A027689 -id:A027689 - cp's OEIS frontend

A022856 a(n) = n-2 + Sum_{i = 1..n-2} (a(i+1) mod a(i)) for n >= 3 with a(1) = a(2) = 1.

1, 1, 1, 2, 3, 5, 8, 12, 17, 23, 30, 38, 47, 57, 68, 80, 93, 107, 122, 138, 155, 173, 192, 212, 233, 255, 278, 302, 327, 353, 380, 408, 437, 467, 498, 530, 563, 597, 632, 668, 705, 743, 782, 822, 863, 905, 948, 992, 1037, 1083, 1130, 1178, 1227

Offset: 1

Clark Kimberling

Essentially triangular numbers + 2, but with three extra initial terms.

G. C. Greubel, Table of n, a(n) for n = 1..5000
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000124, A000217, A007318, A027689.

a[n_] := If[n<4, 1, (n^2-7n+16)/2]; Array[a, 60] (* Jean-François Alcover, Mar 08 2017 *)

for(n=1,100, print1(if(n<4, 1, (n^2 - 7*n +16)/2), ", ")) \\ G. C. Greubel, Jul 13 2017

For n > 3, a(n) = (n^2 - 7*n + 16)/2 = A027689(n-4)/2 = A000217(n-4) + 2 = A000124(n-4) + 1. - Henry Bottomley, Jun 27 2000
a(n) = Sum_{k=0..2} A007318(n-k-2, k) for n > 3. - Johannes W. Meijer, Aug 11 2013
Sum_{n>=1} 1/a(n) = 3 + 2*Pi*tanh(sqrt(15)*Pi/2)/sqrt(15). - Amiram Eldar, Dec 13 2022

A027603 a(n) = n^3 + (n+1)^3 + (n+2)^3 + (n+3)^3.

Original entry on oeis.org

36, 100, 224, 432, 748, 1196, 1800, 2584, 3572, 4788, 6256, 8000, 10044, 12412, 15128, 18216, 21700, 25604, 29952, 34768, 40076, 45900, 52264, 59192, 66708, 74836, 83600, 93024, 103132, 113948, 125496, 137800, 150884, 164772

Offset: 0

Patrick De Geest

Sums of four consecutive cubes. - Wesley Ivan Hurt, Dec 16 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Patrick De Geest, Palindromic Sums of Cubes of Consecutive Integers
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000578, A005898, A027602, A027604.

Cf. A016825, A027689, A173965.

[4*n^3 + 18*n^2 + 42*n + 36: n in [0..40]]; // Vincenzo Librandi, Jun 04 2011

A027603:=n->n^3 + (n+1)^3 + (n+2)^3 + (n+3)^3: seq(A027603(n), n=0..50); # Wesley Ivan Hurt, Dec 16 2015

Table[n^3 +(n+1)^3 +(n+2)^3 +(n+3)^3, {n, 0, 33}] (* or *)
Table[Plus@@(Range[n, n + 3]^3), {n, 0, 33}] (* Alonso del Arte, Jan 24 2011 *)

Vec(-4*(-9+11*x-10*x^2+2*x^3)/(1-x)^4 + O(x^100)) \\ Altug Alkan, Dec 16 2015

A027603_list, m = [], [24, 12, 28, 36]
for _ in range(10**2):
    A027603_list.append(m[-1])
    for i in range(3):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

[n^3+(n+1)^3+(n+2)^3+(n+3)^3 for n in range(0,40)] # Zerinvary Lajos, Jul 03 2008

a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4) for n>=4.
a(n) = 4*n^3 + 18*n^2 + 42*n + 36 = 4*A173965(n+2).
From Bruno Berselli, Jan 24 2011: (Start)
G.f.: 4*(9 - 11*x + 10*x^2 - 2*x^3)/(1-x)^4.
a(n) = A027689(n+1) * A016825(n+1). (End)
E.g.f.: 2*(18 + 32*x + 15*x^2 + 2*x^3)*exp(x). - G. C. Greubel, Aug 24 2022

A163253 An interspersion: the order array of the odd-numbered columns of the double interspersion at A161179.

