A000078 -id:A000078 - cp's OEIS frontend

A337513 G.f. A(x) satisfies: A(x) = 1 - Sum_{k=1..4} (x * A(x))^k.

1, -1, 0, 1, 0, -1, -5, 13, 5, -43, 4, 98, 122, -638, -246, 2912, -537, -9419, -1648, 47005, 2243, -232237, 87988, 904267, -351692, -4123026, 1726126, 20257940, -14035151, -86846040, 73352891, 387126945, -358259621, -1853868355, 2081413376

Offset: 0

Ilya Gutkovskiy, Aug 30 2020

sign

Cf. A000078, A007440, A036766, A337512, A337514.

nmax = 34; A[] = 0; Do[A[x] = 1 - Sum[(x A[x])^k, {k, 1, 4}] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
nmax = 35; CoefficientList[(1/x) InverseSeries[Series[x/(1 - x - x^2 - x^3 - x^4), {x, 0, nmax}], x], x]
b[m_, r_, k_] := b[m, r, k] = If[m + r == 0, 1, Sum[b[m - j, r + j - 1, k], {j, 1, Min[1, m]}] - Sum[b[m + j - 1, r - j, k], {j, 1, Min[k, r]}]]; a[n_] := b[0, n, 4]; Table[a[n], {n, 0, 34}]

G.f.: A(x) = (1/x) * Series_Reversion(x / (1 - x - x^2 - x^3 - x^4)).

A351656 Dirichlet g.f.: Product_{p prime} 1 / (1 - p^(-s) - p^(-2s) - p^(-3s) - p^(-4*s)).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 4, 2, 1, 1, 2, 1, 1, 1, 8, 1, 2, 1, 2, 1, 1, 1, 4, 2, 1, 4, 2, 1, 1, 1, 15, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 2, 2, 1, 1, 8, 2, 2, 1, 2, 1, 4, 1, 4, 1, 1, 1, 2, 1, 1, 2, 29, 1, 1, 1, 2, 1, 1, 1, 8, 1, 1, 2, 2, 1, 1, 1, 8, 8, 1, 1, 2, 1, 1, 1

Offset: 1

Ilya Gutkovskiy, Feb 16 2022

Cf. A000078, A351219, A351346, A351347, A351348, A351655.

t[e_] := t[e] = If[e < 5, 2^(e-1), t[e-1] + t[e-2] + t[e-3] + t[e-4]]; a[1] = 1; a[n_] := Times @@ t /@ Last @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 23 2023 *)

for(n=1, 87, print1(direuler(p=2, n, 1/(1 - X - X^2 - X^3 - X^4))[n], ", "))

Multiplicative with a(p^e) = A000078(e+3).

A101350 Triangle read by rows: T(n,k) = number of k-matchings in the graph obtained by a zig-zag triangulation of a convex n-gon, T(0,0)=T(1,0)=T(2,0)=T(2,1)=1 (n > 2, 0 <= k <= floor(n/2)).

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 5, 2, 1, 7, 7, 1, 9, 16, 3, 1, 11, 29, 15, 1, 13, 46, 43, 5, 1, 15, 67, 95, 30, 1, 17, 92, 179, 104, 8, 1, 19, 121, 303, 271, 58, 1, 21, 154, 475, 591, 235, 13, 1, 23, 191, 703, 1140, 705, 109, 1, 25, 232, 995, 2010, 1746, 506, 21, 1, 27, 277, 1359, 3309, 3780

Offset: 0

Emeric Deutsch, Dec 25 2004

			T(5,2)=7 because in the triangulation of the convex pentagon ABCDEA with diagonals AD and AC we have seven 2-matchings: {AB,CD},{AB,DE},{BC,AD},{BC,DE},{BC,EA},{CD,EA} and {DE,AC}.
Triangle begins:
  1;
  1;
  1,  1;
  1,  3;
  1,  5,  2;
  1,  7,  7;
  1,  9, 16,  3;
  1, 11, 29, 15;
  1, 13, 46, 43, 5;
  ...

Andrew Howroyd, Table of n, a(n) for n = 0..649

Row sums yield A000078 (the tetranacci numbers). T(2n+1, n) = A023610(n) (n > 0). T(2n, n) = A000045(n+1) (the Fibonacci numbers).

