A004524 -id:A004524 - cp's OEIS frontend

A093392 [n/25] + [(n+1)/25] + [(n+2)/25] + [(n+3)/25] + [(n+4)/25].

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 7, 8, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 12, 13, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15

Offset: 0

Reinhard Zumkeller, Mar 28 2004

nonn

Conjectured g.f. confirmed with more terms similar to A093390, A093391, A093393.

Ray Chandler, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1).

Cf. A004524, A093390, A093391, A093393.

Table[Total[Table[Floor[(n+d)/25],{d,0,4}]],{n,0,90}] (* Harvey P. Dale, Sep 04 2024 *)

Empirical g.f.: x^21 / ((x-1)^2*(x^20+x^15+x^10+x^5+1)). - Colin Barker, Apr 01 2013

A093393 [n/9] + [n/4] + [(n+1)/9] + [(n+1)/4] + [(n+2)/9].

Original entry on oeis.org

0, 0, 0, 1, 2, 2, 2, 4, 6, 7, 7, 8, 9, 9, 9, 10, 12, 13, 14, 15, 16, 16, 16, 17, 18, 19, 20, 22, 23, 23, 23, 24, 25, 25, 26, 28, 30, 30, 30, 31, 32, 32, 32, 34, 36, 37, 37, 38, 39, 39, 39, 40, 42, 43, 44, 45, 46, 46, 46, 47, 48, 49, 50, 52, 53, 53, 53, 54, 55, 55, 56, 58, 60, 60

Offset: 0

Reinhard Zumkeller, Mar 28 2004

nonn

a(n) = A004524(n) + A093390(n).

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-2,1,1,-2,1,1,-2,2,-1).

Table[Total[Floor[(n+{0,1,2})/9]]+Total[Floor[(n+{0,1})/4]],{n,0,80}] (* or *) LinearRecurrence[{2,-2,1,1,-2,1,1,-2,2,-1},{0,0,0,1,2,2,2,4,6,7},80] (* Harvey P. Dale, Dec 01 2024 *)

G.f.:(x^3*(2*x^6+x^4+x^3+1))/((x^2+1)*(x^6+x^3+1)*(x-1)^2) [From Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009]

A092038 a(n+1) = a(n) + (a(n) mod 2)^(n mod a(n)), a(1) = 1.

Original entry on oeis.org

1, 2, 2, 3, 4, 4, 4, 5, 6, 6, 6, 7, 8, 8, 8, 9, 10, 10, 10, 11, 12, 12, 12, 13, 14, 14, 14, 15, 16, 16, 16, 17, 18, 18, 18, 19, 20, 20, 20, 21, 22, 22, 22, 23, 24, 24, 24, 25, 26, 26, 26, 27, 28, 28, 28, 29, 30, 30, 30, 31, 32, 32, 32, 33, 34, 34, 34, 35, 36, 36, 36, 37, 38, 38

Offset: 1

Reinhard Zumkeller, Mar 27 2004

nonn

a(n) = floor(n/4) + floor((n-1)/4) + 2 for n>1.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A004524.

a092038 n = a092038_list !! (n-1)
a092038_list = 1 : zipWith (\u v -> v + (v `mod` 2) ^ (u `mod` v))
                           [2..] a092038_list
-- Reinhard Zumkeller, Oct 10 2012

Join[{1},Table[Floor[n/4]+Floor[(n-1)/4]+2,{n,2,80}]] (* Harvey P. Dale, Oct 03 2012 *)

A158856 Triangle T(n, k) = coefficients of p(n, x), where p(n, x) = (1 - x^(2+floor((n-1)/2)))(1 + (-1)^floor(n/2)x^(1+floor(n/2))), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 0, -1, 1, 0, 0, -1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, -1, 0, -1, 1, 0, 1, 0, 0, -1, 0, -1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, -1, 0, -1, 0, -1

Offset: 0

Roger L. Bagula, Mar 28 2009

			Triangle begins as:
  1;
  1,  1;
  1,  0, -1;
  1,  0,  0, -1;
  1,  0,  1,  0,  1;
  1,  0,  1,  1,  0,  1;
  1,  0,  1,  0, -1,  0, -1;
  1,  0,  1,  0,  0, -1,  0, -1;
  1,  0,  1,  0,  1,  0,  1,  0,  1;
  1,  0,  1,  0,  1,  1,  0,  1,  0,  1;
  1,  0,  1,  0,  1,  0, -1,  0, -1,  0, -1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A004524, A154957.