Original entry on oeis.org

1, 4, 2, 9, 5, 3, 16, 10, 7, 6, 25, 17, 13, 11, 8, 36, 26, 21, 18, 14, 12, 49, 37, 31, 27, 22, 19, 15, 64, 50, 43, 38, 32, 28, 23, 20, 81, 65, 57, 51, 44, 39, 33, 29, 24, 100, 82, 73, 66, 58, 52, 45, 40, 34, 30, 121, 101, 91, 83, 74, 67, 59, 53, 46, 41, 35

Offset: 1

Clark Kimberling, Jul 23 2009

A permutation of the natural numbers.
Row 1 consists of the squares.
Beginning with row 5, the columns obey a 3rd-order recurrence:
c(n)=c(n-1)+c(n-2)-c(n-3)+1; thus disregarding row 1, the nonsquares are partitioned by this recurrence.
Except for initial terms, the first ten rows match A000290, A002522, A002061, A059100, A014209, A117950, A027688, A087475, A027689, A117951, and the first column, A035106.

			Corner:
1....4....9...16...25
2....5...10...17...26
3....7...13...21...31
6...11...18...27...38
The double interspersion A161179 begins thus:
1....4....7...12...17...24
2....3....8...11...18...23
5....6...13...16...25...30
9...10...19...22...33...38
Expel the even-numbered columns, leaving
1....7...17...
2....8...18...
5...13...25...
9...19...33...
Then replace each of those numbers by its rank when all the numbers are jointly ranked.

Clark Kimberling, Doubly interspersed sequences, double interspersions and fractal sequences, The Fibonacci Quarterly 48 (2010) 13-20.

Cf. A000037, A000290, A161179, A163254, A163255, A163256, A163257, A163258.

Let S(n,k) denote the k-th term in the n-th row. Three cases:
S(1,k)=k^2;
if n is even, then S(n,k)=k^2+(n-2)k+(n^2-2*n+4)/4;
if n>=3 is odd, then S(n,k)=k^2+(n-2)k+(n^2-2*n+1)/4.

Edited and augmented by Clark Kimberling, Jul 24 2009

A027716 Numbers k such that k^2 + k + 4 is a palindrome.

Original entry on oeis.org

0, 1, 20, 200, 219, 261, 2000, 2234, 2551, 2613, 20000, 20604, 200000, 202885, 205704, 218354, 2000000, 2155139, 2490266, 2620486, 20000000, 25882353, 200000000, 205705704, 2000000000, 2094600194, 20000000000, 20102030400, 20349812814, 21572321960

Offset: 1

Patrick De Geest

Includes 2*10^k for k>=2. - Robert Israel, Apr 16 2019

Giovanni Resta, Table of n, a(n) for n = 1..36
P. De Geest, Palindromic Quasi_Over_Squares of the form n^2+(n+X)

Cf. A027717, A027689, A027714, A027718.

palQ[n_] := Block[{d = IntegerDigits[n]}, d == Reverse[d]]; f[n_] := n^2 + n + 4; Select[Range[0, 10^5], palQ@ f@ # &] (* Giovanni Resta, Aug 29 2018 *)
Select[Range[0,3*10^6],PalindromeQ[#^2+#+4]&] (* The program generates the first 20 terms of the sequence. *) (* Harvey P. Dale, Jul 22 2024 *)

More terms from Giovanni Resta, Aug 29 2018

A027717 Palindromes of form k^2 + k + 4.