Cf. A000078, A023610, A000045.

G:=1/(1-z-t*z^2-t*z^3-t^2*z^4):Gserz:=simplify(series(G,z=0,18)):P[0]:=1: for n from 1 to 16 do P[n]:=sort(coeff(Gserz,z^n)) od:for n from 0 to 16 do seq(coeff(t*P[n],t^k),k=1..1+floor(n/2)) od;# yields the sequence in triangular form

s(n) = 1/(1-x-y*x^2-y*x^3-y^2*x^4) + O(x^n);
my(gf=Pol(s(20))); for(n=0, poldegree(gf), my(p=polcoeff(gf,n)); for(k=0, poldegree(p), print1(polcoeff(p,k), ", ")); print) \\ Andrew Howroyd, Nov 04 2017

G.f.: 1/(1 - z - tz^2 - tz^3 - t^2z^4).

A111427 Tribonacci(tetranacci(n)).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 24, 1705, 8646064, 120879712950776, 6984325661199418257711189170, 2037364946117781220616409939151494028465982215442078400

Offset: 0

Jonathan Vos Post, Nov 13 2005

Alois P. Heinz, Table of n, a(n) for n = 0..16

Cf. A000045, A000073, A000078, A007570.

a:= n-> (<<0|1|0>, <0|0|1>, <1|1|1>>^((<<0|1|0|0>,
        <0|0|1|0>, <0|0|0|1>, <1|1|1|1>>^n)[1, 4]))[1, 3]:
seq(a(n), n=0..12);  # Alois P. Heinz, Nov 07 2018

a(n) = A000073(A000078(n)).

A112677 Sum of digits of the sum of the previous 4 terms.

Original entry on oeis.org

1, 1, 1, 1, 4, 7, 4, 7, 4, 4, 10, 7, 7, 10, 7, 4, 10, 4, 7, 7, 10, 10, 7, 7, 7, 4, 7, 7, 7, 7, 10, 4, 10, 4, 10, 10, 7, 4, 4, 7, 4, 10, 7, 10, 4, 4, 7, 7, 4, 4, 4, 10, 4, 4, 4, 4, 7, 10, 7, 10, 7, 7, 4, 10, 10, 4, 10, 7, 4, 7, 10, 10, 4, 4, 10, 10, 10, 7, 10, 10, 10, 10

Offset: 0

Jonathan Vos Post, Dec 30 2005

This is to the tetranacci sequence as A112661 is to the tribonacci and as A030132 is to Fibonacci. A000288 is the tetranacci sequence (A000078) but starting with values (1,1,1,1). Andrew Carmichael Post (andrewpost(AT)gmail.com) wrote the program that generated this sequence and showed that for any 4 initial integers a(0),a(1),a(2),a(3) the length of the cycle eventually entered is a factor of 312. For instance, starting with (6,6,6,6) continues in a cycle of length 1 since SOD(6+6+6+6) = SOD(24) = 6; and 1 divides 312. For the SOD(tribonacci) which is A112661, the length of any cycle eventually entered is a factor of 78.

All terms for n >= 4 are 4, 7, or 10. The sequence has period 78; the 78 terms after the initial 1,1,1,1 repeat forever. - Nathaniel Johnston, May 04 2011

			a(0)=a(1)=a(2)=a(3)=1.
a(4) = SOD(1+1+1+1) = SOD(4) = 4.
a(5) = SOD(1+1+1+4) = SOD(7) = 7.
a(10) = SOD(4+7+4+4) = SOD(19) = 10, note that we do not iterate SOD to reduce 10 to 1.

Cf. A000078, A000288, A004090, A007953, A010888, A030132, A112661.

