A235700 -id:A235700 - cp's OEIS frontend

A047235 Numbers that are congruent to {2, 4} mod 6.

2, 4, 8, 10, 14, 16, 20, 22, 26, 28, 32, 34, 38, 40, 44, 46, 50, 52, 56, 58, 62, 64, 68, 70, 74, 76, 80, 82, 86, 88, 92, 94, 98, 100, 104, 106, 110, 112, 116, 118, 122, 124, 128, 130, 134, 136, 140, 142, 146, 148, 152, 154, 158, 160, 164, 166, 170, 172, 176, 178, 182, 184, 188, 190, 194, 196, 200, 202, 206

Offset: 1

N. J. A. Sloane

Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 19 ).
Complement of A047273; A093719(a(n)) = 0. - Reinhard Zumkeller, Oct 01 2008
One could prefix an initial term "1" (or not) and define this sequence through a(n+1) = a(n) + (a(n) mod 6). See A001651 for the analog with 3, A235700 (with 5), A047350 (with 7), A007612 (with 9) and A102039 (with 10). Using 4 or 8 yields a constant sequence from that term on. - M. F. Hasler, Jan 14 2014
Nonnegative m such that m^2/6 + 1/3 is an integer. - Bruno Berselli, Apr 13 2017
Numbers divisible by 2 but not by 3. - David James Sycamore, Apr 04 2018
Numbers k for which A276086(k) is of the form 6m+3. - Antti Karttunen, Dec 03 2022

David A. Corneth, Table of n, a(n) for n = 1..10000
Chunhui Lai, A note on potentially K_4-e graphical sequences, arXiv:math/0308105 [math.CO], 2003.
William A. Stein, The modular forms database.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A020760, A020832, A093719, A047273 (complement), A120325 (characteristic function).
Equals 2*A001651.
Cf. A007310 ((6*n+(-1)^n-3)/2). - Bruno Berselli, Jun 24 2010
Positions of 3's in A053669 and in A358840.

[ n eq 1 select 2 else Self(n-1)+2*(1+n mod 2): n in [1..70] ]; // Klaus Brockhaus, Dec 13 2008

seq(6*floor((n+1)/2) + 3 + (-1)^n, n=1..67); # Gary Detlefs, Mar 02 2010

Flatten[Table[{6n - 4, 6n - 2}, {n, 40}]] (* Alonso del Arte, Oct 27 2014 *)

a(n)=(n-1)\2*6+3+(-1)^n \\ Charles R Greathouse IV, Jul 01 2013

first(n) = my(v = vector(n, i, 3*i - 1)); forstep(i = 2, n, 2, v[i]--); v \\ David A. Corneth, Oct 20 2017

a(n) = 2*A001651(n).
n such that phi(3*n) = phi(2*n). - Benoit Cloitre, Aug 06 2003
G.f.: 2*x*(1 + x + x^2)/((1 + x)*(1 - x)^2). a(n) = 3*n - 3/2 - (-1)^n/2. - R. J. Mathar, Nov 22 2008
a(n) = 3*n + 5..n odd, 3*n + 4..n even a(n) = 6*floor((n+1)/2) + 3 + (-1)^n. - Gary Detlefs, Mar 02 2010
a(n) = 6*n - a(n-1) - 6 (with a(1) = 2). - Vincenzo Librandi, Aug 05 2010
a(n+1) = a(n) + (a(n) mod 6). - M. F. Hasler, Jan 14 2014
Sum_{n>=1} 1/a(n)^2 = Pi^2/27. - Dimitris Valianatos, Oct 10 2017
a(n) = (6*n - (-1)^n - 3)/2. - Ammar Khatab, Aug 23 2020
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(6*sqrt(3)). - Amiram Eldar, Dec 11 2021
E.g.f.: 2 + ((6*x - 3)*exp(x) - exp(-x))/2. - David Lovler, Aug 25 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 2/sqrt(3) (10 * A020832).
Product_{n>=1} (1 + (-1)^n/a(n)) = 1/sqrt(3) (A020760). (End)

A102039 a(n) = a(n-1) + last digit of a(n-1), starting at 1.