p[x_, n_]= (1-x^(2+Floor[(n-1)/2]))*(1+(-1)^Floor[n/2]*x^(1+Floor[n/2]))/(1 - x^2);
Table[CoefficientList[p[x, n], x], {n,0,12}]//Flatten (* modified by G. C. Greubel, Mar 07 2022 *)

def p(n,x): return (1-x^(2+((n-1)//2)))*(1+(-1)^(n//2)*x^(1+(n//2)))/(1-x^2)
def A158856(n,k): return ( p(n,x) ).series(x, n+1).list()[k]
flatten([[A158856(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 07 2022

T(n, k) = coefficients of p(n, x), where p(n, x) = (Sum_{j=0..1+floor((n-1)/2)} x^j)*(Sum_{i=0..floor(n/2)} (-x)^i) and p(0, x) = 1.
From G. C. Greubel, Mar 07 2022: (Start)
T(n, k) = coefficients of p(n, x), where p(n, x) = (1 - x^(2+floor((n-1)/2)))*(1 + (-1)^floor(n/2)*x^(1+floor(n/2))).
Sum_{k=0..n} T(n, k) = floor((n+3)/2)*( (1 + floor(n/2)) mod 2 ).
Sum_{k=0..n} abs(T(n, k)) = A004524(n+3).
T(2*n, n) = (1 + (-1)^n)/2.
T(2*n+1, n) = (1 + (-1)^n)/2.
Sum_{k=0..floor(n/2)} T(n, k) = floor((n+4)/4).
T(n, k) = abs(A154957(n,k)). (End)

Edited by G. C. Greubel, Mar 07 2022

A160636 Hankel transform of A114464.

Original entry on oeis.org

1, 0, -1, -2, -8, 0, 128, 1024, 16384, 0, -4194304, -134217728, -8589934592, 0, 35184372088832, 4503599627370496, 1152921504606846976, 0, -75557863725914323419136, -38685626227668133590597632

Offset: 0

Paul Barry, May 21 2009

Hankel transform of A114464(n+1) is A160637.

G. C. Greubel, Table of n, a(n) for n = 0..115

Cf. A004524, A160637.

R:= RealField(); [Round(2^Floor(Binomial(n,2)/2)*((Sqrt(2)/2 -1/2)*Sin(3*Pi(R)*n/4+Pi(R)/4)+(Sqrt(2)/2+1/2)*Cos(Pi(R)*n/4+Pi(R)/4))): n in [0..50]]; // G. C. Greubel, May 03 2018

Table[Round[2^Floor[Binomial[n, 2]/2]*((Sqrt[2]-1)*Sin[(3*n+1)*Pi/4]/2 + (Sqrt[2]+1)*Cos[(n+1)*Pi/4]/2)], {n, 0, 50}] (* G. C. Greubel, May 03 2018 *)
a[ n_] := -Sign[Mod[n - 1, 4]]*(-1)^Quotient[n - 1, 4]*2^Quotient[n (n - 1), 4]; (* Michael Somos, Mar 14 2020 *)

for(n=0,50, print1(round(2^floor(binomial(n,2)/2)*((sqrt(2)-1)*sin((3*n+1)*Pi/4)/2 +(sqrt(2)+1)*cos((n+1)*Pi/4)/2)), ", ")) \\ G. C. Greubel, May 03 2018

A160636(n)=if(n%4!=1,(-1)^((n+2)\4)<<(binomial(n,2)\2),0) \\ M. F. Hasler, May 09 2018

a(n) = 2^floor(C(n,2)/2)*((sqrt(2)-1)*sin((3*n+1)*Pi/4)/2 +(sqrt(2)+1)*cos((n+1)*Pi/4)/2).
a(4k+1) = 0, a(n) = (-1)^floor((n+2)/4) * 2^A011848(n) if n !== 1 (mod 4), where A011848(n) = floor(C(n,2)/2). - M. F. Hasler, May 09 2018
a(n) = -a(2-n) * 2^A004524(n) for all n in Z. - Michael Somos, Mar 14 2020

Comment with an incorrect formula deleted by M. F. Hasler, May 09 2018

A166556 Triangle read by rows, A000012 * A047999.