A112677 := proc(n) option remember: if(n<=3)then return 1:fi: return add(d,d=convert(procname(n-1) + procname(n-2) + procname(n-3) + procname(n-4),base,10)): end: seq(A112677(n),n=0..100); # Nathaniel Johnston, May 04 2011

nxt[{a_,b_,c_,d_}]:={b,c,d,Total[IntegerDigits[a+b+c+d]]}; Transpose[ NestList[ nxt,{1,1,1,1},90]][[1]] (* or *) PadRight[{1,1,1,1},120,{10,10,10,10,4,7,4,7,4,4,10,7,7,10,7,4,10,4,7,7,10,10,7,7,7,4,7,7,7,7,10,4,10,4,10,10,7,4,4,7,4,10,7,10,4,4,7,7,4,4,4,10,4,4,4,4,7,10,7,10,7,7,4,10,10,4,10,7,4,7,10,10,4,4,10,10,10,7}](* Harvey P. Dale, Mar 05 2016 *)

a(0)=a(1)=a(2)=a(3)=1. a(n) = SumDigits(a(n-1) + a(n-2) + a(n-3) + a(n-4)).
a(n) = SumDigits(A000288(n)).
a(n) = A007953(a(n-1) + a(n-2) + a(n-3) + a(n-4)). - Nathaniel Johnston, May 04 2011

Name corrected by Nathaniel Johnston, May 04 2011

A116849 Number of permutations of length n which avoid the patterns 213, 51432.

Original entry on oeis.org

1, 2, 5, 14, 41, 121, 356, 1044, 3057, 8948, 26192, 76674, 224465, 657137, 1923817, 5632105, 16488346, 48270655, 141315320, 413709331, 1211159679, 3545745012, 10380388294, 30389230117, 88966354626, 260454516946, 762496740130

Offset: 1

Lara Pudwell, Feb 26 2006

a(n) is the binomial transform of the tetranacci numbers (A007318 transform of A000078).  For example: a(6) = 121 = 1*1 + 5*1 + 10*2 + 10*4 + 5*8 + 1*15. - Bob Selcoe, Jun 28 2014
From Bob Selcoe, Jul 06 2014: (Start)
In general, the equation for the binomial transform of m-nacci numbers is: a(n) = 2*a(n-1) + sum_({i=0..n-2}, {j=0..m-2}) a(n-i-2)*sum_(i!/(j!*(i-j)!), where i>=j and a(0)=1. So in this sequence, where m=4 (therefore m-2=2) and the offset is 1 (rather than 0): a(n) = 2*a(n-1) + sum_{i=0..n-3} a(n-i-2)*[(i*(i+1)/2)+1].
Another general equation for the binomial transform of m-nacci numbers (when a(0)=1) is: a(n) = 2*a(n-1) + sum_{i=2..m} a(n-i)*2^(i-2) + sum_({i=m+1..n}, {j=0..i-m-1}) a(n-i)*(2^(i-2) - sum_((i-2)!/(j!*(i-2-j)!). Thus, when m=4 and offset is 1: a(n) = 2*a(n-1) + sum_{i=2..4} a(n-i)*2^(i-2) + sum_({i=5..n-1}, {j=0..i-5}) a(n-i)*(2^(i-2) - sum_[(i-2)!/(j!*(i-2-j)!)]). (End)

			From _Bob Selcoe_, Jul 06 2014: (Start)
a(7) = 356 = 2*121 + 41*(0*1/2 + 1) + 14*(1*2/2 + 1) + 5*(2*3/2 + 1) + 2*(3*4/2 + 1) + 1*(4*5/2 + 1)
a(8) = 1044 = 2*356 + 121*2^0 + 41*2^1 + 14*2^2 + 5*(2^3 - 3!/(0!*3!)) + 2*(2^4 - [4!/(0!*4!) + 4!/(1!*3!)]) + 1*(2^5 - [5!/(0!*5!) + 5!/(1!*4!) + 5!/(2!*3!)]). (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Lara Pudwell, Systematic Studies in Pattern Avoidance, 2005.
Index entries for linear recurrences with constant coefficients, signature (5,-8,6,-1).

