A038183 -id:A038183 - cp's OEIS frontend

A100308 Modulo 2 binomial transform of 5^n.

1, 6, 26, 156, 626, 3756, 16276, 97656, 390626, 2343756, 10156276, 60937656, 244531876, 1467191256, 6357828776, 38146972656, 152587890626, 915527343756, 3967285156276, 23803710937656, 95520019531876, 573120117191256

Offset: 0

Paul Barry, Dec 06 2004

5^n may be retrieved through 5^n = Sum_{k=0..n} (-1)^A010060(n-k) * (binomial(n,k) mod 2)*a(k).

Robert Israel, Table of n, a(n) for n = 0..1429
Vladimir Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2 and their polynomial generalization, arXiv:1011.6083 [math.NT], 2010-2012; J. of Algebra Number Theory: Advances and Appl., 7 (2012), no.1, 11-29.

Cf. A001316, A001317, A038183, A100307, A100309, A100310, A100311.

[(&+[5^k*(Binomial(n, k) mod 2): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Feb 03 2023

f:= proc(n) local L,M;
   L:= convert(n,base,2);
   mul(1+5^(2^(k-1)), k = select(t -> L[t]=1, [$1..nops(L)]));
end proc:
map(f, [$0..30]); # Robert Israel, Aug 26 2018

a[n_]:= Sum[Mod[Binomial[n, k], 2] 5^k, {k, 0, n}];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Sep 19 2018 *)

def A100308(n): return sum((bool(~n&n-k)^1)*5**k for k in range(n+1)) # Chai Wah Wu, May 02 2023

def A100308(n): return sum(5^k*(binomial(n, k)%2) for k in range(n+1))
[A100308(n) for n in range(41)] # G. C. Greubel, Feb 03 2023

a(n) = Sum_{k=0..n} (binomial(n, k) mod 2)*5^k.
From Vladimir Shevelev, Dec 26-27 2013: (Start)
Sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(5^(2^k)+1)^r),
Sum_{n>=0} (-1)^A000120(n)/a(n)^r = Product_{k>=0} (1 - 1/(5^(2^k)+1)^r), where r > 0 is a real number.
In particular,
Sum_{n>=0} 1/a(n) = Product_{k>=0} (1 + 1/(5^(2^k)+1)) = 1.2134769...;
Sum_{n>=0} (-1)^A000120(n)/a(n) = 0.8.
a(2^n) = 5^(2^n) + 1, n >= 0.
Note that analogs of Stephan's limit formulas (see Shevelev link) reduce to the relations:
a(2^t*n+2^(t-1)) = 24*(5^(2^(t-1)+1))/(5^(2^(t-1))-1) * a(2^t*n+2^(t-1)-2), t >= 2.
In particular, for t=2,3,4, we have the following formulas:
  a(4*n+2)  =            26 * a(4*n),
  a(8*n+4)  =      (313/13) * a(8*n+2),
  a(16*n+8) = (195313/8138) * a(16*n+6), etc. (End)

A048705 The rule numbers for 1-D CA composed of Rules "90" and "150" so that each direction occurs only once.

Original entry on oeis.org

90, 150, 1721342310, 140117185019831836588493434554119984790, 113427455640312821160607117168492587690

Offset: 1

Antti Karttunen, Mar 09 1999

nonn

The "numerator" (0, 1 and the rest from A020652) is the multiplicity of the "Rule 150" component and the "denominator" (1, 0 and the rest from A020653) is the multiplicity of the "Rule 90" component.
The resulting numbers define one-dimensional linear cellular automata with radius being the sum of the number of the "90" and "150" components.
In hexadecimal the sequence is 5A, 96, 66999966, 69699696969669699696696969699696, 5555555555555555AAAAAAAAAAAAAAAA, ...

Index entries for sequences related to cellular automata

A048706 gives the corresponding "XOR-conjugate" rules.

Cf. A038183, A038184, A048709 (for specific examples). See also A048708, A048720.