Original entry on oeis.org

1, 2, 1, 3, 1, 1, 4, 2, 2, 1, 5, 2, 2, 1, 1, 6, 3, 2, 1, 2, 1, 7, 3, 3, 1, 3, 1, 1, 8, 4, 4, 2, 4, 2, 2, 1, 9, 4, 4, 2, 4, 2, 2, 1, 1, 10, 5, 4, 2, 4, 2, 2, 1, 2, 1, 11, 5, 5, 2, 4, 2, 2, 1, 3, 1, 1, 12, 6, 6, 3, 4, 2, 2, 1, 4, 2, 2, 1

Offset: 0

Gary W. Adamson, Oct 17 2009

			First few rows of the triangle =
   1;
   2, 1;
   3, 1, 1;
   4, 2, 2, 1;
   5, 2, 2, 1, 1;
   6, 3, 2, 1, 2, 1;
   7, 3, 3, 1, 3, 1, 1;
   8, 4, 4, 2, 4, 2, 2, 1;
   9, 4, 4, 2, 4, 2, 2, 1, 1;
  10, 5, 4, 2, 4, 2, 2, 1, 2, 1;
  11, 5, 5, 2, 4, 2, 2, 1, 3, 1, 1;
  12, 6, 6, 3, 4, 2, 2, 1, 4, 2, 2, 1;
  13, 6, 6, 3, 5, 2, 2, 1, 5, 2, 2, 1, 1;
  ...

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Cf. A000027, A004524, A004526, A047999, A080100.

Sums include: A006046 (row), A007729 (diagonal).

A166556:= func< n,k | (&+[(Binomial(j,k) mod 2): j in [k..n]]) >;
[A166556(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Dec 02 2024

A166556 := proc(n,k)
    local j;
    add(A047999(j,k),j=k..n) ;
end proc: # R. J. Mathar, Jul 21 2016

A166556[n_, k_]:= Sum[Mod[Binomial[j,k], 2], {j,k,n}];
Table[A166556[n,k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 02 2024 *)

def A166556(n,k): return sum(binomial(j,k)%2 for j in range(k,n+1))
print(flatten([[A166556(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Dec 02 2024

Triangle read by rows, A000012 * A047999; where A000012 = an infinite lower triangular matrix with all 1's: [1; 1,1; 1,1,1;..]; and A047999 = Sierpinski's gasket.
The operation takes partial sums of Sierpinski's gasket terms, by columns.
From G. C. Greubel, Dec 02 2024: (Start)
T(n, k) = Sum_{j=k..n} (binomial(j,k) mod 2).
T(n, 0) = A000027(n+1).
T(n, 1) = A004526(n+1).
T(n, 2) = A004524(n+1).
T(2*n, n) = A080100(n).
Sum_{k=0..n} T(n, k) = A006046(n+1).
Sum_{k=0..n} (-1)^k*T(n, k) = A006046(floor(n/2)+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A007729(n). (End)

A187480 Rank transform of the sequence round(n/2); complement of A187481.

Original entry on oeis.org

1, 2, 4, 5, 6, 8, 9, 10, 11, 13, 15, 16, 17, 19, 20, 21, 22, 24, 26, 27, 28, 30, 32, 33, 34, 35, 37, 38, 39, 40, 42, 43, 44, 46, 48, 49, 50, 51, 53, 54, 55, 57, 59, 60, 61, 63, 64, 65, 66, 68, 69, 70, 71, 73, 75, 76, 77, 79, 80, 81, 82, 84, 85, 86, 87, 89, 91, 92, 93, 95, 97, 98, 99, 100, 102, 103, 104, 106, 108, 109, 110, 112, 113, 114, 115, 117, 119, 120, 121, 123, 124, 125

Offset: 1

Clark Kimberling, Mar 10 2011

nonn

See A187224. A187480 is the rank transform of (A004525 with initial two zeros removed).

Cf. A187224, A187481, A004524.

seqA = Table[Round[n/2], {n, 1, 180}]  (* A004524 *)
seqB = Table[n, {n, 1, 80}];           (* A000027 *)
jointRank[{seqA_, seqB_}] := {Flatten@Position[#1, {_, 1}],
Flatten@Position[#1, {_, 2}]} &[Sort@Flatten[{{#1, 1} & /@ seqA,
{#1, 2} & /@ seqB}, 1]];
limseqU = FixedPoint[jointRank[{seqA, #1[[1]]}] &, jointRank[{seqA, seqB}]][[1]]                                     (* A187480 *)
Complement[Range[Length[seqA]], limseqU]  (* A187481 *)
(* by Peter J. C. Moses, Mar 10 2011 *)

A209634 Triangle with (1,4,7,10,13,16...,(3*n-2),...) in every column, shifted down twice.