Original entry on oeis.org

1, 2, 4, 8, 16, 22, 24, 28, 36, 42, 44, 48, 56, 62, 64, 68, 76, 82, 84, 88, 96, 102, 104, 108, 116, 122, 124, 128, 136, 142, 144, 148, 156, 162, 164, 168, 176, 182, 184, 188, 196, 202, 204, 208, 216, 222, 224, 228, 236, 242, 244, 248, 256, 262, 264, 268, 276, 282

Offset: 1

Samantha Stones (devilsdaughter2000(AT)hotmail.com), Dec 25 2004

Sequence A001651 is the "base 3" version. In base 4 this rule leads to (1,2,4,4,4...), in base 5 to (1,2,4,8,11,12,14,18,21,22,24,28...) = A235700. - M. F. Hasler, Jan 14 2014

This and the following sequences (none of which is "base"!) could all be defined by a(1) = 1 and a(n+1) = a(n) + (a(n) mod b) with different values of b: A001651 (b=3), A235700 (b=5), A047235 (b=6), A047350 (b=7), A007612 (b=9). Using b=4 or b=8 yields a constant sequence from that term on. - M. F. Hasler, Jan 15 2014

			28 + 8 = 36, 36 + 6 = 42.

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

Apart from initial term, same as A002081.

LinearRecurrence[{2,-2,2,-1},{1,2,4,8,16},60] (* Harvey P. Dale, Jul 02 2022 *)

print1(a=1);for(i=1,99,print1(","a+=a%10)) \\ M. F. Hasler, Jan 14 2014

Vec(x*(5*x^4+2*x^3+2*x^2+1)/((x-1)^2*(x^2+1)) + O(x^100)) \\ Colin Barker, Sep 20 2014

a(n) = if(n==1, 1, (-10-(1-I/2)*(-I)^n-(1+I/2)*I^n+5*n)) \\ Colin Barker, Oct 18 2015

a(n) = 2*a(n-1)-2*a(n-2)+2*a(n-3)-a(n-4) for n>5. G.f.: x*(5*x^4+2*x^3+2*x^2+1) / ((x-1)^2*(x^2+1)). - Colin Barker, Sep 20 2014

a(n) = (-10 - (1-i/2)*(-i)^n - (1+i/2)*i^n + 5*n) for n>1, where i = sqrt(-1). - Colin Barker, Oct 18 2015

A322489 Numbers k such that k^k ends with 4.

Original entry on oeis.org

2, 18, 22, 38, 42, 58, 62, 78, 82, 98, 102, 118, 122, 138, 142, 158, 162, 178, 182, 198, 202, 218, 222, 238, 242, 258, 262, 278, 282, 298, 302, 318, 322, 338, 342, 358, 362, 378, 382, 398, 402, 418, 422, 438, 442, 458, 462, 478, 482, 498, 502, 518, 522, 538, 542, 558

Offset: 1

Bruno Berselli, Dec 12 2018

Also numbers k == 2 (mod 4) such that 2^k and k^2 end with the same digit.

Numbers congruent to {2, 18} mod 20. - Amiram Eldar, Feb 27 2023

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

Cf. A004526, A056849.
Subsequence of A139544, A235700.
Numbers k such that k^k ends with d: A008592 (d=0), A017281 (d=1), A067870 (d=3), this sequence (d=4), A017329 (d=5), A271346 (d=6), A322490 (d=7), A017377 (d=9).

GAP
```
List([1..70], n -> 10*n+3*(-1)^n-5);
```

[10*n+3*(-1)^n-5 for n in 1:70] |> println

Magma
```
[10*n+3*(-1)^n-5: n in [1..70]];
```

select(n->n^n mod 10=4,[$1..558]); # Paolo P. Lava, Dec 18 2018

Mathematica
```
Table[10 n + 3 (-1)^n - 5, {n, 1, 60}]
```
Maxima
```
makelist(10*n+3*(-1)^n-5, n, 1, 70);
```

apply(A322489(n)=10*n+3*(-1)^n-5, [1..70]) \\ M. F. Hasler, Dec 14 2018

Vec(2*x*(1 + 8*x + x^2) / ((1 - x)^2*(1 + x)) + O(x^70)) \\ Colin Barker, Dec 13 2018

[10*n+3*(-1)**n-5 for n in range(1, 70)]

Sage
```
[10*n+3*(-1)^n-5 for n in (1..70)]
```

O.g.f.: 2*x*(1 + 8*x + x^2)/((1 + x)*(1 - x)^2).
E.g.f.: 2 + 3*exp(-x) + 5*(2*x - 1)*exp(x).
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = 10*n + 3*(-1)^n - 5. Therefore:
a(n) = 10*n - 8 for odd n;
a(n) = 10*n - 2 for even n.
a(n+2*k) = a(n) + 20*k.
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(2*Pi/5)*Pi/20 = sqrt(5+2*sqrt(5))*Pi/20. - Amiram Eldar, Feb 27 2023

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A047235 Numbers that are congruent to {2, 4} mod 6.

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A102039 a(n) = a(n-1) + last digit of a(n-1), starting at 1.

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A322489 Numbers k such that k^k ends with 4.

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