A168276 -id:A168276 - cp's OEIS frontend

A227786 Take squares larger than 1, subtract 3 from even squares and 2 from odd squares; a(n) = a(n-1) + A168276(n+1) (with a(1) = 1).

1, 7, 13, 23, 33, 47, 61, 79, 97, 119, 141, 167, 193, 223, 253, 287, 321, 359, 397, 439, 481, 527, 573, 623, 673, 727, 781, 839, 897, 959, 1021, 1087, 1153, 1223, 1293, 1367, 1441, 1519, 1597, 1679, 1761, 1847, 1933, 2023, 2113, 2207, 2301, 2399, 2497, 2599, 2701

Offset: 1

Antti Karttunen, Jul 31 2013

nonn

Conjecture: from n>=2 onward, a(n) gives the positions of 2's in A227761.

a(29) = 897 = 3*13*23 is the first term which is neither prime nor semiprime, that is, has more than two prime divisors.

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

Bisections: A082109, A073577. Cf. also A227761.

a(n) = A000290(n+1) - 2 - (n mod 2).
a(1)=1, and for n>1, a(n) = a(n-1)+A168276(n+1).
a(n) = (1/2) * (2*n^2 + 4*n -3 + (-1)^n) = 2*A116940(n-1) + 1. a(n-1) = 2*ceiling(n^2/2) - 3 = 2*A000985(n) - 3. G.f.: x*(-x^3 - x^2 + 5*x + 1)/((1-x)^3 * (1+x)). - Ralf Stephan, Aug 10 2013

A168277 a(n) = 2*n - (-1)^n - 2.

Original entry on oeis.org

1, 1, 5, 5, 9, 9, 13, 13, 17, 17, 21, 21, 25, 25, 29, 29, 33, 33, 37, 37, 41, 41, 45, 45, 49, 49, 53, 53, 57, 57, 61, 61, 65, 65, 69, 69, 73, 73, 77, 77, 81, 81, 85, 85, 89, 89, 93, 93, 97, 97, 101, 101, 105, 105, 109, 109, 113, 113, 117, 117, 121, 121, 125, 125, 129, 129

Offset: 1

Vincenzo Librandi, Nov 22 2009

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

Cf. A016813, A163980, A168276.

Cf. A006752, A111003 (Pi^2/8).

[n eq 1 select 1 else 4*n-Self(n-1)-6: n in [1..70]]; // Vincenzo Librandi, Sep 16 2013

CoefficientList[Series[(1 + 3 x^2) / ((1 + x) (x - 1)^2), {x, 0, 80}], x] (* Vincenzo Librandi, Sep 16 2013 *)
Table[2 n - (-1)^n - 2, {n, 70}] (* Bruno Berselli, Sep 17 2013 *)
LinearRecurrence[{1,1,-1},{1,1,5},70] (* Harvey P. Dale, Aug 25 2015 *)

a(n)=2*n-(-1)^n-2 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 4*n - a(n-1) - 6, with n>1, a(1)=1.
a(n) = A163980(n-1), n>1. - R. J. Mathar, Nov 25 2009
G.f.: x*(1 + 3*x^2)/( (1+x)*(x-1)^2 ). - R. J. Mathar, Jul 15 2013
a(n) = A168276(n) - 1. - Vincenzo Librandi, Sep 17 2013
a(n) = a(n-1) +a(n-2) -a(n-3). - Vincenzo Librandi, Sep 17 2013
E.g.f.: (-1 + 3*exp(x) + 2*(x - 1)*exp(2*x))*exp(-x). - G. C. Greubel, Jul 16 2016
Sum_{n>=1} 1/a(n)^2 = Pi^2/8 + G, where G is Catalan's constant (A006752). - Amiram Eldar, Aug 21 2022

New definition from Bruno Berselli, Sep 17 2013

A267654 Irregular triangle of palindromic subsequences. Every row has 2n+1 terms. From the second row, there are only two alternated numbers: 2n+4 and 2*n+2.

