A017089 -id:A017089 - cp's OEIS frontend

A278545 Number of neighbors of the n-th term in a full square array read by antidiagonals.

3, 5, 5, 5, 8, 5, 5, 8, 8, 5, 5, 8, 8, 8, 5, 5, 8, 8, 8, 8, 5, 5, 8, 8, 8, 8, 8, 5, 5, 8, 8, 8, 8, 8, 8, 5, 5, 8, 8, 8, 8, 8, 8, 8, 5, 5, 8, 8, 8, 8, 8, 8, 8, 8, 5, 5, 8, 8, 8, 8, 8, 8, 8, 8, 8, 5, 5, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 5, 5, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 5, 5, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 5

Offset: 1

Omar E. Pol, Nov 23 2016

Apart from the first row and the first column, the rest of the elements are 8's.

For the same idea but for a right triangle see A278480; for an isosceles triangle see A278481; for a square spiral see A010731; and for a hexagonal spiral see A010722.

			The corner of the square array begins:
3,5,5,5,5,5,5,5,5,5,...
5,8,8,8,8,8,8,8,8,...
5,8,8,8,8,8,8,8,...
5,8,8,8,8,8,8,...
5,8,8,8,8,8,...
5,8,8,8,8,...
5,8,8,8,...
5,8,8,...
5,8,...
5,...
...

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

Antidiagonal sums give 3 together with the elements > 2 of A017089.

Cf. A010722, A010731, A274912, A274913, A278317, A278290, A278480, A278481.

3, seq(op([5,8$i,5]),i=0..20); # Robert Israel, Dec 04 2016

G.f. 3+x+8*x/(1-x)-3*(1+x)*Theta_2(0,sqrt(x))/(2*x^(1/8)) where Theta_2 is a Jacobi Theta function. - Robert Israel, Dec 04 2016

A017090 a(n) = (8*n + 2)^2.

Original entry on oeis.org

4, 100, 324, 676, 1156, 1764, 2500, 3364, 4356, 5476, 6724, 8100, 9604, 11236, 12996, 14884, 16900, 19044, 21316, 23716, 26244, 28900, 31684, 34596, 37636, 40804, 44100, 47524, 51076, 54756, 58564, 62500, 66564, 70756, 75076, 79524, 84100, 88804, 93636, 98596

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A006752, A016814, A017089 (8n+2), A000290 (n^2).

[(8*n+2)^2: n in [0..35]]; // Vincenzo Librandi, Jul 12 2011

Table[(8*n + 2)^2, {n, 0, 40}] (* Amiram Eldar, Apr 24 2023 *)

a(n)=(8*n+2)^2 \\ Charles R Greathouse IV, Jun 17 2017

G.f.: -4*(1 + 22*x + 9*x^2)/(x-1)^3. - R. J. Mathar, Jul 14 2016
From Amiram Eldar, Apr 24 2023: (Start)
a(n) = A017089(n)^2.
a(n) = 2^2*A016814(n).
Sum_{n>=0} 1/a(n) = Pi^2/64 + G/8, where G is Catalan's constant (A006752). (End)

A047468 Numbers that are congruent to {1, 2} mod 8.

Original entry on oeis.org

1, 2, 9, 10, 17, 18, 25, 26, 33, 34, 41, 42, 49, 50, 57, 58, 65, 66, 73, 74, 81, 82, 89, 90, 97, 98, 105, 106, 113, 114, 121, 122, 129, 130, 137, 138, 145, 146, 153, 154, 161, 162, 169, 170, 177, 178, 185, 186, 193, 194, 201, 202, 209, 210, 217, 218, 225, 226, 233

Offset: 1

N. J. A. Sloane

David Lovler, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Union of A017077 and A017089.

Cf. A047467.

