A102039 -id:A102039 - 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)

A235700 a(n+1) = a(n) + (a(n) mod 5), a(1)=1.

Original entry on oeis.org

1, 2, 4, 8, 11, 12, 14, 18, 21, 22, 24, 28, 31, 32, 34, 38, 41, 42, 44, 48, 51, 52, 54, 58, 61, 62, 64, 68, 71, 72, 74, 78, 81, 82, 84, 88, 91, 92, 94, 98, 101, 102, 104, 108, 111, 112, 114, 118, 121, 122, 124, 128, 131, 132, 134, 138, 141, 142, 144, 148, 151, 152, 154, 158, 161, 162, 164, 168, 171, 172, 174, 178, 181, 182, 184, 188, 191

Offset: 1

M. F. Hasler, Jan 14 2014

Although the present sequence has not been thought of via "writing a(n) in base b", this could be seen as "base 5" version of A102039 (base 10) and A001651 (base 3), A047235 (base 6), A047350 (base 7) and A007612 (base 9). For 4 or 8 one would get a sequence constant from that (3rd resp. 4th) term on.

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

Cf. A001651, A007612, A047235, A047350, A102039.

NestList[#+Mod[#,5]&,1,80] (* Harvey P. Dale, Oct 20 2024 *)

is_A235700(n) = bittest(278,n%10) \\ 278=2^1+2^2+2^4+2^8

PARI
```
A235700 = n -> 2^((n-1)%4)+(n-1)\4*10
```

print1(a=1);for(i=1,99,print1(","a+=a%5))

Vec(x*(2*x^3+2*x^2+1)/((x-1)^2*(x^2+1)) + O(x^100)) \\ Colin Barker, Jan 16 2014

a(n) = 2^(n-1 mod 4) + 10*floor((n-1)/4).
From Colin Barker, Jan 16 2014: (Start)
a(n) = (-10+(1+2*i)*(-i)^n+(1-2*i)*i^n+10*n)/4 where i=sqrt(-1).
a(n) = 2*a(n-1)-2*a(n-2)+2*a(n-3)-a(n-4).
G.f.: x*(2*x^3+2*x^2+1) / ((x-1)^2*(x^2+1)). (End)
E.g.f.: (4 + 5*exp(x)*(x - 1) + cos(x) + 2*sin(x))/2. - Stefano Spezia, Feb 22 2025

A102718 a(n) = a(n-1) + (sum of the last two digits of the sequence so far); a(0)=0; a(1)=1.

Original entry on oeis.org

0, 1, 2, 5, 12, 15, 21, 24, 30, 33, 39, 51, 57, 69, 84, 96, 111, 113, 117, 125, 132, 137, 147, 158, 171, 179, 195, 209, 218, 227, 236, 245, 254, 263, 272, 281, 290, 299, 317, 325, 332, 337, 347, 358, 371, 379, 395, 409, 418, 427, 436, 445, 454, 463, 472, 481, 490, 499, 517, 525, 532, 537, 547, 558, 571

Offset: 0

Eric Angelini, Feb 06 2005

			96 + (9 + 6) = 111;
111 + (1 + 1) = 113;
113 + (1 + 3) = 117.

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A102039.

Join[{0,1,2,5},NestList[#+Total[Take[IntegerDigits[#],-2]]&,12,60]] (* Harvey P. Dale, May 11 2018 *)

first(n) = {n = max(n, 5); res = [0, 1, 2, 5, 12]; res = concat(res, vector(n - 5)); for(i = 6, min(n, 37), res[i] = res[i-1] + vecsum(digits(res[i-1] % 100)));
for(i = 38, n, res[i] = res[i - 20] + 200); res} \\ David A. Corneth, May 11 2018

a(n + 20) = 200 + a(n) for n > 17. - David A. Corneth, May 11 2018

Corrected and extended by Harvey P. Dale, May 11 2018

A102719 a(n) = a(n-1) + (sum of the last three digits of the sequence so far); a(0)=0; a(1)=1; a(2)=2 and a(n) < a(n+1).