Original entry on oeis.org

4, 6, 424, 40204, 48184, 68386, 4002004, 4992994, 6510156, 6830386, 400020004, 424545424, 40000200004, 41162526114, 42314341324, 47678687674, 4000002000004, 4644626264464, 6201427241026, 6866949496686, 400000020000004, 669896222698966, 40000000200000004

Offset: 1

Patrick De Geest

Giovanni Resta, Table of n, a(n) for n = 1..36
P. De Geest, Palindromic Quasi_Over_Squares of the form n^2+(n+X)

Cf. A027716, A027689, A027715, A027728.

Select[Table[n^2+n+4,{n,0,25*10^5}],IntegerDigits[#] == Reverse[ IntegerDigits[ #]]&] (* Harvey P. Dale, Mar 05 2015 *)
Select[Table[n^2+n+4,{n,0,25*10^5}],PalindromeQ] (* Harvey P. Dale, Dec 23 2023 *)

More terms from Giovanni Resta, Aug 29 2018

A213921 Natural numbers placed in table T(n,k) layer by layer. The order of placement: at the beginning filled odd places of layer clockwise, next - even places clockwise. Table T(n,k) read by antidiagonals.

Original entry on oeis.org

1, 2, 3, 5, 4, 7, 10, 8, 9, 13, 17, 14, 6, 16, 21, 26, 22, 11, 12, 25, 31, 37, 32, 18, 15, 20, 36, 43, 50, 44, 27, 23, 24, 30, 49, 57, 65, 58, 38, 33, 19, 35, 42, 64, 73, 82, 74, 51, 45, 28, 29, 48, 56, 81, 91, 101, 92, 66, 59, 39, 34, 41, 63, 72, 100, 111

Offset: 1

Boris Putievskiy, Mar 05 2013

A permutation of the natural numbers.
a(n) is a pairing function: a function that reversibly maps Z^{+} x Z^{+} onto Z^{+}, where Z^{+} is the set of integer positive numbers.
Layer is pair of sides of square from T(1,n) to T(n,n) and from T(n,n) to T(n,1). Enumeration table T(n,k) is layer by layer. The order of the list:
T(1,1)=1;
T(1,2), T(2,1), T(2,2);
. . .
T(1,n), T(3,n), ... T(n,3), T(n,1), T(2,n), T(4,n), ... T(n,4), T(n,2);
...

			The start of the sequence as table:
   1   2   5  10  17  26 ...
   3   4   8  14  22  32 ...
   7   9   6  11  18  27 ...
  13  16  12  15  23  33 ...
  21  25  20  24  19  28 ...
  31  36  30  35  29  34 ...
  ...
The start of the sequence as triangle array read by rows:
   1;
   2,  3;
   5,  4,  7;
  10,  8,  9, 13;
  17, 14,  6, 16, 21;
  26, 22, 11, 12, 25, 31;
  ...

Boris Putievskiy, Rows n = 1..140 of triangle, flattened
Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
Eric W. Weisstein, MathWorld: Pairing functions
Index entries for sequences that are permutations of the natural numbers

Cf. A060734, A060736; table T(n,k) contains: in rows A002522, A014206, A059100, A027688, A117950, A027689, A087475, A027690, A117951, A027691, A114949, A027692, A117619; in columns A002061, A000290, A002378, A005563, A028387, A008865, A028552, A028872, A014209, A028347, A028875.

t=int((math.sqrt(8*n-7) - 1)/ 2)
i=n-t*(t+1)/2
j=(t*t+3*t+4)/2-n
if i > j:
   result=i*i-(j%2)*i+2-int((j+2)/2)
else:
   result=j*j-((i%2)+1)*j + int((i+3)/2)

As a table:
T(n,k) = n*n - (k mod 2)*n + 2 - floor((k+2)/2), if n>k;
T(n,k) = k*k - ((n mod 2)+1)*k + floor((n+3)/2), if n<=k.
As a linear sequence:
a(n) = i*i - (j mod 2)*i + 2 - floor((j+2)/2), if i>j;
a(n) = j*j - ((i mod 2)+1)*j + floor((i+3)/2), if i<=j; where i = n-t*(t+1)/2, j = (t*t+3*t+4)/2-n, t = floor((-1+sqrt(8*n-7))/2).

