A257083 -id:A257083 - cp's OEIS frontend

A246695 Row sums of the triangular array A246694.

1, 3, 9, 18, 35, 57, 91, 132, 189, 255, 341, 438, 559, 693, 855, 1032, 1241, 1467, 1729, 2010, 2331, 2673, 3059, 3468, 3925, 4407, 4941, 5502, 6119, 6765, 7471, 8208, 9009, 9843, 10745, 11682, 12691, 13737, 14859, 16020, 17261, 18543, 19909, 21318, 22815

Offset: 0

Clark Kimberling, Sep 01 2014

Also partial sums of A257083. - Reinhard Zumkeller, Apr 17 2015

			First 5 rows of A246694 preceded by sums
sum = 1: ...... 1
sum = 3: ...... 1 ... 2
sum = 9: ...... 3 ... 2 ... 4
sum = 18: ..... 3 ... 5 ... 4 ... 6
sum = 35: ..... 7 ... 5 ... 8 ... 6 ... 9

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A246694, A246705, A246706.

Cf. A257083, A377802.

a246695 n = a246695_list !! n
a246695_list = scanl1 (+) a257083_list
-- Reinhard Zumkeller, Apr 17 2015

z = 25; t[0, 0] = 1; t[1, 0] = 1; t[1, 1] = 2;
t[n_, 0] := If[OddQ[n], t[n - 1, n - 2] + 1, t[n - 1, n - 1] + 1];
t[n_, 1] := If[OddQ[n], t[n - 1, n - 1] + 1, t[n - 1, n - 2] + 1];
t[n_, k_] := t[n, k - 2] + 1; A246695 = Table[Sum[t[n, k], {k, 0, n}], {n, 0, z}]

Conjectured linear recurrence:  a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6), with a(0) = 1, a(1) = 3, a(2) = 9, a(3) = 18, a(4) = 35, a(5) = 57, a(6) = 91.
Conjectured g.f.: (1 + x + 2*x^2 + x^3 + x^4)/((x - 1)^4*(x + 1)^2).
Conjecture: a(n) = (1/8)*(n + 1)*((-1)^n + 2*n^2 + 4*n + 7). - Eric Simon Jacob, Jul 19 2023 [This conjecture is correct; compare A377802, note offset 1. - Werner Schulte, Nov 22 2024]

Corrected and edited by M. F. Hasler, Nov 17 2014

A271713 Numbers n such that 3*n - 5 is a square.

Original entry on oeis.org

2, 3, 7, 10, 18, 23, 35, 42, 58, 67, 87, 98, 122, 135, 163, 178, 210, 227, 263, 282, 322, 343, 387, 410, 458, 483, 535, 562, 618, 647, 707, 738, 802, 835, 903, 938, 1010, 1047, 1123, 1162, 1242, 1283, 1367, 1410, 1498, 1543, 1635, 1682, 1778, 1827, 1927, 1978, 2082, 2135, 2243, 2298

Offset: 1

Juri-Stepan Gerasimov, Apr 12 2016

Quasipolynomial of order 2 and degree 2. - Charles R Greathouse IV, Apr 12 2016
From Ray Chandler, Apr 13 2016: (Start)
Square roots of resulting squares gives A001651.
Sequence is the union of A141631 and A271740. (End)

			a(3) = 7 because 3*7 - 5 = 16 = 4^2.

Index entries for linear recurrences with constant coefficients, signature (1, 2, -2, -1, 1).

Cf. numbers n such that 3*n + k is a square: A120328 (k=-6), this sequence (k=-5), A056107 (k=-3), A257083 (k=-2), A033428 (k=0), A001082 (k=1), A080663 (k=3), A271675 (k=4), A100536 (k=6).

Cf. A001651, A141631, A271740.

[ n: n in [0..2500] | IsSquare(3*n - 5)];

A271713:=n->`if`(issqr(3*n-5), n, NULL): seq(A271713(n), n=1..5000); # Wesley Ivan Hurt, Apr 13 2016

Select[Range@ 2300, IntegerQ@ Sqrt[3 # - 5] &] (* Michael De Vlieger, Apr 12 2016 *)

is(n)=issquare(3*n-5) \\ Charles R Greathouse IV, Apr 12 2016

a(n)=((n\2*3-(-1)^n)^2+5)/3 \\ Charles R Greathouse IV, Apr 12 2016

from _future_ import division
A271713_list = [(n**2+5)//3 for n in range(10**6) if not (n**2+5) % 3] # Chai Wah Wu, Apr 13 2016

G.f.: x*(2 + x + x^3 + 2*x^4)/((1 - x)^3*(1 + x)^2). - Ilya Gutkovskiy, Apr 12 2016
a(n) = (3/2)*n^2 + O(n). - Charles R Greathouse IV, Apr 12 2016
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>5. - Wesley Ivan Hurt, Apr 13 2016

A257088 a(2n) = 4n if n>0, a(2n + 1) = 2n + 1, a(0) = 1.