Flatten[#+{1,2}&/@(8Range[0,30])] (* or *) LinearRecurrence[{1,1,-1},{1,2,9},60] (* Harvey P. Dale, Mar 26 2013 *)

a(n)=(n-1)\2*8+2-n%2 \\ Charles R Greathouse IV, May 14 2012

a(n) = 8*n - a(n-1) - 13 (with a(1)=1). - Vincenzo Librandi, Aug 06 2010
G.f.: x*(1+x+6*x^2)/((1-x)^2*(1+x)). - Colin Barker, May 13 2012
a(n) = 1 + 8*floor((n-1)/2) + ((n-1) mod 2). - Alois P. Heinz, May 13 2012
a(n) = (-3*(3 + (-1)^n) + 8*n)/2. - Colin Barker, May 14 2012
a(1)=1, a(2)=2, a(3)=9, a(n) = a(n-1) + a(n-2) - a(n-3). - Harvey P. Dale, Mar 26 2013
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(2)*Pi/16 + log(2)/8 + sqrt(2)*log(sqrt(2)+1)/8. - Amiram Eldar, Dec 18 2021
E.g.f.: 6 + ((8*x - 9)*exp(x) - 3*exp(-x))/2. - David Lovler, Sep 02 2022

More terms from Vincenzo Librandi, Aug 06 2010

A101492 Triangle read by rows: T(n,k) = (n-k+1)(4k+1).

Original entry on oeis.org

1, 2, 5, 3, 10, 9, 4, 15, 18, 13, 5, 20, 27, 26, 17, 6, 25, 36, 39, 34, 21, 7, 30, 45, 52, 51, 42, 25, 8, 35, 54, 65, 68, 63, 50, 29, 9, 40, 63, 78, 85, 84, 75, 58, 33, 10, 45, 72, 91, 102, 105, 100, 87, 66, 37, 11, 50, 81, 104, 119, 126, 125, 116, 99, 74, 41, 12, 55, 90, 117

Offset: 0

Lambert Klasen (lambert.klasen(AT)gmx.de) and Gary W. Adamson, Jan 21 2005

The triangle is generated from the product A*B
of the infinite lower triangular matrices A =
1 0 0 0...
1 1 0 0...
1 1 1 0...
1 1 1 1...
... and B =
1 0 0 0...
1 5 0 0...
1 5 9 0...
1 5 9 13...
...
T(n+0,0) = 1*n = A000027(n+1),
T(n+0,1) = 5*n = A008587(n),
T(n+1,2) = 9*n = A008591(n),
T(n+2,3) = 13*n = A008595(n),
so, for example,
T(n,n) = 4*n+1 = A016813(n),
T(n+1,n) = 8*n+2 = A017089(n),
T(n,0)*T(n,1)/10 = A000217(n) (triangular numbers),
T(n,n)*T(n,0) = A001107(n+1) (10-gonal numbers: 4*n^2 - 3*n),
T(n,n)*T(n,1)/5 = A007742(n).

Muniru A Asiru, Rows n=0..150 of triangle, flattened

Row sums give hexagonal pyramidal numbers A002412.

Cf. A101493 for product B*A, A002412.

Flat(List([0..11],n->List([0..n],k->(n+1-k)*(4*k+1)))); # Muniru A Asiru, Mar 07 2019

[[(n+1-k)*(4*k+1): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Mar 07 2019

Flatten[Table[(n+1-k)(4k+1),{n,0,15},{k,0,n}]] (* Harvey P. Dale, Jun 09 2011 *)

T(n, k) = if(k>n,0,(n-k+1)*(4*k+1));
for(i=0,10, for(j=0,i,print1(T(i,j),", "));print())

[[(n-k+1)*(4*k+1) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Mar 07 2019

A103219 Triangle read by rows: T(n,k) = (n+1-k)(4(n+1-k)^2 - 1)/3+2k(n+1-k)^2.

Original entry on oeis.org

1, 10, 3, 35, 18, 5, 84, 53, 26, 7, 165, 116, 71, 34, 9, 286, 215, 148, 89, 42, 11, 455, 358, 265, 180, 107, 50, 13, 680, 553, 430, 315, 212, 125, 58, 15, 969, 808, 651, 502, 365, 244, 143, 66, 17, 1330, 1131, 936, 749, 574, 415, 276, 161, 74, 19, 1771, 1530, 1293

Offset: 0

Lambert Klasen (lambert.klasen(AT)gmx.de) and Gary W. Adamson, Jan 26 2005

The triangle is generated from the product B * A of the infinite lower triangular matrices A =
1 0 0 0...
3 1 0 0...
5 3 1 0...
7 5 3 1...
...
and B =
1 0 0 0...
1 3 0 0...
1 3 5 0...
1 3 5 7...
...