# The definitions of bit_i and floor_log_2 are given in A048700
rule90 := proc(seed,n) option remember: local sl, i: if (0 = n) then (seed) else sl := floor_log_2(seed+1); add(((bit_i(rule90(seed,n-1),i)+bit_i(rule90(seed,n-1),i-2)) mod 2)*(2^i), i=0..(2*n)+sl) fi: end:
rule150 := proc(seed,n) option remember: local sl, i: if (0 = n) then (seed) else sl := floor_log_2(seed+1);
add(((bit_i(rule150(seed,n-1),i)+bit_i(rule150(seed,n-1),i-1)+bit_i(rule150(seed,n-1),i-2)) mod 2)*(2^i), i=0..((2*n)+sl)) fi: end:
# Rule 90 and Rule 150 are commutative in respect to each other:
rule90x150combination := proc(n) local p,q,i; p := extended_A020652[ n ]; # the Rule 150 component [ 0,1,op(A020652) ]
q := extended_A020653[ n ]; # the Rule 90 component [ 1,0,op(A020653) ]
RETURN(sum('bit_i(rule150(rule90(i,q),p),(2*(p+q))) * (2^i)','i'=0..(2^((2*(p+q))+1))-1));
end:

a(n) = rule90x150combination(n) # See the Maple procedures below.

A100309 Modulo 2 binomial transform of 6^n.

Original entry on oeis.org

1, 7, 37, 259, 1297, 9079, 47989, 335923, 1679617, 11757319, 62145829, 435020803, 2178463249, 15249242743, 80603140213, 564221981491, 2821109907457, 19747769352199, 104381066575909, 730667466031363, 3658979549971729

Offset: 0

Paul Barry, Dec 06 2004

6^n may be retrieved through 6^n = Sum_{k=0..n} (-1)^A010060(n-k) * mod(binomial(n,k), 2) * a(k).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Vladimir Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2 and their polynomial generalization, arXiv:1011.6083 [math.NT], 2010-2012; J. of Algebra Number Theory: Advances and Appl., 7 (2012), no.1, 11-29.

Cf. A000120, A000400, A001316, A001317, A010060, A038183.

Cf. A100307, A100308, A100310, A100311.

[(&+[6^k*(Binomial(n,k) mod 2): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Feb 02 2023

a[n_]:= a[n]= Sum[6^k*Mod[Binomial[n,k], 2], {k,0,n}];
Table[a[n], {n,0,40}] (* G. C. Greubel, Feb 02 2023 *)

def A100309(n): return sum((bool(~n&n-k)^1)*6**k for k in range(n+1)) # Chai Wah Wu, May 03 2023

def A100309(n): return sum(6^k*(binomial(n, k)%2) for k in range(n+1))
[A100309(n) for n in range(41)] # G. C. Greubel, Feb 02 2023

a(n) = Sum_{k=0..n} mod(binomial(n, k), 2) * 6^k.
From Vladimir Shevelev, Dec 26-27 2013: (Start)
Sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(6^(2^k)+1)^r),
Sum_{n>=0} (-1)^A000120(n)/a(n)^r = Product_{k>=0} (1 - 1/(6^(2^k)+1)^r), where r>0 is a real number.
In particular,
Sum_{n>=0} 1/a(n) = Product_{k>=0} (1 + 1/(6^(2^k)+1)) = 1.1746508...;
Sum_{n>=0} (-1)^A000120(n)/a(n) = 5/6.
a(2^n) = 6^(2^n) + 1, n>=0.
Note that analogs of Stephan's limit formulas (see Shevelev link) reduce to the relations a(2^t*n+2^(t-1)) = 35*(6^(2^(t-1)+1))/(6^(2^(t-1))-1) * a(2^t*n+2^(t-1)-2), t>=2. In particular, for t=2,3,4, we have the following formulas:
a(4*n+2) = 37 * a(4*n);
a(8*n+4) = 1297/37 * a(8*n+2);
a(16*n+8)= 1679617/47989 * a(16*n+6), etc. (End)

A100310 Modulo 2 binomial transform of 7^n.

Original entry on oeis.org

1, 8, 50, 400, 2402, 19216, 120100, 960800, 5764802, 46118416, 288240100, 2305920800, 13847054404, 110776435232, 692352720200, 5538821761600, 33232930569602, 265863444556816, 1661646528480100, 13293172227840800

Offset: 0

Paul Barry, Dec 06 2004

7^n may be retrieved through 7^n = Sum_{k=0..n} (-1)^A010060(n-k) * mod(binomial(n,k), 2) * a(k).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Vladimir Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2 and their polynomial generalization, arXiv:1011.6083 [math.NT], 2010-2012; J. of Algebra Number Theory: Advances and Appl., 7 (2012), no.1, 11-29.

Cf. A000120, A001316, A001317, A010060, A038183.

Cf. A100307, A100308, A100309, A100311.