Original entry on oeis.org

1, 4, 7, 1, 10, 4, 13, 7, 1, 16, 10, 4, 19, 13, 7, 1, 22, 16, 10, 4, 25, 19, 13, 7, 1, 28, 22, 16, 10, 4, 31, 25, 19, 13, 7, 1, 34, 28, 22, 16, 10, 4, 37, 31, 25, 19, 13, 7, 1, 40, 34, 28, 22, 16, 10, 4, 43, 37, 31, 25, 19, 13, 7, 1, 46, 40, 34, 28, 22, 16, 10

Offset: 1

Ctibor O. Zizka, Mar 11 2012

OEIS contains a lot of similar sequences, for example A152204, A122196, A173284.
Row sums for this sequence gives A006578.
In general, by given triangle with (A-B,2*A-B,...,A*n-B,...) in every column, shifted down K-times, we have the row sum s(n)= A*(n*n+K*n+nmodK)/(2*K) - B*(n+nmodK)/K. In this sequence K=2,A=3,B=2, in A152204 K=2,A=2,B=1.
No triangle with primes in every column, shifted down by K>=2 in OEIS, no row sums of it in OEIS.
From Johannes W. Meijer, Sep 28 2013: (Start)
Triangle read by rows formed from antidiagonals of triangle A143971.
The alternating row sums equal A004524(n+2) + 2*A004524(n+1).
The antidiagonal sums equal A171452(n+1). (End)

			Triangle:
1
4
7,  1
10, 4
13, 7,  1
16, 10, 4
19, 13, 7,  1
22, 16, 10, 4
25, 19, 13, 7,  1
28, 22, 16, 10, 4
...

Cf. A008315, A011973, A102541, A122196, A122197, A128099, A152198, A152204, A173284, A207538.

Cf. (Related to triangle sums) A006578, A000217, A002620, A004524, A171452.

T := (n, k) -> 3*n - 6*k + 4: seq(seq(T(n, k), k=1..floor((n+1)/2)), n=1..15); # Johannes W. Meijer, Sep 28 2013

From Johannes W. Meijer, Sep 28 2013: (Start)
T(n, k) = 3*n - 6*k + 4, n >= 1 and 1 <= k <= floor((n+1)/2).
T(n, k) = A143971(n-k+1, k), n >= 1 and 1 <= k <= floor((n+1)/2). (End)

A260056 Irregular triangle read by rows: coefficients T(n, k) of certain polynomials p(n, x) with exponents in increasing order, n >= 0 and 0 <= k <= 2*n.

Original entry on oeis.org

1, 2, 1, 1, 3, 3, 4, 2, 1, 4, 6, 10, 9, 7, 3, 1, 5, 10, 20, 25, 26, 19, 11, 4, 1, 6, 15, 35, 55, 71, 70, 56, 34, 16, 5, 1, 7, 21, 56, 105, 161, 196, 197, 160, 106, 55, 22, 6, 1, 8, 28, 84, 182, 322, 462, 554, 553, 463, 321, 183, 83, 29, 7, 1, 9, 36, 120, 294, 588, 966, 1338, 1569, 1570, 1337, 967, 587, 295, 119, 37, 8, 1, 10, 45, 165, 450, 1002, 1848, 2892, 3873, 4477, 4476, 3874

Offset: 0

Werner Schulte, Nov 08 2015

The triangle is related to the triangle of trinomial coefficients.

			The irregular triangle T(n,k) begins:
n\k:  0   1   2    3    4    5    6    7    8    9   10  11  12  13  14  ...
0     1;
1     2   1   1;
2     3   3   4    2    1;
3     4   6  10    9    7    3    1;
4     5  10  20   25   26   19   11    4    1;
5     6  15  35   55   71   70   56   34   16    5    1;
6     7  21  56  105  161  196  197  160  106   55   22   6   1;
7     8  28  84  182  322  462  554  553  463  321  183  83  29   7   1;
etc.
The polynomial corresponding to row 2 is p(2,x) = 3+3*x+4*x^2+2*x^3+x^4.

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Cf. A000027 (col 0), A000217 (col 1), A000292 (col 2), A001590, A002426, A004524, A005582 (col 3), A008937, A027907, A095662 (col 5), A113682, A246437.

A027907[n_, k_] := Sum[Binomial[n, j]*Binomial[j, k - j], {j, 0, n}]; Table[ Sum[A027907[j, k], {j, 0, n}], {n,0,10}, {k, 0, 2*n} ] // Flatten (* G. C. Greubel, Mar 07 2017 *)