			Triangle begins:
1,
10,3,
35,18,5,
84,53,26,7,
165,116,71,34,9,
286,215,148,89,42,11,

Row sums give A103220.
T(n, 0) = (n+1)*(4*(n+1)^2 - 1)/3 = A000447(n+1);
T(n+1, n)= 8*n+2 = A017089(n+1);
Cf. A103218 (for product A*B), A103220.

T[n_, k_] := (n + 1 - k)*(4*(n + 1 - k)^2 - 1)/3 + 2*k*(n + 1 - k)^2; Flatten[ Table[ T[n, k], {n, 0, 10}, {k, 0, n}]] (* Robert G. Wilson v, Feb 10 2005 *)

T(n,k)=(n+1-k)*(4*(n+1-k)^2-1)/3+2*k*(n+1-k)^2; for(i=0,10, for(j=0,i,print1(T(i,j),","));print())

A380820 a(0) = 0, a(1) = 1, and for n >= 2, a(n) = a(n-1) + a(n-2) if a(n-1) < n, otherwise a(n-1) - n.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 1, 9, 0, 9, 9, 18, 5, 23, 8, 31, 14, 45, 26, 6, 32, 10, 42, 18, 60, 34, 7, 41, 12, 53, 22, 75, 42, 8, 50, 14, 64, 26, 90, 50, 9, 59, 16, 75, 30, 105, 58, 10, 68, 18, 86, 34, 120, 66, 11, 77, 20, 97, 38, 135, 74, 12, 86, 22, 108, 42, 150

Offset: 0

Ya-Ping Lu, Feb 04 2025

Sequence starts with the first 7 Fibonacci numbers. For n >= 12, a(n) takes the values of (8*n+30)/7, (n+22)/7, (9*n+35)/7, (2*n+26)/7, (11*n+41)/7, (4*n+30)/7, and (15*n+45)/7 sequentially for n = 5, 6, 0, 1, 2, 3, 4 mod 7 (see plot in Links), which correspond to A017089 (n>=2), A000027 (n>=5), A017221 (n>=2), A005843 (n>=4), A017497 (n>=2), A016825 (n>=3), and A008597 (n>=3), respectively.

Terms for n >= 16 are the same as A322558(n) for n >= 17.

Ya-Ping Lu, Plot of A380820
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,2,0,0,0,0,0,0,-1).

Cf. A000027, A000045, A005843, A008597, A016825, A017089, A017221, A017497, A322558.

s={0,1};Do[AppendTo[s,If[s[[-1]]James C. McMahon, Feb 14 2025 *)

def A380820(n): R = [0, 1, 1, 2, 3, 5, 8, 1, 9, 0, 9, 9]; X = [9, 2, 11, 4, 15, 8, 1]; Y = [35, 26, 41, 30, 45, 30, 22]; return R[n] if n < 12 else (X[n%7]*n + Y[n%7])//7

a(n) = A322558(n+1) for n >= 16.

A017091 a(n) = (8*n + 2)^3.

Original entry on oeis.org

8, 1000, 5832, 17576, 39304, 74088, 125000, 195112, 287496, 405224, 551368, 729000, 941192, 1191016, 1481544, 1815848, 2197000, 2628072, 3112136, 3652264, 4251528, 4913000, 5639752, 6434856, 7301384, 8242408, 9261000, 10360232, 11543176, 12812904, 14172488, 15625000

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A002117, A016815, A017089 (8n+2), A000578 (n^3).