A171746 Let f(n) = n + floor(sqrt(n)). Then a(n) is the smallest number of iterations of f on n such that a perfect square is obtained.

Original entry on oeis.org

3, 2, 1, 5, 2, 4, 1, 3, 7, 2, 4, 6, 1, 3, 5, 9, 2, 4, 6, 8, 1, 3, 5, 7, 11, 2, 4, 6, 8, 10, 1, 3, 5, 7, 9, 13, 2, 4, 6, 8, 10, 12, 1, 3, 5, 7, 9, 11, 15, 2, 4, 6, 8, 10, 12, 14, 1, 3, 5, 7, 9, 11, 13, 17, 2, 4, 6, 8, 10, 12, 14, 16, 1, 3, 5, 7, 9, 11, 13, 15, 19, 2, 4, 6, 8, 10, 12, 14, 16, 18, 1, 3, 5

Offset: 1

Neven Juric (neven.juric(AT)apis-it.hr), Oct 07 2010

nonn

Iterate A028392, starting with n: a(n) is the number of steps until a square will be reached. - Reinhard Zumkeller, Feb 23 2012

			f(9)=12, f(12)=15, f(15)=18, f(18)=22, f(22)=26, f(26)=31, f(31)=36. The first square number in this sequence 12,15,18,22,26,31,36 is on the seventh place and therefore a(9)=7.

Matematicko-fizicki list 1/144, problem 2-2, page 29, (1985-1986).

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

Cf. A010052, A028392.

a171746 = (+ 1) . length . takeWhile (== 0) .
                           map a010052 . tail . iterate a028392
-- Reinhard Zumkeller, Feb 23 2012, Oct 14 2010

f[n_] := Length@ NestWhileList[ # + Floor@Sqrt@# &, n, ! IntegerQ@Sqrt@# || # == n &] - 1; Array[f, 93] (* Robert G. Wilson v, Oct 08 2010 *)

f(n) = n + sqrtint(n); \\ A028392
a(n) = my(k=1); while (!issquare(n=f(n)), k++); k; \\ Michel Marcus, Nov 06 2022

From Robert G. Wilson v, Oct 08 2010: (Start)
a(k)=1 for A002061(n): n^2 - n + 1 for n>1;
a(k)=2 for A002522(n): n^2 + 1     for n>1;
a(k)=3 for A014206(n): n^2 + n + 2 for n>1;
a(k)=4 for A059100(n): n^2 + 2     for n>1;
a(k)=5 for A027688(n): n^2 + n + 3 for n>2;
a(k)=6 for A117950(n): n^2 + 3     for n>2;
a(k)=7 for A027689(n): n^2 + n + 4 for n>4;
a(k)=8 for A087475(n): n^2 + 4     for n>3;
a(k)=9 for A027690(n): n^2 + n + 5 for n>4; ... (End)
a(n^2) = 2*n + 1: a(A000290(n)) = A005408(n). - Reinhard Zumkeller, Oct 14 2010

A367964 Triangle of 2-parameter triangular numbers, read by rows. T(n, k) = (n(n + 1) + k(k + 1)) / 2.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 6, 7, 9, 12, 10, 11, 13, 16, 20, 15, 16, 18, 21, 25, 30, 21, 22, 24, 27, 31, 36, 42, 28, 29, 31, 34, 38, 43, 49, 56, 36, 37, 39, 42, 46, 51, 57, 64, 72, 45, 46, 48, 51, 55, 60, 66, 73, 81, 90, 55, 56, 58, 61, 65, 70, 76, 83, 91, 100, 110

Offset: 0

Peter Luschny, Dec 07 2023

If the rows of the triangle are extended for k > n, the array A144216 is created, which is symmetrical to the main diagonal and therefore contains no new information compared to this triangle.

