A027170 -id:A027170 - cp's OEIS frontend

A027179 a(n) = T(n,0) + T(n,1) + ... + T(n,[ n/2 ]), T given by A027170.

1, 3, 15, 26, 81, 130, 350, 558, 1417, 2282, 5632, 9190, 22296, 36834, 88280, 147422, 349929, 589786, 1388672, 2359254, 5516356, 9437138, 21931190, 37748686, 87250096, 150994890, 347306730, 603979718, 1383135310, 2415919042

Offset: 0

Clark Kimberling

nonn

A027182 a(n) = self-convolution of row n of array T given by A027170.

Original entry on oeis.org

18, 150, 820, 3726, 15492, 61660, 240328, 928110, 3572692, 13749580, 52977216, 204478276, 790747504, 3063728856, 11891504944, 46231328174, 180005765748, 701823431740, 2739725128448, 10707198384348, 41888225645152, 164027934617024

Offset: 0

Clark Kimberling

nonn

A027184 a(n) = (1/2)*(n-th largest even number in array T given by A027170).

Original entry on oeis.org

5, 15, 21, 29, 38, 47, 69, 78, 95, 125, 132, 159, 181, 197, 236, 239, 285, 335, 370, 389, 395, 447, 492, 509, 523, 553, 575, 645, 719, 797, 874, 879, 895, 965, 1035, 1048, 1055, 1109, 1149, 1247, 1349, 1450, 1455, 1506, 1565, 1633, 1679, 1797, 1874, 1919, 1945, 2000

Offset: 1

Clark Kimberling

nonn

a(n) = A027183(n) / 2. - Sean A. Irvine, Oct 24 2019

Corrected and extended by Sean A. Irvine, Oct 24 2019

A026907 Triangular array T read by rows (9-diamondization of Pascal's triangle). Step 1: t(n,k) = sum of 9 entries in diamond-shaped subarray of Pascal's triangle having vertices C(n,k), C(n+4,k+2), C(n+2,k), C(n+2,k+2). Step 2: T(n,k) = t(n,k) - t(0,0) + 1.

Original entry on oeis.org

1, 13, 13, 28, 44, 28, 46, 90, 90, 46, 67, 154, 198, 154, 67, 91, 239, 370, 370, 239, 91, 118, 348, 627, 758, 627, 348, 118, 148, 484, 993, 1403, 1403, 993, 484, 148, 181, 650, 1495, 2414, 2824, 2414, 1495, 650, 181, 217, 849, 2163, 3927, 5256, 5256, 3927, 2163, 849, 217

Offset: 0

Clark Kimberling

			Triangle starts:
   1;
  13,  13;
  28,  44,  28;
  46,  90,  90,  46;
  67, 154, 198, 154,  67;
  91, 239, 370, 370, 239,  91;
  ...

Indranil Ghosh, Rows 0..125, flattened

Cf. A007318, A027170.
Sums: A026915 (row), A026916, A026917, A026918 (diagonal).
Diagonals: A026908, A026909, A026910, A026911, A026912, A026913, A026914.

A026907:= func< n,k | Binomial(n,k) + 3*Binomial(n+4,k+2) - 18 >;
[A026907(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 21 2025

t[n_, k_]:=Binomial[n + 4, k + 2 ] + Binomial[n + 3, k + 1] + Binomial[n + 3, k + 2] + Binomial[n + 2, k] + Binomial[n + 2, k + 1] + Binomial[n + 2, k + 2] + Binomial[n + 1, k] + Binomial[n + 1, k + 1] + Binomial[n, k] ; T[n_, k_]:=t[n,k] - t[0, 0] + 1; Flatten[Table[T[n, k], {n, 0, 9},{k, 0, n}]] (* Indranil Ghosh, Mar 13 2017 *)

alias(C, binomial);
t(n,k) = C(n+4,k+2) + C(n+3,k+1) + C(n+3,k+2) + C(n+2,k) + C(n+2,k+1) + C(n+2,k+2) + C(n+1,k) + C(n+1,k+1) + C(n,k);
T(n,k) = t(n,k)-t(0,0)+1;
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print());
\\ Michel Marcus, Mar 13 2017

def A026907(n,k): return binomial(n,k) +3*binomial(n+4,k+2) -18
print(flatten([[A026907(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Aug 21 2025

