 0  -2 0
  1 0  -4 0  1
  1 0  -6 0  5 0
  1 0  -8 0 13 0   -2
  1 0 -10 0 25 0  -12 0
  1 0 -12 0 41 0  -38 0   4
  1 0 -14 0 61 0  -88 0  28 0
  1 0 -16 0 85 0 -170 0 104 0 -8
5th column: A001844; 7th column: -A035597=-2*A005900(n+1); 9th column: 4*A006325(n+2); 11th column: -8*(1,8,34,104) (from columns 4,5,6,7 of F(n,m)).
As a triangular array, this is the Riordan array ((1+x)^2, x/(1-x)). - Philippe Deléham, Feb 21 2012

			Triangle T(n,k) begins:
  1
  2, 1
  1, 3,  1
  0, 4,  4,  1
  0, 4,  8,  5,   1
  0, 4, 12, 13,   6,   1
  0, 4, 16, 25,  19,   7,   1
  0, 4, 20, 41,  44,  26,   8,  1
  0, 4, 24, 61,  85,  70,  34,  9,  1
  0, 4, 28, 85, 146, 155, 104, 43, 10, 1
- _Philippe Deléham_, Feb 21 2012

Muniru A Asiru, Table of n, a(n) for n = 0..5151

Cf. A130713 (column 0), A113311 (column 1), A008574 (column 2), A001844 (column 3), A005900 (column 4), A006325 (column 5), A033455 (column 6).

Cf. A267633.

Flat(List([0..12],n->List([0..n],k->Binomial(n,n-k)+Binomial(n-1,n-k-1)-Binomial(n-2,n-k-2)-Binomial(n-3,n-k-3)))); # Muniru A Asiru, Mar 22 2018

A130713 := proc(n)
    if n <= 2 and n >=0 then
        op(n+1,[1,2,1]) ;
    else
        0;
    end if;
end proc:
A202241 := proc(n,m)
    option remember;
    if n < 0 then
        0 ;
    elif m = 0 then
        A130713(n);
    else
        procname(n,m-1)+procname(n-1,m) ;
    end if;
end proc:
for d from 0 to 12 do
    for m from 0 to d do
        printf("%d,",A202241(d-m,m)) ;
    end do:
end do: # R. J. Mathar, Dec 22 2011
C := proc (n, k) if 0 <= k and k <= n then factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if end proc:
for n from 0 to 10 do
     seq(C(n, n-k) + C(n-1, n-k-1) - C(n-2, n-k-2) - C(n-3, n-k-3), k = 0..n);
end do; # Peter Bala, Mar 20 2018

rows = 12;
T[0] = PadRight[{1, 2, 1}, rows];
T[n_ /; nJean-François Alcover, Jun 29 2019 *)

def Trow(n): return [binomial(n, n-k) + binomial(n-1, n-k-1) - binomial(n-2, n-k-2) - binomial(n-3, n-k-3) for k in (0..n)]
for n in (0..9): print(Trow(n)) # Peter Luschny, Mar 21 2018

F(1,m) = m+2.
F(2,m) = A034856(m+1).
F(3,m) = A000297(m-1).
Sum_{m=0..d} F(d-m,m) = A116453(d-3), d >= 3 (antidiagonal sums).
As a triangular array T(n,k), 0 <= k <= n, satisfies: T(n,k) = T(n-1,k) + T(n-1,k-1) with T(0,0) = 1, T(1,0) = 2, T(2,0) = 1, T(3,0) = 0. - Philippe Deléham, Feb 21 2012
Unsigned diagonals of A267633 (beginning with its main diagonal) appear to be the reverse rows of this entry's triangle beginning with the fourth row. - Tom Copeland, Jan 26 2016
T(n,k) = C(n, n-k) + C(n-1, n-k-1) - C(n-2, n-k-2) - C(n-3, n-k-3), where C(n, k) = n!/(k!*(n-k)!) if 0 <= k <= n, otherwise 0. - Peter Bala, Mar 20 2018

cp's OEIS Frontend

A213500 Rectangular array T(n,k): (row n) = b**c, where b(h) = h, c(h) = h + n - 1, n >= 1, h >= 1, and ** = convolution.

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A001249 Squares of tetrahedral numbers: a(n) = binomial(n+3,n)^2.

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A061167 a(n) = n^5 - n.

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A098359 Multiplication table of the square numbers read by antidiagonals.

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A130713 a(0)=a(2)=1, a(1)=2, a(n)=0 for n > 2.

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A213505 Rectangular array: (row n) = b**c, where b(h) = h^2, c(h) = (n-1+h)^2, n>=1, h>=1, and ** = convolution.

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A271662 Convolution of nonzero pentagonal numbers (A000326) with themselves.

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A213500 Rectangular array T(n,k): (row n) = bc, where b(h) = h, c(h) = h + n - 1, n >= 1, h >= 1, and = convolution.

A213505 Rectangular array: (row n) = bc, where b(h) = h^2, c(h) = (n-1+h)^2, n>=1, h>=1, and = convolution.