A116849 := proc(n) coeftayl( -(x*(x-1)^3)/(-6*x^3+x^4+8*x^2-5*x+1), x=0, n); end proc: seq(A116849(n), n=1..30); # Wesley Ivan Hurt, Jul 06 2014

Rest[CoefficientList[Series[-(x (x - 1)^3)/(- 6 x^3 + x^4 + 8 x^2 - 5 x + 1), {x, 0, 40}], x] ](* Vincenzo Librandi, Jun 29 2014 *)

G.f.: -(x*(x-1)^3)/(-6*x^3+x^4+8*x^2-5*x+1).
a(n) = 5*a(n-1) -8*a(n-2) +6*a(n-3) -a(n-4) for n>3. - Vincenzo Librandi, Jun 29 2014
From Bob Selcoe, Jul 06 2014: (Start)
a(n) = 2*a(n-1) + sum_{i=1..n-3} a(n-i-2)*[(i*(i+1)/2)+1].
a(n) = 2*a(n-1) + sum_{i=2..4} a(n-i)*2^(i-2) + sum_({i=5..n-1}, {j=0..i-5}) a(n-i)*(2^(i-2) - sum_[(i-2)!/(j!*(i-2-j)!)]). (End)

A181190 Maximal length of chain-addition sequence mod 10 with window of size n.

Original entry on oeis.org

1, 60, 124, 1560, 4686, 1456, 18744, 585936, 4882810, 212784

Offset: 1

Alexander Dashevsky (atanvarnoalda(AT)gmail.com), Oct 10 2010

Chain addition mod 10 with window n: take an n-digit 'seed'. Take the sum of its digits mod 10 and append to the seed. Repeat with the last n digits of the string, until the seed appears again.
This sequence shows the lengths of the longest sequences for different window sizes.
a(1)-a(10) all occur for seed 1 (among others). If this is always true, the sequence continues: 406224, 12695306, 4272460934, 380859180, 122070312496, 518798826, 3433227539058. - Lars Blomberg, Feb 12 2013
Comment from Michel Lagneau, Jan 20 2017, edited by N. J. A. Sloane, Jan 24 2017: (Start)
If seed 1 is always as good as or better than any other, as seems likely, then this sequence has the following alternative description.
Consider the n initial terms of an infinite sequence S(k, n) of decimal digits given by 0, 0,..., 0, 1. The succeeding terms are given by the final digits in the sum of the n immediately preceding terms. The sequence lists the period of each sequence corresponding to n = 2, 3, ...
a(2) = period of A000045 mod 10 (Fibonacci numbers mod 10) = A001175(10).
a(3) = period of A000073 mod 10 (tribonacci numbers mod 10) = A046738(10).
a(4) = period of A000078 mod 10 (tetranacci numbers mod 10) = A106295(10).
a(5) = period of A001591 mod 10 (pentanacci numbers mod 10) = A106303(10).
a(6) = period of A001592 mod 10 (hexanacci numbers mod 10).
a(7) = period of A122189 mod 10 (heptanacci numbers mod 10).
a(8) = period of A079262 mod 10 (octanacci numbers mod 10).
a(4) = 1560 because the four initial terms 0, 0, 0, 1 => S(k, 4) = 0, 0, 0, 1, 1, 2, 4, 8, 5, 9, 6, 8, 8, 1, 3, 0, 2, 6, 1, 9, 8, ... (tetranacci numbers mod 10). This sequence is periodic with period 1560:
S(1560 + 1, 4) = S(1, 4) = 0,
S(1560 + 2, 4) = S(2, 4) = 0,
S(1560 + 3, 4) = S(3, 4) = 0,
S(1560 + 4, 4) = S(4, 4) = 1.
(End)

			For n=2, the longest sequence begins with '01' (among others):
01123583145943707741561785381909987527965167303369549325729101.
It is 60 digits long (not counting the second '01' at the end).
For n=3, one of the longest sequences begins again with '001':
00112473441944756893025746770415061742394699425184352079627546556679289964992013
48570291225960516297849144970639807524172091001 (124 digits long without the second '001').

Cf. A000045, A000073, A000078, A001591, A001592, A003893, A079262, A122189

a(8)-a(10) from Lars Blomberg, Feb 12 2013

A196875 a(n) = a(n-4) + a(n-3) + a(n-2) + a(n-1) + (n-5).

Original entry on oeis.org

1, 1, 1, 1, 4, 8, 16, 32, 64, 125, 243, 471, 911, 1759, 3394, 6546, 12622, 24334, 46910, 90427, 174309, 335997, 647661, 1248413, 2406400, 4638492, 8940988, 17234316, 33220220, 64034041, 123429591, 237918195, 458602075, 883983931, 1703933822, 3284438054

Offset: 1

Aditya Subramanian, Oct 07 2011

Alois P. Heinz, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2,0,0,-1,1).