Original entry on oeis.org

0, 1, 2, 5, 13, 22, 29, 42, 57, 71, 86, 101, 103, 107, 115, 122, 127, 137, 148, 161, 169, 185, 199, 218, 229, 242, 250, 257, 271, 281, 292, 305, 313, 320, 325, 335, 346, 359, 376, 392, 406, 416, 427, 440, 448, 464, 478, 497, 517, 530, 538, 554, 568, 587, 607

Offset: 0

Eric Angelini, Feb 06 2005

			86+(1+8+6) = 101.
101+(1+0+1) = 103.
103+(1+0+3) = 107.

Alois P. Heinz, Table of n, a(n) for n = 0..10000

Cf. A102039.

a:= proc(n) option remember; local k, i; if n<2 then n elif n=2 then b(2):= [2,1,0]; 2 else k:= a(n-1) +add (i, i=b(n-1)); b(n):= [convert (k, base, 10)[], b(n-1)[]][1..3]; k fi end: seq (a(n), n=0..80); # Alois P. Heinz, Aug 13 2009

nxt[{a_,b_,c_}]:={b,c,c+Total[Take[Flatten[Join[IntegerDigits[a],IntegerDigits[b],IntegerDigits[c]]],-3]]}; NestList[nxt,{0,1,2},60][[;;,1]] (* Harvey P. Dale, Mar 13 2025 *)

Corrected and extended by Alois P. Heinz, Aug 13 2009

A235699 a(n+1) = a(n) + (a(n) mod 10) + 1, a(0) = 0.

Original entry on oeis.org

0, 1, 3, 7, 15, 21, 23, 27, 35, 41, 43, 47, 55, 61, 63, 67, 75, 81, 83, 87, 95, 101, 103, 107, 115, 121, 123, 127, 135, 141, 143, 147, 155, 161, 163, 167, 175, 181, 183, 187, 195, 201, 203, 207, 215, 221, 223, 227, 235, 241, 243, 247, 255, 261, 263, 267, 275, 281, 283, 287, 295, 301, 303, 307, 315, 321, 323, 327, 335, 341, 343, 347, 355

Offset: 0

M. F. Hasler, Jan 14 2014

Instead of (a(n) mod 10) one might say "the last (decimal) digit of a(n)".
Apart from the initial term, the first differences form the periodic sequence (2,4,8,6)[repeated].
Without the final "+ 1" and starting with 1, one gets A102039: Indeed, the last digit cycles through 2,4,8,6 and therefore the sequence never becomes constant.

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

Cf. A102039, A235698, A045844.

NestList[#+Mod[#,10]+1&,0,80] (* or *) Join[{0},LinearRecurrence[{2,-2,2,-1},{1,3,7,15},80]] (* Harvey P. Dale, Dec 21 2014 *)

print1(a=0);for(i=1,99,print1(","a+=a%10+1))

a(n) = 5*n-6+cos(n*Pi/2)+2*sin(n*Pi/2), for n>0. - Giovanni Resta, Jan 15 2014
From Colin Barker, Jan 16 2014: (Start)
a(n) = -6+(1/2+i)*(-i)^n+(1/2-i)*i^n+5*n for n>0 where i=sqrt(-1).
a(n) = 2*a(n-1)-2*a(n-2)+2*a(n-3)-a(n-4) for n>4.
G.f.: x*(5*x^3+3*x^2+x+1) / ((x-1)^2*(x^2+1)). (End)

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

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A235700 a(n+1) = a(n) + (a(n) mod 5), a(1)=1.

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A102718 a(n) = a(n-1) + (sum of the last two digits of the sequence so far); a(0)=0; a(1)=1.

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A102719 a(n) = a(n-1) + (sum of the last three digits of the sequence so far); a(0)=0; a(1)=1; a(2)=2 and a(n) < a(n+1).

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A235699 a(n+1) = a(n) + (a(n) mod 10) + 1, a(0) = 0.

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