[(&+[7^k*(Binomial(n,k) mod 2): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Feb 01 2023

a[n_]:= a[n]= Sum[7^k*Mod[Binomial[n,k], 2], {k,0,n}];
Table[a[n], {n,0,40}] (* G. C. Greubel, Feb 01 2023 *)

def A100310(n): return sum((bool(~n&n-k)^1)*7**k for k in range(n+1)) # Chai Wah Wu, May 03 2023

def A100310(n): return sum(7^k*(binomial(n, k)%2) for k in range(n+1))
[A100310(n) for n in range(41)] # G. C. Greubel, Feb 01 2023

a(n) = Sum_{k=0..n} mod(binomial(n, k), 2)*7^k.
From Vladimir Shevelev, Dec 26-27 2013: (Start)
Sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(7^(2^k)+1)^r),
Sum_{n>=0} (-1)^A000120(n)/a(n)^r = Product_{k>=0} (1 - 1/(7^(2^k)+1)^r), where r>0 is a real number.
In particular,
Sum_{n>=0} 1/a(n) = Product_{k>=0} (1 + 1/(7^(2^k)+1)) = 1.1479779...;
Sum_{n>=0} (-1)^A000120(n)/a(n) = 6/7.
a(2^n) = 7^(2^n)+1, n>=0.
Note that analogs of Stephan's limit formulas (see Shevelev link) reduce to the relations:
a(2^t*n+2^(t-1)) = 48*(7^(2^(t-1)+1))/(7^(2^(t-1))-1) * a(2^t*n+2^(t-1)-2), t>=2.
In particular, for t=2,3,4, we have the following formulas:
a(4*n+2) = 50 * a(4*n);
a(8*n+4) = 1201/25 * a(8*n+2);
a(16*n+8)= 2882401/60050 * a(16*n+6), etc. (End)

A048710 Family 1 "Rule 90 x Rule 150 Array" read by antidiagonals.

Original entry on oeis.org

1, 5, 7, 17, 27, 21, 85, 119, 65, 107, 257, 427, 325, 455, 273, 1285, 1799, 1105, 1755, 1365, 1911, 4369, 6939, 5397, 7607, 4097, 6827, 5189, 21845, 30583, 16705, 27499, 20485, 28679, 17745, 28123, 65537

Offset: 0

Antti Karttunen, Mar 18 1999

Infinitely many one-dimensional cellular automaton rules (given in sequence A048705) occur in this array, as combinations of CA-rules "90" (generates rows) and "150" (generates columns).
No pattern occurs twice in such arrays.
Each row/column can be generated from its predecessor row/column with SHIFTXORADJ transformation, given in A048711.

			   1  5  17   85  257 1105 ... [ beginning of array ]
   7 27 119  427 1799 ...
  21 65 325 1105 5397 ...

Rows = A038183, A048711, A048713, columns = A038184, A048712, A048713, diagonal = A048709. Cf. A048720.

trinv := n -> floor((1+sqrt(1+8*n))/2); # Gives integral inverses of the triangular numbers

a(n) = rule150(rule90(1, (((trinv(n)-1)*(((1/2)*trinv(n))+1))-n)), (n-((trinv(n)*(trinv(n)-1))/2))).

A048711 2nd row of Family 1 "90 X 150 array": generations 0 .. n of Rule 90 starting from seed pattern 7.

Original entry on oeis.org

7, 27, 119, 427, 1799, 6939, 30583, 109227, 458759, 1769499, 7798903, 27984299, 117901063, 454761243, 2004318071, 7158278827, 30064771079, 115964117019, 511101108343, 1833951035819, 7726646167303

Offset: 0

Antti Karttunen

nonn

Also generated by applying one generation of "Rule 150" to each term of A038183 or by doing a transformation SHIFTXORADJ(A038183)

N. J. A. Sloane, Transforms: Maple implementation of binary eXclusive OR (XORnos).

A048713.

# Maple procedure for doing Shift XOR adjacent terms transformation:
SHIFTXORADJ := proc(a) local b,i:
if whattype(a) <> list then RETURN([ ]); fi: if nops(a) <= 1 then RETURN([ ]); fi: b := [ ]:
for i from 2 to nops(a) do b := [ op(b), XORnos((a[ i-1 ]*2),a[ i ]) ]: od: RETURN(b); end:

a(n) = product('((bit_i((n+1), i)*(2^(2^(i+1))))+1)', 'i'=0..floor_log_2(n+2)) + 2*product('((bit_i(n, i)*(2^(2^(i+1))))+1)', 'i'=0..floor_log_2(n+1));

A051656 Sum_{i=0..n} (C(n,i) mod 2)Fibonacci(2i).