Cf. A000126, A196787.

a:= n-> (Matrix(6, (i, j)-> `if`(i=j-1, 1, `if`(i=6, [1, -1, 0, 0, -2, 3][j], 0)))^n. <<-1, 1, 1, 1, 1, 4>>)[1, 1]: seq(a(n), n=1..50); # Alois P. Heinz, Oct 15 2011

nn = 40; a[1] = a[2] = a[3] = a[4] = 1; Do[a[n] = a[n - 1] + a[n - 2] + a[n - 3] + a[n - 4] + (n - 5), {n, 5, nn}]; Table[a[n], {n, nn}] (* T. D. Noe, Oct 07 2011 *)
RecurrenceTable[{a[1]==a[2]==a[3]==a[4]==1,a[n]==a[n-1]+a[n-2]+a[n-3]+a[n-4]+(n-5)},a,{n,40}] (* or *) LinearRecurrence[{3,-2,0,0,-1,1},{1,1,1,1,4,8},40] (* Harvey P. Dale, Aug 25 2014 *)

G.f.: (x^5-3*x^4+2*x-1)*x / ((x^4+x^3+x^2+x-1)*(x-1)^2 ).
a(n) = +3*a(n-1) -2*a(n-2) -a(n-5) +a(n-6).
a(n) = 5/9-n/3 +(10*A000078(n) +17*A000078(n+1) +21*A000078(n+2) -14*A000078(n+3))/9. - R. J. Mathar, Oct 16 2011

A202193 Triangle read by rows: T(n,m) = coefficient of x^n in expansion of (x/(1 - x - x^2 - x^3 - x^4))^m.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 4, 5, 3, 1, 8, 12, 9, 4, 1, 15, 28, 25, 14, 5, 1, 29, 62, 66, 44, 20, 6, 1, 56, 136, 165, 129, 70, 27, 7, 1, 108, 294, 401, 356, 225, 104, 35, 8, 1, 208, 628, 951, 944, 676, 363, 147, 44, 9, 1, 401, 1328, 2211, 2424, 1935, 1176, 553, 200, 54, 10, 1

Offset: 1

Vladimir Kruchinin, Dec 14 2011

From Philippe Deléham, Feb 16 2014: (Start)
As a Riordan array, this is (1/(1 - x - x^2 - x^3 - x^4), x/(1 - x - x^2 - x^3 - x^4)).
T(n,0) = A000078(n+3); T(n+1,1) = A118898(n+4).
Row sums are A103142(n).
Diagonal sums are A077926(n)*(-1)^n.
Tetranacci convolution triangle. (End)

			Triangle begins:
   1;
   1,  1;
   2,  2,  1;
   4,  5,  3,  1;
   8, 12,  9,  4,  1;
  15, 28, 25, 14,  5,  1;
  29, 62, 66, 44, 20,  6,  1;

Similar sequences : A037027 (Fibonacci convolution triangle), A104580 (tribonacci convolution triangle). - Philippe Deléham, Feb 16 2014

T(n,m):=if n=m then 1 else sum(sum((-1)^i*binomial(k,k-i)*binomial(n-m-4*i-1,k-1),i,0,(n-m-k)/4)*binomial(k+m-1,m-1),k,1,n-m);

T(n,m) = Sum_{k=1..n-m} (Sum_{i=0..floor((n-m-k)/4)} (-1)^i*binomial(k,k-i)*binomial(n-m-4*i-1,k-1))*binomial(k+m-1,m-1), n > m, T(n,n)=1.
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k) + T(n-3,k) + T(n-4,k), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Feb 16 2014
G.f. for column m: (x/(1 - x - x^2 - x^3 - x^4))^m. - Jason Yuen, Feb 17 2025

A202805 a(n) is the largest k in an n_nacci(k) sequence (Fibonacci(k) for n=2, tribonacci(k) for n=3, etc.) such that n_nacci(k) >= 2^(k-n-1).