Original entry on oeis.org

2, 4, 2, 4, 6, 4, 6, 4, 6, 8, 6, 8, 6, 8, 6, 8, 10, 8, 10, 8, 10, 8, 10, 8, 10, 12, 10, 12, 10, 12, 10, 12, 10, 12, 10, 12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 16, 14, 16, 14, 16, 14, 16, 14, 16, 14, 16, 14, 16, 14, 16

Offset: 0

Paul Curtz, Jan 19 2016

Row sums = 2, 10, 26, 50, ... = A069894(n).
Starting from A053186(n) =
0,                            for b(n)
0, 1, 2,                              for c(n)
0, 1, 2, 3, 4,                                for d(n)
0, 1, 2, 3, 4, 5, 6,
etc,
a(n) is used for
1) b(n+1) = b(n) + (a(0)=2) i.e. 0, 2, 4, 6, ... = A005843(n).
2) c(n+3) = c(n) + (period 3:repeat 4, 2, 4) i.e. 0, 1, 2, 4, 3, 6, 8, ... =  A265667(n).
3) d(n+5) = d(n) + (period 5:repeat 6, 4, 6, 4, 6) i.e. 0, 1, 2, 3, 4, 6, 5, 8, 7, 10, ... = A265734(n).
Etc.
a(n) has a companion with the same terms,differently distributed,yielding permutations of the nonnegative numbers. See A265672.
a(n) other writing (by pairs):
2,   4,  2,  4,
6,   4,  6,  4,
6,   8,  6,  8,  6,  8,  6,  8,
10   8, 10,  8, 10,  8, 10,  8,
10, 12, 10, 12, 10, 12, 10, 12, 10, 12, 10, 12,
14, 12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 12,
etc.
First column: A168276(n+2). Second column: A168273(n+2).
Row sums: 12, 20, 56, 72, ... = 4*A074378(n+1).
The last term of the successive rows is the number of their terms.
Main diagonal: A005843(n+1).

			The triangle is
2,
4, 2, 4,
6, 4, 6, 4, 6,
8, 6, 8, 6, 8, 6, 8,
etc.

Cf. A007395, A005843, A008586, A016825, A053186, A069894, A074378, A086520, A168273, A168276, A265667, A265672, A265734.

Table[2 (n - 1) + 2 (Boole@ OddQ@ k + 1), {n, 0, 7}, {k, 2 n + 1}] // Flatten (* Michael De Vlieger, Jan 19 2016 *)

a(n) = 2 * A086520(n+2).
a(2n) = 4*n + 2 times 4*n + 2 = 2, 2, 6, 6, 6, 6, 6, 6, 10,....
a(2n+1) = 4*(n+1) times 4*(n+1) = 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, 12, ....

cp's OEIS Frontend

A227786 Take squares larger than 1, subtract 3 from even squares and 2 from odd squares; a(n) = a(n-1) + A168276(n+1) (with a(1) = 1).

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A168277 a(n) = 2*n - (-1)^n - 2.

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A267654 Irregular triangle of palindromic subsequences. Every row has 2n+1 terms. From the second row, there are only two alternated numbers: 2n+4 and 2*n+2.

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cp's OEIS Frontend

A227786 Take squares larger than 1, subtract 3 from even squares and 2 from odd squares; a(n) = a(n-1) + A168276(n+1) (with a(1) = 1).

Views

Author

Keywords

Comments

Links

Crossrefs

Formula

A168277 a(n) = 2*n - (-1)^n - 2.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A267654 Irregular triangle of palindromic subsequences. Every row has 2*n+1 terms. From the second row, there are only two alternated numbers: 2*n+4 and 2*n+2.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

A267654 Irregular triangle of palindromic subsequences. Every row has 2n+1 terms. From the second row, there are only two alternated numbers: 2n+4 and 2*n+2.