Original entry on oeis.org

1, 1, 4, 3, 8, 5, 12, 7, 16, 9, 20, 11, 24, 13, 28, 15, 32, 17, 36, 19, 40, 21, 44, 23, 48, 25, 52, 27, 56, 29, 60, 31, 64, 33, 68, 35, 72, 37, 76, 39, 80, 41, 84, 43, 88, 45, 92, 47, 96, 49, 100, 51, 104, 53, 108, 55, 112, 57, 116, 59, 120, 61, 124, 63, 128

Offset: 0

Michael Somos, Apr 16 2015

			G.f. = 1 + x + 4*x^2 + 3*x^3 + 8*x^4 + 5*x^5 + 12*x^6 + 7*x^7 + 16*x^8 + ...

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

Cf. A001082, A005408, A007310, A008574, A040001, A215947.

CF. A257083 (partial sums), A246695.

import Data.List (transpose)
a257088 n = a257088_list !! n
a257088_list = concat $ transpose [a008574_list, a005408_list]
-- Reinhard Zumkeller, Apr 17 2015

a[ n_] := Which[ n < 1, Boole[n == 0], OddQ[n], n, True, 2 n];
a[ n_] := SeriesCoefficient[ (1 + x + 2*x^2 + x^3 + x^4) / (1 - 2*x^2 + x^4), {x, 0, n}];

PARI
```
{a(n) = if( n<1, n==0, n%2, n, 2*n)};
```

{a(n) = if( n<0, 0, polcoeff( (1 + x + 2*x^2 + x^3 + x^4) / (1 - 2*x^2 + x^4) + x * O(x^n), n))};

Euler transform of length 4 sequence [ 1, 3, -1, -1].
a(n) is multiplicative with a(2^e) = 2^(e+1) if e>0, otherwise a(p^e) = p^e.
G.f.: (1 + x + 2*x^2 + x^3 + x^4) / (1 - 2*x^2 + x^4).
G.f.: (1 - x^3) * (1 - x^4) / ((1 - x) * (1 - x^2)^3).
MOBIUS transform of A215947 is [1, 4, 3, 8, 5, ...].
a(n) = n * A040001(n) if n>0.
a(n) + a(n-1) = A007310(n) if n>0.
a(n) = A001082(n+1) - A001082(n) if n>0.
Binomial transform with a(0)=0 is A128543 if n>0.
a(2*n) = A008574(n). a(2*n + 1) = A005408(n).
a(n) = A022998(n) if n>0. - R. J. Mathar, Apr 19 2015
From Amiram Eldar, Jan 28 2025: (Start)
Dirichlet g.f.: (1+2^(1-s)) * zeta(s-1).
Sum_{k=1..n} a(k) ~ (3/4) * n^2. (End)
a(n) = gcd(n^n, 2*n). - Mia Boudreau, Jun 27 2025

A271675 Numbers m such that 3*m + 4 is a square.

Original entry on oeis.org

0, 4, 7, 15, 20, 32, 39, 55, 64, 84, 95, 119, 132, 160, 175, 207, 224, 260, 279, 319, 340, 384, 407, 455, 480, 532, 559, 615, 644, 704, 735, 799, 832, 900, 935, 1007, 1044, 1120, 1159, 1239, 1280, 1364, 1407, 1495, 1540, 1632, 1679, 1775, 1824, 1924, 1975, 2079, 2132, 2240, 2295, 2407

Offset: 1

Juri-Stepan Gerasimov, Apr 12 2016

7 is the unique prime in this sequence. If m is in this sequence, then 3*m + 4 = k^2 for k is nonzero integer, that is, m = (k^2 - 4)/3 = (k-2)*(k+2)/3. So m can be only prime if one of divisors is prime and another one is 1. Otherwise there should be more than 1 prime divisors, that is n must be composite. - Altug Alkan, Apr 12 2016
From Ray Chandler, Apr 12 2016: (Start)
Square roots of resulting squares gives A001651 (with a different starting point).
Sequence is the union of (positive terms) in A140676 and A270710. (End)
The sequence terms are the exponents in the expansion of Sum_{n >= 0} q^n*(1 - q)*(1 - q^3)*...*(1 - q^(2*n+1)) = 1 - q^4 - q^7 + q^15 + q^20 - q^32 - q^50 + + - - .... - Peter Bala, Dec 19 2024

			a(4) = 32 because 3*32 + 4 = 100 = 10*10.

Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A001651, A140676, A270710.

Cf. numbers n such that 3*n + k is a square: A120328 (k=-6), A271713 (k=-5), A056107 (k=-3), A257083 (k=-2), A033428 (k=0), A001082 (k=1), A080663 (k=3), this sequence (k=4), A100536 (k=6).

Magma
```
[n: n in [0..4000] | IsSquare(3*n+4)];
```

Select[Range[0,2500], IntegerQ@ Sqrt[3 # + 4] &] (* Michael De Vlieger, Apr 12 2016 *)
LinearRecurrence[{1,2,-2,-1,1},{0,4,7,15,20},60] (* Harvey P. Dale, Dec 09 2016 *)

from gmpy2 import is_square
for n in range(0,10**5):
    if(is_square(3*n+4)):print(n)
# Soumil Mandal, Apr 12 2016

O.g.f.: x^2*(4 + 3*x - x^3)/((1 + x)^2*(1 - x)^3).
E.g.f.: 1 + (1 - 2*x)*exp(-x)/8 - 3*(3 - 4*x - 2*x^2)*exp(x)/8.
a(n) = A001082(n+1) - 1 = (6*n*(n+1) + (2*n + 1)*(-1)^n - 1)/8 - 1. Therefore: a(2*k+1) = k*(3*k+4), a(2*k) = (k+1)*(3*k-1).
Sum_{n>=2} 1/a(n) = 19/16 - Pi/(4*sqrt(3)). - Amiram Eldar, Jul 26 2024

Edited and extended by Bruno Berselli, Apr 12 2016

A271723 Numbers k such that 3*k - 8 is a square.

Original entry on oeis.org

3, 4, 8, 11, 19, 24, 36, 43, 59, 68, 88, 99, 123, 136, 164, 179, 211, 228, 264, 283, 323, 344, 388, 411, 459, 484, 536, 563, 619, 648, 708, 739, 803, 836, 904, 939, 1011, 1048, 1124, 1163, 1243, 1284, 1368, 1411, 1499, 1544, 1636, 1683, 1779, 1828, 1928, 1979, 2083, 2136, 2244, 2299

Offset: 1

Juri-Stepan Gerasimov, Apr 13 2016

Square roots of resulting squares gives A001651. - Ray Chandler, Apr 14 2016

			a(1) = 3 because 3*3 - 8 = 1^2.

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

Cf. A001651.

Cf. numbers n such that 3*n + k is a square: this sequence (k=-8), A120328 (k=-6), A271713 (k=-5), A056107 (k=-3), A257083 (k=-2), A033428 (k=0), A001082 (k=1), A080663 (k=3), A271675 (k=4), A100536 (k=6), A271741 (k=7), A067725 (k=9).

Magma
```
[n: n in [1..2400] | IsSquare(3*n-8)];
```

seq(seq(((3*m+k)^2+8)/3, k=1..2),m=0..50); # Robert Israel, Dec 05 2016

Select[Range@ 2400, IntegerQ@ Sqrt[3 # - 8] &] (* Bruno Berselli, Apr 14 2016 *)
LinearRecurrence[{1,2,-2,-1,1},{3,4,8,11,19},60] (* Harvey P. Dale, Oct 02 2020 *)

from gmpy2 import is_square
[n for n in range(3000) if is_square(3*n-8)] # Bruno Berselli, Dec 05 2016

[(6*(n-1)*n-(2*n-1)*(-1)**n+23)/8 for n in range(1, 60)] # Bruno Berselli, Dec 05 2016

From Ilya Gutkovskiy, Apr 13 2016: (Start)
G.f.: x*(3 + x - 2*x^2 + x^3 + 3*x^4)/((1 - x)^3*(1 + x)^2).
a(n) = (6*(n - 1)*n - (2*n - 1)*(-1)^n + 23)/8. (End)

cp's OEIS Frontend

A246695 Row sums of the triangular array A246694.

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A271713 Numbers n such that 3*n - 5 is a square.

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A257088 a(2*n) = 4*n if n>0, a(2*n + 1) = 2*n + 1, a(0) = 1.

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A271675 Numbers m such that 3*m + 4 is a square.

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Magma

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A271723 Numbers k such that 3*k - 8 is a square.

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Magma

Maple

Mathematica

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Python

Formula

A257088 a(2n) = 4n if n>0, a(2n + 1) = 2n + 1, a(0) = 1.