Original entry on oeis.org

0, 1, 3, 12, 21, 77, 168, 609, 987, 3572, 7755, 28059, 47376, 171409, 372099, 1346268, 2178309, 7881197, 17108664, 61899729, 104512485, 378129724, 820851717, 2969869413, 4809706272, 17401680769, 37775923491, 136674575148

Offset: 0

Antti Karttunen, Nov 30 1999

Positions in the first column (A003622) of Wythoff array of the terms which have their Zeckendorf Expansion patterned as row[2n+1] in Pascal's Triangle computed modulo 2 (A047999)

Proof in preparation, to be published (see A048757).

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Cf. A048757, A047999, A035513, A038183, A051256. First row of A050609, First column of A050610.
a(n) = A019586[A048757[n]]. A048757[n] = SS(Athis_sequence[n])+1, where SSx means the second Fibonacci Successor of x (= x's Z.E. shifted left twice).
Cf. A001906.

a051656 = sum . zipWith (*) a001906_list . a047999_row
-- Reinhard Zumkeller, Feb 27 2011

Table[Sum[Mod[Binomial[n,i],2]Fibonacci[2i],{i,0,n}],{n,0,30}] (* Harvey P. Dale, Oct 30 2011 *)

a(n)=sum(i=0,n,if(!bitand(i,n-i),fibonacci(2*i))) \\ Charles R Greathouse IV, Jan 04 2013

a(n) = sum_{i=0..n} (C(2n, 2i) mod 2)*F(2*i) = FL(n)product_{i=0..inf} L(2^i)^bit(n, i) where L is n-th Lucas number (A000032) and FL is defined as in A048757: FL(n) = n-th fibonacci number if n even, n-th Lucas number if n odd.

A089893 a(n) = (A001317(2n)-1)/4.

Original entry on oeis.org

0, 1, 4, 21, 64, 321, 1092, 5461, 16384, 81921, 278532, 1392661, 4210752, 21053761, 71582788, 357913941, 1073741824, 5368709121, 18253611012, 91268055061, 275951648832, 1379758244161, 4691178030148, 23455890150741

Offset: 0

Ralf Stephan, Jan 10 2004

nonn

a(n) = (A038192(n) - 3)/12 = (A038183(n) - 1)/4, reflecting the fact that, if you look at each row of the Pascal triangle's parity as a binary number (cf. A001317), then the numbers in odd rows are thrice the numbers in even rows.
Conjecture: a(2^k) = 2^(2^(k+1)-2). [This conjecture is true. - Vladimir Shevelev, Nov 28 2010]
Conjectures: lim(n->inf, a(2n+1)/a(2n)) = 5, lim(n->inf, a(4n+2)/a(4n+1)) = 17/5, lim(n->inf, a(8n+4)/a(8n+3)) = 257/85 etc. [This follows from the formula, for n>=0, t>=1: ( 4*a(2^t*n+2^(t-1))+1 )/( 4*a(2^t*n+2^(t-1)-1)+1 ) = 3*F_t/(F_t-2), where F_t= A000215(t) - Vladimir Shevelev, Nov 28 2010]

Vladimir Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2, arXiv:1011.6083 [math.NT], 2010-2012.

Cf. A000215, A001317, A038183, A038192.

a1317[n_] := Sum[2^k Mod[Binomial[n, k], 2] , {k, 0, n}];
a[n_] := (a1317[2n] - 1)/4;
Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Jan 18 2019 *)

a(n)=(sum(k=0,2*n+1,(binomial(2*n+1,k)%2)*2^k)-3)/12

def A089893(n): return sum((bool(~(m:=n<<1)&m-k)^1)<>2 # Chai Wah Wu, May 02 2023

A038185 One-dimensional cellular automaton 'sigma' (Rule 150).

Original entry on oeis.org

1, 3, 5, 13, 17, 59, 81, 219, 257, 899, 1349, 3437, 4353, 15235, 20805, 56173, 65537, 229379, 344069, 876557, 1118225, 3913787, 5313617, 14399195, 16842753, 58949635, 88424453, 225271821, 285282321

Offset: 0

Antti Karttunen, Feb 09 1999

nonn

Generation n (starting from the generation 0: 1) cut after the central 1-column and interpreted as a binary number.

Decimal representation of the x-axis, from the left edge to the origin, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 518", based on the 5-celled von Neumann neighborhood. Initialized with a single black (ON) cell at stage zero. - Robert Price, Feb 22 2017

N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
S. Wolfram, A New Kind of Science
Wolfram Research, Wolfram Atlas of Simple Programs
Index to 2D 5-Neighbor Cellular Automata
Index to Elementary Cellular Automata
Index entries for sequences related to cellular automata

Cf. A006977, A006978, A038183, a(n) = floor(A038184[ n ]/2^n)

bit_n := (x,n) -> `mod`(floor(x/(2^n)),2);
sigmacut := proc(n): if (0 = n) then (1)
else sum('((bit_n(sigmagen(n-1),i+1+n-1)+bit_n(sigmagen(n-1),i+n-1)+bit_n(sigmagen(n-1),i-1+n-1)) mod 2)*(2^i)', 'i'=0..(n)) fi: end:

A166555 Triangle read by rows, T(n, k) = 2^k * A047999(n, k).