			Triangle T(n, k) starts:
  0 |  0;
  1 |  1,  2;
  2 |  3,  4,  6;
  3 |  6,  7,  9, 12;
  4 | 10, 11, 13, 16, 20;
  5 | 15, 16, 18, 21, 25, 30;
  6 | 21, 22, 24, 27, 31, 36, 42;
  7 | 28, 29, 31, 34, 38, 43, 49, 56;
  8 | 36, 37, 39, 42, 46, 51, 57, 64, 72;
  9 | 45, 46, 48, 51, 55, 60, 66, 73, 81,  90;
 10 | 55, 56, 58, 61, 65, 70, 76, 83, 91, 100, 110;
.
Start at row 0, column 0 with 0. Go down by adding the column index in step n. At row n, restart the counting and go n steps right by adding the row index in step n, then change direction and go down again by adding the column index. After 3*n steps on this path you are at T(2*n, n) which is 2*triangular(n) + (triangular(2*n) - triangular(n)) = (5*n^2 + 3*n)/2. These are the sliced heptagonal numbers A147875 (see the illustration of Leo Tavares).
.
The equation T(n, k) = (n*(n + 1) + k*(k + 1))/2 can be extended to all n, k in ZZ.
  [n\k] ... -6  -5  -4  -3  -2  -1   0   1   2   3   4   5  ...
  -------------------------------------------------------------
  [-5] ..., 25, 20, 16, 13, 11, 10, 10, 11, 13, 16, 20, 25, ...
  [-4] ..., 21, 16, 12,  9,  7,  6,  6,  7,  9, 12, 16, 21, ...
  [-3] ..., 18, 13,  9,  6,  4,  3,  3,  4,  6,  9, 13, 18, ...
  [-2] ..., 16, 11,  7,  4,  2,  1,  1,  2,  4,  7, 11, 16, ...
  [-1] ..., 15, 10,  6,  3,  1,  0,  0,  1,  3,  6, 10, 15, ...
  [ 0] ..., 15, 10,  6,  3,  1,  0,  0,  1,  3,  6, 10, 15, ...
  [ 1] ..., 16, 11,  7,  4,  2,  1,  1,  2,  4,  7, 11, 16, ...
  [ 2] ..., 18, 13,  9,  6,  4,  3,  3,  4,  6,  9, 13, 18, ...
  [ 3] ..., 21, 16, 12,  9,  7,  6,  6,  7,  9, 12, 16, 21, ...
  [ 4] ..., 25, 20, 16, 13, 11, 10, 10, 11, 13, 16, 20, 25, ...

Columns: A000217, A000124, A152950, A166136, A121263.
Diagonals: A002378, A000290, A002061, A059100, A027689.
Cf. A147875 (T(2*n, n)), A016061 (row sums), A367965 (alternating row sums), A143216 (the multiplicative equivalent), A144216 (extended array).

T := (n, k) -> (n*(n + 1) + k*(k + 1)) / 2:
for n from 0 to 10 do seq(T(n, k), k = 0..n) od;

Module[{n=1},NestList[Append[#+n,n*++n]&,{0},10]] (* or *)
Table[(n(n+1)+k(k+1))/2,{n,0,10},{k,0,n}] (* Paolo Xausa, Dec 07 2023 *)

# A purely additive construction:
from functools import cache
@cache
def a_row(n: int) -> list[int]:
    if n == 0: return [0]
    row = a_row(n - 1) + [0]
    for k in range(n): row[k] += n
    row[n] = row[n - 1] + n
    return row

Recurrence: T(n, n) = n + T(n, n-1) starting with T(0, 0) = 0.
For k <> n: T(n, k) = n + T(n-1, k).
T(n, k) = t(n) + t(k), where t(n) are the triangular numbers A000217.
G.f.: (x + x*(2 - 5*x + x^2)*y + x^4*y^2)/((1 - x)^3*(1 - x*y)^3). - Stefano Spezia, Dec 07 2023

A214870 Natural numbers placed in table T(n,k) layer by layer. The order of placement: at the beginning filled odd places of layer clockwise, next - even places counterclockwise. T(n,k) read by antidiagonals.

