A033994 -id:A033994 - cp's OEIS frontend

A198061 Array read by antidiagonals, m>=0, n>=0, A(m,n) = sum{k=0..n} sum{j=0..m} sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

1, 0, 2, 0, 2, 3, 0, 2, 6, 4, 0, 2, 11, 12, 5, 0, 2, 20, 32, 20, 6, 0, 2, 37, 84, 70, 30, 7, 0, 2, 70, 224, 240, 130, 42, 8, 0, 2, 135, 612, 834, 550, 217, 56, 9, 0, 2, 264, 1712, 2968, 2354, 1092, 336, 72, 10, 0, 2, 521, 4884, 10826, 10310, 5551, 1960, 492

Offset: 0

Peter Luschny, Nov 02 2011

			m\n  [0] [1]  [2]   [3]    [4]     [5]    [6]
----------------------------------------------
[0]   1   2    3     4      5       6       7    A000027
[1]   0   2    6    12     20      30      42    A002378
[2]   0   2   11    32     70     130     217    A033994
[3]   0   2   20    84    240     550    1092    A098077
[4]   0   2   37   224    834    2354    5551
[5]   0   2   70   612   2968   10310   28854

Cf. A198060.

A198061 := proc(m, n) local i,j,k,pow;
pow := (a,b) -> if a=0 and b=0 then 1 else a^b fi;
add(add(add((-1)^(j+i)*binomial(i,j)*pow(n,j)*pow(k,m-j),i=0..m),j=0..m),k=0..n) end:
for m from 0 to 8 do lprint(seq(A198061(m,n), n=0..6)) od;

Unprotect[Power]; 0^0 = 1; Protect[Power]; a[m_, n_] :=  Sum[(-1)^(j+i)*Binomial[i, j]*n^j*k^(m-j) , {i, 0, m}, {j, 0, m}, {k, 0, n}]; Table[a[m-n, n], {m, 0, 10}, {n, 0, m}] // Flatten (* Jean-François Alcover, Jul 26 2013 *)

A198061(n,2) = A006127(n+1)

A174814 a(n) = n(n+1)(5*n+1)/3.

Original entry on oeis.org

0, 4, 22, 64, 140, 260, 434, 672, 984, 1380, 1870, 2464, 3172, 4004, 4970, 6080, 7344, 8772, 10374, 12160, 14140, 16324, 18722, 21344, 24200, 27300, 30654, 34272, 38164, 42340, 46810, 51584, 56672, 62084, 67830, 73920, 80364, 87172, 94354, 101920, 109880

Offset: 0

Bruno Berselli, Dec 01 2010 - Dec 02 2010

Also zero followed by bisection (even part) of A088003.

Numbers ending in 0, 2 or 4 (cf. 2*A053796(n)). Therefore we can easily see that a(m)^(2*k+1)==-1 (mod 5) only for m in A047219, while a(m)^(2*k)==-1 (mod 5) only for m in A016873 and k odd.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

[n*(n+1)*(5*n+1)/3: n in [0..50]]; // Vincenzo Librandi, May 30 2011

Table[1/3 n (1+n) (1+5 n), {n,0,50}]  (* Harvey P. Dale, Feb 25 2011 *)

a(n) = n*(n+1)*(5*n+1)/3 \\ Charles R Greathouse IV, May 30 2011

G.f.: 2*x*(2+3*x)/(1-x)^4.
a(n) = 2*A033994(n) for n>0.
a(n) = n*A147875(n+1)-sum(k=1..n, A147875(k)) for n>0.
a(-n) = -A144945(n).

A212501 Number of (w,x,y,z) with all terms in {1,...,n} and w > x < y >= z.

Original entry on oeis.org

0, 0, 2, 13, 45, 115, 245, 462, 798, 1290, 1980, 2915, 4147, 5733, 7735, 10220, 13260, 16932, 21318, 26505, 32585, 39655, 47817, 57178, 67850, 79950, 93600, 108927, 126063, 145145, 166315, 189720, 215512, 243848, 274890, 308805

Offset: 0

Clark Kimberling, May 19 2012

For a guide to related sequences, see A211795.

Partial sums of A033994. - J. M. Bergot, Jun 14 2013

			a(8)=798 which results from the following: 1*(8+9+10+11+12+13+14) + 2*(8+9+10+11+12+13) + 3*(8+9+10+11+12) + 4*(8+9+10+11) + 5*(8+9+10) + 6*(8+9) + 7*(8) = 798 = 77+126+150+152+135+102+56. - _J. M. Bergot_, Aug 23 2022

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A211795.

A212501:=n->(n-1)*n*(n+1)*(5*n-2)/24: seq(A212501(n), n=0..60); # Wesley Ivan Hurt, Oct 07 2017

t = Compile[{{n, _Integer}}, Module[{s = 0},
(Do[If[w > x < y >= z, s = s + 1],
{w, 1, #}, {x, 1, #}, {y, 1, #}, {z, 1, #}] &[n]; s)]];
Map[t[#] &, Range[0, 40]]   (* A212501 *)
LinearRecurrence[{5,-10,10,-5,1},{0,0,2,13,45},50] (* Harvey P. Dale, May 01 2023 *)

a(n)=n*(n-1)*(n+1)*(5*n-2)/24 \\ Charles R Greathouse IV, Jun 14 2013

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 4.
G.f.: x^2*(2+3*x)/(1-x)^5. - Bruno Berselli, May 31 2012
a(n) = (n-1)*n*(n+1)*(5*n-2)/24. - Bruno Berselli, May 31 2012

A337574 a(n) is the dot product of the vectors of the first n primes and the next n primes.