[(8*n+2)^3: n in [0..35]]; // Vincenzo Librandi, Jul 12 2011

(8*Range[0,30]+2)^3 (* or *) LinearRecurrence[{4,-6,4,-1},{8,1000,5832,17576},30] (* Harvey P. Dale, Dec 30 2019 *)

G.f.: 8*(1 + 121*x + 235*x^2 + 27*x^3)/(x-1)^4. - R. J. Mathar, Jul 14 2016
From Amiram Eldar, Apr 24 2023: (Start)
a(n) = A017089(n)^3.
a(n) = 2^3*A016815(n).
Sum_{n>=0} 1/a(n) = Pi^3/512 + 7*zeta(3)/128. (End)

A017092 a(n) = (8*n + 2)^4.

Original entry on oeis.org

16, 10000, 104976, 456976, 1336336, 3111696, 6250000, 11316496, 18974736, 29986576, 45212176, 65610000, 92236816, 126247696, 168896016, 221533456, 285610000, 362673936, 454371856, 562448656

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1)

Cf. A017089 (8n+2), A000583 (n^4).

[(8*n+2)^4: n in [0..30]]; // Vincenzo Librandi, Jul 12 2011

(8*Range[0,20]+2)^4 (* or *) LinearRecurrence[{5,-10,10,-5,1},{16,10000,104976,456976,1336336},20] (* Harvey P. Dale, Dec 16 2017 *)

G.f.: -16*(1 + 620*x + 3446*x^2 + 1996*x^3 + 81*x^4)/(x-1)^5. - R. J. Mathar, Jul 14 2016

A017093 a(n) = (8*n + 2)^5.

Original entry on oeis.org

32, 100000, 1889568, 11881376, 45435424, 130691232, 312500000, 656356768, 1252332576, 2219006624, 3707398432, 5904900000, 9039207968, 13382255776, 19254145824, 27027081632, 37129300000, 50049003168, 66338290976, 86617093024, 111577100832, 141985700000, 178689902368

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A013663, A016817, A017089 (8n+2), A000584 (n^5).

[(8*n+2)^5: n in [0..30]]; // Vincenzo Librandi, Jul 12 2011

(8Range[0,20]+2)^5  (* Harvey P. Dale, Feb 24 2011 *)

G.f.: 32*(1 + 3119*x + 40314*x^2 + 63854*x^3 + 15349*x^4 + 243*x^5)/(x-1)^6. - R. J. Mathar, Jul 14 2016
From Amiram Eldar, Apr 24 2023: (Start)
a(n) = A017089(n)^5.
a(n) = 2^5*A016817(n).
Sum_{n>=0} 1/a(n) = 5*Pi^5/98304 + 31*zeta(5)/2048. (End)

A017094 a(n) = (8*n + 2)^6.

Original entry on oeis.org

64, 1000000, 34012224, 308915776, 1544804416, 5489031744, 15625000000, 38068692544, 82653950016, 164206490176, 304006671424, 531441000000, 885842380864, 1418519112256, 2194972623936, 3297303959104

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (7, -21, 35, -35, 21, -7, 1).

Cf. A017089 (8n+2), A001014 (n^6).

[(8*n+2)^6: n in [0..30]]; // Vincenzo Librandi, Jul 12 2011

cp's OEIS Frontend

A278545 Number of neighbors of the n-th term in a full square array read by antidiagonals.

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A017090 a(n) = (8*n + 2)^2.

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A047468 Numbers that are congruent to {1, 2} mod 8.

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A101492 Triangle read by rows: T(n,k) = (n-k+1)*(4*k+1).

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A103219 Triangle read by rows: T(n,k) = (n+1-k)*(4*(n+1-k)^2 - 1)/3+2*k*(n+1-k)^2.

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A380820 a(0) = 0, a(1) = 1, and for n >= 2, a(n) = a(n-1) + a(n-2) if a(n-1) < n, otherwise a(n-1) - n.

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A017091 a(n) = (8*n + 2)^3.

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A017092 a(n) = (8*n + 2)^4.

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A101492 Triangle read by rows: T(n,k) = (n-k+1)(4k+1).

A103219 Triangle read by rows: T(n,k) = (n+1-k)(4(n+1-k)^2 - 1)/3+2k(n+1-k)^2.