T(n,0) = n+1, and T(n,k) = 0 for k < 0 or k > 2*n, and T(n+1,k) = T(n,k) + T(n,k-1) + T(n,k-2) for k > 0.
T(n,k) = Sum_{j=0..n} A027907(j,k) for 0 <= k <= 2*n.
T(n,k) = Sum_{j=0..k} (-1)^(k-j)*A027907(n+1,j+1) for 0 <= k <= 2*n.
T(n,k) = T(n,2*n-1-k) + (-1)^k for 0 <= k < 2*n.
p(n,x) = Sum_{k=0..2*n} T(n,k)*x^k = Sum_{k=0..n} (1+x+x^2)^k for n >= 0.
p(n,x) = ((1+x+x^2)^(n+1)-1)/(x+x^2), p(n,0) = p(n,-1) = n+1 for n >= 0.
p(n+1,x) = (1+x+x^2)*p(n,x)+1 for n >= 0.
Sum_{n>=0} p(n,x)*t^n = 1/((1-t)*(1-t*(1+x+x^2))).
T(n,2*n) = 1, and T(n,n) = A113682(n) for n >= 0.
T(n,n-1) = A246437(n+1), and T(n,n-1)+T(n,n) = A002426(n+1) for n > 0.
If d(n) is n-th antidiagonal sum of the triangle then: d(n) = A008937(n+1), and d(n+2)-d(n) = A001590(n+5) for n >= 0.
Conjecture: If a(n) is n-th antidiagonal alternating sum of the triangle then: a(n) = A004524(n+3).
Sum_{k=0..2*n} (-1)^k*T(n,k)^2 = (3^(n+1)-1)/2 for n >= 0.
Sum_{k=0..2*n} (-1)^k*(y*k+1)*T(n,k) = Sum{k=0..n} y*k+1 = (n+1)*(y*n+2)/2 for real y and n >= 0.
Conjecture of linear recurrence for column k: Sum_{m=0..k+2} (-1)^m*T(n+m,k)* binomial(k+2,m) = 0 for k >= 0 and n >= 0.

A290561 a(n) = n + cos(n*Pi/2).

Original entry on oeis.org

1, 1, 1, 3, 5, 5, 5, 7, 9, 9, 9, 11, 13, 13, 13, 15, 17, 17, 17, 19, 21, 21, 21, 23, 25, 25, 25, 27, 29, 29, 29, 31, 33, 33, 33, 35, 37, 37, 37, 39, 41, 41, 41, 43, 45, 45, 45, 47, 49, 49, 49, 51, 53, 53, 53, 55, 57, 57, 57, 59, 61, 61, 61, 63, 65, 65, 65

Offset: 0

Jean-François Alcover and Paul Curtz, Aug 06 2017

a(n) divides A289296(n).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Cf. A004524, A085046, A289296, A290562.

A290561:=n->n+cos(n*Pi/2): seq(A290561(n), n=0..150); # Wesley Ivan Hurt, Aug 06 2017

a[n_] := n + Cos[n*Pi/2]; Table[a[n], {n, 0, 60}]

a(n) = n + round(cos(n*Pi/2)); \\ Michel Marcus, Aug 06 2017

Vec((x^3 + x^2 - x + 1)/((x - 1)^2*(x^2 + 1)) + O(x^100)) \\ Colin Barker, Aug 06 2017

G.f.: (x^3 + x^2 - x + 1)/((x - 1)^2*(x^2 + 1)).
a(n) = n if n == 3 (mod 4), and a(n) = a(n-4) + 4 otherwise, for n>2.
a(n) = a(n+20) - 20.
a(n) = 2*A004524(n) + 1.
a(n) + A290562(n) = 2*n.
a(n) * A290562(n) = n^2 - cos(n*Pi/2)^2 = A085046(n) for n>0.
A290562(n) = -a(-n).
From Colin Barker, Aug 06 2017: (Start)
a(n) = ((-i)^n + i^n)/2 + n where i=sqrt(-1).
a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - a(n-4) for n>3. (End)

cp's OEIS Frontend

A093392 [n/25] + [(n+1)/25] + [(n+2)/25] + [(n+3)/25] + [(n+4)/25].

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A093393 [n/9] + [n/4] + [(n+1)/9] + [(n+1)/4] + [(n+2)/9].

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A092038 a(n+1) = a(n) + (a(n) mod 2)^(n mod a(n)), a(1) = 1.

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A158856 Triangle T(n, k) = coefficients of p(n, x), where p(n, x) = (1 - x^(2+floor((n-1)/2)))*(1 + (-1)^floor(n/2)*x^(1+floor(n/2))), read by rows.

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A160636 Hankel transform of A114464.

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A166556 Triangle read by rows, A000012 * A047999.

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A187480 Rank transform of the sequence round(n/2); complement of A187481.

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A209634 Triangle with (1,4,7,10,13,16...,(3*n-2),...) in every column, shifted down twice.

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A158856 Triangle T(n, k) = coefficients of p(n, x), where p(n, x) = (1 - x^(2+floor((n-1)/2)))(1 + (-1)^floor(n/2)x^(1+floor(n/2))), read by rows.