Original entry on oeis.org

1, 2, 3, 5, 4, 7, 10, 9, 8, 13, 17, 16, 6, 14, 21, 26, 25, 11, 12, 22, 31, 37, 36, 18, 15, 20, 32, 43, 50, 49, 27, 24, 23, 30, 44, 57, 65, 64, 38, 35, 19, 33, 42, 58, 73, 82, 81, 51, 48, 28, 29, 45, 56, 74, 91, 101, 100, 66, 63, 39, 34, 41, 59, 72, 92, 111

Offset: 1

Boris Putievskiy, Mar 11 2013

Permutation of the natural numbers.
a(n) is a pairing function: a function that reversibly maps Z^{+} x Z^{+} onto Z^{+}, where Z^{+} is the set of integer positive numbers.
Layer is pair of sides of square from T(1,n) to T(n,n) and from T(n,n) to T(n,1).
Enumeration table T(n,k) layer by layer. The order of the list:
  T(1,1)=1;
  T(1,2), T(2,1), T(2,2);
  . . .
  T(1,n), T(3,n), ... T(n,3), T(n,1); T(n,2), T(n,4), ... T(4,n), T(2,n);
  . . .

			The start of the sequence as table:
   1   2   5  10  17  26 ...
   3   4   9  16  25  36 ...
   7   8   6  11  18  27 ...
  13  14  12  15  24  35 ...
  21  22  20  23  19  28 ...
  31  32  30  33  29  34 ...
  ...
The start of the sequence as triangle array read by rows:
   1;
   2,  3;
   5,  4,  7;
  10,  9,  8, 13;
  17, 16,  6, 14, 21;
  26, 25, 11, 12, 22, 31;
  ...

Boris Putievskiy, Rows n = 1..140 of triangle, flattened
Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
Eric Weisstein's World of Mathematics, Pairing functions
Index entries for sequences that are permutations of the natural numbers

Cf. A060734, A060736, A185725, A213921, A213922; table T(n,k) contains: in rows A002522, A000290, A059100, A005563, A117950, A008865, A087475, A028872, A117951, A028347, A114949, A028875, A117619, A028878, A189833, A028881, A189834, A028884, A114948, A028560, A189836; in columns A002061, A014206, A002378, A027688, A028387, A027689, A028552, A027690, A014209, A027691, A027692, A082111, A027693, A028557, A027694, A108195, A187710, A048058, A048840.

t=int((math.sqrt(8*n-7) - 1)/ 2)
i=n-t*(t+1)/2
j=(t*t+3*t+4)/2-n
if i > j:
   result=i*i-i+(j%2)*(2-(j+1)/2)+((j+1)%2)*(j/2+1)
else:
   result=j*j-2*(i%2)*j + (i%2)*((i+1)/2+1) + ((i+1)%2)*(-i/2+1)

As table
T(n,k) = k*k-2*(n mod 2)*k+(n mod 2)*((n+1)/2+1)+((n+1) mod 2)*(-n/2+1), if n<=k;
T(n,k) = n*n-n+(k mod 2)*(2-(k+1)/2)+((k+1) mod 2)*(k/2+1), if n>k.
As linear sequence
a(n) = j*j-2*(i mod 2)*j+(i mod 2)*((i+1)/2+1)+((i+1) mod 2)*(-i/2+1), if i<=j;
a(n) = i*i-i+(j mod 2)*(2-(j+1)/2)+((j+1) mod 2)*(j/2+1), if i>j; where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor((-1+sqrt(8*n-7))/2).

A294070 a(n) = (1/4)(n^2 - 2n)^2 + (9/4)(n^2 - 2n) + 6.