From G. C. Greubel, Aug 21 2025: (Start)
T(n, k) = binomial(n,k) + 3*binomial(n+4, k+2) - 18.
Sum_{k=0..n} (-1)^k*T(n, k) = 6*(1+(-1)^n)* floor((n+1)/2) + [n=0]. (End)

A027181 a(n) = Lucas(n+4) - 2*(n+3).

Original entry on oeis.org

1, 3, 8, 17, 33, 60, 105, 179, 300, 497, 817, 1336, 2177, 3539, 5744, 9313, 15089, 24436, 39561, 64035, 103636, 167713, 271393, 439152, 710593, 1149795, 1860440, 3010289, 4870785, 7881132, 12751977, 20633171, 33385212, 54018449, 87403729, 141422248

Offset: 0

Clark Kimberling

Let F be a homogeneous polynomial in n + 4 variables f0, f1, f2, g0, g1, ..., gn, defined as the determinant of a Sylvester matrix of polynomials f2*x^2 + f1*x + f0 and gn*x^n + ... + g1*x + g0. It appears that a(n) is equal to the l1-norm of F, i.e., the sum of absolute values of coefficients of F. - Anton Mosunov, Apr 13 2019

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Cf. A000032, A000045, A027170, A192978.

List([0..40], n-> Lucas(1,-1,n+4)[2] -2*(n+3)); # G. C. Greubel, Jul 24 2019

[Lucas(n+4) - (2*n+6): n in [0..40]]; // Vincenzo Librandi, Apr 16 2011

LinearRecurrence[{3,-2,-1,1},{1,3,8,17},40] (* Vladimir Joseph Stephan Orlovsky, Jan 25 2012 *)
Table[LucasL[n+4]-2*(n+3), {n,0,40}] (* G. C. Greubel, Jul 24 2019 *)

Vec((1 + x^2)/((1 - x)^2*(1 - x - x^2)) + O(x^40)) \\ Colin Barker, Mar 10 2017

vector(40, n, n--; f=fibonacci; f(n+5)+f(n+3)-2*(n+3)) \\ G. C. Greubel, Jul 24 2019

[lucas_number2(n+4,1,-1) -2*(n+3) for n in range(40)] # G. C. Greubel, Apr 14 2019

a(n) = Sum_{k=0..floor(n/2)} A027170(n-k, k).
G.f.: (1 + x^2)/((1 - x)^2*(1 - x - x^2)).
From Colin Barker, Mar 10 2017: (Start)
a(n) = -4 + (2^(-1-n)*((1-sqrt(5))^n*(-15+7*sqrt(5)) + (1+sqrt(5))^n*(15+7*sqrt(5))))/sqrt(5) - 2*(1+n).
a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4) for n>3.
(End)

cp's OEIS Frontend

A027179 a(n) = T(n,0) + T(n,1) + ... + T(n,[ n/2 ]), T given by A027170.

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A027182 a(n) = self-convolution of row n of array T given by A027170.

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A027184 a(n) = (1/2)*(n-th largest even number in array T given by A027170).

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Extensions

A026907 Triangular array T read by rows (9-diamondization of Pascal's triangle). Step 1: t(n,k) = sum of 9 entries in diamond-shaped subarray of Pascal's triangle having vertices C(n,k), C(n+4,k+2), C(n+2,k), C(n+2,k+2). Step 2: T(n,k) = t(n,k) - t(0,0) + 1.

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Magma

Mathematica

PARI

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Formula

A027181 a(n) = Lucas(n+4) - 2*(n+3).

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GAP

Magma

Mathematica

PARI

PARI

Sage

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