Original entry on oeis.org

6, 31, 112, 279, 652, 1231, 2140, 3363, 5132, 7647, 10600, 14583, 19754, 25435, 31894, 40617, 50866, 62583, 76174, 91431, 108124, 127319, 147868, 172493, 200392, 230281, 262140, 297413, 334756, 374607, 419958, 471113, 524892, 583853, 649458, 717339, 790760, 868997, 951672, 1039871, 1134792

Offset: 1

Robert Israel, Sep 01 2020

nonn

			a(3) = 2*7 + 3*11 + 5*13 = 112.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000040, A337573 (n such that a(n) is prime).

Cf. A033994 (similar when prime(n) is replaced with n).

P:= :
[seq(P[1..t]^%T . P[t+1..2*t],t=1..250)];

Table[Prime[Range[n]].Prime[Range[n+1,2n]],{n,50}] (* Harvey P. Dale, Mar 30 2024 *)

a(n) = sum(k=1, n, prime(k)*prime(n+k)); \\ Michel Marcus, Sep 02 2020

a(n) = Sum_{k=1..n} prime(k)*prime(n+k).

A257093 a(n) = n(n+1)(13*n+2)/6.

Original entry on oeis.org

0, 5, 28, 82, 180, 335, 560, 868, 1272, 1785, 2420, 3190, 4108, 5187, 6440, 7880, 9520, 11373, 13452, 15770, 18340, 21175, 24288, 27692, 31400, 35425, 39780, 44478, 49532, 54955, 60760, 66960, 73568, 80597, 88060, 95970, 104340, 113183, 122512, 132340

Offset: 0

Luce ETIENNE, Apr 16 2015

This sequence gives the number of triangles of all sizes in (5*n^2)-polyiamonds in a tetragonal or hexagonal or heptagonal configuration.
It is the sum of (1/2)*Sum_{j=0..n-1} (n-j)*(5*n+1-j) triangles oriented in one direction and (1/2)*Sum_{j-0..n-1} (n-j)*(5*n-1-3*j) oriented in the opposite direction.
Shäfli's notation: 3.3.3.3.3 for a(1).
The difference between this sequence and A050409(n) equals A000292(n-1).
Also, (1/3)*(A002717(2*n) + A255211(n) - 2*A000330(n)) gives A033994(n): a (5*n^2)-polyiamond in pentagonal configuration that does not belong to this sequence because a(1)=6.
a(n) is odd only when n mod 4 = 1.

			Second comment a(0) = 0; a(1) = 3 + 2; a(2) = 16 + 12; a(3) = 46 + 36; a(4) = 100 + 80; a(5) = 185 + 150; a(6) = 308 + 252.

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

Cf. A002411, A011379, A033429, A050409, A000292, A002717, A000330, A033994, A255211.

[n*(n+1)*(13*n+2)/6: n in [0..40]]; // Vincenzo Librandi, Apr 16 2015

Table[n (n + 1) (13 n + 2)/6, {n, 0, 40}] (* Vincenzo Librandi, Apr 16 2015 *)
CoefficientList[Series[x (5+8x)/(1-x)^4,{x,0,50}],x] (* or *) LinearRecurrence[{4,-6,4,-1},{0,5,28,82},60] (* Harvey P. Dale, Feb 12 2023 *)

a(n) = Sum_{j=0..n-1} (n-j)*(5*n-2*j).
From Vincenzo Librandi, Apr 16 2015: (Start)
G.f.: x*(5+8*x)/(1-x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). (End)
E.g.f.: exp(x)*x*(30 + 54*x + 13*x^2)/6. - Stefano Spezia, Mar 02 2025

Corrected by Harvey P. Dale, Feb 12 2023

cp's OEIS Frontend

A198061 Array read by antidiagonals, m>=0, n>=0, A(m,n) = sum{k=0..n} sum{j=0..m} sum{i=0..m} (-1)^(j+i)*C(i,j)*n^j*k^(m-j).

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A174814 a(n) = n*(n+1)*(5*n+1)/3.

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A212501 Number of (w,x,y,z) with all terms in {1,...,n} and w > x < y >= z.

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A337574 a(n) is the dot product of the vectors of the first n primes and the next n primes.

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Examples

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Programs

Maple

Mathematica

PARI

Formula

A257093 a(n) = n*(n+1)*(13*n+2)/6.

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Magma

Mathematica

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A198061 Array read by antidiagonals, m>=0, n>=0, A(m,n) = sum{k=0..n} sum{j=0..m} sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

A174814 a(n) = n(n+1)(5*n+1)/3.

A257093 a(n) = n(n+1)(13*n+2)/6.