Original entry on oeis.org

4, 6, 15, 40, 96, 204, 391, 690, 1140, 1786, 2679, 3876, 5440, 7440, 9951, 13054, 16836, 21390, 26815, 33216, 40704, 49396, 59415, 70890, 83956, 98754, 115431, 134140, 155040, 178296, 204079, 232566, 263940, 298390, 336111, 377304, 422176, 470940, 523815

Offset: 1

Jan Lakota Nono, Aug 14 2018

			2*2, 2*3, 3*5, 5*8, 8*12, 12*17, 17*23, 23*30, 30*38, ...

Colin Barker, Table of n, a(n) for n = 1..1000
SESC NSU Correspondence School, First assignments for 2018/2019 (in Russian), Kvant, 2018, No. 7, p. 42, Mathematics section, 6th grade, exercise no. 2. "Calculate and show in a reduced fraction form the following sum: 1/(2*3) + 2/(3*5) + 3/(5*8) + 4/(8*12) + 5/(12*17)."
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A027689, A152948.

List([1..40],n->(n^2-3*n+6)*(n^2-n+4)/4); # Muniru A Asiru, Aug 16 2018

[(n^2-3*n+6)*(n^2-n+4)/4: n in [1..40]]; // Vincenzo Librandi, Aug 30 2018

b:=n->(n^2-3*n+6)/2: seq(b(n)*b(n+1),n=1..40); # Muniru A Asiru, Aug 16 2018

Times@@@Partition[Array[(#^2 -3# +6)/2 &, 40], 2, 1] (* Michael De Vlieger, Sep 24 2018 *)
LinearRecurrence[{5,-10,10,-5,1}, {4,6,15,40,96}, 40] (* G. C. Greubel, Feb 10 2019 *)

Vec(x*(4 - 14*x + 25*x^2 - 15*x^3 + 6*x^4)/(1-x)^5 + O(x^40)) \\ Colin Barker, Nov 26 2018

[(n^2-3*n+6)*(n^2-n+4)/4 for n in (1..40)] # G. C. Greubel, Feb 10 2019

a(n) = A152948(n) * A152948(n+1).
From Muniru A Asiru, Aug 16 2018: (Start)
a(n) = (n^2 - 3*n + 6)*(n^2 - n + 4)/4.
a(n) = A152948(n)*A027689(n-1)/2. (End)
a(n) = A266883(A061925(n-1)). - Bruno Berselli, Aug 30 2018
From Colin Barker, Nov 26 2018: (Start)
G.f.: x*(4 - 14*x + 25*x^2 - 15*x^3 + 6*x^4)/(1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5.
a(n) = (24 - 18*n + 13*n^2 - 4*n^3 + n^4)/4. (End)
E.g.f.: (1/4)*exp(x)*(16 + 8*x + 14*x^2 + 6*x^3 + x^4). - Stefano Spezia, Nov 30 2018

cp's OEIS Frontend

A022856 a(n) = n-2 + Sum_{i = 1..n-2} (a(i+1) mod a(i)) for n >= 3 with a(1) = a(2) = 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A027603 a(n) = n^3 + (n+1)^3 + (n+2)^3 + (n+3)^3.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Python

Sage

Formula

A163253 An interspersion: the order array of the odd-numbered columns of the double interspersion at A161179.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

Extensions

A027716 Numbers k such that k^2 + k + 4 is a palindrome.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Extensions

A027717 Palindromes of form k^2 + k + 4.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Extensions

A213921 Natural numbers placed in table T(n,k) layer by layer. The order of placement: at the beginning filled odd places of layer clockwise, next - even places clockwise. Table T(n,k) read by antidiagonals.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python

Formula

A171746 Let f(n) = n + floor(sqrt(n)). Then a(n) is the smallest number of iterations of f on n such that a perfect square is obtained.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

A367964 Triangle of 2-parameter triangular numbers, read by rows. T(n, k) = (n(n + 1) + k(k + 1)) / 2.

A294070 a(n) = (1/4)(n^2 - 2n)^2 + (9/4)(n^2 - 2n) + 6.