Original entry on oeis.org

1, 1, 2, 1, 0, 4, 1, 2, 4, 8, 1, 0, 0, 0, 16, 1, 2, 0, 0, 16, 32, 1, 0, 4, 0, 16, 0, 64, 1, 2, 4, 8, 16, 32, 64, 128, 1, 0, 0, 0, 0, 0, 0, 0, 256, 1, 2, 0, 0, 0, 0, 0, 0, 256, 512, 1, 0, 4, 0, 0, 0, 0, 0, 256, 0, 1024, 1, 2, 4, 8, 0, 0, 0, 0, 256, 512, 1024, 2048, 1, 0, 0, 0, 16, 0, 0, 0, 256, 0, 0, 0, 4096

Offset: 0

Gary W. Adamson, Oct 17 2009

Number of positive terms in n-th row (n>=0) equals to A000120(n). - Vladimir Shevelev, Oct 25 2010

Former name: Triangle read by rows, Sierpinski's gasket, A047999 * (1,2,4,8,...) diagonalized. - G. C. Greubel, Dec 02 2024

			First few rows of the triangle are:
  1;
  1, 2;
  1, 0, 4;
  1, 2, 4, 8;
  1, 0, 0, 0, 16;
  1, 2, 0, 0, 16,.32;
  1, 0, 4, 0, 16,..0,..64;
  1, 2, 4, 8, 16,.32,..64,..128;
  1, 0, 0, 0,..0,..0,...0,....0,..256;
  1, 2, 0, 0,..0,..0,...0,....0,..256,...512;
  1, 0, 4, 0,..0,..0,...0,....0,..256,.....0,...1024;
  1, 2, 4, 8,..0,..0,...0,....0,..256,...512,...1024,...2048;
  1, 0, 0, 0, 16,..0,...0,....0,..256,.....0,......0,......0,..4096;
  ...

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Cf. A000007, A000079, A000120, A038183, A147999.

Sums include: A001317 (row), A101624 (diagonal), A101625 (odd rows of signed diagonal).

A166555:= func< n,k | 2^k*( Binomial(n,k) mod 2) >;
[A166555(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Dec 01 2024

A166555[n_, k_]:= 2^k*Mod[Binomial[n, k], 2];
Table[A166555[n,k], {n,0,14}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 01 2024 *)

def A166555(n,k): return 2^k*int(not ~n & k) if kA166555(n,k) for k in range(n+1)] for n in range(15)])) # G. C. Greubel, Dec 01 2024

Triangle read by rows, A047999 * Q. A047999 = Sierpinski's gasket, Q = an infinite lower triangular matrix with (1,2,4,8,...) as the main diagonal and the rest zeros.
Sum_{k=0..n} T(n, k) = A001317(n).
From G. C. Greubel, Dec 02 2024: (Start)
T(n, k) = 2^k * (binomial(n,k) mod 2).
T(n, n) = A000079(n).
T(2*n, n) = A000007(n).
Sum_{k=0..n} (-1)^k*T(n, k) = (1/2)*( (1+(-1)^n)*A038183(n/2) - (1-(-1)^n) *A038183((n-1)/2) ).
Sum_{k=0..floor(n/2)} T(n-k, k) = A101624(n).
Sum_{k=0..floor((2*m+1)/2)} T(2*m-k+1, k) = A101625(m+1), m >= 0. (End)

New name by G. C. Greubel, Dec 02 2024

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A100308 Modulo 2 binomial transform of 5^n.

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A048705 The rule numbers for 1-D CA composed of Rules "90" and "150" so that each direction occurs only once.

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A100309 Modulo 2 binomial transform of 6^n.

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A100310 Modulo 2 binomial transform of 7^n.

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A048710 Family 1 "Rule 90 x Rule 150 Array" read by antidiagonals.

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A048711 2nd row of Family 1 "90 X 150 array": generations 0 .. n of Rule 90 starting from seed pattern 7.

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A051656 Sum_{i=0..n} (C(n,i) mod 2)*Fibonacci(2*i).

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A051656 Sum_{i=0..n} (C(n,i) mod 2)Fibonacci(2i).