Original entry on oeis.org

6, 12, 25, 48, 94, 184, 363, 719, 1430, 2851, 5691, 11371, 22728, 45443, 90870, 181724, 363429, 726839, 1453658, 2907295, 5814566, 11629107

Offset: 2

Frank M Jackson, Dec 24 2011

From Frank M Jackson, Jul 02 2023: (Start)
Define the n_nacci sequence, basically row n in A092921, with an offset of 0, n_nacci(k) = 0 for 0 <= k <= n-2 and n_nacci(n-1) = 1. Thereafter, n_nacci(k) for k >= n continues as the sum of its previous n terms.
This means that n_nacci(k) = 2^(k-n) for n <= k <= 2n-1. In the limit as n tends to infinity the n_nacci sequence after an initial large set of zeros followed by 1 has successive terms of ascending powers of 2.
As the n-acci constants, (A001622, A058265, A086088, A103814,...) are smaller than 2, for each n_nacci sequence there is a largest k such that n_nacci(k) >= 2^(k-n-1). (End)

			For n=3, the tribonacci sequence is 0,0,1,1,2,4,7,...,149,274,504,... and the 13th term is 504 < 512 so a(n)=12 because 274 is greatest term >= 2^(12-3-1) = 256.

Cf. A000045, A000073, A000078.

nAcci := proc(n,k)
    option remember ;
    if k <= n-2 then
        0;
    elif k = n-1 then
        1;
    else
        add( procname(n,i),i=k-n..k-1) ;
    end if;
end proc:
A202805 := proc(n)
    local k ;
    for k from n do
        if nAcci(n,k) < 2^(k-n-1) then
            return k-1;
        end if;
    end do:
end proc:
for n from 2 do
    print(n,A202805(n)) ;
end do: # R. J. Mathar, Mar 11 2024

fib[n_, m_] := (Block[{nacci}, (Do[nacci[g]=0, {g, 0, m - 2}];
nacci[m-1]=1;nacci[p_] := (nacci[p]=Sum[nacci[h], {h, p-m, p-1}]);nacci[n])]);
crossover[q_] := (Block[{$RecursionLimit=Infinity}, (k=0;While[fib[k+q+1, q]>=2^k, k++];k+q)]);
Table[crossover[j], {j, 2, 12}]

def nacci(n): # generator of n_nacci terms
    window = [0]*(n-1) + [1]
    yield from window
    while True:
        an = sum(window)
        yield an
        window = window[1:] + [an]
def a(n):
    pow2 = 1
    for k, t in enumerate(nacci(n)):
        if k > n + 1: pow2 <<= 1
        if 0 < t < pow2: return k-1
print([a(n) for n in range(2, 12)]) # Michael S. Branicky, Jan 29 2025

Edited by N. J. A. Sloane, May 20 2023
There seems to be an error in the Comment. See "History" tab. - N. J. A. Sloane, Jun 24 2023
Removed musing about what might define "complete" sequences. - R. J. Mathar, Mar 11 2024
a(17)-a(23) from Michael S. Branicky, Jan 29 2025

cp's OEIS Frontend

A337513 G.f. A(x) satisfies: A(x) = 1 - Sum_{k=1..4} (x * A(x))^k.

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A351656 Dirichlet g.f.: Product_{p prime} 1 / (1 - p^(-s) - p^(-2*s) - p^(-3*s) - p^(-4*s)).

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A101350 Triangle read by rows: T(n,k) = number of k-matchings in the graph obtained by a zig-zag triangulation of a convex n-gon, T(0,0)=T(1,0)=T(2,0)=T(2,1)=1 (n > 2, 0 <= k <= floor(n/2)).

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A111427 Tribonacci(tetranacci(n)).

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A112677 Sum of digits of the sum of the previous 4 terms.

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A116849 Number of permutations of length n which avoid the patterns 213, 51432.

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A181190 Maximal length of chain-addition sequence mod 10 with window of size n.

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A196875 a(n) = a(n-4) + a(n-3) + a(n-2) + a(n-1) + (n-5).

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A351656 Dirichlet g.f.: Product_{p prime} 1 / (1 - p^(-s) - p^(-2s) - p^(-3s) - p^(-4*s)).